chapter #6 functions of random variables question #2reeder/5080/5080solutions/elird haxhiu...

Chapter #6 – Functions of Random Variables

Question #2: Let 𝑋~𝑈𝑁𝐼𝐹(0,1). Use the CDF technique to find the PDF of the following

random variables: a) 𝑌 = 𝑋1/4, b) 𝑊 = 𝑒−𝑋, c) 𝑍 = 1 − 𝑒−𝑋, and d) 𝑈 = 𝑋(1 − 𝑋).

a) Since 𝑋~𝑈𝑁𝐼𝐹(0,1), we know that the density function is 𝑓𝑋(𝑥) = {1 𝑖𝑓 𝑥 ∈ (0,1)0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

while the distribution function is 𝐹𝑋(𝑥) = {

0 𝑖𝑓 𝑥 ∈ (−∞, 0]

𝑥 𝑖𝑓 𝑥 ∈ (0,1)

1 𝑖𝑓 𝑥 ∈ [1, ∞) . We can then use the

CDF technique to find 𝐹𝑌(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(𝑋1/4 ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑦4) = 𝐹𝑋(𝑦4), so

that 𝑓𝑌(𝑦) =𝑑

𝑑𝑦𝐹𝑌(𝑦) =

𝑑

𝑑𝑦𝐹𝑋(𝑦4) = 𝑓𝑋(𝑦4)4𝑦3 = (1)4𝑦3. Since 0 < 𝑥 < 1, the

bounds are 0 < 𝑦4 < 1 or 0 < 𝑦 < 1. Therefore, 𝑓𝑌(𝑦) = {4𝑦3 𝑖𝑓 𝑦 ∈ (0,1)0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

.

b) We have 𝐹𝑊(𝑤) = 𝑃(𝑊 ≤ 𝑤) = 𝑃(𝑒−𝑋 ≤ 𝑤) = 𝑃(𝑋 ≥ −𝑙𝑛 (𝑤)) = 1 − 𝐹𝑋(−𝑙𝑛 (𝑤)),

so that 𝑓𝑊(𝑤) =𝑑

𝑑𝑤𝐹𝑊(𝑤) =

𝑑

𝑑𝑤[1 − 𝐹𝑋(− 𝑙𝑛(𝑤))] = −𝑓𝑋(− 𝑙𝑛(𝑤)) (−

1

𝑤) =

1

𝑤. Since

0 < 𝑥 < 1, the bounds are 0 < − 𝑙𝑛(𝑤) < 1 or 𝑒−1 < 𝑤 < 1. The probability density

function of the random variable 𝑊 is therefore 𝑓𝑊(𝑤) = {1

𝑤 𝑖𝑓 𝑤 ∈ (𝑒−1, 1)

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 .

c) By the CDF technique, we have that the 𝐹𝑍(𝑧) = 𝑃(𝑍 ≤ 𝑧) = 𝑃(1 − 𝑒−𝑋 ≤ 𝑧) =

𝑃(𝑒−𝑋 ≥ 1 − 𝑧) = 𝑃(𝑋 ≤ −𝑙 𝑛(1 − 𝑧)) = 𝐹𝑋(−𝑙 𝑛(1 − 𝑧)), so we have that 𝑓𝑍(𝑧) =

𝑑

𝑑𝑧𝐹𝑍(𝑧) =

𝑑

𝑑𝑧𝐹𝑋(−𝑙 𝑛(1 − 𝑧)) = 𝑓𝑋(− 𝑙𝑛(1 − 𝑧)) (−

−1

1−𝑧) =

1

1−𝑧. Since 0 < 𝑥 < 1, the

bounds are 0 < −𝑙 𝑛(1 − 𝑧) < 1 or 0 < 𝑧 < 1 − 𝑒−1. Therefore, the probability

density function of this random variable is 𝑓𝑍(𝑧) = {1

1−𝑧 𝑖𝑓 𝑧 ∈ (0,1 − 𝑒−1)


d) We will use Theorem 6.3.2 for this question. Suppose that 𝑋 is a continuous random

variable with density 𝑓𝑋(𝑥) and assume that 𝑌 = 𝑢(𝑋) is a one-to-one transformation

with inverse 𝑥 = 𝑤(𝑦). If 𝑑

𝑑𝑦𝑤(𝑦) is continuous and nonzero, the density of 𝑌 is given

by 𝑓𝑌(𝑦) = 𝑓𝑋(𝑤(𝑦)) |𝑑

𝑑𝑦𝑤(𝑦)|. Here, the transformation is 𝑢 = 𝑥(1 − 𝑥) = −𝑥2 + 𝑥

and, based on the y-values of the graph over the x-values of (0,1), the range over

which the density of 𝑈 is defined is (0,1

4). Since this transformation is not one-to-one,

we must partition the interval (0,1) into two parts: (0,1

2) and (

1

2, 1). Then solve the

transformation by completing the square to obtain 𝑢 = −𝑥2 + 𝑥 → −𝑢 = 𝑥2 − 𝑥 →

1

4− 𝑢 = 𝑥2 − 𝑥 +

1

4 →

1

4− 𝑢 = (𝑥 −

1

2)

2

→ 𝑥 −1

2= ±√

1

4− 𝑢 → 𝑥 =

1

2± √

1

4− 𝑢.

Since 𝑥 = 𝑤(𝑢) =1

2± √

1

4− 𝑢, we have 𝑤′(𝑢) = ±

1

2(

1

4− 𝑢)

−1

2 (−1) =±1

2√1

4−𝑢

. Thus,

𝑓𝑈(𝑢) = 𝑓𝑋 (1

2+ √

1

4− 𝑢) |𝑤′(𝑢)| + 𝑓𝑋 (

1

2− √

1

4− 𝑢) |𝑤′(𝑢)| = (1 + 1)|𝑥′(𝑢)| =

1

√1

4−𝑢

,

so we can conclude that the density function is 𝑓𝑈(𝑢) = {(1

4− 𝑢)

−1

2 𝑖𝑓 𝑢 ∈ (0,

1

4)

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

.

Question #3: The measured radius 𝑅 of a circle has PDF 𝑓𝑅(𝑟) = 6𝑟(1 − 𝑟) if 𝑟 ∈ (0,1) and

𝑓𝑅(𝑟) = 0 otherwise. Find the distribution of a) the circumference and b) area of the circle.

a) The circumference of a circle is given by 𝐶 = 2𝜋𝑅, so by the CDF technique we have

that 𝐹𝐶(𝑐) = 𝑃(𝐶 ≤ 𝑐) = 𝑃(2𝜋𝑅 ≤ 𝑐) = 𝑃 (𝑅 ≤𝑐

2𝜋) = 𝐹𝑅 (

𝑐

2𝜋). The density is thus

𝑓𝐶(𝑐) =𝑑

𝑑𝑐𝐹𝐶(𝑐) =

𝑑

𝑑𝑐𝐹𝑅 (

𝑐

2𝜋) = 𝑓𝑅 (

𝑐

2𝜋) (

1

2𝜋) = 6 (

𝑐

2𝜋) (1 −

𝑐

2𝜋) (

1

2𝜋) = ⋯ =

6𝑐(2𝜋−𝑐)

(2𝜋)3 .

Since 0 < 𝑟 < 1 we have 0 <𝑐

2𝜋< 1 so that 0 < 𝑐 < 2𝜋. Therefore, the probability

density function of the circumference is given by 𝑓𝐶(𝑐) = {6𝑐(2𝜋−𝑐)

(2𝜋)3 𝑖𝑓 𝑐 ∈ (0,2𝜋)


b) The area is given by 𝐴 = 𝜋𝑅2, so we have 𝐹𝐴(𝑎) = 𝑃(𝐴 ≤ 𝑎) = 𝑃(𝜋𝑅2 ≤ 𝑎) =

𝑃 (𝑅2 ≤𝑎

𝜋) = 𝑃 (|𝑅| ≤ ±√

𝑎

𝜋) = 𝑃 (−√

𝑎

𝜋≤ 𝑅 ≤ √

𝑎

𝜋) = 𝐹𝑅 (√

𝑎

𝜋) − 𝐹𝑅 (−√

𝑎

𝜋). Thus,

we have 𝑓𝐴(𝑎) =𝑑

𝑑𝑎𝐹𝐴(𝑎) =

𝑑

𝑑𝑎[𝐹𝑅 (√

𝑎

𝜋) − 𝐹𝑅 (−√

𝑎

𝜋)] = ⋯ =

3(√𝜋−√𝑎)

𝜋3/2. Since we have

0 < 𝑟 < 1, then 0 < √𝑎

𝜋< 1 so that 0 < 𝑎 < 𝜋 and 𝑓𝐴(𝑎) = {

3(√𝜋−√𝑎)

𝜋3/2 𝑖𝑓 𝑎 ∈ (0, 𝜋)


Question #10: Suppose 𝑋 has density 𝑓𝑋(𝑥) =1

2𝑒−|𝑥| for all 𝑥 ∈ ℝ. a) Find the density of the

random variable 𝑌 = |𝑋|. b) If 𝑊 = 0 when 𝑋 ≤ 0 and 𝑊 = 1 when 𝑋 > 0 find the CDF of 𝑊.

a) We have 𝐹𝑌(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(|𝑋| < 𝑦) = 𝑃(−𝑦 ≤ 𝑋 ≤ 𝑦) = 𝐹𝑋(𝑦) − 𝐹𝑋(−𝑦), so

that we obtain 𝑓𝑌(𝑦) =𝑑


𝑑

𝑑𝑦[𝐹𝑋(𝑦) − 𝐹𝑋(−𝑦)] = 𝑓𝑋(𝑦)(1) − 𝑓𝑋(−𝑦)(−1) =

1

2𝑒−|𝑦| +

1

2𝑒−|−𝑦| = 𝑒−|𝑦|. Since the transformation was an absolute value function and

we have the bounds −∞ < 𝑥 < ∞, the bounds become 0 < 𝑦 < ∞. This allows us to

write the probability density function of 𝑌 as 𝑓𝑌(𝑦) = {𝑒−𝑦 𝑖𝑓 𝑦 ∈ (0, ∞)


b) We see that 𝑃(𝑊 = 0) =1

2 and 𝑃(𝑊 = 1) =

1

2 since 𝑓𝑋(𝑥) is symmetric. This allows us

to write the cumulative distribution function as 𝐹𝑊(𝑤) = {

0 𝑖𝑓 𝑤 ∈ (−∞, 0)1

2 𝑖𝑓 𝑤 ∈ [0,1)

1 𝑖𝑓 𝑤 ∈ [1, ∞)

.

Question #13: Suppose 𝑋 has density 𝑓𝑋(𝑥) =1

24𝑥2 for 𝑥 ∈ (−2,4) and 𝑓𝑋(𝑥) = 0 otherwise.

Find the probability density function of the random variable 𝑌 = 𝑋2.

We will use Theorem 6.3.2 for this question: Suppose that 𝑋 is a continuous random

variable with density 𝑓𝑋(𝑥) and assume that 𝑌 = 𝑢(𝑋) is a one-to-one transformation

with inverse 𝑥 = 𝑤(𝑦). If 𝑑

𝑑𝑦𝑤(𝑦) is continuous and nonzero, the density of 𝑌 is given

by 𝑓𝑌(𝑦) = 𝑓𝑋(𝑤(𝑦)) |𝑑

𝑑𝑦𝑤(𝑦)|. Here, the transformation is 𝑢 = 𝑥2 and, based on the

y-values of the graph over the x-values of (0,1), the domain over which the density of

𝑌 is defined is (0,16). Solving the transformation then gives that 𝑥 = 𝑤(𝑦) = ±√𝑦 so

that 𝑤′(𝑦) = ±1

2√𝑦. We must consider two cases in the interval (0,16): over (0,4) the

transformation is not one-to-one and over (4,16) it is one-to-one. The density is thus

𝑓𝑌(𝑦) = {

𝑓𝑋(−√𝑦)|𝑤′(𝑦)| + 𝑓𝑋(√𝑦)|𝑤′(𝑦)| 𝑖𝑓 𝑦 ∈ (0,4)

𝑓𝑋(√𝑦)|𝑤′(𝑦)| 𝑖𝑓 𝑦 ∈ [4,16)


= {

√𝑦

24 𝑖𝑓 𝑦 ∈ (0,4)

√𝑦

48 𝑖𝑓 𝑦 ∈ [4,16)


.

Question #16: Let 𝑋1 and 𝑋2 be independent random variables each having density function

𝑓𝑋𝑖(𝑥) =

1

𝑥2 for 𝑥𝑖 ∈ [1, ∞) and 𝑓𝑋𝑖(𝑥) = 0 otherwise. a) Find the joint PDF of 𝑈 = 𝑋1𝑋2 and

𝑉 = 𝑋1. b) Find the marginal probability density function of the random variable 𝑈.

a) Since 𝑋1 and 𝑋2 are independent, their joint density is simply the product of their

marginal densities, so 𝑓𝑋1𝑋2(𝑥1, 𝑥2) = (

1

𝑥12) (

1

𝑥22) =

1

(𝑥1𝑥2)2 whenever we have that

(𝑥1, 𝑥2) ∈ [1, ∞) × [1, ∞) and zero otherwise. We will use Theorem 6.3.6, which says

that if 𝑢 = 𝑓(𝑥1, 𝑥2) and 𝑣 = 𝑓(𝑥1, 𝑥2) and we can solve uniquely for 𝑥1 and 𝑥2, then

we have 𝑓𝑈𝑉(𝑢, 𝑣) = 𝑓𝑋1𝑋2(𝑥1(𝑢, 𝑣), 𝑥2(𝑢, 𝑣))|𝐽|, where 𝐽 = 𝑑𝑒𝑡 [

𝜕𝑥1

𝜕𝑢

𝜕𝑥1

𝜕𝑣𝜕𝑥2

𝜕𝑢

𝜕𝑥2

𝜕𝑣

]. Here, we

have that 𝑣 = 𝑥1 so 𝑥1 = 𝑣 and 𝑢 = 𝑥1𝑥2 so 𝑥2 =𝑢

𝑥1=

𝑢

𝑣, so we can calculate the

Jacobian as 𝐽 = 𝑑𝑒𝑡 [0 11

𝑣−

𝑢

𝑣2

] = −1

𝑣. We can therefore find the joint density as

𝑓𝑈𝑉(𝑢, 𝑣) = 𝑓𝑋1𝑋2(𝑣,

𝑢

𝑣) |−

1

𝑣| = (

1

(𝑣⋅𝑢

𝑣)

2) (1

𝑣) =

1

𝑢2𝑣 if 1 < 𝑣 < 𝑢 < ∞. We can find this

region of integration by substituting into the constraints of 𝑓𝑋1𝑋2(𝑥1, 𝑥2). The first is

that 1 < 𝑥1 < ∞ so 1 < 𝑣 < ∞ while the second is 1 < 𝑥2 < ∞ so 1 <𝑢

𝑣< ∞, which

reduces to 𝑣 < 𝑢 < ∞. Combining these gives the required bounds 1 < 𝑣 < 𝑢 < ∞.

b) We have 𝑓𝑈(𝑢) = ∫ 𝑓𝑈𝑉(𝑢, 𝑣)𝑑𝑣𝑢

1= ∫

1

𝑢2𝑣𝑑𝑣

𝑢

1= [

1

𝑢2𝑙𝑛(𝑣)]

1

𝑢

=𝑙𝑛(𝑢)

𝑢2 if 1 < 𝑢 < ∞.

Question #18: Let 𝑋 and 𝑌 have joint density function 𝑓𝑋𝑌(𝑥, 𝑦) = 𝑒−𝑦 for 0 < 𝑥 < 𝑦 < ∞

and 𝑓𝑋𝑌(𝑥, 𝑦) = 0 otherwise. a) Then find the joint density function of 𝑆 = 𝑋 + 𝑌 and 𝑇 = 𝑋.

b) Find the marginal density function of 𝑆. c) Find the marginal density function of 𝑇.

a) We have that 𝑡 = 𝑥 so 𝑥 = 𝑡 and 𝑠 = 𝑥 + 𝑦 so 𝑦 = 𝑠 − 𝑥 = 𝑠 − 𝑡, so we can calculate

the Jacobian as 𝐽 = 𝑑𝑒𝑡 [

𝜕𝑥

𝜕𝑠

𝜕𝑥

𝜕𝑡𝜕𝑦

𝜕𝑠

𝜕𝑦

𝜕𝑡

] = 𝑑𝑒𝑡 [0 11 −1

] = −1. We can therefore find the joint

density as 𝑓𝑆𝑇(𝑠, 𝑡) = 𝑓𝑋𝑌(𝑡, 𝑠 − 𝑡)|−1| = (𝑒−(𝑠−𝑡))(1) = 𝑒𝑡−𝑠 if 0 < 𝑡 <𝑠

2< ∞. We

can find these bounds by substituting into the bounds 0 < 𝑥 < 𝑦 < ∞ with our solved

transformations, which gives 0 < 𝑡 < 𝑠 − 𝑡 < ∞ or 0 < 2𝑡 < 𝑠 < ∞ or 0 < 𝑡 <𝑠

2< ∞.

b) We have that 𝑓𝑆(𝑠) = ∫ 𝑓𝑆𝑇(𝑠, 𝑡)𝑑𝑡𝑠/2

0= ∫ 𝑒𝑡−𝑠𝑑𝑡

𝑠/2

0= ∫ 𝑒𝑡𝑒−𝑠𝑑𝑡

𝑠/2

0= [𝑒𝑡𝑒−𝑠]0

𝑠/2=

𝑒−𝑠[𝑒𝑡]0𝑠/2

= 𝑒−𝑠(𝑒𝑠/2 − 1) = if 0 < 𝑠 < ∞.

c) We have that 𝑓𝑇(𝑡) = ∫ 𝑓𝑆𝑇(𝑠, 𝑡)𝑑𝑠∞

2𝑡= ∫ 𝑒𝑡−𝑠𝑑𝑠

∞

2𝑡= ∫ 𝑒𝑡𝑒−𝑠𝑑𝑠

∞

2𝑡= [−𝑒𝑡𝑒−𝑠]2𝑡

∞ =

−𝑒𝑡[𝑒−𝑠]2𝑡∞ = −𝑒𝑡(0 − 𝑒−2𝑡) = 𝑒−𝑡 if 0 < 𝑡 < ∞. Note that we have omitted the steps

where the infinite limit of integration is replaced by a parameter and a limit to infinity

with that parameter is evaluated to show that it goes to zero.

Question #21: Let 𝑋 and 𝑌 have joint density 𝑓𝑋𝑌(𝑥, 𝑦) = 2(𝑥 + 𝑦) for 0 < 𝑥 < 𝑦 < 1 and

𝑓𝑋𝑌(𝑥, 𝑦) = 0 otherwise. a) Find the joint probability density function of 𝑆 = 𝑋 and 𝑇 = 𝑋𝑌.

b) Find the marginal probability density function of the random variable 𝑇.

a) We have that 𝑠 = 𝑥 so 𝑥 = 𝑠 and 𝑡 = 𝑥𝑦 so 𝑦 =𝑡

𝑥=

𝑡

𝑠, so we can calculate the Jacobian

as 𝐽 = 𝑑𝑒𝑡 [

𝜕𝑥

𝜕𝑠

𝜕𝑥

𝜕𝑡𝜕𝑦

𝜕𝑠

𝜕𝑦

𝜕𝑡

] = 𝑑𝑒𝑡 [1 0

−𝑡

𝑠

1

𝑠

] =1

𝑠. The joint probability density function is thus

𝑓𝑆𝑇(𝑠, 𝑡) = 𝑓𝑋𝑌 (𝑠,𝑡

𝑠) |

1

𝑠| = 2 (𝑠 +

𝑡

𝑠) (

1

𝑠) = 2 (1 +

𝑡

𝑠2). The region is then found by

substituting to 0 < 𝑥 < 𝑦 < 1, so we have 0 < 𝑠 <𝑡

𝑠< 1 or 0 < 𝑠2 < 𝑡 < 𝑠 < 1. This

can be visualized as the region between 𝑡 = 𝑠 and 𝑡 = 𝑠2 on the 𝑠𝑡 plane.

b) The marginal probability density function is given by 𝑓𝑇(𝑡) = ∫ 𝑓𝑆𝑇(𝑠, 𝑡)𝑑𝑠√𝑡

𝑡=

∫ 2 (1 +𝑡

𝑠2) 𝑑𝑠

√𝑡

𝑡= [2𝑠 −

2𝑡

𝑠]

𝑡

√𝑡

= (2√𝑡 − 2√𝑡) − (2𝑡 − 2) = 2 − 2𝑡 if 0 < 𝑡 < 1.

Question #25: Let 𝑋1, 𝑋2, 𝑋3,𝑋4 be independent random variables. Assume that 𝑋2, 𝑋3, 𝑋4 are

each distributed Poisson with parameter 5 and the random variable 𝑌 = 𝑋1 + 𝑋2 + 𝑋3 + 𝑋4

is distributed Poisson with parameter 25. a) What is the distribution of the random variable

𝑋1? b) What is the distribution of the random variable 𝑊 = 𝑋1 + 𝑋2?

a) We first note that while 𝑋1,𝑋2, 𝑋3, 𝑋4 are independent, they are not 𝑖𝑖𝑑 since only

𝑋2, 𝑋3, 𝑋4~𝑃𝑂𝐼(5) with 𝑋1 not being listed. Thus, we must use the unsimplified

formula 6.4.4, which says if 𝑋1, … , 𝑋𝑛 are independent random variables with moment

generating functions 𝑀𝑋𝑖(𝑡) and 𝑌 = 𝑋1 + ⋯ + 𝑋𝑛, then the moment generating

function of 𝑌 is 𝑀𝑌(𝑡) = [𝑀𝑋1(𝑡)] … [𝑀𝑋𝑛

(𝑡)]. We use the fact that if some 𝑋~𝑃𝑂𝐼(𝜇),

then 𝑀𝑋(𝑡) = 𝑒𝜇(𝑒𝑡−1) to solve this problem. If we let 𝑌 = 𝑋1 + 𝑋2 + 𝑋3 + 𝑋4, we have

that 𝑀𝑌(𝑡) = [𝑀𝑋1(𝑡)][𝑀𝑋2

(𝑡)][𝑀𝑋3(𝑡)][𝑀𝑋4

(𝑡)] = 𝑒25(𝑒𝑡−1) since 𝑌~𝑃𝑂𝐼(25).

Substituting gives [𝑀𝑋1(𝑡)][𝑒5(𝑒𝑡−1)][𝑒5(𝑒𝑡−1)][𝑒5(𝑒𝑡−1)] = 𝑒25(𝑒𝑡−1), which reduces to

[𝑀𝑋1(𝑡)][𝑒5(𝑒𝑡−1)]

3= 𝑒5(𝑒𝑡−1) so that [𝑀𝑋1

(𝑡)] =𝑒25(𝑒𝑡−1)

[𝑒5(𝑒𝑡−1)]3 =

𝑒25(𝑒𝑡−1)

𝑒15(𝑒𝑡−1)= 𝑒10(𝑒𝑡−1). This

is the moment generating function of a poisson 10 random variable, so 𝑋1~𝑃𝑂𝐼(10).

b) We have 𝑀𝑊(𝑡) = [𝑀𝑋1(𝑡)][𝑀𝑋2

(𝑡)] = [𝑒10(𝑒𝑡−1)][𝑒5(𝑒𝑡−1)] = 𝑒15(𝑒𝑡−1), which is the

moment generating function of a poisson 15 random variable, so 𝑊~𝑃𝑂𝐼(15). We can

see a general pattern here; if 𝑋𝑖~𝑃𝑂𝐼(𝜇) for 𝑖 = 1, … , 𝑛 are independent random

variables and we define some 𝑍 = ∑ 𝑋𝑖𝑘𝑖=1 for 𝑘 ≤ 𝑛, then we have that 𝑍~𝑃𝑂𝐼(𝑘𝜇).

Chapter #6 – Functions of Random Variables

Question #17: Suppose that 𝑋1 and 𝑋2 denote a random sample of size 2 from a gamma

distribution such that 𝑋𝑖~𝐺𝐴𝑀 (2,1

2). Find the PDF of a) 𝑌 = √𝑋1 + 𝑋2 and b) 𝑊 =

𝑋1

𝑋2.

a) We know that if some 𝑋~𝐺𝐴𝑀(𝜃, 𝜅), then 𝑓𝑋(𝑥) =1

𝜃𝜅Γ(𝜅)𝑥𝜅−1𝑒−

𝑥

𝜃 if 𝑥 > 0. Since 𝑋1

and 𝑋2 are independent, their joint density function is given by 𝑓𝑋1𝑋2(𝑥1, 𝑥2) =

𝑓𝑋1(𝑥1)𝑓𝑋2(𝑥2) = (1

√2Γ(1

2)𝑥1−1

2𝑒−𝑥12 )(

1

√2Γ(1

2)𝑥2−1

2𝑒−𝑥22 ) =

1

2𝜋

1

√𝑥1

1

√𝑥2𝑒−

(𝑥1+𝑥2)

2 if 𝑥1, 𝑥2 > 0,

since Γ (1

2) = √𝜋. We have the transformation 𝑦 = √𝑥1 + 𝑥2 and generate another

transformation 𝑤 = 𝑥1, so we have 𝑥1 = 𝑤 and 𝑥2 = 𝑦2 − 𝑤, which allows us to find

𝐽 = 𝑑𝑒𝑡 [1 0−1 2𝑦

] = 2𝑦. Then the joint density of 𝑊 and 𝑌 is given by 𝑓𝑊𝑌(𝑤, 𝑦) =

𝑓𝑋1𝑋2(𝑤, 𝑦2 − 𝑤)|𝐽| =

1

𝜋

1

√𝑤

1

√𝑦2−𝑤𝑒−

𝑦2

2 𝑦 if 𝑤 > 0, 𝑦2 − 𝑤 > 0; these bounds can be

combined to give 0 < 𝑤 < 𝑦2. Finally, the density of 𝑌 is 𝑓𝑌(𝑦) = ∫ 𝑓𝑊𝑌(𝑤, 𝑦)∞

−∞𝑑𝑤 =

1

𝜋∫ 𝑦

1

√𝑤

1

√𝑦2−𝑤𝑒−

𝑦2

2𝑦2

0𝑑𝑤 = ⋯ = 𝑦𝑒−

𝑦2

2 if 𝑦 > 0 and zero otherwise. The evaluation of

this integral has been omitted, but can be computed by two substitutions.

b) We have 𝑤 =𝑥1

𝑥2 and generate 𝑧 = 𝑥1 so that 𝑥1 = 𝑧 and 𝑥2 =

𝑧

𝑤, which allows us to

calculate 𝐽 = det [1 01

𝑤−

𝑧

𝑤2

] = −𝑧

𝑤2. Then we have 𝑓𝑍𝑊(𝑧, 𝑤) = 𝑓𝑋1𝑋2 (𝑧,

𝑧

𝑤) |𝐽| =

1

2𝜋

1

√𝑧

1

√𝑧/𝑤𝑒−

𝑧+𝑧/𝑤

2𝑧

𝑤2=

1

2𝜋

1

𝑤3/2𝑒−

(𝑧+𝑧/𝑤)

2 if 𝑧, 𝑤 > 0, so that the density of 𝑊 is given by

𝑓𝑊(𝑤) = ∫ 𝑓𝑊𝑍(𝑤, 𝑧)∞

−∞𝑑𝑧 =

1

2𝜋∫

1

𝑤3/2 𝑒−(𝑧+𝑧/𝑤)

2∞

0𝑑𝑧 = ⋯ =

1

𝜋

1

(𝑤+1)√𝑤 if 𝑤 > 0. The

evaluation of this integral has been omitted, but can be computed by substitution.

Question #26: Let 𝑋1 and 𝑋2 be independent negative binomial random variables such that

𝑋1~𝑁𝐵(𝑟1, 𝑝) and 𝑋2~𝑁𝐵(𝑟2, 𝑝). a) Find the MGF and distribution of 𝑌 = 𝑋1 + 𝑋2.

We use Theorem 6.4.3, which says that if the random variables 𝑋𝑖 are independent

with respective MGFs 𝑀𝑋𝑖(𝑡), then the MGF of the random variable that is their sum

is simply the product of their respective MGFs. Also, if some discrete random variable

𝑋~𝑁𝐵(𝑟, 𝑝), then 𝑀𝑋(𝑡) = (𝑝𝑒𝑡

1−𝑞𝑒𝑡)𝑟

. Therefore, the moment generating function of 𝑌

is 𝑀𝑌(𝑡) = [𝑀𝑋1(𝑡)][𝑀𝑋2(𝑡)] = (𝑝𝑒𝑡

1−𝑞𝑒𝑡)𝑟1(

𝑝𝑒𝑡

1−𝑞𝑒𝑡)𝑟2= (

𝑝𝑒𝑡

1−𝑞𝑒𝑡)𝑟1+𝑟2

. This then allows to

determine the distribution of 𝑌 = 𝑋1 + 𝑋2, namely that 𝑌~𝑁𝐵(𝑟1 + 𝑟2, 𝑝).

Question #27: Recall that 𝑌~𝐿𝑂𝐺𝑁(𝜇, 𝜎2) if 𝑙𝑛(𝑌)~𝑁(𝜇, 𝜎2). Assume that 𝑌𝑖~𝐿𝑂𝐺𝑁(𝜇𝑖, 𝜎𝑖2)

for 𝑖 = 1,… , 𝑛 are independent. Find the distribution functions of the following random

variables a) 𝐴 = ∏ 𝑌𝑖𝑛𝑖=1 , b) 𝐵 = ∏ 𝑌𝑖

𝑛𝑖=1

𝑎, c) 𝐶 =

𝑌1

𝑌2, d) find 𝐸(𝐴) = 𝐸(∏ 𝑌𝑖

𝑛𝑖=1 ).

a) We have that 𝑙𝑛(𝐴) = 𝑙𝑛(∏ 𝑌𝑖𝑛𝑖=1 ) = ∑ 𝑙𝑛(𝑌𝑖)

𝑛𝑖=1 = 𝑙𝑛(𝑌1) + ⋯+ 𝑙𝑛(𝑌𝑛), so the

random variable 𝑙𝑛(𝐴) is the sum of 𝑛 normally distributed random variables. This

implies that 𝑙𝑛(𝐴)~𝑁(∑ 𝜇𝑖𝑛𝑖=1 , ∑ 𝜎𝑖

2𝑛𝑖=1 ), which means 𝐴~𝐿𝑂𝐺𝑁(∑ 𝜇𝑖

𝑛𝑖=1 , ∑ 𝜎𝑖

2𝑛𝑖=1 ).

b) The random variable 𝑙𝑛(𝐵) = 𝑙𝑛(∏ 𝑌𝑖𝑎𝑛

𝑖=1 ) = ∑ 𝑙𝑛(𝑌𝑖𝑎)𝑛

𝑖=1 = ∑ 𝑎𝑖𝑙𝑛(𝑌𝑖)𝑛𝑖=1 =

𝑎1 𝑙𝑛(𝑌1) + ⋯+ 𝑎2𝑙𝑛(𝑌𝑛). We use that if some 𝑋~𝑁(𝜇, 𝜎2), then 𝑎𝑋~𝑁(𝑎𝜇, 𝑎2𝜎2) to

conclude that 𝑙𝑛(𝐵)~𝑁(∑ 𝑎𝑖𝜇𝑖𝑛𝑖=1 , ∑ 𝑎𝑖

2𝜎𝑖2𝑛

𝑖=1 ) so 𝐵~𝐿𝑂𝐺𝑁(∑ 𝑎𝑖𝜇𝑖𝑛𝑖=1 , ∑ 𝑎𝑖

2𝜎𝑖2𝑛

𝑖=1 ).

c) We have that 𝑙𝑛(𝐶) = 𝑙𝑛 (𝑌1

𝑌2) = 𝑙𝑛(𝑌1) − 𝑙𝑛(𝑌2), so the random variable 𝑙𝑛(𝐶) is the

sum of two normally distributed random variables. Thus, 𝑙𝑛(𝐶)~𝑁(𝜇1 − 𝜇2, 𝜎12 + 𝜎2

2)

which implies that the distribution of 𝐶 =𝑌1

𝑌2 is 𝐶~𝐿𝑂𝐺𝑁(𝜇1 − 𝜇2, 𝜎1

2 + 𝜎22).

d) For 𝑋~𝑁(𝜇, 𝜎2), we have 𝑀𝑋(𝑡) = 𝐸(𝑒𝑡𝑋) = 𝑒𝜇𝑡+𝜎2𝑡2/2 and for 𝑍~𝑁(0,1), we have

𝑀𝑍(𝑡) = 𝐸(𝑒𝑡𝑍) = 𝑒𝑡2/2. Thus, the expected value is given by 𝐸(𝑌𝑖) = 𝐸(𝑒𝜇𝑖+𝜎𝑖𝑍) =

𝑒𝜇𝑖+𝜎𝑖2/2. Since the random variables 𝑌𝑖 are all independent, we therefore have that

𝐸(𝐴) = 𝐸(∏ 𝑌𝑖𝑛𝑖=1 ) = ∏ 𝐸(𝑌𝑖)

𝑛𝑖=1 = ∏ (𝑒𝜇𝑖+𝜎𝑖

2/2)𝑛𝑖=1 = 𝑒𝑥𝑝{∑ 𝜇𝑖

𝑛𝑖=1 +∑ 𝜎𝑖

2/2𝑛𝑖=1 }.

Question #28: Let 𝑋1 and 𝑋2 be a random sample of size 2 from a continuous distribution

with PDF of the form 𝑓𝑋(𝑥) = 2𝑥 if 0 < 𝑥 < 1 and zero otherwise. a) Find the marginal

densities of 𝑌1 and 𝑌2, the smallest and largest order statistics, b) find the joint probability

density function of 𝑌1 and 𝑌2, and c) find the density of the sample range 𝑅 = 𝑌2 − 𝑌1.

a) Since 𝑓𝑋(𝑥) = {2𝑥𝑖𝑓𝑥 ∈ (0,1)0𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

, we know that 𝐹𝑋(𝑥) = {

0𝑖𝑓𝑥 ∈ (−∞, 0]

𝑥2𝑖𝑓𝑥 ∈ (0,1)

1𝑖𝑓𝑥 ∈ [1,∞)

. Then

from Theorem 6.5.2, we have that 𝑔1(𝑦1) = 𝑛𝑓𝑋(𝑦1)[1 − 𝐹𝑋(𝑦1)]𝑛−1 so we can

calculate the smallest order statistic as 𝑔1(𝑦1) = 2[2𝑦1][1 − 𝑦12]2−1 = 4𝑦1 − 4𝑦1

3

whenever 𝑦1 ∈ (0,1). Similarly, 𝑔𝑛(𝑦𝑛) = 𝑛𝑓𝑋(𝑦𝑛)[𝐹𝑋(𝑦𝑛)]𝑛−1 so we can calculate the

largest order statistic as 𝑔2(𝑦2) = 2[2𝑦2][𝑦22]2−1 = 4𝑦2

3 whenever 𝑦2 ∈ (0,1).

b) From Theorem 6.5.1, the joint probability density function of the order statistics is

𝑔𝒀(𝑦1, … , 𝑦𝑛) = 𝑛! 𝑓𝑋(𝑦1)…𝑓𝑋(𝑦𝑛). In this question, we have that 𝑔𝑌1𝑌2(𝑦1, 𝑦2) =

2! 𝑓𝑋(𝑦1)𝑓𝑋(𝑦2) = 2(2𝑦1)(2𝑦2) = 8𝑦1𝑦2 whenever we have 0 < 𝑦1 < 𝑦2 < 1.

c) We first find the joint density of the smallest and largest order statistics in order to

make a transformation to get the marginal density of the sample range. From the

work we did above, we have that 𝑔𝑌1𝑌2(𝑦1, 𝑦2) = 8𝑦1𝑦2. We have the transformation

𝑟 = 𝑦2 − 𝑦1 and generate 𝑠 = 𝑦1, so we have 𝑦1 = 𝑠 and 𝑦2 = 𝑟 + 𝑠 which allows us to

calculate 𝐽 = 𝑑𝑒𝑡 [1 01 1

] = 1. The joint density of 𝑆 and 𝑅 is therefore 𝑓𝑆𝑅(𝑠, 𝑟) =

𝑔𝑌1𝑌2(𝑠, 𝑟 + 𝑠)|𝐽| = 8𝑠(𝑟 + 𝑠)(1) = 8𝑠2 + 8𝑠𝑟 if 0 < 𝑠 < 𝑟 + 𝑠 < 1, which can also be

written as 0 < 𝑠 < 1 − 𝑟. The marginal density is thus 𝑓𝑅(𝑟) = ∫ 𝑓𝑆𝑅(𝑠, 𝑟)∞

−∞𝑑𝑠 =

∫ (8𝑠2 + 8𝑠𝑟)1−𝑟

0𝑑𝑠 = [

8

3𝑠3 + 4𝑟𝑠2]

0

1−𝑟

=8

3(1 − 𝑟)3 + 4𝑟(1 − 𝑟)2 if 0 < 𝑟 < 1.

Question #31: Consider a random sample of size 𝑛 from an exponential distribution such

that 𝑋𝑖~𝐸𝑋𝑃(1). Give the density of a) the smallest order statistic denoted by 𝑌1, b) the

largest order statistic denoted by 𝑌𝑛, c) the sample range of the order statistics 𝑅 = 𝑌𝑛 − 𝑌1.

a) Since 𝑓𝑋𝑖(𝑥𝑖) = {𝑒−𝑥𝑖 𝑖𝑓𝑥 ∈ (0,∞)0𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

, we have 𝐹𝑋𝑖(𝑥𝑖) = {0𝑖𝑓𝑥 ∈ (−∞, 0]

1 − 𝑒−𝑥𝑖 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.

Then we have that 𝑔1(𝑦1) = 𝑛𝑓𝑋(𝑦1)[1 − 𝐹𝑋(𝑦1)]𝑛−1 = 𝑛𝑒−𝑦1[1 − (1 − 𝑒−𝑦1)]𝑛−1 =

𝑛𝑒−𝑦1(𝑒−𝑦1)𝑛−1 = 𝑛𝑒−𝑛𝑦1 if 𝑦1 > 0 and zero otherwise.

b) Similarly, we have 𝑔𝑛(𝑦𝑛) = 𝑛𝑓𝑋(𝑦𝑛)[𝐹𝑋(𝑦𝑛)]𝑛−1 = 𝑛𝑒−𝑦𝑛[1 − 𝑒−𝑦𝑛]𝑛−1 for 𝑦𝑛 > 0.

c) Since the exponential distribution has the memoryless property, the difference of

𝑅 = 𝑌𝑛 − 𝑌1 will not be conditional on the value of 𝑌1. This allows us to treat 𝑌1 = 0,

so that 𝑅 = 𝑌𝑛 − 𝑌1 = 𝑌𝑛 − 0 = 𝑌𝑛. We then use the fact that the range of a set of 𝑛

order statistics from an exponential distribution is the same as the largest order

statistic from a set of 𝑛 − 1 order statistics. From above, we have that 𝑔𝑛(𝑦𝑛) =

𝑛𝑒−𝑦𝑛[1 − 𝑒−𝑦𝑛]𝑛−1, so substituting 𝑛 − 1 gives 𝑔𝑅(𝑟) = (𝑟 − 1)[1 − 𝑒−𝑟]𝑟−2𝑒−𝑟.

Question #32: A system is composed of five independent components connected in series

one after the other. a) If the PDF of the time to failure of each component is 𝑋𝑖~𝐸𝑋𝑃(1), then

give the PDF of the time to failure of the system; b) if the components are connected in

parallel so that all must fail before the system fails, give the PDF of the time to failure.

a) Since 𝑋𝑖~𝐸𝑋𝑃(1), we know that the density is given by 𝑓𝑋𝑖(𝑥𝑖) = 𝑒−𝑥𝑖 for 𝑥𝑖 > 0 and

𝐹𝑋𝑖(𝑥𝑖) = {0𝑖𝑓𝑥 ∈ (−∞, 0]

1 − 𝑒−𝑥𝑖 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒. The system in the series fails whenever the

earliest component fails, which happens at time 𝑋(1) = 𝑌1, the first order statistic.

Thus, the probability density function of the time to failure is therefore given by

𝑓𝑌1(𝑦1) = 𝑛𝑓𝑋(𝑦1)[1 − 𝐹𝑋(𝑦1)]𝑛−1 = 5𝑒−𝑦1[𝑒−𝑦1]4 = 5𝑒−5𝑦1 whenever 𝑦1 > 0.

b) For the system in parallel, the system in the series fails whenever the last component

fails, which happens at time 𝑋(5) = 𝑌5, the greatest order statistic. Thus, the density is

𝑓𝑌5(𝑦5) = 𝑛𝑓𝑋(𝑦𝑛)[𝐹𝑋(𝑦𝑛)]𝑛−1 = 5𝑒−𝑦5[1 − 𝑒−𝑦5]4 whenever 𝑦5 > 0.

Question #33: Consider a random sample of size 𝑛 from a geometric distribution such that

𝑋𝑖~𝐺𝐸𝑂(𝑝). Give the CDF of a) the minimum 𝑌1, b) the 𝑘𝑡ℎ smallest 𝑌𝑘, c) the maximum 𝑌𝑛.

a) If some 𝑋𝑖~𝐺𝐸𝑂(𝑝), then 𝑓𝑋𝑖(𝑥) = 𝑝(1 − 𝑝)𝑥 and 𝐹𝑋𝑖(𝑥) = 1 − (1 − 𝑝)𝑥+1. Consider

the event 𝑋(1) > 𝑚, which happens if and only if 𝑋(𝑖) > 𝑚 for all 𝑖 = 1,… , 𝑛. Therefore,

we have 𝑃(𝑋(1) > 𝑚) = 𝑃(𝑋(𝑖) > 𝑚)𝑛= [1 − 𝐹𝑋𝑖(𝑚)]

𝑛= [(1 − 𝑝)𝑚]𝑛 = (1 − 𝑝)𝑚𝑛,

which implies that the CDF is given by 𝐹𝑋(1)(𝑚) = 𝑃(𝑋(1) ≤ 𝑚) = 1 − (1 − 𝑝)𝑚𝑛.

b) The event 𝑋(𝑘) ≤ 𝑚 happens when 𝑘 of the 𝑋(𝑖) satisfy 𝑋(𝑖) ≤ 𝑚 and the other 𝑛 − 𝑘

satisfy 𝑋(𝑖) > 𝑚. Thus, we have 𝑃(𝑋(𝑘) ≤ 𝑚) = (𝑛𝑘)𝑃(𝑋(𝑖) ≤ 𝑚)

𝑛𝑃(𝑋(𝑖) > 𝑚)

𝑛−𝑘=

(𝑛𝑘)[1 − (1 − 𝑝)𝑚]𝑘[(1 − 𝑝)𝑚]𝑛−𝑘, which is the distribution function of 𝑌𝑘.

c) The event 𝑋(𝑛) ≤ 𝑚 happens when 𝑋(𝑖) ≤ 𝑚 for all 𝑖 = 1,… , 𝑛. Thus, 𝑃(𝑋(𝑛) ≤ 𝑚) =

𝑃(𝑋(𝑖) ≤ 𝑚)𝑛= (∑ [𝑝(1 − 𝑝)𝑗−1]𝑛

𝑗=1 )𝑛= (𝑝

1−(1−𝑝)𝑚

1−(1−𝑝))𝑛

= [1 − (1 − 𝑝)𝑚]𝑛.

Chapter #7 – Limiting Distributions

Question #30: If 𝑋~𝑃𝐴𝑅(𝜃, 𝜅), then 𝑓𝑋(𝑥) =𝜅

𝜃(1+𝑥

𝜃)

𝜅+1 if 𝑥 > 0 and zero otherwise. Consider

a random sample of size 𝑛 = 5 from a Pareto distribution where 𝑋~𝑃𝐴𝑅(1,2); that is,

suppose that 𝑋1, … , 𝑋5 are drawn from the given Pareto distribution above. a) Find the joint

PDF of the second and fourth order statistics given by 𝑌2 = 𝑋(2) and 𝑌4 = 𝑋(4), and b) find the

joint PDF of the first three order statistics given by 𝑌1 = 𝑋(2), 𝑌2 = 𝑋(2) and 𝑌3 = 𝑋(2).

a) The CDF of the population is given by 𝐹𝑋(𝑥) = ∫ 𝑓𝑋(𝑡)𝑥

0𝑑𝑡 = ∫ [

2

(1+𝑡)3]𝑥

0𝑑𝑡, so that we

can calculate the joint density using Corollary 6.5.1 as 𝑓𝑌2𝑌4(𝑦2, 𝑦4) =

5!

(2−1)!(4−2−1)!(5−4)![𝐹𝑋(𝑦2)]2−1𝑓𝑋(𝑦2)[𝐹𝑋(𝑦4) − 𝐹𝑋(𝑦2)]4−2−1𝑓𝑋(𝑦4)[1 − 𝐹𝑋(𝑦4)]5−4 =

5! 𝑓𝑋(𝑦2)𝑓𝑋(𝑦4)[𝐹𝑋(𝑦2)][𝐹𝑋(𝑦4) − 𝐹𝑋(𝑦2)][1 − 𝐹𝑋(𝑦4)] if 0 < 𝑦2 < 𝑦4 < ∞.

b) From Theorem 6.5.4, we have 𝑔(𝑦1, … , 𝑦𝑟) =𝑛!

(𝑛−𝑟)![1 − 𝐹𝑋(𝑦𝑟)]𝑛−𝑟[𝑓𝑋(𝑦1) … 𝑓𝑋(𝑦𝑟)],

so we may calculate that 𝑓𝑌1𝑌2𝑌3(𝑦1, 𝑦2, 𝑦3) = 60[1 − 𝐹𝑋(𝑦3)]2[𝑓𝑋(𝑦1)𝑓𝑋(𝑦2)𝑓𝑋(𝑦3)]

whenever 0 < 𝑦1 < 𝑦2 < 𝑦3 < ∞, since we have that 𝑛!

(𝑛−𝑟)!=

5!

2!= 60.

Question #1: Consider a random sample of size 𝑛 from a distribution with cumulative

distribution function 𝐹𝑋(𝑥) = 1 −1

𝑥 whenever 1 ≤ 𝑥 < ∞ and zero otherwise. That is, let the

random variables 𝑋1, … , 𝑋𝑛 be ~𝑖𝑖𝑑 from the distribution with CDF 𝐹𝑋(𝑥). a) Derive the CDF

of the smallest order statistic given by 𝑋(1) = 𝑋1:𝑛, b) find the limiting distribution of 𝑋1:𝑛;

that is, if 𝐺𝑛(𝑦) denotes the order statistic from above, find lim𝑛→∞

𝐺𝑛(𝑦), c) find the limiting

distribution of 𝑋1:𝑛𝑛 ; that is, find the CDF of 𝑋(1)

𝑛 and its limit as 𝑛 → ∞.

a) We can compute that 𝐹𝑋1:𝑛(𝑦) = 𝑃(𝑋1:𝑛 ≤ 𝑦) = 1 − 𝑃(𝑋1:𝑛 > 𝑦) = 1 − 𝑃(𝑋𝑖 > 𝑦)𝑛 =

1 − [1 − (1 −1

𝑦)]

𝑛

= 1 − (1

𝑦)

𝑛

whenever 1 ≤ 𝑦 < ∞. We thus have that the CDF of

the smallest order statistic is 𝐹𝑋1:𝑛(𝑦) = {

1 − (1/𝑦)𝑛 𝑖𝑓 𝑦 ≥ 1 0 𝑖𝑓 𝑦 < 1

. Finally, we note

that 𝑃(𝑋1:𝑛 > 𝑦) ≡ 𝑃(𝑋𝑖 > 𝑦)𝑛 since the smallest order statistic is greater than some

𝑦 if and only if all 𝑛 of the independent samples are also greater than 𝑦. We can use

this approach for any order statistic, including the largest, by changing the exponent.

b) We have that lim𝑛→∞

𝐺𝑛(𝑦) = {lim

𝑛→∞[1 − (1/𝑦)𝑛] 𝑖𝑓 𝑦 ≥ 1

lim𝑛→∞

0 𝑖𝑓 𝑦 < 1 = {

0 𝑖𝑓 𝑦 ≤ 1 1 𝑖𝑓 𝑦 > 1

= 𝐺(𝑦),

so the limiting distribution of 𝑋1:𝑛 is degenerate at 𝑦 = 1. From Definition 7.2.2, this

means that 𝐺(𝑦) is the cumulative distribution function of some discrete distribution

𝑔(𝑦) that assigns probability one at 𝑦 = 1 and zero otherwise.

c) As before, we have 𝐹𝑋1:𝑛𝑛 (𝑦) = 𝑃(𝑋1:𝑛

𝑛 ≤ 𝑦) = 𝑃 (𝑋1:𝑛 ≤ 𝑦1

𝑛) = 1 − 𝑃 (𝑋1:𝑛 > 𝑦1

𝑛) =

1 − 𝑃 (𝑋𝑖 > 𝑦1

𝑛)𝑛

= 1 − [1 − (1 −1

𝑦1𝑛

)]

𝑛

= 1 − (1

𝑦) whenever 𝑦 ≥ 1. Therefore, it is

clear that the limiting distribution of this sequence of random variables is given by

lim𝑛→∞

𝐹𝑋1:𝑛𝑛 (𝑦) = {

1 − 1/𝑦 𝑖𝑓 𝑦 ≥ 1 0 𝑖𝑓 𝑦 < 1

= 𝐺(𝑦) since there is no dependence on 𝑛.

Question #2: Consider a random sample of size 𝑛 from a distribution with CDF given by

𝐹(𝑥) =1

1+𝑒−𝑥 for all 𝑥 ∈ ℝ. Find the limiting distribution of a) 𝑋𝑛:𝑛 and b) 𝑋𝑛:𝑛 − 𝑙𝑛 (𝑛).

a) We have 𝐹𝑋𝑛:𝑛(𝑦) = 𝑃(𝑋𝑛:𝑛 ≤ 𝑦) = 𝑃(𝑋𝑖 ≤ 𝑦)𝑛 = (

1

1+𝑒−𝑦)𝑛

for all 𝑦 ∈ ℝ. Since

lim𝑛→∞

[(1

1+𝑒−𝑦)𝑛

] = 0, we conclude that 𝑋𝑛:𝑛 does not have a limiting distribution.

b) We calculate that 𝐹𝑋𝑛:𝑛−𝑙𝑛 (𝑛)(𝑦) = 𝑃(𝑋𝑛:𝑛 − 𝑙𝑛 (𝑛) ≤ 𝑦) = 𝑃(𝑋𝑛:𝑛 ≤ 𝑦 + 𝑙 𝑛(𝑛)) =

𝐹𝑋𝑛:𝑛(𝑦 + 𝑙 𝑛(𝑛)) = (

1

1+𝑒−(𝑦+𝑙 𝑛(𝑛)))𝑛

= (1

1+𝑒−𝑦𝑒−𝑙𝑛 (𝑛))

𝑛

= (1

1+𝑒−𝑦

𝑛

)

𝑛

. Evaluating this

limit gives lim𝑛→∞

[𝐹𝑋𝑛:𝑛(𝑦 + 𝑙 𝑛(𝑛))] = lim

𝑛→∞[(

1

1+𝑒−𝑦

𝑛

)

𝑛

] = 𝑒𝑒−𝑦 for all 𝑦 ∈ ℝ.

Question #3: Consider a random sample of size 𝑛 from the distribution 𝐹(𝑥) = 1 − 𝑥−2 if

𝑥 > 1 and zero otherwise. Find the limiting distribution of a) 𝑋1:𝑛, b) 𝑋𝑛:𝑛 and c) 1

√𝑛𝑋𝑛:𝑛.

a) We can compute that 𝐹𝑋1:𝑛(𝑦) = 𝑃(𝑋1:𝑛 ≤ 𝑦) = 1 − 𝑃(𝑋1:𝑛 > 𝑦) = 1 − 𝑃(𝑋𝑖 > 𝑦)𝑛 =

1 − [1 − (1 −1

𝑦2)]𝑛

= 1 − (1

𝑦2𝑛) if 𝑦 > 1. Thus, 𝐹𝑋1:𝑛(𝑦) = {

1 −1

𝑦2𝑛 𝑖𝑓 𝑦 > 1

0 𝑖𝑓 𝑦 ≤ 1 so

the limiting distribution is lim𝑛→∞

𝐹𝑋1:𝑛(𝑦) = {

lim𝑛→∞

[1 −1

𝑦2𝑛] 𝑖𝑓 𝑦 > 1

lim𝑛→∞

[0] 𝑖𝑓 𝑦 ≤ 1 = {

1 𝑖𝑓 𝑦 > 10 𝑖𝑓 𝑦 ≤ 1

.

We therefore say that the limiting distribution is degenerate at 𝑦 = 1.

b) We have 𝐹𝑋𝑛:𝑛(𝑦) = 𝑃(𝑋𝑛:𝑛 ≤ 𝑦) = 𝑃(𝑋𝑖 ≤ 𝑦)𝑛 = (1 −

1

𝑦2)𝑛

whenever 𝑦 > 1. Thus,

𝐹𝑋𝑛:𝑛(𝑦) = {

(1 −1

𝑦2)𝑛

𝑖𝑓 𝑦 > 1

0 𝑖𝑓 𝑦 ≤ 1 so lim

𝑛→∞𝐹𝑋𝑛:𝑛

(𝑦) = {lim

𝑛→∞[(1 −

1

𝑦2)𝑛

] 𝑖𝑓 𝑦 > 1

lim𝑛→∞

[0] 𝑖𝑓 𝑦 ≤ 1 = 0.

We would therefore conclude that there is no limiting distribution for 𝑋𝑛:𝑛.

c) We compute that 𝐹 1

√𝑛𝑋𝑛:𝑛

(𝑦) = 𝑃 (1

√𝑛𝑋𝑛:𝑛 ≤ 𝑦) = 𝑃(𝑋𝑛:𝑛 ≤ √𝑛𝑦) = 𝐹𝑋𝑛:𝑛

(√𝑛𝑦) =

(1 −1

(√𝑛𝑦)2)𝑛

= (1 −1

𝑛𝑦2)𝑛

whenever √𝑛𝑦 > 1 or 𝑦 >1

√𝑛. We can therefore compute

the limit as lim𝑛→∞

𝐹 1

√𝑛𝑋𝑛:𝑛

(𝑦) = {lim

𝑛→∞[(1 −

1

𝑛𝑦2)

𝑛

] 𝑖𝑓 𝑦 >1

√𝑛

lim𝑛→∞

[0] 𝑖𝑓 𝑦 ≤1

√𝑛

= {𝑒−

1

𝑦2 𝑖𝑓 𝑦 > 0 0 𝑖𝑓 𝑦 ≤ 0

.

Question #5: Suppose that 𝑍𝑖~𝑁(0,1) and that the 𝑍𝑖 are all independent. Use moment

generating functions to find the limiting distribution of 𝐴𝑛 =∑ (𝑍𝑖+

1

𝑛)𝑛

𝑖=1

√𝑛 as 𝑛 → ∞.

We have 𝐴𝑛 =∑ (𝑍𝑖+

1

𝑛)𝑛

𝑖=1

√𝑛=

(∑ 𝑍𝑖𝑛𝑖=1 )+(∑ (

1

𝑛)𝑛

𝑖=1 )

√𝑛=

(∑ 𝑍𝑖𝑛𝑖=1 )+𝑛

1

𝑛

√𝑛=

∑ 𝑍𝑖𝑛𝑖=1

√𝑛+

1

√𝑛, so the MGF is

𝑀𝐴𝑛(𝑡) = [𝑀1(𝑡)]𝑛[𝑀2(𝑡)] = [𝑀1(𝑡)]𝑛 [𝐸 (𝑒

1

√𝑛𝑡)] since 𝐴𝑛 is the sum of two parts so

we can multiply their respective MGFs. The MGF of a standard normal random

variable with 𝜇 = 0 and 𝜎2 = 1 is given by 𝑀𝑍(𝑡) = 𝑒𝑡2/2, which allows us to calculate

that 𝑀1(𝑡) = 𝑒(

𝑡

√𝑛)

2/2

= 𝑒𝑡2

2𝑛. Also, we have that 𝐸 (𝑒1

√𝑛𝑡) = 𝑒

𝑡

√𝑛 so combining these

gives 𝑀𝐴𝑛(𝑡) = [𝑀1(𝑡)]𝑛 [𝐸 (𝑒

1

√𝑛𝑡)] = [𝑒

𝑡2

2𝑛]𝑛

[𝑒𝑡

√𝑛] = [𝑒𝑡2

2 ] [𝑒𝑡

√𝑛]. Then we can use

Theorem 7.3.1 to calculate lim𝑛→∞

𝑀𝐴𝑛(𝑡) = lim

𝑛→∞[𝑒

𝑡2

2 ] [𝑒𝑡

√𝑛] = 𝑒𝑡2

2 = 𝑀(𝑡), which we

know is the MGF of a standard normal, so the limiting distribution is 𝐴~𝑁(0,1). Note

that this is also a direct consequence of the Central Limit Theorem.

Question #9: Let 𝑋1, 𝑋2, … , 𝑋100 be a random sample of size 𝑛 = 100 from an exponential

distribution such that each 𝑋𝑖~𝐸𝑋𝑃(1) and let 𝑌 = 𝑋1 + 𝑋2 + ⋯ + 𝑋100. a) Give a normal

approximation for the probability 𝑃(𝑌 > 110), and b) if �̅� =𝑌

100 is the sample mean, then

give a normal approximation to the probability 𝑃(1.1 < �̅� < 1.2).

a) Since each 𝑋𝑖~𝐸𝑋𝑃(1), we know that 𝐸(𝑋𝑖) = 𝜃 = 1 while 𝑉𝑎𝑟(𝑋𝑖) = 𝜃2 = 1. Due to

the independence of the 𝑋𝑖s, we have that 𝐸(𝑌) = 𝐸(∑ 𝑋𝑖100𝑖=1 ) = ∑ 𝐸(𝑋𝑖)

100𝑖=1 = 100 and

𝑉𝑎𝑟(𝑌) = 𝑉𝑎𝑟(∑ 𝑋𝑖100𝑖=1 ) = ∑ 𝑉𝑎𝑟(𝑋𝑖)

100𝑖=1 = 100 so 𝑠𝑑(𝑌) = √100 = 10. We can

therefore calculate that 𝑃(𝑌 > 110) = 1 − 𝑃(𝑌 ≤ 110) = 1 − 𝑃(∑ 𝑋𝑖100𝑖=1 ≤ 110) =

1 − 𝑃 (∑ 𝑋𝑖

100𝑖=1 −100

10≤

110−100

10) ≈ 1 − 𝑃(𝑍 ≤ 1) = 1 − Φ(1) = 1 − 0.8413 = 0.1587,

where 𝑍 denotes the standard normal distribution with 𝜇 = 0 and 𝜎2 = 1.

b) We know that 𝑍𝑛 =�̅�−𝜇

𝜎/√𝑛→𝑑 𝑁(0,1) by the Central Limit Theorem. We then have that

𝐸(�̅�) = 𝐸 (𝑌

100) =

1

100𝐸(𝑌) = 1 and 𝑉𝑎𝑟(�̅�) = 𝑉𝑎𝑟 (

𝑌

100) =

1

10,000𝑉𝑎𝑟(𝑌) =

1

100 so

𝑠𝑑(𝑌) =1

10, which allows us to find 𝑃(1.1 < �̅� < 1.2) = 𝑃 (

1.1−1

1/10<

�̅�−1

1/10<

1.2−1

1/10) ≈

𝑃(1 < 𝑍𝑛 < 2) = Φ(2) − Φ(1) = 0.9772 − 0.8413 = 0.1359. Here, we have used the

fact that 𝜇 = 1 and 𝜎 = 1 which come from the population distribution 𝑋𝑖~𝐸𝑋𝑃(1).

Question #11: Let 𝑋𝑖~𝑈𝑁𝐼𝐹(0,1) where 𝑋1, 𝑋2, … , 𝑋20 are all independent. Find a normal

approximation for the probability 𝑃(∑ 𝑋𝑖20𝑖=1 ≤ 12).

Since each 𝑋𝑖~𝑈𝑁𝐼𝐹(0,1), we know that 𝐸(𝑋𝑖) = 1/2 while 𝑉𝑎𝑟(𝑋𝑖) = 1/12. Due to

the independence of the 𝑋𝑖s, we have that 𝐸(∑ 𝑋𝑖20𝑖=1 ) = ∑ 𝐸(𝑋𝑖)

20𝑖=1 = 10 and

𝑉𝑎𝑟(∑ 𝑋𝑖20𝑖=1 ) = ∑ 𝑉𝑎𝑟(𝑋𝑖)

20𝑖=1 = 5/3, so that 𝑠𝑑(∑ 𝑋𝑖

20𝑖=1 ) = √5/3. This allows us to

find 𝑃(∑ 𝑋𝑖20𝑖=1 ≤ 12) = 𝑃 (

∑ 𝑋𝑖20𝑖=1 −10

√5/3≤

12−10

√5/3) ≈ 𝑃(𝑍 ≤ 1.55) = Φ(1.55) = 0.9394.

Chapter #8 – Statistics and Sampling Distributions

Question #1: Let 𝑋 denote the weight in pounds of a single bag of feed where 𝑋~𝑁(101,4).

What is the probability that 20 bags will weigh at least 2,000 pounds?

Let 𝑌 = ∑ 𝑋𝑖20𝑖=1 where 𝑋𝑖~𝑁(101,4). We have that 𝐸(𝑌) = 𝐸(∑ 𝑋𝑖

20𝑖=1 ) = ∑ 𝐸(𝑋𝑖)

20𝑖=1 =

20(101) = 2,020 and 𝑉𝑎𝑟(𝑌) = 𝑉𝑎𝑟(∑ 𝑋𝑖20𝑖=1 ) = ∑ 𝑉𝑎𝑟(𝑋𝑖)

20𝑖=1 = 20(4) = 80 such

that 𝑠𝑑(𝑌) = √80 = 4√5. We can thus calculate the probability 𝑃(𝑌 ≥ 2,000) =

𝑃(∑ 𝑋𝑖20𝑖=1 ≥ 2,000) = 𝑃 (

∑ 𝑋𝑖20𝑖=1 −𝐸(𝑌)

𝑠𝑑(𝑌)≥

2,000−𝐸(𝑌)

𝑠𝑑(𝑌)) = 𝑃 (

∑ 𝑋𝑖20𝑖=1 −2,020

4√5≥

2,000−2,020

4√5) ≈

𝑃 (𝑍 ≥ −20

4√5) = 𝑃(𝑍 ≥ −2.24) = 1 − Φ(−2.24) = 0.987, where 𝑍~𝑁(0,1).

Question #2: Let 𝑆 denote the diameter of a shaft and 𝐵 the diameter of a bearing, where

both 𝑆 and 𝐵 are independent and 𝑆~𝑁(1,0.0004) and 𝐵~𝑁(1.01,0.0009). a) If a shaft and

bearing are selected at random, what is the probability that the shaft diameter will exceed

the bearing diameter? b) Now assume equal variances (𝜎𝑆2 = 𝜎𝐵

2 = 𝜎2) such that we have

𝑆~𝑁(1, 𝜎2) and 𝐵~𝑁(1.01, 𝜎2). Find the value of 𝜎 that will yield a probability of

noninterference of 0.95 (which means the shaft diameter exceeds the bearing diameter).

a) Define 𝑌 = 𝑆 − 𝐷, since we wish to find 𝑃(𝑆 > 𝐷) = 𝑃(𝑆 − 𝐷 > 0) = 𝑃(𝑌 > 0). We

have that 𝐸(𝑌) = 𝐸(𝑆 − 𝐷) = 𝐸(𝑆) − 𝐸(𝐷) = 1 − 1.01 = −0.01 and 𝑉𝑎𝑟(𝑌) =

𝑉𝑎𝑟(𝑆 − 𝐷) = 𝑉𝑎𝑟(𝑆) + 𝑉𝑎𝑟(𝐷) = 0.0004 + 0.0009 = 0.0013 such that 𝑠𝑑(𝑌) =

√0.0013 = 0.036. Thus, we have 𝑃(𝑆 > 𝐷) = 𝑃(𝑆 − 𝐷 > 0) = 𝑃(𝑌 > 0) =

𝑃 (𝑌−𝐸(𝑌)

𝑠𝑑(𝑌)>

0−𝐸(𝑌)

𝑠𝑑(𝑌)) = 𝑃 (

𝑌+0.01

0.036>

0.01

0.036) ≈ 𝑃(𝑍 > 0.28) = 1 − Φ(0.28) = 0.39.

b) For 𝑌 = 𝑆 − 𝐷, we have that 𝐸(𝑌) = −0.01 but 𝑉𝑎𝑟(𝑌) = 2𝜎2 so 𝑠𝑑(𝑌) = √2𝜎. We

wish to find 𝜎 so that 𝑃(𝑌 > 0) = 0.95 → 1 − 𝑃 (𝑍 ≤0.01

√2𝜎) = 0.95 → Φ (

0.01

√2𝜎) = 0.05.

Since only the critical value 𝑍𝛼 = −1.645 ensures that Φ(−1.645) = 0.05, we must

solve 0.01

√2𝜎= −1.645 → 𝜎 = −0.004. But since we must have 𝜎 ≥ 0, no such 𝜎 exists.

Question #3: Let 𝑋1, … , 𝑋𝑛 be a random sample of size 𝑛 where they are ~𝑖𝑖𝑑 such that

𝑋𝑖~𝑁(𝜇, 𝜎2) and define 𝑈 = ∑ 𝑋𝑖𝑛𝑖=1 and 𝑊 = ∑ 𝑋𝑖

2𝑛𝑖=1 . a) Find a statistic that is a function of

𝑈 and 𝑊 and unbiased for the parameter 𝜃 = 2𝜇 − 5𝜎2. b) Find a statistic that is unbiased

for 𝛾 = 𝜎2 + 𝜇2. c) If 𝑐 is a constant and 𝑌𝑖 = 1 if 𝑋𝑖 ≤ 𝑐 and zero otherwise, find a statistic

that is a function of 𝑌1, … , 𝑌𝑛 and is unbiased for 𝐹𝑋(𝑐) = Φ (𝑐−𝜇

𝜎) = ∫

1

√2𝜋𝑒−

𝑡2

2(𝑐−𝜇)/𝜎

−∞𝑑𝑡.

a) We first find an estimator for 𝜇 = 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ) =

1

𝑛𝐸(∑ 𝑋𝑖

𝑛𝑖=1 ) =

1

𝑛𝐸(𝑈) =

𝑈

𝑛

and then for 𝜎2 = 𝐸(𝑆2) = 𝐸 (1

𝑛−1∑ (𝑋𝑖 − �̅�)2𝑛

𝑖=1 ) = 𝐸 (1

𝑛−1[∑ (𝑋𝑖

2)𝑛𝑖=1 − 𝑛�̅�2]) =

1

𝑛−1𝐸(∑ (𝑋𝑖

2)𝑛𝑖=1 − 𝑛�̅�2) =

1

𝑛−1𝐸(∑ 𝑋𝑖

2𝑛𝑖=1 − 𝑛�̅�2) =

1

𝑛−1𝐸 [𝑊 − 𝑛 (

1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 )

2

] =

1

𝑛−1𝐸 [𝑊 − 𝑛

(∑ 𝑋𝑖𝑛𝑖=1 )

2

𝑛2 ] =1

𝑛−1𝐸 [𝑊 −

(∑ 𝑋𝑖𝑛𝑖=1 )

2

𝑛] =

1

𝑛−1𝐸 [𝑊 −

1

𝑛𝑈2] =

1

𝑛−1[𝑊 −

𝑈2

𝑛].

We thus have 𝜃 = 2𝜇 − 5𝜎2 = 2 [𝑈

𝑛] − 5 [

1

𝑛−1(𝑊 −

𝑈2

𝑛)] =

2𝑈

𝑛−

5

𝑛−1(𝑊 −

1

𝑛𝑈2),

which is an unbiased estimator of 𝜃 since 𝐸(𝜃) = ⋯ = 𝜃.

b) Since we found that 𝜇 = 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ), then we have 𝜇2 = [𝐸 (

1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 )]

2

=

𝐸 [(1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 )

2

− 𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 )] = 𝐸 [

1

𝑛2(∑ 𝑋𝑖

𝑛𝑖=1 )2 −

1

𝑛2 𝑉𝑎𝑟(∑ 𝑋𝑖𝑛𝑖=1 )] = 𝐸 [

1

𝑛2 𝑈2 −

1

𝑛2 𝑛𝜎2] = 𝐸 [𝑈2

𝑛2 −𝜎2

𝑛] =

𝑈2

𝑛2 −𝜎2

𝑛. We previously found that 𝜎2 =

1

𝑛−1[𝑊 −

𝑈2

𝑛], so

combining these we find that 𝛾 = 𝜎2 + 𝜇2 = 𝜎2 +𝑈2

𝑛2−

𝜎2

𝑛=

(1−𝑛)

𝑛𝜎2 +

𝑈2

𝑛2=

(1−𝑛)

𝑛[

1

𝑛−1(𝑊 −

𝑈2

𝑛)] +

𝑈2

𝑛2 =1

𝑛(𝑊 −

𝑈2

𝑛) +

𝑈2

𝑛2 =𝑊

𝑛−

𝑈2

𝑛2 +𝑈2

𝑛2 =𝑊

𝑛, which is an

unbiased estimator of 𝛾 since 𝐸(𝛾) = ⋯ = 𝛾.

c) We have 𝑃(𝑌𝑖 = 1) = 𝑃(𝑋𝑖 ≤ 𝑐) = 𝑃 (𝑋𝑖−𝜇

𝜎≤

𝑐−𝜇

𝜎) = 𝑃 (𝑍 ≤

𝑐−𝜇

𝜎) = Φ (

𝑐−𝜇

𝜎) = 𝐹𝑋(𝑐)

and 𝐸(𝑌𝑖) = 1 ∙ 𝑃(𝑌𝑖 = 1) + 0 ∙ 𝑃(𝑌𝑖 = 0) = 𝑃(𝑌𝑖 = 1) = Φ (𝑐−𝜇

𝜎) = 𝐹𝑋(𝑐). Then,

𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑌𝑖

𝑛𝑖=1 ) =

1

𝑛𝐸(∑ 𝑌𝑖

𝑛𝑖=1 ) =

1

𝑛∑ 𝐸(𝑌𝑖)

𝑛𝑖=1 =

1

𝑛𝑛Φ (

𝑐−𝜇

𝜎) = Φ (

𝑐−𝜇

𝜎) = 𝐹𝑋(𝑐),

which means that �̅� is an unbiased estimator of 𝐹𝑋(𝑐) = Φ((𝑐 − 𝜇)/𝜎).

Question #4: Assume that 𝑋1 and 𝑋2 are independent normal random variables such that

each 𝑋𝑖~𝑁(𝜇, 𝜎2) and define 𝑌1 = 𝑋1 + 𝑋2 and 𝑌2 = 𝑋1 − 𝑋2. Show that the random variables

𝑌1 and 𝑌2 are independent and normally distributed.

Since 𝑋1 and 𝑋2 are independent normal random variables, we know that their joint

density function is 𝑓𝑋1𝑋2(𝑥1, 𝑥2) = 𝑓𝑋1

(𝑥1)𝑓𝑋2(𝑥2) = [

1

√2𝜋𝜎𝑒

−(𝑥1−𝜇)2

2𝜎2 ] [1

√2𝜋𝜎𝑒

−(𝑥2−𝜇)2

2𝜎2 ] =

1

2𝜋𝜎2 𝑒−

1

2𝜎2[(𝑥1−𝜇)2+(𝑥2−𝜇)2]. We have the transformation 𝑦1 = 𝑥1 + 𝑥2 and 𝑦2 = 𝑥1 − 𝑥2,

which can be solved to obtain 𝑥1 =𝑦1+𝑦2

2 and 𝑥2 =

𝑦1−𝑦2

2. This allows us to calculate

the Jacobian 𝐽 = 𝑑𝑒𝑡 [1/2 1/21/2 −1/2

] = −1

2, so we can compute the joint density

𝑓𝑌1𝑌2(𝑦1, 𝑦2) = 𝑓𝑋1𝑋2

(𝑦1+𝑦2

2,

𝑦1−𝑦2

2) |𝐽| =

1

2[

1

2𝜋𝜎2 𝑒−

1

2𝜎2((𝑦1+𝑦2

2−𝜇)

2+(

𝑦1−𝑦22

−𝜇)2

)]. After

simplifying this expression, we have 𝑓𝑌1𝑌2(𝑦1, 𝑦2) =

1

4𝜋𝜎2 𝑒−

1

4𝜎2[𝑦1−2𝜇]2

𝑒−

1

4𝜎2[𝑦2]2

. Since

the marginal densities can be separated, this shows that 𝑌1 and 𝑌2 are independent

and normally distributed. Moreover, we see that 𝑌1~𝑁(2𝜇, 2𝜎2) and 𝑌2~𝑁(0,2𝜎2).

Question #12: The distance in feet by which a parachutist misses a target is 𝐷 = √𝑋12 + 𝑋2

2

where 𝑋1 and 𝑋2 are independent with each 𝑋𝑖~𝑁(0,25). Find the probability 𝑃(𝐷 ≤ 12.25).

We wish to find 𝑃(𝐷 ≤ 12.25) = 𝑃 (√𝑋12 + 𝑋2

2 ≤ 12.25) = 𝑃[𝑋12 + 𝑋2

2 ≤ (12.25)2] =

𝑃[(𝑋1 − 0)2 + (𝑋2 − 0)2 ≤ (12.25)2] = 𝑃[(𝑋1 − 𝜇)2 + (𝑋2 − 𝜇)2 ≤ (12.25)2] =

𝑃 [(𝑋1−𝜇)2

𝜎2 +(𝑋2−𝜇)2

𝜎2 ≤(12.25)2

𝜎2 ] = 𝑃 [∑(𝑋𝑖−𝜇)2

𝜎22𝑖=1 ≤

(12.25)2

𝜎2 ] = 𝑃 [∑(𝑋𝑖−0)2

𝜎22𝑖=1 ≤

(12.25)2

𝜎2 ].

Since 𝜇 = 0 and 𝜎2 = 25, we have 𝑃 [∑(𝑋𝑖−0)2

25

2𝑖=1 ≤

(12.25)2

25] ≈ 𝑃 [𝜒2(2) ≤

(12.25)2

25] ≈

𝑃[𝜒2(2) ≤ 6] = 0.95. Note that we have used Corollary 8.3.4 to transform the

question into one using the chi-square distribution, since ∑ (𝑋𝑖−𝜇

𝜎)

2𝑛𝑖=1 ~𝜒2(𝑛). This is

because 𝑋𝑖~𝑁(𝜇, 𝜎2) implies that 𝑋𝑖−𝜇

𝜎~𝑁(0,1) so (

𝑋𝑖−𝜇

𝜎)

2

~𝜒2(1) and that the sum of

𝑛 independent chi-square distributed random variables is distributed 𝜒2(𝑛).


Question #8: Suppose that 𝑋 and 𝑌 are independent and distributed 𝑋~𝜒2(𝑚) and 𝑌~𝜒2(𝑛).

Is the random variable 𝑍 = 𝑌 − 𝑋 distributed chi-square if we have 𝑛 > 𝑚?

No. The random variable 𝑍 = 𝑌 − 𝑋 can clearly take on negative values, whereas as a

random variable following the chi-square distribution must be positive.

Question #9: Suppose that 𝑋~𝜒2(𝑚), 𝑆 = 𝑋 + 𝑌~𝜒2(𝑚 + 𝑛) and that 𝑋 and 𝑌 are

independent random variables. Use moment generating functions to show that 𝑆 − 𝑋~𝜒2(𝑛).

We know that if some 𝐴~𝜒2(𝑣), then its MGF is given by 𝑀𝑌(𝑡) = (1 − 2𝑡)−𝑣/2. We

thus have 𝑀𝑋(𝑡) = (1 − 2𝑡)−𝑚/2 and 𝑀𝑆(𝑡) = 𝑀𝑋+𝑌(𝑡) = (1 − 2𝑡)−(𝑚+𝑛)/2. Since 𝑋

and 𝑌 are independent, we know that 𝑀𝑋+𝑌(𝑡) = 𝑀𝑋(𝑡)𝑀𝑌(𝑡), which implies that

𝑀𝑆−𝑋(𝑡) = 𝑀(𝑋+𝑌)−𝑋(𝑡) = 𝑀𝑌(𝑡) =𝑀𝑋+𝑌(𝑡)

𝑀𝑋(𝑡)=

(1−2𝑡)−(𝑚+𝑛)/2

(1−2𝑡)−𝑚/2 = (1 − 2𝑡)−𝑛/2. Thus, we

have that 𝑌 = 𝑆 − 𝑋 is distributed chi-square with 𝑛 degrees of freedom.

Question #14: If 𝑇~𝑡(𝜈), find the distribution of the random variable 𝑇2.

We know that if 𝑍~𝑁(0,1) and 𝑉~𝜒2(𝜈) are independent random variables, then the

distribution of 𝑇 =𝑍

√𝑉/𝜈 is Student’s t distribution. But then we can square this to

produce 𝑇2 =𝑍2

𝑉/𝜈=

𝑍2/1

𝑉/𝜈, which makes it clear that 𝑇2~𝐹(1, 𝜈). The reason for this is

that we know if some 𝑍~𝑁(0,1), then 𝑍2~𝜒2(1). Moreover, we are already given that

𝑉~𝜒2(𝜈). Combining these results with the fact that if some 𝑉1~𝜒2(𝜈1) and 𝑉2~𝜒

2(𝜈2)

are independent, then the random variable 𝑋 =𝑉1/𝜈1

𝑉2/𝜈2~𝐹(𝜈1, 𝜈2). Therefore, 𝑇2 follows

the F distribution with 1 and 𝜈 degrees of freedom whenever 𝑇~𝑡(𝜈).

Question #15: Suppose that 𝑋𝑖~𝑁(𝜇, 𝜎2) for 𝑖 = 1,… , 𝑛 and 𝑍𝑖~𝑁(0,1) for 𝑖 = 1, . . , 𝑘 and

that all variables are independent. Find the distribution of the following random variables.

a) 𝑋1 − 𝑋2~𝑁(𝜇 − 𝜇, 𝜎2 + 𝜎2) ≡ 𝑁(0,2𝜎2)

b) 𝑋2 + 2𝑋3~𝑁(𝜇 + 2𝜇, 𝜎2 + 4𝜎2) ≡ 𝑁(3𝜇, 5𝜎2)

c) 𝑍12~𝜒2(1) since the square of a standard normal random variable is chi-square.

d) 𝑋1−𝑋2

𝜎𝑆𝑧√2~𝑡(𝑘 − 1) since 𝑋1 − 𝑋2~𝑁(0,2𝜎

2) implies that 𝑋1−𝑋2

𝜎√2~𝑁(0,1) and dividing this

by the sample standard deviation 𝑆𝑧 of the Z sample makes it clear that ~𝑡(𝑘 − 1).

e) √𝑛(�̅�−𝜇)

𝜎𝑆𝑧~𝑡(𝑘 − 1) since 𝑍 =

�̅�−𝜇

𝜎/√𝑛~𝑁(0,1), 𝑉𝑧 =

(𝑘−1)𝑆𝑧2

𝜎2~𝜒2(𝑘 − 1) and we can write

√𝑛(�̅�−𝜇)

𝜎𝑆𝑧=

𝑍

𝑆𝑧=

𝑍

√𝑉𝑧/(𝑘−1)~𝑡(𝑘 − 1) by the definition of the t distribution (see above).

f) 𝑍12 + 𝑍2

2 = 𝜒2(1) + 𝜒2(1)~𝜒2(1 + 1) ≡ 𝜒2(2) since we can simply add the

parameters for a sum of independent chi-square random variables.

g) 𝑍12 − 𝑍2

2 → the distribution is unknown.

h) 𝑍1

√𝑍22~𝑡(1) since 𝑉 = 𝑍2

2~𝜒2(1) and we can write 𝑍1

√𝑍22=

𝑍1

√𝑍22/1

=𝑍1

√𝑉/1~𝑡(1).

i) 𝑍12

𝑍22~𝐹(1,1) since 𝑉1 = 𝑍1

2~𝜒2(1), 𝑉2 = 𝑍22~𝜒2(1) and we have

𝑍12

𝑍22 =

𝑉1/1

𝑉2/1= ~𝐹(1,1).

j) 𝑍1

𝑍2~𝐶𝐴𝑈(1,0) since we can generate the joint transformation 𝑢 =

𝑧1

𝑧2 and 𝑣 = 𝑧2,

calculate the joint density 𝑓𝑈𝑉(𝑢, 𝑣) and integrate out 𝑑𝑣 to find 𝑓𝑈(𝑢) =1

𝜋(𝑢2+1).

k) �̅�

𝑍→ the distribution is unknown.

l) √𝑛𝑘(�̅�−𝜇)

𝜎√∑ 𝑍𝑖2𝑘

𝑖=1

~𝑡(𝑘) since 𝑊 =�̅�−𝜇

𝜎/√𝑛~𝑁(0,1) and 𝑉 = ∑ 𝑍𝑖

2𝑘𝑖=1 ~𝜒2(𝑘) and we can write the

expression √𝑛𝑘(�̅�−𝜇)

𝜎√∑ 𝑍𝑖2𝑘

𝑖=1

=√𝑛(�̅�−𝜇)

𝜎

√∑ 𝑍𝑖2𝑘

𝑖=1𝑘

=𝑊

√𝑉/𝑘~𝑡(𝑘) by the definition of the distribution.

m) ∑(𝑋𝑖−𝜇)

2

𝜎2𝑛𝑖=1 + ∑ (𝑍𝑖 − �̅�)2𝑘

𝑖=1 ~𝜒2(𝑛 + 𝑘 − 1) since ∑(𝑋𝑖−𝜇)

2

𝜎2𝑛𝑖=1 ~𝜒2(𝑛) by Corollary

8.3.4 and ∑ (𝑍𝑖 − �̅�)2𝑘𝑖=1 = (𝑘 − 1)𝑆𝑧

2 =(𝑘−1)𝑆𝑧

2

12~𝜒2(𝑘 − 1) by Theorem 8.3.6. Thus,

we have the sum of two chi-square random variables so we sum the parameters.

n) �̅�

𝜎2+

1

𝑘∑ 𝑍𝑖𝑘𝑖=1 ~𝑁 (

𝜇

𝜎2,1

𝑛𝜎2+

1

𝑘) since �̅�~𝑁 (𝜇,

𝜎2

𝑛) implies that the random variable

1

𝜎2�̅� = ∑ (

1

𝜎2)𝑋𝑖

𝑛

𝑛𝑖=1 ~𝑁 (∑ (

1

𝑛𝜎2)𝑛

𝑘=1 𝜇, ∑ (1

𝑛𝜎2)2

𝑛𝑘=1 𝜎2) ≡ 𝑁 (

𝜇

𝜎2,1

𝑛𝜎2). Also, we have

1

𝑘∑ 𝑍𝑖𝑘𝑖=1 = �̅�~𝑁 (0,

1

𝑘) so the distribution of their sum is normal and we sum their

respective means and variances to conclude that �̅�

𝜎2+

1

𝑘∑ 𝑍𝑖𝑘𝑖=1 ~𝑁 (

𝜇

𝜎2,1

𝑛𝜎2+

1

𝑘).

o) 𝑘�̅�2~𝜒2(1) since √𝑘�̅�~𝑁(0,1), so it must be that (√𝑘�̅�)2= 𝑘�̅�2~𝜒2(1).

p) (𝑘−1)∑ (𝑋𝑖−�̅�)

2𝑛𝑖=1

(𝑛−1)𝜎2∑ (𝑍𝑖−𝑍)2𝑘

𝑖=1

~𝐹(𝑛 − 1, 𝑘 − 1) since we can simplify the random variable as

(𝑘−1)∑ (𝑋𝑖−�̅�)2𝑛

𝑖=1

(𝑛−1)𝜎2∑ (𝑍𝑖−𝑍)2𝑘

𝑖=1

=

12 ∑ (𝑋𝑖−�̅�)2𝑛

𝑖=1𝑛−1

𝜎2 ∑ (𝑍𝑖−�̅�)2𝑘

𝑖=1𝑘−1

=12𝑆𝑋

2

𝜎2𝑆𝑍2 and 12𝑆𝑋

2~𝜒2(𝑛 − 1) and 𝜎2𝑆𝑍2~𝜒2(𝑘 − 1). We

thus have the ratio of two chi-square random variables over their respective degrees

of freedom, which we know follows the F distribution.

Question #18: Assume that 𝑍~𝑁(0,1), 𝑉1~𝜒2(5) and 𝑉2~𝜒

2(9) are all independent. Then

compute the probability that a) 𝑃(𝑉1 + 𝑉2 < 8.6), b) 𝑃 (𝑍

√𝑉1/5< 2.015), c) 𝑃(𝑍 > 0.611√𝑉2),

d) 𝑃 (𝑉1

𝑉2< 1.45) and e) find the value of 𝑏 such that 𝑃 (

𝑉1

𝑉1+𝑉2< 𝑏) = 0.9.

a) Since 𝑉1~𝜒2(5) and 𝑉2~𝜒

2(9), we know that 𝑉1 + 𝑉2~𝜒2(14). This allows us to

compute 𝑃(𝑉1 + 𝑉2 < 8.6) = 0.144 using the tables for the chi-square distribution.

b) We know that if 𝑍~𝑁(0,1) and some 𝑉~𝜒2(𝜈) are independent random variables,

then 𝑇 =𝑍

√𝑉/𝜈 follows the t distribution with 𝜈 degrees of freedom. We thus have that

𝑇 =𝑍

√𝑉1/5~𝑡(5), so we can compute 𝑃 (

𝑍

√𝑉1/5< 2.015) = 0.95 using the t-table.

c) We wish to compute 𝑃(𝑍 > 0.611√𝑉2) = 𝑃 (𝑍

√𝑉2> 0.611) = 𝑃 (

𝑍

√𝑉23 > 0.611(3)) =

𝑃 (𝑍

√𝑉2/9> 1.833) = 0.05, from using the t-table since we know that

𝑍

√𝑉2/9~𝑡(9).

d) We wish to compute 𝑃 (𝑉1

𝑉2< 1.45) = 𝑃 (

𝑉1/5

𝑉2/9< 1.45 (

9

5)) = 𝑃 (

𝑉1/5

𝑉2/9< 2.61). We know

that if some 𝑉1~𝜒2(𝜈1) and 𝑉2~𝜒

2(𝜈2) are independent, then the random variable

given by 𝑋 =𝑉1/𝜈1

𝑉2/𝜈2~𝐹(𝜈1, 𝜈2). We thus have that

𝑉1/5

𝑉2/9~𝐹(5,9), so we can therefore use

the F-table to compute the desired probability as 𝑃 (𝑉1/5

𝑉2/9< 2.61) = 0.9.

e) We wish to compute 𝑏 sich that 𝑃 (𝑉1

𝑉1+𝑉2< 𝑏) = 𝑃 (

𝑉1+𝑉2

𝑉1>

1

𝑏) = 𝑃 (1 +

𝑉2

𝑉1>

1

𝑏) =

𝑃 (𝑉2

𝑉1>

1

𝑏− 1) = 𝑃 (

𝑉2/9

𝑉1/5>

5

9(1

𝑏− 1)) = 0.9. But we know that 𝐹 =

𝑉2/9

𝑉1/5~𝐹(9,5) so we

can use tables to find that 𝑃(𝐹 > 0.383) = 0.9. This means that we must solve the

equation 5

9(1

𝑏− 1) = 0.383 →

1

𝑏=

9

5(0.383) + 1 → 𝑏 =

19

5(0.383)+1

= 0.592.

Question #19: Suppose that 𝑇~𝑡(1). a) Show that the CDF of 𝑇 is 𝐹𝑇(𝑡) =1

2+

1

𝜋𝑎𝑟𝑐𝑡𝑎𝑛(𝑡)

and b) show that the 100 ⋅ 𝛾𝑡ℎ percentile is given by 𝑡𝛾(1) = 𝑡𝑎𝑛 [𝜋 (𝛾 −1

2)].

a) If some 𝑇~𝑡(𝜈), then its density is given by 𝑓𝑇(𝑡) =Γ(

𝜈+1

2)

Γ(𝜈

2)√𝜈𝜋

(1 +𝑡2

𝜈)−𝜈+1

2. When 𝜈 = 1,

we have 𝑓𝑇(𝑡) =Γ(1)

Γ(1

2)√𝜋

(1 + 𝑡2)−1 =1

𝜋

1

1+𝑡2 since Γ(1) = 1 and Γ (

1

2) = √𝜋. We thus

have that 𝑓𝑇(𝑡) =1

𝜋

1

1+𝑡2 when 𝜈 = 1, which is the density of a Cauchy random variable.

To find the cumulative distribution, we simply compute 𝐹𝑇(𝑡) =1

𝜋∫

1

1+𝑥2

𝑡

−∞𝑑𝑥 =

1

𝜋[𝑎𝑟𝑐𝑡𝑎𝑛(𝑥)]−∞

𝑡 =1

𝜋(𝑎𝑟𝑐𝑡𝑎𝑛(𝑡) − (−

𝜋

2)) =

1

𝜋𝑎𝑟𝑐𝑡𝑎𝑛(𝑡) +

1

2.

b) The 100 ⋅ 𝛾𝑡ℎ percentile is the value of 𝑡 such that 𝐹𝑇(𝑡) = 𝛾. From the work above,

we have 1

𝜋𝑎𝑟𝑐𝑡𝑎𝑛(𝑡) +

1

2= 𝛾 → 𝑎𝑟𝑐𝑡𝑎𝑛(𝑡) = (𝛾 −

1

2) 𝜋 → 𝑡 = 𝑡𝑎𝑛 [(𝛾 −

1

2) 𝜋]. This

proves that the 100 ⋅ 𝛾𝑡ℎ percentile is given by 𝑡𝛾(1) = 𝑡𝑎𝑛 [(𝛾 −1

2)𝜋].


Question #22: Compute 𝐸(𝑋𝑝) for 𝑝 > 0 if we have that 𝑋~𝐵𝐸𝑇𝐴(𝑎, 𝑏).

Since 𝑋~𝐵𝐸𝑇𝐴(𝑎, 𝑏), its PDF is 𝑓𝑋(𝑥) =Γ(𝑎+𝑏)

Γ(𝑎)Γ(𝑏)𝑥𝑎−1(1 − 𝑥)𝑏−1 whenever 0 < 𝑥 < 1

and 𝑎 > 0, 𝑏 > 0. Then using the definition of expected value, we can compute

𝐸(𝑋𝑝) = ∫ 𝑥𝑝𝑓𝑋(𝑥)∞

−∞𝑑𝑥 =

Γ(𝑎+𝑏)

Γ(𝑎)Γ(𝑏)∫ 𝑥𝑝𝑥𝑎−1(1 − 𝑥)𝑏−11

0𝑑𝑥 =

Γ(𝑎+𝑏)

Γ(𝑎)Γ(𝑏)

Γ(𝑎+𝑝)Γ(𝑏)

Γ(𝑎+𝑏+𝑝)=

Γ(𝑎+𝑏)Γ(𝑎+𝑝)

Γ(𝑎)Γ(𝑎+𝑏+𝑝) since we have

Γ(𝑎+𝑏)

Γ(𝑎)Γ(𝑏)∫ 𝑥𝑎−1(1 − 𝑥)𝑏−11

0𝑑𝑥 = 1 so we can solve for the

integral to conclude that Γ(𝑎)Γ(𝑏)

Γ(𝑎+𝑏)= ∫ 𝑥𝑎−1(1 − 𝑥)𝑏−11

0𝑑𝑥. In this case, we are solving

that ∫ 𝑥𝑝𝑥𝑎−1(1 − 𝑥)𝑏−11

0𝑑𝑥 = ∫ 𝑥𝑝+𝑎−1(1 − 𝑥)𝑏−11

0𝑑𝑥 =

Γ(𝑎+𝑝)Γ(𝑏)

Γ(𝑎+𝑏+𝑝). Therefore, all of

the moments of the beta distribution for some fixed 𝑝 > 0 can be written in terms of

the gamma function, which can be evaluated numerically.

Question #24: Suppose that 𝑌𝜈~𝜒2(𝜈). Use moment generating functions to find the limiting

distribution of the transformed random variable 𝑌𝜈−𝜈

√2𝜈 as 𝜈 → ∞.

This result follows directly from the Central Limit Theorem. If we let 𝑌𝜈 = ∑ 𝑋𝑖𝑛𝑖=1

where 𝑋𝑖~χ2(1) for 𝑖 = 1, … , 𝑛, then 𝑌𝜈~𝜒2(𝑛) so that 𝐸(𝑌𝜈) = 𝑛, 𝑉𝑎𝑟(𝑌𝜈) = 2𝑛 and

𝑠𝑑(𝑌𝜈) = √2𝑛. Therefore, 𝑌𝜈−𝐸(𝑌𝜈)

𝑠𝑑(𝑌𝜈)=

𝑌𝜈−𝑛

√2𝑛→ 𝑍~𝑁(0,1) as 𝑛 → ∞. We will now prove

this result using moment generating functions. By the definition of MGFs, we have

𝑀[

𝑌𝜈−𝜈

√2𝜈 ]

(𝑡) = 𝐸 [𝑒𝑡(

𝑌𝜈−𝜈

√2𝜈 )

] = 𝐸 [𝑒−𝑡√

𝜈

2𝑒𝑡

𝑌𝜈

√2𝜈] = 𝑒−𝑡

√2

√𝜈𝐸 [𝑒𝑡

𝑌𝜈

√2𝜈] = 𝑒−𝑡

√2

√𝜈𝑀𝑌𝜈(

𝑡

√2𝜈) =

𝑒−𝑡

√2

√𝜈 (1 −2𝑡

√2𝜈)

−𝜈

2= 𝑒

−𝑡√2

√𝜈 (1 − 𝑡√2

𝜈)

−𝜈

2

. In order to evaluate lim𝜈→∞

𝑀[

𝑌𝜈−𝜈

√2𝜈 ]

(𝑡), we first

take logarithms and then exponentiate the result. This implies that ln [M[

Yν−ν

√2ν ]

(t)] =

ln [𝑒−𝑡

√2

√𝜈 (1 − 𝑡√2

√𝜈)

−𝜈

2] = ln [e

−𝑡√2

√𝜈] + ln [(1 − 𝑡√2

√𝜈)

−𝜈

2] = −𝑡

√2

√𝜈−

ν

2ln (1 − 𝑡

√2

√𝜈). From

here, we use the Taylor series ln(1 − 𝑧) = −𝑧 −𝑧2

2−

𝑧3

3− ⋯ for 𝑧 = 𝑡

√2

√𝜈 to evaluate

the limit, which then gives lim𝜈→∞

ln [M[

Yν−ν

√2ν ]

(t)] = lim𝜈→∞

[−𝑡√2

√𝜈−

ν

2ln (1 − 𝑡

√2

√𝜈)] ≈

lim𝜈→∞

[−𝑡√2

√𝜈−

ν

2(−𝑡

√2

√𝜈−

𝑡2

𝜈−

𝑡3232

3𝜈32

− ⋯ )] = lim𝜈→∞

[−𝑡√2

√𝜈+ 𝑡

√2

√𝜈+

𝑡2

2+

𝑡3√2

3√𝜈+ ⋯ ] =

lim𝜈→∞

[𝑡2

2+

𝑡3√2

3√𝜈+ ⋯ ] =

𝑡2

2+ 0 + ⋯. This result therefore implies that the limit

lim𝜈→∞

𝑀[

𝑌𝜈−𝜈

√2𝜈 ]

(𝑡) = 𝑒lim

𝜈→∞ln[M

[Yν−ν

√2ν ]

(t)]

= 𝑒𝑡2

2 , which is the moment generating function of

a random variable that follows a standard normal distribution. This proves that the

random variable 𝑌𝜈−𝑛

√2𝑛→ 𝑍~𝑁(0,1) as 𝑛 → ∞, just as is guaranteed by the CLT.

Chapter #9 – Point Estimation

Question #1: Assume that 𝑋1, … , 𝑋𝑛 are independent and identically distributed with

common density 𝑓(𝑥; 𝜃), where 𝜃 > 0 is an unknown parameter. Find the method of

moments estimator (MME) of 𝜃 if the density function is a) 𝑓(𝑥; 𝜃) = 𝜃𝑥𝜃−1 for 0 < 𝑥 < 1,

b) 𝑓(𝑥; 𝜃) = (𝜃 + 1)𝑥−𝜃−2 whenever 𝑥 > 1, and c) 𝑓(𝑥; 𝜃) = 𝜃2𝑥𝑒−𝜃𝑥 whenever 𝑥 > 0.

a) We begin by computing the first population moment, so 𝐸(𝑋) = ∫ 𝑥𝑓(𝑥; 𝜃)1

0𝑑𝑥 =

∫ 𝑥(𝜃𝑥𝜃−1)1

0𝑑𝑥 = 𝜃 ∫ 𝑥𝜃1

0𝑑𝑥 =

𝜃

𝜃+1[𝑥𝜃+1]

0

1=

𝜃

𝜃+1(1 − 0) =

𝜃

𝜃+1. We therefore have

𝐸(𝑋) =𝜃

𝜃+1. Next, we equate the first population moment with the first sample

moment, which gives 𝜇1′ = 𝑀1

′ →𝜃

𝜃+1=

1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 →

𝜃

𝜃+1= �̅�. Finally, we replace 𝜃

by 𝜃 and solve the equation �̂�

�̂�+1= �̅� for 𝜃, which implies that 𝜃𝑀𝑀𝐸 =

�̅�

1−�̅�.

b) Just as above, we first compute 𝐸(𝑋) = ∫ 𝑥𝑓(𝑥; 𝜃)∞

1𝑑𝑥 = ∫ 𝑥[(𝜃 + 1)𝑥−𝜃−2]

∞

1𝑑𝑥 =

(𝜃 + 1) ∫ 𝑥−𝜃−1∞

1𝑑𝑥 =

𝜃+1

−𝜃[𝑥−𝜃]

1

∞= −

𝜃+1

𝜃[0 − 1] =

𝜃+1

𝜃. Thus, we have 𝐸(𝑋) =

𝜃+1

𝜃

which means that 𝜇1′ = 𝑀1

′ →𝜃+1

𝜃=

1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 →

𝜃+1

𝜃= �̅� and 𝜃𝑀𝑀𝐸 =

1

�̅�−1.

c) We have 𝐸(𝑋) = ∫ 𝑥𝑓(𝑥; 𝜃)∞

0𝑑𝑥 = ∫ 𝑥[𝜃2𝑥𝑒−𝜃𝑥]

∞

0𝑑𝑥 = 𝜃2 ∫ 𝑥2𝑒−𝜃𝑥∞

0𝑑𝑥 = ⋯ =

2

𝜃

after doing integration by parts. We can also find this directly by noting that the

density 𝑓(𝑥, 𝜃) = 𝜃2𝑥𝑒−𝜃𝑥 suggests that 𝑋~𝐺𝐴𝑀𝑀𝐴 (1

𝜃, 2). This then implies that

𝐸(𝑋) = 𝜅𝜃 =1

𝜃2 =

2

𝜃. We therefore set 𝜇1

′ = 𝑀1′ such that

2

𝜃=

1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 or

2

𝜃= �̅�, and

then solve for the method of moments estimator, which is given by 𝜃𝑀𝑀𝐸 =2

�̅�.

Question #2: Assume that 𝑋1, … , 𝑋𝑛 are independent and identically distributed. Find the

method of moments estimator (MME) of the unknown parameters if the random sample

comes from a) 𝑋~𝑁𝐵(3, 𝑝), b) 𝑋~𝐺𝐴𝑀𝑀𝐴(2, 𝜅), c) 𝑋~𝑊𝐸𝐼 (𝜃,1

2), and d) 𝑋~𝑃𝐴𝑅(𝜃, 𝜅).

a) Since 𝑋~𝑁𝐵(3, 𝑝), we know that 𝐸(𝑋) =𝑟

𝑝=

3

𝑝. Equating this with the first sample

moment gives 𝜇1′ = 𝑀1

′ →3

𝑝= �̅�, so the estimator is �̂�𝑀𝑀𝐸 =

3

�̅�.

b) Since 𝑋~𝐺𝐴𝑀𝑀𝐴(2, 𝜅), we know that 𝐸(𝑋) = 𝜅𝜃 = 2𝜅. Equating this with the first

sample moment gives 𝜇1′ = 𝑀1

′ → 2𝜅 = �̅�, so the estimator is �̂�𝑀𝑀𝐸 =�̅�

2.

c) Since 𝑋~𝑊𝐸𝐼 (𝜃,1

2), we know that 𝐸(𝑋) = 𝜃Γ (1 +

1

𝛽) = 𝜃Γ (1 +

1

1/2) = 𝜃Γ(3) =

𝜃(3 − 1)! = 2𝜃. Thus, we have 𝜇1′ = 𝑀1

′ → 2𝜃 = �̅�, so the estimator is 𝜃𝑀𝑀𝐸 =�̅�

2.

d) Since 𝑋~𝑃𝐴𝑅(𝜃, 𝜅), we have 𝜇1 =𝜃

𝜅−1 and 𝜇2 = 𝜎2 + 𝜇1

2 =𝜃2𝜅

(𝜅−2)(𝜅−1)2 +𝜃2

(𝜅−1)2. This

means that 𝜇1 =𝜃

𝜅−1= 𝑀1

′ = �̅� and 𝜇2 =𝜃2𝜅

(𝜅−2)(𝜅−1)2 +𝜃2

(𝜅−1)2 = 𝑀2′ =

1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 . We

must solve for the unknown parameters 𝜃 and 𝜅 in terms of the two sample moments

�̅� and 1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 . From the first equation, we can solve to find 𝜃 = (𝜅 − 1)�̅� and

substitute into the second equation to find �̅�2(𝜅−1)2𝜅

(𝜅−2)(𝜅−1)2 +�̅�2(𝜅−1)2

(𝜅−1)2 =1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 →

�̅�2𝜅

(𝜅−2)+ �̅�2 =

1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 → �̅�2 (

𝜅

𝜅−2+ 1) =

1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 →

2𝜅−2

𝜅−2=

∑ 𝑋𝑖2𝑛

𝑖=1

𝑛�̅�2 . But this means

𝑛�̅�2(2𝜅 − 2) = (𝜅 − 2) ∑ 𝑋𝑖2𝑛

𝑖=1 → 2𝑛�̅�2𝜅 − 2𝑛�̅�2 = 𝜅 ∑ 𝑋𝑖2𝑛

𝑖=1 − 2 ∑ 𝑋𝑖2𝑛

𝑖=1 , so that

2𝑛�̅�2𝜅 − 𝜅 ∑ 𝑋𝑖2𝑛

𝑖=1 = 2𝑛�̅�2 − 2 ∑ 𝑋𝑖2𝑛

𝑖=1 → 𝜅(2𝑛�̅�2 − ∑ 𝑋𝑖2𝑛

𝑖=1 ) = 2𝑛�̅�2 − 2 ∑ 𝑋𝑖2𝑛

𝑖=1 .

Finally, we divide through to find 𝜅 =2𝑛�̅�2−2 ∑ 𝑋𝑖

2𝑛𝑖=1

2𝑛�̅�2−∑ 𝑋𝑖2𝑛

𝑖=1

. Plugging in to the other equation

implies that 𝜃 = (𝜅 − 1)�̅� = (2𝑛�̅�2−2 ∑ 𝑋𝑖

2𝑛𝑖=1

2𝑛�̅�2−∑ 𝑋𝑖2𝑛

𝑖=1

− 1) �̅�, so that the two method of

moments estimators are �̂�𝑀𝑀𝐸 =2𝑛�̅�2−2 ∑ 𝑋𝑖

2𝑛𝑖=1

2𝑛�̅�2−∑ 𝑋𝑖2𝑛

𝑖=1

and 𝜃𝑀𝑀𝐸 = (2𝑛�̅�2−2 ∑ 𝑋𝑖

2𝑛𝑖=1

2𝑛�̅�2−∑ 𝑋𝑖2𝑛

𝑖=1

− 1) �̅�.

Question #3: Assume that 𝑋1, … , 𝑋𝑛 are independent and identically distributed with

common density 𝑓(𝑥; 𝜃), where 𝜃 > 0 is an unknown parameter. Find the maximum

likelihood estimator (MLE) for 𝜃 when the PDF is a) 𝑓(𝑥; 𝜃) = 𝜃𝑥𝜃−1 whenever 0 < 𝑥 < 1,

b) 𝑓(𝑥; 𝜃) = (𝜃 + 1)𝑥−𝜃−2 whenever 𝑥 > 1, and c) 𝑓(𝑥, 𝜃) = 𝜃2𝑥𝑒−𝜃𝑥 whenever 𝑥 > 0.

a) We first find the likelihood function based on the joint density of 𝑋1, … , 𝑋𝑛, which is

𝐿(𝜃) = 𝑓(𝑥1; 𝜃) … 𝑓(𝑥𝑛; 𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ 𝜃𝑥𝑖

𝜃−1𝑛𝑖=1 = 𝜃𝑛(𝑥1 … 𝑥𝑛)𝜃−1. Next, we

construct the log likelihood function, since it is easier to differentiate and achieves a

maximum at the same point as the likelihood function. This gives ln[𝐿(𝜃)] =

ln[𝜃𝑛(𝑥1 … 𝑥𝑛)𝜃−1] = 𝑛 ln(𝜃) + (𝜃 − 1)[ln(𝑥1) + ⋯ + ln (𝑥𝑛)], which we differentiate

so 𝜕

𝜕𝜃ln[𝐿(𝜃)] =

𝜕

𝜕𝜃[𝑛 ln(𝜃) + (𝜃 − 1) ∑ ln(𝑥𝑖)

𝑛𝑖=1 ] =

𝑛

𝜃+ ∑ ln (𝑥𝑖)

𝑛𝑖=1 . We then solve

for the value of 𝜃 which makes the derivative equal zero, so 𝑛

𝜃+ ∑ ln (𝑥𝑖)

𝑛𝑖=1 = 0 → 𝜃 =

−𝑛

∑ ln (𝑥𝑖)𝑛𝑖=1

. Since it is clear that the second derivative of ln[𝐿(𝜃)] is negative, we have

found that the maximum likelihood estimator is 𝜃𝑀𝐿𝐸 = −𝑛

∑ ln (𝑋𝑖)𝑛𝑖=1

. (Note that we

must capitalize the 𝑋𝑖 from 𝑥𝑖 when presenting the estimator.)

b) We have 𝐿(𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ (𝜃 + 1)𝑥𝑖

−𝜃−2𝑛𝑖=1 = (𝜃 + 1)𝑛(𝑥1 … 𝑥𝑛)−𝜃−2 so that

ln[𝐿(𝜃)] = ln[(𝜃 + 1)𝑛(𝑥1 … 𝑥𝑛)−𝜃−2] = 𝑛 ln(𝜃 + 1) − (𝜃 + 2) ∑ ln (𝑥𝑖)𝑛𝑖=1 . Then we

find 𝜕


𝜕

𝜕𝜃[𝑛 ln(𝜃 + 1) − (𝜃 + 2) ∑ ln(𝑥𝑖)

𝑛𝑖=1 ] =

𝑛

𝜃+1− ∑ ln(𝑥𝑖)

𝑛𝑖=1 . Finally,

we must solve 𝑛

𝜃+1− ∑ ln(𝑥𝑖)

𝑛𝑖=1 = 0 → 𝜃 =

𝑛

∑ ln(𝑥𝑖)𝑛𝑖=1

− 1. Since the second derivative

of ln[𝐿(𝜃)] will be negative, we have found that 𝜃𝑀𝐿𝐸 =𝑛

∑ ln(𝑋𝑖)𝑛𝑖=1

− 1.

c) We have 𝐿(𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ 𝜃2𝑥𝑖𝑒−𝜃𝑥𝑖𝑛

𝑖=1 = 𝜃2𝑛(𝑥1 … 𝑥𝑛)𝑒−𝜃(𝑥1+⋯+𝑥𝑛) so that

ln[𝐿(𝜃)] = ln[𝜃2𝑛(𝑥1 … 𝑥𝑛)𝑒−𝜃(∑ 𝑥𝑖𝑛𝑖=1 )] = 2𝑛 ln(𝜃) + ∑ ln(𝑥𝑖)

𝑛𝑖=1 − 𝜃 ∑ 𝑥𝑖

𝑛𝑖=1 . Then we

have 𝜕


𝜕

𝜕𝜃[2𝑛 ln(𝜃) + ∑ ln(𝑥𝑖)

𝑛𝑖=1 − 𝜃 ∑ 𝑥𝑖

𝑛𝑖=1 ] =

2𝑛

𝜃− ∑ 𝑥𝑖

𝑛𝑖=1 . Finally, we

must solve 2𝑛

𝜃− ∑ 𝑥𝑖

𝑛𝑖=1 = 0 → 𝜃 =

2𝑛

∑ 𝑥𝑖𝑛𝑖=1

=2

�̅�, which implies that 𝜃𝑀𝐿𝐸 =

2

�̅�.

Question #4: Assume that 𝑋1, … , 𝑋𝑛 are independent and identically distributed. Find the

maximum likelihood estimator (MLE) of the parameter if the distribution is a) 𝑋𝑖~𝐵𝐼𝑁(1, 𝑝),

b) 𝑋𝑖~𝐺𝐸𝑂(𝑝) , c) 𝑋𝑖~𝑁𝐵(3, 𝑝), d) 𝑋𝑖~𝐺𝐴𝑀𝑀𝐴(𝜃, 2), e) 𝑋𝑖~𝑊𝐸𝐼 (𝜃,1

2), and f) 𝑋𝑖~𝑃𝐴𝑅(1, 𝜅).

a) Since the density of 𝑋~𝐵𝐼𝑁(1, 𝑝) is 𝑓(𝑥; 𝑝) = (1𝑥)𝑝𝑥(1 − 𝑝)1−𝑥 = 𝑝𝑥(1 − 𝑝)1−𝑥, we

have 𝐿(𝑝) = ∏ 𝑓(𝑥𝑖; 𝑝)𝑛𝑖=1 = ∏ 𝑝𝑥𝑖(1 − 𝑝)1−𝑥𝑖𝑛

𝑖=1 = 𝑝∑ 𝑥𝑖𝑛𝑖=1 (1 − 𝑝)𝑛−∑ 𝑥𝑖

𝑛𝑖=1 and then

ln[𝐿(𝑝)] = ln[𝑝∑ 𝑥𝑖𝑛𝑖=1 (1 − 𝑝)𝑛−∑ 𝑥𝑖

𝑛𝑖=1 ] = (∑ 𝑥𝑖

𝑛𝑖=1 ) ln(𝑝) + (𝑛 − ∑ 𝑥𝑖

𝑛𝑖=1 )ln (1 − 𝑝).

Differentiating gives 𝜕

𝜕𝑝ln[𝐿(𝑝)] =

𝜕

𝜕𝑝[(∑ 𝑥𝑖

𝑛𝑖=1 ) ln(𝑝) + (𝑛 − ∑ 𝑥𝑖

𝑛𝑖=1 )ln (1 − 𝑝)] =


𝑝−

𝑛−∑ 𝑥𝑖𝑛𝑖=1

1−𝑝= 0 →


𝑝=

𝑛−∑ 𝑥𝑖𝑛𝑖=1

1−𝑝→ (1 − 𝑝) ∑ 𝑥𝑖

𝑛𝑖=1 = 𝑝(𝑛 − ∑ 𝑥𝑖

𝑛𝑖=1 ) →

∑ 𝑥𝑖𝑛𝑖=1 − 𝑝 ∑ 𝑥𝑖

𝑛𝑖=1 = 𝑝𝑛 − 𝑝 ∑ 𝑥𝑖

𝑛𝑖=1 → ∑ 𝑥𝑖

𝑛𝑖=1 = 𝑝𝑛 → 𝑝 =


𝑛= �̅�. Since the

second derivative will be negative, we have found that �̂�𝑀𝐿𝐸 = �̅�.

b) Since 𝑓(𝑥; 𝑝) = 𝑝(1 − 𝑝)𝑥−1, we have 𝐿(𝑝) = ∏ 𝑓(𝑥𝑖; 𝑝)𝑛𝑖=1 = ∏ 𝑝(1 − 𝑝)𝑥𝑖−1𝑛

𝑖=1 =

𝑝𝑛(1 − 𝑝)[∑ 𝑥𝑖𝑛𝑖=1 ]−𝑛 and then the log likelihood function becomes ln[𝐿(𝑝)] =

ln[𝑝𝑛(1 − 𝑝)[∑ 𝑥𝑖𝑛𝑖=1 ]−𝑛] = 𝑛 ln(𝑝) + {[∑ 𝑥𝑖

𝑛𝑖=1 ] − 𝑛}ln (1 − 𝑝). Differentiating gives

𝜕


𝜕

𝜕𝑝[𝑛 ln(𝑝) + {[∑ 𝑥𝑖

𝑛𝑖=1 ] − 𝑛} ln(1 − 𝑝)] =

𝑛

𝑝−

[∑ 𝑥𝑖𝑛𝑖=1 ]−𝑛

1−𝑝. Equating this

with zero implies 𝑛

𝑝−

[∑ 𝑥𝑖𝑛𝑖=1 ]−𝑛

1−𝑝= 0 →

𝑛

𝑝=

[∑ 𝑥𝑖𝑛𝑖=1 ]−𝑛

1−𝑝→ (1 − 𝑝)𝑛 = 𝑝[∑ 𝑥𝑖

𝑛𝑖=1 − 𝑛] →

𝑛 − 𝑛𝑝 = 𝑝 ∑ 𝑥𝑖𝑛𝑖=1 − 𝑛𝑝 → 𝑛 = 𝑝 ∑ 𝑥𝑖

𝑛𝑖=1 → 𝑝 =

𝑛


=1

1

𝑛∑ 𝑥𝑖

𝑛𝑖=1

=1

�̅�. Since the second

derivative will be negative, we have found that �̂�𝑀𝐿𝐸 =1

�̅�.

c) Since 𝑋~𝑁𝐵(3, 𝑝), we have 𝑓(𝑥; 𝑝) = (𝑥−13−1

)𝑝3(1 − 𝑝)𝑥−3 =(𝑥−1)!

2(𝑥−3)!𝑝3(1 − 𝑝)𝑥−3 =

(𝑥−1)(𝑥−2)

2𝑝3(1 − 𝑝)𝑥−3 =

1

2(𝑥2 − 3𝑥 + 2)𝑝3(1 − 𝑝)𝑥−3. This implies that the

likelihood function 𝐿(𝑝) = ∏ 𝑓(𝑥𝑖; 𝑝)𝑛𝑖=1 = ∏ [

1

2(𝑥𝑖

2 − 3𝑥𝑖 + 2)𝑝3(1 − 𝑝)𝑥𝑖−3]𝑛𝑖=1 =

2−𝑛(𝑥𝑖2 − 3𝑥𝑖 + 2)𝑛𝑝3𝑛(1 − 𝑝)[∑ 𝑥𝑖

𝑛𝑖=1 ]−3𝑛, so the log likelihood function ln[𝐿(𝑝)] =

ln[2−𝑛(𝑥𝑖2 − 3𝑥𝑖 + 2)𝑛𝑝3𝑛(1 − 𝑝)[∑ 𝑥𝑖

𝑛𝑖=1 ]−3𝑛] = −𝑛 ln(2) + 𝑛 ln(𝑥𝑖

2 − 3𝑥𝑖 + 2) +

3𝑛 ln(𝑝) + {[∑ 𝑥𝑖𝑛𝑖=1 ] − 3𝑛} ln(1 − 𝑝). Differentiating this then gives

𝜕


𝜕

𝜕𝑝[−𝑛 ln(2) + 𝑛 ln(𝑥𝑖

2 − 3𝑥𝑖 + 2) + 3𝑛 ln(𝑝) + {[∑ 𝑥𝑖𝑛𝑖=1 ] − 3𝑛} ln(1 − 𝑝)] =

3𝑛

𝑝−

[∑ 𝑥𝑖𝑛𝑖=1 ]−3𝑛

1−𝑝= 0 → ⋯ → 𝑝 =

3

�̅�. Therefore, we have that �̂�𝑀𝐿𝐸 =

3

�̅�.

d) Since 𝑋~𝐺𝐴𝑀𝑀𝐴(𝜃, 2), we have 𝑓(𝑥; 𝜃) =1

𝜃2Γ(2)𝑥2−1𝑒−

𝑥

𝜃 =1

𝜃2 𝑥𝑒−𝑥

𝜃. This means

𝐿(𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ [

1

𝜃2 𝑥𝑖𝑒−

𝑥𝑖𝜃 ]𝑛

𝑖=1 =1

𝜃2𝑛(𝑥1 … 𝑥𝑛)𝑒−

1

𝜃∑ 𝑥𝑖

𝑛𝑖=1 so that ln[𝐿(𝜃)] =

ln [1

𝜃2𝑛(𝑥1 … 𝑥𝑛)𝑒−

1

𝜃∑ 𝑥𝑖

𝑛𝑖=1 ] = −2𝑛 ln(𝜃) + ∑ ln (𝑥𝑖)

𝑛𝑖=1 −

1

𝜃∑ 𝑥𝑖

𝑛𝑖=1 . Differentiating

gives 𝜕


𝜕

𝜕𝜃[−2𝑛 ln(𝜃) + ∑ ln (𝑥𝑖)

𝑛𝑖=1 −

1

𝜃∑ 𝑥𝑖

𝑛𝑖=1 ] = −

2𝑛

𝜃+

1

𝜃2∑ 𝑥𝑖

𝑛𝑖=1 . Then

we solve 1

𝜃2∑ 𝑥𝑖

𝑛𝑖=1 −

2𝑛

𝜃= 0 →

1

𝜃2∑ 𝑥𝑖

𝑛𝑖=1 =

2𝑛

𝜃→ 𝜃 ∑ 𝑥𝑖

𝑛𝑖=1 = 2𝑛𝜃2 → 𝜃 =


2𝑛=

�̅�

2.

Since the second derivative will be zero, we have found that 𝜃𝑀𝐿𝐸 =�̅�

2.

e) Since 𝑋~𝑊𝐸𝐼 (𝜃,1

2), we have 𝑓(𝑥; 𝜃) =

1

2√𝜃𝑥

1

2−1𝑒

−√𝑥

𝜃 =1

2√𝜃𝑥−

1

2𝑒−√

𝑥

𝜃. Thus, we have

𝐿(𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ [

1

2√𝜃𝑥

𝑖

−1

2𝑒−

√𝑥𝑖

√𝜃 ]𝑛𝑖=1 =

1

2𝑛𝜃𝑛2

(𝑥1

−1

2 … 𝑥𝑛

−1

2) 𝑒−

1

√𝜃∑ 𝑥

𝑖

12𝑛

𝑖=1 so that the

log of the likelihood function is ln[𝐿(𝜃)] = ln [1

2𝑛𝜃𝑛2

(𝑥1

−1

2 … 𝑥𝑛

−1

2) 𝑒−

1

√𝜃∑ 𝑥

𝑖

12𝑛

𝑖=1 ] =

−𝑛 ln(2) −𝑛

2ln(𝜃) + ∑ 𝑥

𝑖

−1

2𝑛𝑖=1 −

1

√𝜃∑ 𝑥

𝑖

1

2𝑛𝑖=1 . Differentiating this gives

𝜕


𝜕

𝜕𝜃[−𝑛 ln(2) −

𝑛

2ln(𝜃) + ∑ 𝑥

𝑖

−1

2𝑛𝑖=1 −

1

√𝜃∑ 𝑥

𝑖

1

2𝑛𝑖=1 ] = −

𝑛

2𝜃+

∑ √𝑥𝑖𝑛𝑖=1

2𝜃32

. Setting this equal to

zero and solving implies −𝑛

2𝜃+


2𝜃32

= 0 →∑ √𝑥𝑖

𝑛𝑖=1

2𝜃32

=𝑛

2𝜃→ 2𝜃 ∑ √𝑥𝑖

𝑛𝑖=1 = 2𝑛𝜃

3

2 →

√𝜃 =∑ √𝑥𝑖

𝑛𝑖=1

𝑛→ 𝜃 = [


𝑛]

2

. Therefore, we have found 𝜃𝑀𝐿𝐸 = [∑ √𝑋𝑖

𝑛𝑖=1

𝑛]

2

.

f) Since 𝑋~𝑃𝐴𝑅(1, 𝜅), we have 𝑓(𝑥; 𝜅) =𝜅

(1+𝑥)𝜅+1 so the likelihood function is 𝐿(𝜅) =

∏ 𝑓(𝑥𝑖; 𝜅)𝑛𝑖=1 = ∏ [𝜅(1 + 𝑥𝑖)−𝜅−1]𝑛

𝑖=1 = 𝜅𝑛 ∏ (1 + 𝑥𝑖)−𝜅−1𝑛

𝑖=1 . Then we have that

ln[𝐿(𝜅)] = ln[𝜅𝑛 ∏ (1 + 𝑥𝑖)−𝜅−1𝑛𝑖=1 ] = 𝑛 ln(𝜅) − (𝜅 + 1) ∑ ln(1 + 𝑥𝑖)𝑛

𝑖=1 . Next, we

compute the derivative so that 𝜕

𝜕𝜅ln[𝐿(𝜅)] =

𝜕

𝜕𝜅[𝑛 ln(𝜅) − (𝜅 + 1) ∑ ln(1 + 𝑥𝑖)

𝑛𝑖=1 ] =

𝑛

𝜅− ∑ ln(1 + 𝑥𝑖)

𝑛𝑖=1 . Finally, we set this result equal to zero and solve for 𝜅 to find that

𝑛

𝜅− ∑ ln(1 + 𝑥𝑖)

𝑛𝑖=1 = 0 →

𝑛

𝜅= ∑ ln(1 + 𝑥𝑖)

𝑛𝑖=1 → 𝜅 =

𝑛

∑ ln(1+𝑥𝑖)𝑛𝑖=1

. Since the second

derivative will be negative, we have found that �̂�𝑀𝐿𝐸 =𝑛

∑ ln(1+𝑋𝑖)𝑛𝑖=1

.


Question #7: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝐺𝐸𝑂(𝑝). Find the Maximum

Likelihood Estimator (MLE) for a) 𝐸(𝑋) =1

𝑝, b) 𝑉𝑎𝑟(𝑋) =

1−𝑝

𝑝2 , and c) 𝑃(𝑋 > 𝑘) = (1 − 𝑝)𝑘

where 𝑘 ∈ {1,2, … }. Do it both ways for each part to verify the Invariance Property.

a) We begin by computing �̂�𝑀𝐿𝐸 by first calculating the likelihood function 𝐿(𝑝) =

∏ 𝑓(𝑥𝑖, 𝑝)𝑛𝑖=1 = ∏ (1 − 𝑝)𝑥𝑖−1𝑝𝑛

𝑖=1 = 𝑝𝑛(1 − 𝑝)∑ (𝑥𝑖−1)𝑛𝑖=1 . Then we can compute

ln[𝐿(𝑝)] = ln[𝑝𝑛(1 − 𝑝)∑ (𝑥𝑖−1)𝑛𝑖=1 ] = 𝑛 ln(𝑝) + [∑ (𝑥𝑖 − 1)𝑛

𝑖=1 ] ln(1 − 𝑝) and then

differentiate 𝜕


𝜕

𝜕𝑝[𝑛 ln(𝑝) + [∑ (𝑥𝑖 − 1)𝑛

𝑖=1 ] ln(1 − 𝑝)] =𝑛

𝑝−

∑ (𝑥𝑖−1)𝑛𝑖=1

1−𝑝.

Setting equal to zero and solving for 𝑝 gives 𝑛

𝑝−

∑ 𝑥𝑖𝑛𝑖=1 −𝑛

1−𝑝= 0 →

𝑛

𝑝=

∑ 𝑥𝑖𝑛𝑖=1 −𝑛

1−𝑝→

(1 − 𝑝)𝑛 = 𝑝∑ 𝑥𝑖𝑛𝑖=1 − 𝑛𝑝 → 𝑛 − 𝑝𝑛 = 𝑝∑ 𝑥𝑖

𝑛𝑖=1 − 𝑝𝑛 → 𝑛 = 𝑝∑ 𝑥𝑖

𝑛𝑖=1 . This then

implies that 𝑝 =𝑛


=1

�̅�. Since the second derivative will be negative, we have

found that �̂�𝑀𝐿𝐸 =1

�̅�. By the Invariance Property of the Maximum Likelihood

Estimator, we have that 𝜏(�̂�) =1

𝑝𝑀𝐿𝐸=

1

1/�̅�= �̅� as the MLE for 𝜏(𝑝) = 𝐸(𝑋) =

1

𝑝.

b) Since �̂�𝑀𝐿𝐸 =1

�̅� and 𝜏(𝑝) = 𝑉𝑎𝑟(𝑋) =

1−𝑝

𝑝2, then 𝜏(�̂�) =

1−𝑝𝑀𝐿𝐸

𝑝𝑀𝐿𝐸2 =

1−1/�̅�

(1/�̅�)2= �̅�(�̅� − 1), by

the Invariance Property of the Maximum Likelihood Estimator.

c) Since �̂�𝑀𝐿𝐸 =1

�̅� and 𝜏(𝑝) = (1 − 𝑝)𝑘, then 𝜏(�̂�) = (1 − �̂�𝑀𝐿𝐸)

𝑘 = (1 − 1/�̅�)𝑘, by the

Invariance Property of the Maximum Likelihood Estimator.

Question #12: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝐿𝑂𝐺𝑁(𝜇, 𝜎2). Find the Maximum

Likelihood Estimator (MLE) for a) the parameters 𝜇 and 𝜎2, and b) 𝜏(𝜇, 𝜎2) = 𝐸(𝑋).

a) We have that the density function of 𝑋 is 𝑓𝑋(𝑥; 𝜇, 𝜎2) =

1

√2𝜋𝜎2

1

𝑥𝑒−

1

2𝜎2(ln𝑥−𝜇)2

, so that

the likelihood function of the sample is given by 𝐿(𝜇, 𝜎2) = ∏ 𝑓(𝑥𝑖, 𝑝)𝑛𝑖=1 =

∏ [1

√2𝜋𝜎2

1

𝑥𝑖𝑒−

1

2𝜎2(ln𝑥𝑖−𝜇)

2

]𝑛𝑖=1 = (2𝜋𝜎2)−

𝑛

2(𝑥1…𝑥𝑛)−1𝑒

−1

2𝜎2∑ (ln𝑥𝑖−𝜇)

2𝑛𝑖=1 . Then the log

likelihood function is ln[𝐿(𝜇, 𝜎2)] = ln [(2𝜋𝜎2)−𝑛

2(𝑥1…𝑥𝑛)−1𝑒

−1

2𝜎2∑ (ln𝑥𝑖−𝜇)

2𝑛𝑖=1 ] =

−𝑛

2ln(2𝜋𝜎2) − ∑ ln(𝑥𝑖)

𝑛𝑖=1 −

1

2𝜎2∑ (ln 𝑥𝑖 − 𝜇)2𝑛𝑖=1 . We differentiate this with respect

to both parameters and set the resulting expressions equal to zero so we can

simultaneously solve for the parameters, so 𝜕

𝜕𝜇ln[𝐿(𝜇, 𝜎2)] =

1

𝜎2∑ (ln 𝑥𝑖 − 𝜇)𝑛𝑖=1 = 0

and 𝜕

𝜕𝜎2ln[𝐿(𝜇, 𝜎2)] = −

𝑛

2𝜎2+

1

2𝜎4∑ (ln 𝑥𝑖 − 𝜇)2𝑛𝑖=1 = 0. The first equation implies

1

𝜎2∑ (ln 𝑥𝑖 − 𝜇)𝑛𝑖=1 = 0 → ∑ (ln 𝑥𝑖 − 𝜇)𝑛

𝑖=1 = 0 → ∑ (ln 𝑥𝑖)𝑛𝑖=1 − 𝑛𝜇 = 0 → 𝜇 =

∑ (ln𝑥𝑖)𝑛𝑖=1

𝑛

and the second −𝑛

2𝜎2+

1

2𝜎4∑ (ln 𝑥𝑖 − 𝜇)2𝑛𝑖=1 = 0 →

1

2𝜎4∑ (ln 𝑥𝑖 − 𝜇)2𝑛𝑖=1 =

𝑛

2𝜎2→

1

𝜎2∑ (ln 𝑥𝑖 − 𝜇)2𝑛𝑖=1 = 𝑛 → 𝜎2 =

1

𝑛∑ (ln 𝑥𝑖 − 𝜇)2𝑛𝑖=1 . Thus, we have that the maximum

likelihood estimators are �̂�𝑀𝐿𝐸 =1

𝑛∑ (ln 𝑥𝑖)𝑛𝑖=1 and �̂�𝑀𝐿𝐸

2 =1

𝑛∑ (ln 𝑥𝑖 − �̂�𝑀𝐿𝐸)

2𝑛𝑖=1 .

b) We know that 𝑋~𝐿𝑂𝐺𝑁(𝜇, 𝜎2) if and only if 𝑌 = ln(𝑋)~𝑁(𝜇, 𝜎2). But 𝑌 = ln(𝑋) if and

only if 𝑋 = 𝑒𝑌 implies that 𝐸(𝑋) = 𝐸(𝑒𝑌) = 𝑀𝑌(1) = 𝑒𝜇(1)+

𝜎212

2 = 𝑒𝜇+𝜎2

2 . By the

Invariance Property of the Maximum Likelihood Estimator, we can conclude that

𝜏(�̂�, �̂�2)𝑀𝐿𝐸 = 𝜏(�̂�𝑀𝐿𝐸 , �̂�𝑀𝐿𝐸2 ) = 𝑒�̂�𝑀𝐿𝐸+

1

2�̂�𝑀𝐿𝐸2

is the MLE for 𝜏(𝜇, 𝜎2) = 𝐸(𝑋) = 𝑒𝜇+𝜎2

2 .

Question #17: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝑈𝑁𝐼𝐹(𝜃 − 1, 𝜃 + 1). a) Show that

the sample mean �̅� is an unbiased estimator for 𝜃; b) show that the midrange 𝑀 =𝑋(1)+𝑋(𝑛)

2

is also an unbiased estimator for the parameter 𝜃; c) which one has a smaller variance?

a) To show that �̅� is an unbiased estimator for 𝜃, we must verify that 𝐸(�̅�) = 𝜃. But we

see that 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) =

1


𝑛𝑖=1 ) =

1

𝑛∑ 𝐸(𝑋𝑖)𝑛𝑖=1 =

1

𝑛∑ [

(𝜃+1)+(𝜃−1)

2]𝑛

𝑖=1 =

1

𝑛∑ 𝜃𝑛𝑖=1 =

1

𝑛𝑛𝜃 = 𝜃, so it is clear that the sample mean is an unbiased estimator for 𝜃.

b) We have that 𝐸(𝑀) = 𝐸 (𝑋(1)+𝑋(𝑛)

2) =

1

2𝐸(𝑋(1) + 𝑋(𝑛)) =

1

2[𝐸(𝑋(1)) + 𝐸(𝑋(𝑛))]. We

must therefore compute the mean of the smallest and largest order statistics, which

we can do by first finding their density functions. We first note that since

𝑋~𝑈𝑁𝐼𝐹(𝜃 − 1, 𝜃 + 1), then 𝑓𝑋(𝑡) =1

(𝜃+1)−(𝜃−1)=

1

2 whenever 𝑡 ∈ (𝜃 − 1, 𝜃 + 1) and

𝐹𝑋(𝑡) = ∫1

2

𝑡

𝜃−1𝑑𝑥 =

1

2[𝑥]𝜃−1

𝑡 =𝑡−(𝜃−1)

2 whenever 𝑡 ∈ (𝜃 − 1, 𝜃 + 1). Then the

distribution function of 𝑋(𝑛) is given by 𝐹𝑛(𝑡) = 𝑃(𝑋(𝑛) ≤ 𝑡) = 𝑃(𝑋𝑖 ≤ 𝑡)𝑛 =

(𝑡−(𝜃−1)

2)𝑛

=(𝑡−𝜃+1)𝑛

2𝑛 so the density function of 𝑋(𝑛) is 𝑓𝑛(𝑡) =

𝑑

𝑑𝑡𝐹𝑛(𝑡) =

𝑛(𝑡−𝜃+1)𝑛−1

2𝑛.

We can then compute the mean of 𝑋(𝑛) as 𝐸(𝑋(𝑛)) = ∫ 𝑡𝑓𝑛(𝑡)𝜃+1

𝜃−1𝑑𝑡 =

∫ 𝑡𝑛(𝑡−𝜃+1)𝑛−1

2𝑛

𝜃+1

𝜃−1𝑑𝑡. This integral can be calculated by completing the substitution

𝑢 = 𝑡 − 𝜃 + 1 so that 𝑑𝑢 = 𝑑𝑡 and 𝑡 = 𝑢 + 𝜃 − 1. This then implies

∫ 𝑡𝑛(𝑡−𝜃+1)𝑛−1

2𝑛

𝜃+1

𝜃−1𝑑𝑡 =

𝑛

2𝑛∫ (𝑢 + 𝜃 − 1)𝑢𝑛−12

0𝑑𝑢 =

𝑛

2𝑛∫ 𝑢𝑛 + 𝜃𝑢𝑛−1 − 𝑢𝑛−12

0𝑑𝑢 =

𝑛

2𝑛[𝑢𝑛+1

𝑛+1+

𝜃𝑢𝑛

𝑛−

𝑢𝑛

𝑛]0

2

=𝑛

2𝑛[2𝑛+1

𝑛+1+

𝜃2𝑛

𝑛−

2𝑛

𝑛] =

2𝑛

𝑛+1+ 𝜃 − 1. We can similarly compute

that the expected value of the first order statistic is 𝐸(𝑋(1)) = −2𝑛

𝑛+1+ 𝜃 + 1. Thus, we

have that 𝐸(𝑀) =1

2[𝐸(𝑋(1)) + 𝐸(𝑋(𝑛))] =

1

2[(𝜃 −

2𝑛

𝑛+1+ 1) + (𝜃 +

2𝑛

𝑛+1− 1)] =

1

2[2𝜃] = 𝜃, so the midrange is also an unbiased estimator for the parameter 𝜃.

c) We have that 𝑉𝑎𝑟(�̅�) = 𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) =

1

𝑛2𝑉𝑎𝑟(∑ 𝑋𝑖

𝑛𝑖=1 ) =

1

𝑛2∑ 𝑉𝑎𝑟(𝑋𝑖)𝑛𝑖=1 =

1

𝑛2∑

[(𝜃+1)−(𝜃−1)]2

12

𝑛𝑖=1 =

1

𝑛2∑

4

12

𝑛𝑖=1 =

1

𝑛2∑

1

3

𝑛𝑖=1 =

1

𝑛2𝑛

3=

1

3𝑛. Similarly, we can calculate

that 𝑉𝑎𝑟(𝑀) = 𝑉𝑎𝑟 (𝑋(1)+𝑋(𝑛)

2) =

1

4𝑉𝑎𝑟(𝑋(1) + 𝑋(𝑛)).

Question #21: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝐵𝐼𝑁(1, 𝑝). a) Find the Cramer-Rao

lower bound for the variances of all unbiased estimators of 𝑝; b) Find the Cramer-Rao lower

bound for the variances of unbiased estimators of 𝑝(1 − 𝑝); c) Find a UMVUE of 𝑝.

a) We have that 𝐶𝑅𝐿𝐵 =[𝜏′(𝑝)]

2

𝑛𝐸[(𝜕

𝜕𝑝ln𝑓(𝑋;𝑝))

2], so we compute each of these parts individually.

First, we have 𝜏(𝑝) = 𝑝, so 𝜏′(𝑝) = 1 and [𝜏′(𝑝)]2 = 1. Next, since 𝑋~𝐵𝐼𝑁(1, 𝑝) we

know that 𝑓𝑋(𝑥) = (1𝑥)𝑝𝑥(1 − 𝑝)1−𝑥 = 𝑝𝑥(1 − 𝑝)1−𝑥 and so 𝑓(𝑋; 𝑝) = 𝑝𝑋(1 − 𝑝)1−𝑋,

which means that ln 𝑓(𝑋; 𝑝) = 𝑋 ln(𝑝) + (1 − 𝑋) ln(1 − 𝑝). Taking the derivative and

squaring gives 𝜕

𝜕𝑝ln 𝑓(𝑋; 𝑝) =

𝑋

𝑝−

1−𝑋

1−𝑝=

(1−𝑝)𝑋−𝑝(1−𝑋)

𝑝(1−𝑝)=

𝑋−𝑝𝑋−𝑝+𝑝𝑋

𝑝(1−𝑝)=

𝑋−𝑝

𝑝(1−𝑝)→

(𝜕

𝜕𝑝ln 𝑓(𝑋; 𝑝))

2

= (𝑋−𝑝

𝑝(1−𝑝))2

=𝑋2−2𝑝𝑋+𝑝2

𝑝2(1−𝑝)2. Finally, we compute 𝐸 [(

𝜕

𝜕𝑝ln 𝑓(𝑋; 𝑝))

2

] =

𝐸 [𝑋2−2𝑝𝑋+𝑝2

𝑝2(1−𝑝)2] =

1

𝑝2(1−𝑝)2𝐸(𝑋2 − 2𝑝𝑋 + 𝑝2) =

1

𝑝2(1−𝑝)2[𝐸(𝑋2) − 2𝑝𝐸(𝑋) + 𝑝2] =

1

𝑝2(1−𝑝)2[(𝑝(1 − 𝑝) + 𝑝2) − 2𝑝(𝑝) + 𝑝2] =

1

𝑝2(1−𝑝)2[𝑝 − 𝑝2 + 𝑝2 − 2𝑝2 + 𝑝2] =

1

𝑝2(1−𝑝)2[𝑝 − 𝑝2] =

𝑝(1−𝑝)

𝑝2(1−𝑝)2=

1

𝑝(1−𝑝). Thus, we have found that 𝐶𝑅𝐿𝐵 =

𝑝(1−𝑝)

𝑛.

b) Now, 𝜏(𝑝) = 𝑝(1 − 𝑝) = 𝑝 − 𝑝2, so [𝜏′(𝑝)]2 = [1 − 2𝑝]2 = 1 − 4𝑝 + 4𝑝2, so the

Cramer-Rao Lower Bound becomes 𝐶𝑅𝐿𝐵 =(1−4𝑝+4𝑝2)𝑝(1−𝑝)

𝑛.

c) Since for the estimator �̂� = �̅�, we have 𝐸(�̂�) = 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) =

1


𝑛𝑖=1 ) =

1


1

𝑛∑ 𝑝𝑛𝑖=1 =

1

𝑛𝑛𝑝 = 𝑝 and then 𝑉𝑎𝑟(�̂�) = 𝑉𝑎𝑟(�̅�) = 𝑉𝑎𝑟 (

1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) =

1


𝑛𝑖=1 ) =

1


1

𝑛2∑ 𝑝(1 − 𝑝)𝑛𝑖=1 =

1

𝑛2𝑛𝑝(1 − 𝑝) =

𝑝(1−𝑝)

𝑛=

𝐶𝑅𝐿𝐵, we can conclude that �̂� = �̅� is a Uniform Minimum Variance Unbiased

Estimator (UMVUE) for the parameter 𝑝 in 𝑋~𝐵𝐼𝑁(1, 𝑝).

Question #22: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝑁(𝜇, 9). a) Find the Cramer-Rao

lower bound for the variances of unbiased estimators of 𝜇; b) is the Maximum Likelihood

Estimator �̂�𝑀𝐿𝐸 = �̅� a UMVUE for the parameter 𝜇?

a) We have 𝜏(𝜇) = 𝜇, so 𝜏′(𝜇) = 1 and [𝜏′(𝜇)]2 = 1. Next, since 𝑋~𝑁(𝜇, 9) we know that

the density is 𝑓𝑋(𝑥) =1

√18𝜋𝑒−

1

18(𝑥−𝜇)2 , so that we have 𝑓(𝑋; 𝜇) =

1

√18𝜋𝑒−

1

18(𝑋−𝜇)2 and

ln 𝑓(𝑋; 𝜇) = −1

2ln(18𝜋) −

1

18(𝑋 − 𝜇)2. We then differentiate twice to obtain

𝜕

𝜕𝜇ln 𝑓(𝑋; 𝜇) =

1

9(𝑋 − 𝜇) →

𝜕2

𝜕𝜇2ln 𝑓(𝑋; 𝜇) = −

1

9. Since we have shown than the

expression 𝐸 [(𝜕

𝜕𝜇ln 𝑓(𝑋; 𝑝))

2

] = −𝐸 [𝜕2

𝜕𝜇2ln 𝑓(𝑋; 𝜇)] and −𝐸 (−

1

9) =

1

9, we can

conclude that the Cramer-Rao Lower Bound is 𝐶𝑅𝐿𝐵 =9

𝑛. This then means that

𝑉𝑎𝑟(𝑇) ≥9

𝑛 for any unbiased estimator 𝑇 of the parameter 𝜇 in 𝑋~𝑁(𝜇, 9).

b) We first verify that 𝐸(�̂�𝑀𝐿𝐸) = 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) =

1


𝑛𝑖=1 ) =

1


1

𝑛∑ 𝜇𝑛𝑖=1 =

1

𝑛𝑛𝜇 = 𝜇, so �̂�𝑀𝐿𝐸 = �̅� is an unbiased estimator for 𝜇. Then we compute

𝑉𝑎𝑟(�̂�𝑀𝐿𝐸) = 𝑉𝑎𝑟(�̅�) = 𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) =

1


𝑛𝑖=1 ) =

1


1

𝑛2∑ 9𝑛𝑖=1 =

1

𝑛29𝑛 =

9

𝑛= 𝐶𝑅𝐿𝐵, so that �̂�𝑀𝐿𝐸 = �̅� a UMVUE for the parameter 𝜇.

Question #23: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝑁(0, 𝜃). a) Is the Maximum

Likelihood Estimator (MLE) for 𝜃 unbiased?; b) is the MLE also a UMVUE for 𝜃?

a) We first find 𝜃𝑀𝐿𝐸 by noting that since 𝑋~𝑁(0, 𝜃), then its density function is 𝑓𝑋(𝑥) =

1

√2𝜋𝜃𝑒−

1

2𝜃𝑥2 so the likelihood function is 𝐿(𝜃) = ∏ 𝑓𝑋(𝑥; 𝜃)

𝑛𝑖=1 = ∏

1

√2𝜋𝜃𝑒−

1

2𝜃𝑥𝑖2

𝑛𝑖=1 =

(2𝜋𝜃)−𝑛

2𝑒−1

2𝜃∑ 𝑥𝑖

2𝑛𝑖=1 and then ln[𝐿(𝜃)] = −

𝑛

2ln(2𝜋𝜃) −

1

2𝜃∑ 𝑥𝑖

2𝑛𝑖=1 . Next, we

differentiate so that 𝜕

𝜕𝜃ln[𝐿(𝜃)] = −

𝑛2𝜋

4𝜋𝜃+

1

2𝜃2∑ 𝑥𝑖

2𝑛𝑖=1 = 0 →

1

2𝜃2∑ 𝑥𝑖

2𝑛𝑖=1 =

𝑛

2𝜃→

∑ 𝑥𝑖2𝑛

𝑖=1 = 𝑛𝜃 → 𝜃 =1

𝑛∑ 𝑥𝑖

2𝑛𝑖=1 . Since the second derivative is negative, we have

𝜃𝑀𝐿𝐸 =1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 . We verify unbiasedness by computing 𝐸(𝜃𝑀𝐿𝐸) = 𝐸 (

1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 ) =

1


2𝑛𝑖=1 ) =

1

𝑛∑ 𝐸(𝑋𝑖

2)𝑛𝑖=1 =

1

𝑛∑ (𝜃 + 02)𝑛𝑖=1 =

1

𝑛∑ 𝜃𝑛𝑖=1 =

1

𝑛𝑛𝜃 = 𝜃.

b) The estimator 𝜃𝑀𝐿𝐸 will be a UMVUE for 𝜃 if 𝑉𝑎𝑟(𝜃𝑀𝐿𝐸) = 𝐶𝑅𝐿𝐵. We therefore begin

by computing the Cramer-Rao Lower Bound. First, we have 𝜏(𝜃) = 𝜃, so 𝜏′(𝜃) = 1

and [𝜏′(𝜃)]2 = 1. Next, since we previously found that 𝑓(𝑋; 𝜃) =1

√2𝜋𝜃𝑒−

1

2𝜃𝑋2

, then we

have ln 𝑓(𝑋; 𝜃) = −1

2ln(2𝜋𝜃) −

1

2𝜃𝑋2 so that

𝜕

𝜕𝜃ln 𝑓(𝑋; 𝜃) = −

1

2𝜃+

𝑋2

2𝜃2. We then find

𝜕2

𝜕𝜃2ln 𝑓(𝑋; 𝜃) =

1

2𝜃2−

2𝑋2

2𝜃3=

1

2𝜃2−

𝑋2

𝜃3 and take the negative of its expected value to

obtain −𝐸 [1

2𝜃2−

𝑋2

𝜃3] = − [

1

2𝜃2−

1

𝜃3𝐸(𝑋2)] =

1

𝜃3(𝜃 + 02) −

1

2𝜃2=

1

𝜃2−

1

2𝜃2=

1

2𝜃2. This

implies that 𝐶𝑅𝐿𝐵 =2𝜃2

𝑛. We must verify that the variance of our estimator is equal

to this lower bound, so we compute 𝑉𝑎𝑟(𝜃𝑀𝐿𝐸) = 𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 ) =

1


2𝑛𝑖=1 ) =

1

𝑛2∑ 𝑉𝑎𝑟(𝑋𝑖

2)𝑛𝑖=1 . In order to compute 𝑉𝑎𝑟(𝑋𝑖

2), we use the formula

𝑉𝑎𝑟(𝑋𝑖2) = 𝐸(𝑋𝑖

4) − [𝐸(𝑋𝑖2)]2 = ⋯ = 3𝜃2 − 𝜃2 = 2𝜃2 by finding the moments of 𝑋𝑖

using the derivatives of the Moment Generating Function at 𝑡 = 0. Then we have that

𝑉𝑎𝑟(𝜃𝑀𝐿𝐸) =1

𝑛2∑ 𝑉𝑎𝑟(𝑋𝑖

2)𝑛𝑖=1 =

1

𝑛2∑ 2𝜃2𝑛𝑖=1 =

1

𝑛2𝑛2𝜃2 =

2𝜃2

𝑛= 𝐶𝑅𝐿𝐵, which verifies

that the Maximum Likelihood Estimator is a UMVUE for the parameter 𝜃.


Question #31: Let 𝜃 and �̃� be the MLE and MME estimators for the parameter 𝜃, where

𝑋1, … , 𝑋𝑛 is a random sample of size 𝑛 from a Uniform distribution such that 𝑋𝑖~𝑈𝑁𝐼𝐹(0, 𝜃).

Show that a) 𝜃 is MSE consistent, and b) �̃� is MSE consistent.

a) We first derive the MLE 𝜃 for 𝜃. Since 𝑋~𝑈𝑁𝐼𝐹(0, 𝜃), we know that the density

function is 𝑓(𝑥; 𝜃) =1

𝜃 for 𝑥 ∈ (0, 𝜃). This allows us to construct the likelihood

function 𝐿(𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ 𝜃−1𝑛

𝑖=1 = 𝜃−𝑛 whenever 𝑥1:𝑛 ≥ 0 and 𝑥𝑛:𝑛 ≤ 𝜃 and

zero otherwise. Then the log likelihood function is ln[𝐿(𝜃)] = −𝑛 ln(𝜃) so that

𝜕

𝜕𝜃ln[𝐿(𝜃)] = −

𝑛

𝜃< 0 for all 𝑛 and 𝜃. This means that 𝐿(𝜃) = 𝜃−𝑛 is a decreasing

function of 𝜃 for 𝑥𝑛:𝑛 ≤ 𝜃 since its first derivative is always negative, so we can

conclude that the MLE is the largest order statistic, so 𝜃 = 𝑋𝑛:𝑛. Next, we show that

this estimator is MSE consistent, which means verifying that lim𝑛→∞

𝐸[𝑋𝑛:𝑛 − 𝜃]2 = 0.

But then we can see that lim𝑛→∞

𝐸[𝑋𝑛:𝑛 − 𝜃]2 = lim𝑛→∞

𝐸[𝑋𝑛:𝑛2 − 2𝜃𝑋𝑛:𝑛 + 𝜃2] =

lim𝑛→∞

[𝐸(𝑋𝑛:𝑛2 ) − 2𝜃𝐸(𝑋𝑛:𝑛) + 𝜃2]. In order to compute this limit, we must find the first

and second moments of the largest order statistic. But we already know that 𝑓𝑛(𝑦) =

𝑛𝑓(𝑦; 𝜃)𝐹(𝑦; 𝜃)𝑛−1 =𝑛

𝜃(

𝑦

𝜃)

𝑛−1

=𝑛𝑦𝑛−1

𝜃𝑛 , so we can calculate 𝐸(𝑋𝑛:𝑛) = ∫ 𝑦𝑓𝑛(𝑦)𝜃

0𝑑𝑦 =

∫ 𝑦𝑛𝑦𝑛−1

𝜃𝑛

𝜃

0𝑑𝑦 =

𝑛

𝜃𝑛 ∫ 𝑦𝑛𝜃

0𝑑𝑦 =

𝑛

𝜃𝑛 [𝑦𝑛+1

𝑛+1]

0

𝜃

=𝑛

𝜃𝑛 (𝜃𝑛+1

𝑛+1− 0) =

𝑛

𝑛+1𝜃 and 𝐸(𝑋𝑛:𝑛

2 ) =

∫ 𝑦2𝑓𝑛(𝑦)𝜃

0𝑑𝑦 = ∫ 𝑦2 𝑛𝑦𝑛−1

𝜃𝑛

𝜃

0𝑑𝑦 =

𝑛

𝜃𝑛 ∫ 𝑦𝑛+1𝜃

0𝑑𝑦 =

𝑛

𝜃𝑛 [𝑦𝑛+2

𝑛+2]

0

𝜃

=𝑛

𝜃𝑛 (𝜃𝑛+2

𝑛+2) =

𝑛

𝑛+2𝜃2.

Thus, we have lim𝑛→∞

[𝐸(𝑋𝑛:𝑛2 ) − 2𝜃𝐸(𝑋𝑛:𝑛) + 𝜃2] = lim

𝑛→∞[

𝑛

𝑛+2𝜃2 − 2𝜃

𝑛

𝑛+1𝜃 + 𝜃2] =

lim𝑛→∞

[𝑛

𝑛+2𝜃2 −

2𝑛

𝑛+1𝜃2 + 𝜃2] = 𝜃2 − 2𝜃2 + 𝜃2 = 0. That this limit is zero verifies that

the maximum likelihood estimator 𝜃 = 𝑋𝑛:𝑛 is mean square error (MSE) consistent.

b) We first derive the MME �̃� for 𝜃. Since 𝑋~𝑈𝑁𝐼𝐹(0, 𝜃), we know that 𝐸(𝑋) =𝜃

2 so we

can equate 𝜇1′ = 𝑀1

′ →𝜃

2=

1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 →

𝜃

2= �̅� → 𝜃 = 2�̅�. This means that �̃� = 2�̅�.

Next, we show that this estimator is MSE consistent, which means verifying that

lim𝑛→∞

𝐸[2�̅� − 𝜃]2 = 0. But we have lim𝑛→∞

𝐸[2�̅� − 𝜃]2 = lim𝑛→∞

𝐸[4�̅�2 − 4𝜃�̅� + 𝜃2] =

lim𝑛→∞

[4𝐸(�̅�2) − 4𝜃𝐸(�̅�) + 𝜃2]. We therefore compute 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ) =

1


𝑛𝑖=1 ) =

1

𝑛∑ 𝐸(𝑋𝑖)

𝑛𝑖=1 =

1

𝑛∑

𝜃

2

𝑛𝑖=1 =

1

𝑛𝑛

𝜃

2=

𝜃

2 and 𝐸(�̅�2) = 𝑉𝑎𝑟(�̅�) + 𝐸(�̅�)2 =

𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ) + (

𝜃

2)

2

=1

𝑛2∑ 𝑉𝑎𝑟(𝑋𝑖)

𝑛𝑖=1 +

𝜃2

4=

1

𝑛2∑

𝜃2

12

𝑛𝑖=1 +

𝜃2

4=

1

𝑛2 𝑛𝜃2

12+

𝜃2

4=

𝜃2

12𝑛+

𝜃2

4=

(3𝑛+1)𝜃2

12𝑛. Thus, we can compute that lim

𝑛→∞[4𝐸(�̅�2) − 4𝜃𝐸(�̅�) + 𝜃2] =

lim𝑛→∞

[4(3𝑛+1)𝜃2

12𝑛− 4𝜃

𝜃

2+ 𝜃2] = lim

𝑛→∞[

(3𝑛+1)𝜃2

3𝑛− 2𝜃2 + 𝜃2] = 𝜃2 − 2𝜃2 + 𝜃2 = 0. That

this limit is zero verifies that the MME �̃� = 2�̅� is mean square error (MSE) consistent.

Question #29: Let 𝑋1, … , 𝑋𝑛 be a random sample of size 𝑛 from a Bernoulli distribution such

that 𝑋𝑖~𝐵𝐼𝑁(1, 𝑝). For a Uniform prior density 𝑝~𝑈𝑁𝐼𝐹(0,1) and a squared error loss

function 𝐿(𝑡; 𝑝) = (𝑡 − 𝑝)2, a) find the posterior distribution of the unknown parameter 𝑝,

b) find the Bayes estimator of 𝑝, and c) find the Bayes risk for the Bayes estimator of 𝑝 above.

a) We have that the posterior density is given by 𝑓P|𝑥(𝑝) =𝑓(𝑥1,…,𝑥𝑛;𝑝)𝑝(𝑝)

∫ 𝑓(𝑥1,…,𝑥𝑛;𝑝)𝑝(𝑝)𝑑𝑝, where

𝑓(𝑥1, … , 𝑥𝑛; 𝑝) = ∏ 𝑓(𝑥𝑖; 𝑝)𝑛𝑖=1 = ∏ 𝑝𝑥(1 − 𝑝)1−𝑥𝑛

𝑖=1 = 𝑝∑ 𝑥𝑖𝑛𝑖=1 (1 − 𝑝)𝑛−∑ 𝑥𝑖

𝑛𝑖=1 since

the random variables are independent and identically distributed and 𝑝(𝑝) = 1 since

the prior density is uniform. We then express ∫ 𝑝∑ 𝑥𝑖𝑛𝑖=1 (1 − 𝑝)𝑛−∑ 𝑥𝑖

𝑛𝑖=1

1

0𝑑𝑝 in terms of

the beta distribution. Recall that if 𝑌~𝐵𝐸𝑇𝐴(𝑎, 𝑏), then its density is 𝑓(𝑦; 𝑎, 𝑏) =

1

𝐵(𝑎,𝑏)𝑦𝑎−1(1 − 𝑦)𝑏−1 where 𝐵(𝑎, 𝑏) =

Γ(𝑎)Γ(𝑏)

Γ(𝑎+𝑏). Next, we must define 𝑎 = ∑ 𝑥𝑖

𝑛𝑖=1 and

𝑏 = 𝑛 − ∑ 𝑥𝑖𝑛𝑖=1 , so we can write ∫ 𝑝∑ 𝑥𝑖

𝑛𝑖=1 (1 − 𝑝)𝑛−∑ 𝑥𝑖

𝑛𝑖=1

1

0𝑑𝑝 = 𝐵(𝑎 + 1, 𝑏 + 1) =

𝐵(∑ 𝑥𝑖𝑛𝑖=1 + 1, 𝑛 − ∑ 𝑥𝑖

𝑛𝑖=1 + 1). Thus, we have 𝑓P|𝑥(𝑝) =

𝑝∑ 𝑥𝑖𝑛𝑖=1 (1−𝑝)𝑛−∑ 𝑥𝑖

𝑛𝑖=1

𝐵(∑ 𝑥𝑖𝑛𝑖=1 +1,𝑛−∑ 𝑥𝑖

𝑛𝑖=1 +1)

=

1

𝐵(𝑎+1,𝑏+1)𝑝𝑎(1 − 𝑝)𝑏, which verifies that the random variable given by

P|𝑥~𝐵𝐸𝑇𝐴(∑ 𝑥𝑖𝑛𝑖=1 + 1, 𝑛 − ∑ 𝑥𝑖

𝑛𝑖=1 + 1) ≡ 𝐵𝐸𝑇𝐴(𝑎 + 1, 𝑏 + 1).

b) For some random variable 𝑌~𝐵𝐸𝑇𝐴(𝑎, 𝑏), we know that 𝐸(𝑌) =𝑎

𝑎+𝑏. Moreover,

Theorem 9.5.2 states that when we have a squared error loss function, the Bayes

estimator is simply the expected value of the posterior distribution. This implies that

the Bayes estimator of 𝑝 is given by �̂�𝐵𝐸 =∑ 𝑥𝑖

𝑛𝑖=1 +1

∑ 𝑥𝑖𝑛𝑖=1 +1+𝑛−∑ 𝑥𝑖

𝑛𝑖=1 +1

=∑ 𝑥𝑖

𝑛𝑖=1 +1

𝑛+2.

c) The risk function in this case is 𝑅𝑇(𝑝) = 𝐸[(𝑇 − 𝑝)2], where 𝑇 =∑ 𝑥𝑖

𝑛𝑖=1 +1

𝑛+2 is the Bayes

Estimator derived above. We would therefore substitute for 𝑇 in the risk function,

evaluate the expected value of that expression and then compute ∫ 𝐸[(𝑇 − 𝑝)2]1

0𝑑𝑝.

Question #34: Consider a random sample of size 𝑛 from a distribution with discrete

probability mass function 𝑓𝑋(𝑥; 𝑝) = (1 − 𝑝)𝑥𝑝 for 𝑥 ∈ {0,1,2, … }. a) Find the MLE of the

unknown parameter 𝑝. b) Find the MLE of 𝜃 =1−𝑝

𝑝. c) Find the CRLB for variances of all

unbiased estimators of the parameter 𝜃 above. d) Is the MLE of 𝜃 =1−𝑝

𝑝 a UMVUE? e) Is the

MLE of 𝜃 =1−𝑝

𝑝 also MSE consistent? f) Compute the asymptotic distribution of the MLE of

𝜃 =1−𝑝

𝑝. g) If we have the estimator 𝜃 =

𝑛

𝑛+1�̅�, then find the risk functions of both 𝜃 and �̅�

using the loss function given by 𝐿(𝑡; 𝜃) =(𝑡−𝜃)2

𝜃2+𝜃.

a) We have 𝐿(𝑝) = ∏ 𝑓(𝑥𝑖; 𝑝)𝑝𝑖=1 = ∏ (1 − 𝑝)𝑥𝑖𝑝 𝑝

𝑖=1 = 𝑝𝑛(1 − 𝑝)∑ 𝑥𝑖𝑛𝑖=1 , so that

ln[𝐿(𝑝)] = 𝑛 ln(𝑝) + ∑ 𝑥𝑖𝑛𝑖=1 ln (1 − 𝑝). Then we have that

𝜕


𝑛

𝑝−


1−𝑝.

Setting this equal to zero and solving for 𝑝 gives the estimator �̂�𝑀𝐿𝐸 =1

1+�̅�.

b) By the Invariance Property, we have that the estimator is 𝜃𝑀𝐿𝐸 =1−𝑝𝑀𝐿𝐸

𝑝𝑀𝐿𝐸= �̅�.

c) Since 𝜃 = 𝜏(𝑝) =1−𝑝

𝑝=

1

𝑝− 1, then 𝜏′(𝑝) = −

1

𝑝2 and [𝜏′(𝑝)]2 =1

𝑝4. Then since

𝑓(𝑋; 𝑝) = (1 − 𝑝)𝑋𝑝, we can compute ln 𝑓(𝑋; 𝑝) = ln(𝑝) + 𝑋ln (1 − 𝑝) so that

𝜕

𝜕𝑝ln 𝑓(𝑋; 𝑝) =

1

𝑝−

𝑋

1−𝑝 and

𝜕2

𝜕𝑝2ln 𝑓(𝑋; 𝑝) = −

1

𝑝2−

𝑋

(1−𝑝)2. We can then compute the

negative of the expected value of this second derivative so that −𝐸 [𝜕2

𝜕𝑝2 ln 𝑓(𝑋; 𝑝)] =

1

𝑝2+

1

(1−𝑝)2𝐸(𝑋) =

1

𝑝2+

1−𝑝

𝑝(1−𝑝)2=

(1−𝑝)2+𝑝(1−𝑝)

𝑝2(1−𝑝)2=

1−2𝑝+𝑝2+𝑝−𝑝2

𝑝2(1−𝑝)2=

1−𝑝

𝑝2(1−𝑝)2=

1

𝑝2(1−𝑝).

These results imply that 𝐶𝑅𝐿𝐵 =𝑝2(1−𝑝)

𝑛𝑝4 =1−𝑝

𝑛𝑝2 .

d) We first verify that 𝐸(𝜃𝑀𝐿𝐸) = 𝐸(�̅�) = 𝐸 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ) =

1

𝑛∑ 𝐸(𝑋𝑖)

𝑛𝑖=1 =

1

𝑛𝑛

1−𝑝

𝑝=

1−𝑝

𝑝,

so that the MLE is an unbiased estimator of 𝜃 =1−𝑝

𝑝. Next, we compute 𝑉𝑎𝑟(𝜃𝑀𝐿𝐸) =

𝑉𝑎𝑟(�̅�) = 𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ) =

1


𝑛𝑖=1 =

1

𝑛2 𝑛1−𝑝

𝑝2 =1−𝑝

𝑛𝑝2 = 𝐶𝑅𝐿𝐵, which verifies

that 𝜃𝑀𝐿𝐸 = �̅� is the UMVUE for the parameter 𝜃 =1−𝑝

𝑝.

e) To verify that 𝜃𝑀𝐿𝐸 = �̅� is MSE consistent, we must show that lim𝑛→∞

𝐸 [�̅� −1−𝑝

𝑝]

2

= 0.

But we can see that we have lim𝑛→∞

𝐸 [�̅� −1−𝑝

𝑝]

2

= lim𝑛→∞

𝐸 [�̅�2 −2(1−𝑝)

𝑝�̅� +

(1−𝑝)2

𝑝2 ] =

lim𝑛→∞

[𝐸(�̅�2) −2(1−𝑝)

𝑝𝐸(�̅�) +

(1−𝑝)2

𝑝2 ], so we must compute the expectation of both the

mean and the mean squared. However, we already know that 𝐸(�̅�) =1−𝑝

𝑝= 𝜃 since

𝜃𝑀𝐿𝐸 is unbiased. Then 𝐸(�̅�2) = 𝑉𝑎𝑟(�̅�) + 𝐸(�̅�)2 = 𝑉𝑎𝑟 (1

𝑛∑ 𝑋𝑖

𝑛𝑖=1 ) + (

1−𝑝

𝑝)

2

=

1


𝑛𝑖=1 +

(1−𝑝)2

𝑝2=

1

𝑛2∑

1−𝑝

𝑝2𝑛𝑖=1 +

(1−𝑝)2

𝑝2=

1

𝑛2𝑛

1−𝑝

𝑝2+

(1−𝑝)2

𝑝2=

1−𝑝

𝑛𝑝2+

(1−𝑝)2

𝑝2.

Thus lim𝑛→∞

[𝐸(�̅�2) −2(1−𝑝)

𝑝𝐸(�̅�) +

(1−𝑝)2

𝑝2] = lim

𝑛→∞[

1−𝑝

𝑛𝑝2+

(1−𝑝)2

𝑝2−

2(1−𝑝)

𝑝

1−𝑝

𝑝+

(1−𝑝)2

𝑝2] =

(1−𝑝)2

𝑝2 −2(1−𝑝)2

𝑝2 +(1−𝑝)2

𝑝2 = 0. This shows that 𝜃𝑀𝐿𝐸 = �̅� is MSE consistent.

f) We use Definition 9.4.5, which states that for large values of 𝑛, the MLE estimator is

distributed normal with mean 𝜃 =1−𝑝

𝑝 and variance 𝐶𝑅𝐿𝐵. Since we previously found

that 𝐶𝑅𝐿𝐵 =1−𝑝

𝑛𝑝2 , we can conclude that 𝜃𝑀𝐿𝐸~𝑁 (1−𝑝

𝑝,

1−𝑝

𝑛𝑝2).

g) Definition 9.5.2 states that the risk function is the expected loss 𝑅𝑇(𝜃) = 𝐸[𝐿(𝑇; 𝜃)].

In this case, the loss function is 𝐿(𝑡; 𝜃) =(𝑡−𝜃)2

𝜃2+𝜃=

𝑡2−2𝜃𝑡+𝜃2

𝜃2+𝜃. Therefore, for the

estimator �̅� we compute 𝑅�̅�(𝜃) = 𝐸 [�̅�2−2𝜃�̅�+𝜃2

𝜃2+𝜃] =

1

𝜃2+𝜃[𝐸(�̅�2) − 2𝜃𝐸(�̅�) + 𝜃2] =

1

𝜃2+𝜃[

1−𝑝

𝑛𝑝2 +(1−𝑝)2

𝑝2 − 2𝜃1−𝑝

𝑝+ 𝜃2] =

1

𝜃2+𝜃[

1

𝑛𝑝𝜃2 + 𝜃2 − 2𝜃2 + 𝜃2] =

𝜃2

(𝜃2+𝜃)𝑛𝑝=

𝜃

(𝜃+1)𝑛𝑝=

(1−𝑝)/𝑝

[(1−𝑝)/𝑝+1]𝑛𝑝=

1−𝑝

[1/𝑝]𝑛𝑝2 =1−𝑝

𝑛𝑝=

𝜃

𝑛. Similarly, for the estimator 𝜃 =

𝑛

𝑛+1�̅� we

can compute 𝑅�̂�(𝜃) = 𝐸 [(

𝑛

𝑛+1�̅�)

2−2𝜃(

𝑛

𝑛+1�̅�)+𝜃2

𝜃2+𝜃] = ⋯ =

𝑛+𝜃

(𝜃+1)(𝑛+1)2.

Question #36: Let 𝑋1, … , 𝑋𝑛 be a random sample of size 𝑛 from a Normal distribution such

that each 𝑋𝑖~𝑁(0, 𝜃). Find the asymptotic distribution of the MLE of the parameter 𝜃.

From the previous assignment, we know that 𝜃𝑀𝐿𝐸 =1

𝑛∑ 𝑋𝑖

2𝑛𝑖=1 . We then use

Definition 9.4.5, which states that for large values of 𝑛, the MLE estimator is

distributed normal with mean 𝜃 and variance 𝐶𝑅𝐿𝐵. That is, we have that

𝜃𝑀𝐿𝐸~𝑁(𝜃, 𝐶𝑅𝐿𝐵). This means that we must compute the Cramer-Rao Lower Bound.

Since 𝜏(𝜃) = 𝜃, then 𝜏′(𝜃) = 1 and [𝜏′(𝜃)]2 = 1. Next, since we previously found that

𝑓(𝑋; 𝜃) =1

√2𝜋𝜃𝑒−

1

2𝜃𝑋2

, then we have ln 𝑓(𝑋; 𝜃) = −1

2ln(2𝜋𝜃) −

1

2𝜃𝑋2 so that

𝜕

𝜕𝜃ln 𝑓(𝑋; 𝜃) = −

1

2𝜃+

𝑋2

2𝜃2. We then find 𝜕2

𝜕𝜃2 ln 𝑓(𝑋; 𝜃) =1

2𝜃2 −2𝑋2

2𝜃3 =1

2𝜃2 −𝑋2

𝜃3 and take

the negative of its expected value to obtain −𝐸 [1

2𝜃2 −𝑋2

𝜃3] = − [1

2𝜃2 −1

𝜃3 𝐸(𝑋2)] =

1

𝜃3(𝜃 + 02) −

1

2𝜃2 =1

𝜃2 −1

2𝜃2 =1

2𝜃2. This implies that 𝐶𝑅𝐿𝐵 =2𝜃2

𝑛. Combining these

facts reveals that the asymptotic distribution of the MLE is 𝜃𝑀𝐿𝐸~𝑁 (𝜃,2𝜃2

𝑛). We can

transform this to get a standard normal distribution by noting that the random

variable �̂�𝑀𝐿𝐸−𝜃

𝜃√2/𝑛~𝑁(0,1) or large values of 𝑛. We could further reduce this by

multiplying through by the constant 𝜃 so that �̂�𝑀𝐿𝐸−𝜃

√2/𝑛~𝑁(0, 𝜃2).

Chapter #10 – Sufficiency and Completeness

Question #6: Let 𝑋1, … , 𝑋𝑛 be independent and each 𝑋𝑖~𝐵𝐼𝑁(𝑚𝑖, 𝑝). Use the Factorization

Criterion to show that 𝑆 = ∑ 𝑋𝑖𝑛𝑖=1 is sufficient for the unknown parameter 𝑝.

Since each 𝑋𝑖~𝐵𝐼𝑁(𝑚𝑖, 𝑝), we know that the probability mass function is given by

𝑓(𝑥;𝑚𝑖, 𝑝) = (𝑚𝑖𝑥)𝑝𝑥(1 − 𝑝)𝑚𝑖−𝑥1{𝑥 = 0,… ,𝑚𝑖}. We can then construct their joint

probability mass function due to the fact that they are independent as

𝑓(𝑥1, … , 𝑥𝑛; 𝑚𝑖 , 𝑝) = ∏ 𝑓(𝑥𝑖; 𝑚𝑖 , 𝑝)𝑛𝑖=1 = ∏ (𝑚𝑖

𝑥𝑖) 𝑝𝑥𝑖(1 − 𝑝)𝑚𝑖−𝑥𝑖1{𝑥𝑖 = 0,… ,𝑚𝑖}

𝑛𝑖=1 =

[∏ (𝑚𝑖𝑥𝑖)𝑛

𝑖=1 ] 𝑝∑ 𝑥𝑖𝑛𝑖=1 (1 − 𝑝)∑ 𝑚𝑖

𝑛𝑖=1 −∑ 𝑥𝑖

𝑛𝑖=1 1{𝑥𝑖 = 0,… ,𝑚𝑖}. If 𝐶 = [∏ (𝑚𝑖

𝑥𝑖)𝑛

𝑖=1 ], then we

have that 𝐶𝑝∑ 𝑥𝑖𝑛𝑖=1 𝑞∑ 𝑚𝑖

𝑛𝑖=1 −∑ 𝑥𝑖

𝑛𝑖=1 1{𝑥𝑖 = 0,… ,𝑚𝑖} = 𝐶

𝑝∑ 𝑥𝑖𝑛𝑖=1 𝑞∑ 𝑚𝑖

𝑛𝑖=1

𝑞∑ 𝑥𝑖𝑛𝑖=1

1{𝑥𝑖 = 0,… ,𝑚𝑖} =

𝐶 (𝑝

𝑞)∑ 𝑥𝑖𝑛𝑖=1

𝑞∑ 𝑚𝑖𝑛𝑖=1 1{𝑥𝑖 = 0,… ,𝑚𝑖}. But then if we define 𝑠 = ∑ 𝑥𝑖

𝑛𝑖=1 , we have that

𝑓(𝑥1, … , 𝑥𝑛; 𝑚𝑖 , 𝑝) = 𝐶 (𝑝

𝑞)𝑠

𝑞∑ 𝑚𝑖𝑛𝑖=1 1{𝑥𝑖 = 0,… ,𝑚𝑖} = 𝑔(𝑠;𝑚𝑖 , 𝑝)ℎ(𝑥1, … , 𝑥𝑛). Since

𝑔(𝑠;𝑚𝑖, 𝑝) = (𝑝

𝑞)𝑠

𝑞∑ 𝑚𝑖𝑛𝑖=1 does not depend on 𝑥1, … , 𝑥𝑛 except through 𝑠 = ∑ 𝑥𝑖

𝑛𝑖=1

and ℎ(𝑥1, … , 𝑥𝑛) = 𝐶1{𝑥𝑖 = 0,… ,𝑚𝑖} does not involve 𝑝, the Factorization Criterion

guarantees that 𝑆 = ∑ 𝑋𝑖𝑛𝑖=1 is sufficient for the unknown parameter 𝑝.

Question #7: Let 𝑋1, … , 𝑋𝑛 be independent and each 𝑋𝑖~𝑁𝐵(𝑟𝑖, 𝑝). This means that each 𝑋𝑖

has probability mass function 𝑃(𝑋𝑖 = 𝑥) = (𝑥−1𝑟𝑖−1

) 𝑝𝑟𝑖(1 − 𝑝)𝑥−𝑟𝑖 for 𝑥 = 𝑟𝑖, 𝑟𝑖 + 1, 𝑟𝑖 + 2,…

Find a sufficient statistic for the unknown parameter 𝑝 using the Factorization Criterion.

As in the question above, we have that 𝑓(𝑥1, … , 𝑥𝑛; 𝑟𝑖, 𝑝) = ∏ 𝑓(𝑥𝑖; 𝑟𝑖, 𝑝)𝑛𝑖=1 =

∏ (𝑥𝑖−1𝑟𝑖−1

) 𝑝𝑟𝑖(1 − 𝑝)𝑥𝑖−𝑟𝑖1{𝑥𝑖 = 𝑟𝑖, 𝑟𝑖 + 1,… }𝑛𝑖=1 . After applying the product operator,

this becomes [∏ (𝑥𝑖−1𝑟𝑖−1

)𝑛𝑖=1 ] 𝑝∑ 𝑟𝑖

𝑛𝑖=1 𝑞∑ 𝑥𝑖

𝑛𝑖=1 −∑ 𝑟𝑖

𝑛𝑖=1 1{𝑥𝑖 = 𝑟𝑖, 𝑟𝑖 + 1,… }. Then if we define

𝐶 = [∏ (𝑥𝑖−1𝑟𝑖−1

)𝑛𝑖=1 ], this expression becomes 𝐶

𝑝∑ 𝑟𝑖𝑛𝑖=1 𝑞∑ 𝑥𝑖

𝑛𝑖=1

𝑞∑ 𝑟𝑖𝑛𝑖=1

1{𝑥𝑖 = 𝑟𝑖, 𝑟𝑖 + 1,… } =

𝐶 (𝑝

𝑞)∑ 𝑟𝑖𝑛𝑖=1

𝑞∑ 𝑥𝑖𝑛𝑖=1 1{𝑥𝑖 = 𝑟𝑖, 𝑟𝑖 + 1,… }. Finally, if we let 𝑠 = ∑ 𝑥𝑖

𝑛𝑖=1 , we have that the

joint mass function is 𝑓(𝑥1, … , 𝑥𝑛; 𝑟𝑖 , 𝑝) = 𝐶 (𝑝


𝑞𝑠1{𝑥𝑖 = 𝑟𝑖, 𝑟𝑖 + 1,… } =

𝑔(𝑠; 𝑟𝑖, 𝑝)ℎ(𝑥1, … , 𝑥𝑛). Since 𝑔(𝑠; 𝑟𝑖, 𝑝) = (𝑝


𝑞𝑠 does not depend on 𝑥1, … , 𝑥𝑛

except through 𝑠 = ∑ 𝑥𝑖𝑛𝑖=1 and ℎ(𝑥1, … , 𝑥𝑛) = 𝐶1{𝑥𝑖 = 𝑟𝑖, 𝑟𝑖 + 1,… } does not involve

𝑝, the Factorization Criterion guarantees that 𝑆 = ∑ 𝑋𝑖𝑛𝑖=1 is sufficient for 𝑝.

Question #16: Let 𝑋1, … , 𝑋𝑛 be independent and each 𝑋𝑖~𝑁𝐵(𝑟𝑖, 𝑝). This means that each 𝑋𝑖

has mass function 𝑃(𝑋𝑖 = 𝑥) = (𝑥−1𝑟𝑖−1

)𝑝𝑟𝑖(1 − 𝑝)𝑥−𝑟𝑖 for 𝑥 = 𝑟𝑖, 𝑟𝑖 + 1, 𝑟𝑖 + 2,… Find the

Maximum Likelihood Estimator (MLE) of 𝑝 by maximizing the MLE of the sufficient statistic.

In the previous question, we found that 𝐿(𝑝) = 𝑓(𝑥1, … , 𝑥𝑛; 𝑟𝑖, 𝑝) =

[∏ (𝑥𝑖−1𝑟𝑖−1

)𝑛𝑖=1 ] 𝑝∑ 𝑟𝑖

𝑛𝑖=1 (1 − 𝑝)∑ 𝑥𝑖

𝑛𝑖=1 (1 − 𝑝)−∑ 𝑟𝑖

𝑛𝑖=1 . Taking the natural logarithm gives

ln[𝐿(𝑝)] = ∑ (𝑥𝑖−1𝑟𝑖−1

)𝑛𝑖=1 + ∑ 𝑟𝑖

𝑛𝑖=1 ln(𝑝) + ∑ 𝑥𝑖

𝑛𝑖=1 ln(1 − 𝑝) − ∑ 𝑟𝑖

𝑛𝑖=1 ln(1 − 𝑝). Then

differentiating the log likelihood function and equating to zero implies that

𝜕


∑ 𝑟𝑖𝑛𝑖=1

𝑝−


1−𝑝+

∑ 𝑟𝑖𝑛𝑖=1

1−𝑝= 0 → (1 − 𝑝)∑ 𝑟𝑖

𝑛𝑖=1 − 𝑝∑ 𝑥𝑖

𝑛𝑖=1 + 𝑝∑ 𝑟𝑖

𝑛𝑖=1 = 0.

Then we have ∑ 𝑟𝑖𝑛𝑖=1 − 𝑝∑ 𝑟𝑖


𝑛𝑖=1 + 𝑝∑ 𝑟𝑖

𝑛𝑖=1 = 0 → ∑ 𝑟𝑖


𝑛𝑖=1 = 0.

This implies that the Maximum Likelihood Estimator of 𝑝 is �̂�𝑀𝐿𝐸 =∑ 𝑟𝑖𝑛𝑖=1


.

Question #12: Let 𝑋1, … , 𝑋𝑛 be independent and identically distributed from a two

parameter exponential distribution 𝐸𝑋𝑃(𝜃, 𝜂) such that the probability density function is

𝑓(𝑥; 𝜃, 𝜂) =1

𝜃𝑒−𝑥+𝜂

𝜃 1{𝑥 > 𝜂}. Find jointly sufficient statistics for the parameters 𝜃 and 𝜂.

Since the random variables are iid, their joint probability density function is thus

given by 𝑓(𝑥1, … , 𝑥𝑛; 𝜃, 𝜂) = ∏ 𝑓(𝑥𝑖; 𝜃, 𝜂)𝑛𝑖=1 1{𝑥𝑖 > 𝜂} = ∏ 𝜃−1𝑒

−𝑥𝑖+𝜂

𝜃𝑛𝑖=1 1{𝑥𝑖 > 𝜂} =

𝜃−𝑛𝑒1

𝜃(𝑛𝜂−∑ 𝑥𝑖

𝑛𝑖=1 )[∏ 1{𝑥𝑖 > 𝜂}𝑛

𝑖=1 ]. This then shows that 𝑆1 = ∑ 𝑋𝑖𝑛𝑖=1 and 𝑆2 = 𝑋1:𝑛 are

jointly sufficient for 𝜃 and 𝜂 by the Factorization Criterion with ℎ(𝑥1, … , 𝑥𝑛) = 1 being

independent of the unknown parameters 𝜃 and 𝜂 and 𝑔(𝑠1, 𝑠2; 𝜃, 𝜂) =

𝜃−𝑛𝑒1

𝜃(𝑛𝜂−𝑠1)∏ 1{𝑠2 > 𝜂}𝑛

𝑖=1 depending on 𝑥1, … , 𝑥𝑛 only through 𝑆1 and 𝑆2.

Question #13: Let 𝑋1, … , 𝑋𝑛 be independent and identically distributed from a beta

distribution 𝐵𝐸𝑇𝐴(𝜃1, 𝜃2) such that the probability density function of each of these random

variables is given by 𝑓(𝑥; 𝜃1, 𝜃2) =Γ(𝜃1+𝜃2)

Γ(𝜃1)Γ(𝜃2)𝑥𝜃1−1(1 − 𝑥)𝜃2−1 whenever 0 < 𝑥 < 1. Find

jointly sufficient statistics for the unknown parameters 𝜃1 and 𝜃2.

Since the random variables are iid, their joint density is given by 𝑓(𝑥1, … , 𝑥𝑛; 𝜃1, 𝜃2) =

∏ 𝑓(𝑥𝑖; 𝜃1, 𝜃2)𝑛𝑖=1 1{0 < 𝑥𝑖 < 1} = ∏

Γ(𝜃1+𝜃2)

Γ(𝜃1)Γ(𝜃2)𝑥𝑖𝜃1−1(1 − 𝑥𝑖)

𝜃2−1𝑛𝑖=1 1{0 < 𝑥𝑖 < 1} =

[Γ(𝜃1+𝜃2)

Γ(𝜃1)Γ(𝜃2)]𝑛[𝑥1…𝑥𝑛]

𝜃1−1[(1 − 𝑥1)… (1 − 𝑥𝑛)]𝜃2−1∏ 1{0 < 𝑥𝑖 < 1}𝑛

𝑖=1 . This then

shows that 𝑆1 = ∏ 𝑋𝑖𝑛𝑖=1 and 𝑆2 = ∏ (1 − 𝑋𝑖)

𝑛𝑖=1 are jointly sufficient for 𝜃1 and 𝜃2 by

the Factorization Criterion with ℎ(𝑥1, … , 𝑥𝑛) = ∏ 1{0 < 𝑥𝑖 < 1}𝑛𝑖=1 being

independent of the unknown parameters and 𝑔(𝑠1, 𝑠2; 𝜃1, 𝜃2) =

[Γ(𝜃1+𝜃2)

Γ(𝜃1)Γ(𝜃2)]𝑛[𝑠1]

𝜃1−1[𝑠2]𝜃2−1 depending on the observations only through 𝑆1 and 𝑆2.

Question #18: Let 𝑋~𝑁(0, 𝜃) for 𝜃 > 0. a) Show that 𝑋2 is complete and sufficient for the

unknown parameter 𝜃, and b) show that 𝑁(0, 𝜃) is not a complete family.

a) Since 𝑋~𝑁(0, 𝜃), we know that 𝑓𝑋(𝑥; 𝜃) =1

√2𝜋𝜃𝑒−

𝑥2

2𝜃 for 𝑥 ∈ ℝ. Therefore, by the

Regular Exponential Class (REC) Theorem, 𝑋2 is complete and sufficient for 𝜃.

b) Since 𝑋~𝑁(0, 𝜃), we know that 𝐸(𝑋) = 0 for all 𝜃 > 0. Therefore, completeness fails

because we have a nontrivial unbiased estimator of 𝐸(𝑋) = 0.

Chapter #10 – Sufficiency and Completeness

Question #21: If 𝑋1, … , 𝑋𝑛 is a random sample from a Bernoulli distribution such that each

𝑋𝑖~𝐵𝐸𝑅𝑁(𝑝) ≡ 𝐵𝐼𝑁(1, 𝑝) where 𝑝 is the unknown parameter to be estimated, find the

UMVUE for a) 𝜏(𝑝) = 𝑉𝑎𝑟(𝑋) = 𝑝(1 − 𝑝), and b) 𝜏(𝑝) = 𝑝2.

a) We first verify that the Bernoulli distribution is a member of the Regular Exponential

Class (REC) by noting that its density can be written as 𝑓(𝑥; 𝑝) = 𝑝𝑥(1 − 𝑝)1−𝑥 =

𝑝𝑥

(1−𝑝)𝑥−1=

𝑝𝑥

(1−𝑝)𝑥(1 − 𝑝) = (

𝑝

1−𝑝)𝑥(1 − 𝑝) = exp {ln [(

𝑝

1−𝑝)𝑥(1 − 𝑝)]}. This equality

implies 𝑓(𝑥; 𝑝) = exp {𝑥 ln (𝑝

1−𝑝) + ln(1 − 𝑝)} = exp {𝑥 ln (

𝑝

1−𝑝)} exp{ln(1 − 𝑝)} =

(1 − 𝑝) exp {𝑥 ln (𝑝

1−𝑝)} = 𝑐(𝑝) exp{𝑡1(𝑥)𝑞1(𝑝)}, so the Bernoulli distribution is a

member of the REC by Definition 10.4.2. We then use Theorem 10.4.2, which

guarantees the existence of sufficient statistics for distribution from the REC, to

construct the sufficient statistic 𝑆1 = ∑ 𝑡1(𝑋𝑖)𝑛𝑖=1 = ∑ 𝑋𝑖

𝑛𝑖=1 . Next, we appeal to the

Rao-Blackwell Theorem in justifying the use of 𝑆1 (or any one-to-one function of it) in

our search for a UMVUE for 𝑉𝑎𝑟(𝑋) = 𝑝(1 − 𝑝). Our initial guess for an estimator is

𝑇 = �̅�(1 − �̅�), so we first compute that 𝐸(𝑇) = 𝐸[�̅�(1 − �̅�)] = 𝐸(�̅�) − 𝐸(�̅�2) =

𝐸 (1

𝑛∑ 𝑋𝑖𝑛𝑖=1 ) − [𝑉𝑎𝑟(�̅�) + 𝐸(�̅�)2] =

1

𝑛∑ 𝐸(𝑋𝑖)𝑛𝑖=1 −

1

𝑛2∑ 𝑉𝑎𝑟(𝑋𝑖)𝑛𝑖=1 + 𝐸(�̅�)2. Thus,

we have calculated that 𝐸(𝑇) =1

𝑛𝑛𝑝 −

1

𝑛2𝑛𝑝(1 − 𝑝) − (

1

𝑛𝑛𝑝)

2

= 𝑝 −𝑝(1−𝑝)

𝑛− 𝑝2 =

𝑝(1 − 𝑝) (1 −1

𝑛) = 𝑝(1 − 𝑝) (

𝑛−1

𝑛), which implies that 𝑇∗ =

𝑛

𝑛−1𝑇 =

𝑛

𝑛−1[�̅�(1 − �̅�]

will have expected value equal to 𝑉𝑎𝑟(𝑋) = 𝑝(1 − 𝑝). The Lehman-Scheffe Theorem

finally guarantees that 𝑇∗ is a UMVUE for 𝑉𝑎𝑟(𝑋) = 𝑝(1 − 𝑝) since it states that any

unbiased estimator which is a function of complete sufficient statistics is a UMVUE.

b) We note that for the complete sufficient statistic 𝑆1 = ∑ 𝑋𝑖𝑛𝑖=1 , we have 𝐸(𝑆1) = 𝑛𝑝

and 𝑉𝑎𝑟(𝑆12) = 𝑛𝑝(1 − 𝑝) since 𝑆1~𝐵𝐼𝑁(𝑛, 𝑝), which is true because it is the sum of 𝑛

independent Bernoulli random variables. This implies 𝐸(𝑆12) = 𝑉𝑎𝑟(𝑆1

2) + 𝐸(𝑆1)2 =

𝑛𝑝(1 − 𝑝) + (𝑛𝑝)2 = 𝑛𝑝(1 − 𝑝) + 𝑛2𝑝2. By the Lehman-Scheffe Theorem, we know

that we must use some function of the complete sufficient statistic 𝑆1 to construct a

UMVUE for the unknown parameter 𝑝2. We note that for 𝑇 = 𝑆1 − 𝑆12, we have 𝐸(𝑇) =

𝐸(𝑆1) − 𝐸(𝑆12) = 𝑛𝑝 − 𝑛𝑝(1 − 𝑝) + 𝑛2𝑝2 = 𝑛𝑝 − 𝑛𝑝 + 𝑛𝑝2 + 𝑛2𝑝2 = 𝑝2(𝑛 + 𝑛2).

This implies that the statistic 𝑇∗ =1

𝑛+𝑛2𝑇 =

𝑆1−𝑆12

𝑛+𝑛2=

∑ 𝑋𝑖𝑛𝑖=1 +(∑ 𝑋𝑖

𝑛𝑖=1 )

2

𝑛+𝑛2 will have

expected value equal to 𝑝2, so it is a UMVUE by the Lehman-Scheffe Theorem.

Question #23: If 𝑋1, … , 𝑋𝑛 is a random sample from a Normal distribution such that each

𝑋𝑖~𝑁(𝜇, 9) where 𝜇 is unknown, find the UMVUE for a) the 95th percentile, and b) 𝑃(𝑋1 ≤ 𝑐),

where 𝑐 is a known constant. Hint: find the conditional distribution of 𝑋1 given �̅� = 𝑥, and

apply the Rao-Blackwell Theorem with 𝑇 = 𝑢(𝑋1), where define 𝑢(𝑥) = 1{𝑥 ≤ 𝑐}.

a) The 95th percentile of a random variable 𝑋 from a 𝑁(𝜇, 9) distribution is the value of

𝑘 such that 𝑃(𝑋 ≤ 𝑘) = 0.95 → 𝑃 (𝑋−𝜇

3≤

𝑘−𝜇

3) = 0.95 → 𝑃 (𝑍 ≤

𝑘−𝜇

3) = 0.95 where

𝑍~𝑁(0,1). From tabulations of the standard normal distribution function Φ(𝑧), we

know that 𝑃(𝑍 ≤ 1.645) = 0.95, so we equate 𝑘−𝜇

3= 1.645 → 𝑘 = 4.935 + 𝜇 = 𝜏(𝜇).

This is what we wish to find a UMVUE for, but since the expectation of a constant is

that constant itself, we simply need to find a UMVUE for 𝜇. We begin by verifying that

the Normal distribution is a member of the Regular Exponential Class (REC) by noting

that the density of 𝑋~𝑁(𝜇, 9) can be written as 𝑓(𝑥; 𝜇) =1

√18𝜋exp {−

1

18(𝑥 − 𝜇)2} =

1

√18𝜋exp {−

𝑥2

18+

𝜇𝑥

9−

𝜇2

18}, where we have that 𝑡1(𝑥) = 𝑥2 and 𝑡2(𝑥) = 𝑥. Thus, the

Normal distribution is a member of the REC by Definition 10.4.2. We then use

Theorem 10.4.2, which guarantees the existence of sufficient statistics for

distribution from the REC, to construct the sufficient statistics 𝑆1 = ∑ 𝑡1(𝑋𝑖)𝑛𝑖=1 =

∑ 𝑋𝑖2𝑛

𝑖=1 and 𝑆2 = ∑ 𝑡2(𝑋𝑖)𝑛𝑖=1 = ∑ 𝑋𝑖

𝑛𝑖=1 . Since the sample mean is an unbiased

estimator for the population mean, we have that 𝐸(𝑇) = 𝐸(�̅�) = 𝐸 (𝑆2

𝑛) = 𝜇. Thus, an

unbiased estimator for 𝜏(𝜇) = 𝑘 = 4.935 + 𝜇 is given by 𝑇∗ = 4.935 + �̅�, which is

also a UMVUE of 𝜏(𝜇) by the Lehmann-Scheffe Theorem.

b) Note that we are trying to estimate 𝑃(𝑋1 ≤ 𝑐) = 𝑃 (𝑋1−𝜇

3≤

𝑐−𝜇

3) = Φ(

𝑐−𝜇

3) = 𝜏(𝜇),

where Φ:ℝ → (0,1) is the cumulative distribution function of 𝑍~𝑁(0,1). Since 𝜏(𝜇) is

a nonlinear function of 𝜇, we cannot simply insert �̅� to obtain a UMVUE. To find an

unbiased estimator, we note that 𝑢(𝑋1) = 1{𝑋1 ≤ 𝑐} is unbiased for 𝜏(𝜇) since we

have 𝐸[𝑢(𝑋1)] = 𝐸[1{𝑋1 ≤ 𝑐}] = 𝑃(𝑋1 ≤ 𝑐) = 𝜏(𝜇). But since it is not a function of

the complete sufficient statistic 𝑆2 = ∑ 𝑋𝑖𝑛𝑖=1 , this estimator cannot be a UMVUE.

However, the Rao-Blackwell Theorem states that 𝐸[𝑢(𝑋1)|𝑆2] = 𝐸[1{𝑋1 ≤ 𝑐}|𝑆2] will

also be unbiased and will be a function of 𝑆2 = ∑ 𝑋𝑖𝑛𝑖=1 . The Lehmann-Scheffe

Theorem then guarantees that 𝐸[1{𝑋1 ≤ 𝑐}|𝑆2] will be a UMVUE. In order to find this,

we must compute the conditional distribution of 𝑋1 given 𝑆2. We know that the

random variable 𝑆2 = ∑ 𝑋𝑖𝑛𝑖=1 ~𝑁(𝑛𝜇, 9𝑛) and that 𝑋1 = 𝑥, 𝑆2 = 𝑠 is equivalent to

𝑋1 = 𝑥,∑ 𝑋𝑖𝑛𝑖=2 = 𝑠 − 𝑥. This implies that 𝑓𝑋1|𝑆2(𝑥|𝑠) = ⋯ =

exp{−(𝑥−𝜇′)2/2(𝜎′)

2}

√2𝜋𝜎′,

where 𝜇′ =𝑠

𝑛 and (𝜎′)2 =

9(𝑛−1)

𝑛. Therefore, if we let 𝐴~𝑁 (

𝑠

𝑛,9(𝑛−1)

𝑛) we have that

𝐸[1{𝑋1 ≤ 𝑐}|𝑆2] = 𝑃(𝐴 ≤ 𝑐) = Φ(𝑐−𝑠/𝑛

3√(𝑛−1)/𝑛), which is a UMVUE for Φ(

𝑐−𝜇

3) = 𝜏(𝜇).

Question #25: If 𝑋1, … , 𝑋𝑛 is a random sample from the probability density function

𝑓(𝑥; 𝜃) = 𝜃𝑥𝜃−11{0 < 𝑥 < 1} where 𝜃 > 0 is the unknown parameter, find the UMVUE for

a) 𝜏(𝜃) =1

𝜃 by using the fact that 𝐸[− ln(𝑋)] =

1

𝜃, and b) the unknown parameter 𝜃.

a) We first verify that the density is a member of the REC by nothing that it can be

written as 𝑓(𝑥; 𝜃) = 𝜃𝑥𝜃−1 = exp{ln[𝜃𝑥𝜃−1]} = exp{ln(𝜃) + (𝜃 − 1) ln(𝑥)} =

exp{ln(𝜃)} exp{(𝜃 − 1) ln(𝑥)} = 𝜃 exp{(𝜃 − 1) ln(𝑥)}, where 𝑡1(𝑥) = ln(𝑥). We then

use Theorem 10.4.2, which guarantees the existence of sufficient statistics for REC

distributions, to construct the sufficient statistic 𝑆1 = ∑ 𝑡1(𝑋𝑖)𝑛𝑖=1 = ∑ ln(𝑋𝑖)

𝑛𝑖=1 . Next,

we appeal to the Rao-Blackwell Theorem in justifying the use of 𝑆1 (or any one-to-one

function of it) in our search for a UMVUE for 1

𝜃. From the hint provided, we initially

guess that 𝑇 =−𝑆1

𝑛=

∑ −ln(𝑋𝑖)𝑛𝑖=1

𝑛 and check that 𝐸(𝑇) = 𝐸 [

∑ − ln(𝑋𝑖)𝑛𝑖=1

𝑛] =

1

𝑛∑ 𝐸[− ln(𝑋𝑖)]𝑛𝑖=1 =

1

𝑛𝑛1

𝜃=

1

𝜃. The Lehman-Scheffe Theorem finally guarantees that

𝑇 = −1

𝑛∑ ln(𝑋𝑖)𝑛𝑖=1 is a UMVUE for

1

𝜃 since it states that any unbiased estimator which

is a function of complete sufficient statistics is a UMVUE.

b) Any UMVUE of the unknown parameter 𝜃 must be a function of the complete and

sufficient statistic 𝑆1 = ∑ ln(𝑋𝑖)𝑛𝑖=1 by the Lehman-Scheffe Theorem. We begin by

noting that 𝐸(𝑆1) = 𝐸[∑ ln(𝑋𝑖)𝑛𝑖=1 ] = ∑ 𝐸[ln(𝑋𝑖)]

𝑛𝑖=1 = −∑ 𝐸[− ln(𝑋𝑖)]

𝑛𝑖=1 = −

𝑛

𝜃, so

we would like to be able to compute 𝐸 (1

𝑆1) ≠

1

𝐸(𝑆1). However, this involves finding

𝐸 [−1

ln(𝑋)] since we know that 𝐸[− ln(𝑋)] =

1

𝜃. We do this by finding the distribution

of 𝑌 = −ln(𝑋) using the CDF technique, which shows that 𝑌~𝐸𝑋𝑃 (1

𝜃) with density

𝑓(𝑦; 𝜃) = 𝜃𝑒−𝜃𝑥1{𝑥 > 0}. This is equivalent to 𝑌~𝐺𝐴𝑀𝑀𝐴 (1

𝜃, 1), so by the Moment

Generating Function technique, we see that 𝑆1 = ∑ ln(𝑋𝑖)𝑛𝑖=1 ~𝐺𝐴𝑀𝑀𝐴 (

1

𝜃, 𝑛). We can

thus calculate 𝐸 [−1

ln(𝑋)] =

𝜃𝑛

Γ(𝑛)∫

1

𝑥𝑥𝑛−1𝑒−𝜃𝑥

∞

0𝑑𝑥 = ⋯ =

𝜃

𝑛−1, which implies that 𝑇 =

𝑛−1

𝑆=

𝑛−1

∑ ln(𝑋𝑖)𝑛𝑖=1

is an unbiased estimator of 𝜃. Then the Lehmann-Scheffe Theorem

guarantees that it is also a UMVUE for the unknown parameter 𝜃.

Question #31: If 𝑋1, … , 𝑋𝑛 is a random sample from the probability density function

𝑓(𝑥; 𝜃) = 𝜃(1 + 𝑥)−(1+𝜃)1{𝑥 > 0} for unknown 𝜃 > 0, find a) the MLE of 𝜃, b) a complete and

sufficient statistic for 𝜃, c) the CRLB for 𝜏(𝜃) =1

𝜃, d) the UMVUE for 𝜏(𝜃) =

1

𝜃, e) the mean

and variance of the asymptotic normal distribution of the MLE, and f) the UMVUE for 𝜃.

a) We have 𝐿(𝜃) = ∏ 𝑓(𝑥𝑖; 𝜃)𝑛𝑖=1 = ∏ 𝜃(1 + 𝑥𝑖)

−(1+𝜃)𝑛𝑖=1 = 𝜃𝑛[∏ (1 + 𝑥𝑖)

𝑛𝑖=1 ]−(1+𝜃) so

that ln[𝐿(𝜃)] = 𝑛 ln(𝜃) − (1 + 𝜃)∑ ln(1 + 𝑥𝑖)𝑛𝑖=1 . Then we have that

𝜕


𝑛

𝜃− ∑ ln(1 + 𝑥𝑖)

𝑛𝑖=1 = 0 → 𝜃 =

𝑛

∑ ln(1+𝑥𝑖)𝑛𝑖=1

so that 𝜃𝑀𝐿𝐸 =𝑛


.

b) To check that it is a member of the REC, we verify that we can write the probability

density function of 𝑋 as 𝑓(𝑥; 𝜃) = 𝜃(1 + 𝑥)−(1+𝜃) = exp{ln[𝜃(1 + 𝑥)−(1+𝜃)]} =

exp{ln(𝜃) − (1 + 𝜃) ln(1 + 𝑥)}, where 𝑡1(𝑥) = 1 and 𝑡2(𝑥) = ln(1 + 𝑥). Thus, 𝑓(𝑥; 𝜃)

is a member of the REC and 𝑆2 = ∑ 𝑡2(𝑋𝑖)𝑛𝑖=1 = ∑ ln(1 + 𝑋𝑖)

𝑛𝑖=1 is a complete and

sufficient statistic for the unknown parameter 𝜃 to be estimated.

c) Since 𝜏(𝜃) =1

𝜃, we have [𝜏′(𝜃)]2 = [−

1

𝜃2]2

=1

𝜃4. Then we have that 𝑓(𝑋; 𝜃) =

𝜃(1 + 𝑋)−(1+𝜃) so its log is ln 𝑓(𝑋; 𝜃) = ln(𝜃) − (1 + 𝜃) ln(1 + 𝑋) and 𝜕

𝜕𝜃ln 𝑓(𝑋; 𝜃) =

1

𝜃− ln(1 + 𝑋). Finally,

𝜕2

𝜕𝜃2ln 𝑓(𝑋; 𝜃) = −

1

𝜃2 so that −𝐸 [

𝜕2

𝜕𝜃2ln 𝑓(𝑋; 𝜃)] =

1

𝜃2. These

results combined allow us to conclude that 𝐶𝑅𝐿𝐵 =(1/𝜃4)

𝑛(1/𝜃2)=

1

𝑛𝜃2.

d) We previously verified that this density is a member of the REC and that the statistic

𝑆2 = ∑ ln(1 + 𝑋𝑖)𝑛𝑖=1 is complete and sufficient for 𝜃. Next, we use the Rao-Blackwell

Theorem in justifying the use of 𝑆2 (or any one-to-one function of it) in our search for

a UMVUE for 1

𝜃. In order to compute 𝐸(𝑆2), we need to find the distribution of the

random variable 𝑌 = ln(1 + 𝑋), which we do using the CDF technique. We thus have

that 𝐹𝑌(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(ln(1 + 𝑋) ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑒𝑦 − 1) = 𝐹𝑋(𝑒𝑦 − 1), so that

then 𝑓𝑌(𝑦) =𝑑


𝑑

𝑑𝑦𝐹𝑋(𝑒

𝑦 − 1) = 𝑒𝑦𝑓𝑋(𝑒𝑦 − 1) = 𝑒𝑦[𝜃(1 + 𝑒𝑦 − 1)−(1+𝜃)] =

𝜃𝑒𝑦(𝑒𝑦)−(1+𝜃) = 𝜃𝑒𝑦𝑒−(𝑦+𝜃𝑦) = 𝜃𝑒𝑦−𝑦−𝜃𝑦 = 𝜃𝑒−𝜃𝑦 whenever 𝑦 > 0. It is

immediately clear that 𝑌~𝐸𝑋𝑃(𝜃), so that 𝐸(𝑌) = 𝐸[ln(1 + 𝑋)] =1

𝜃. This allows us to

find 𝐸(𝑆2) = 𝐸[∑ ln(1 + 𝑋𝑖)𝑛𝑖=1 ] = ∑ 𝐸[ln(1 + 𝑋𝑖)]

𝑛𝑖=1 =

𝑛

𝜃. Since we want an unbiased

estimator for 1

𝜃, it is clear that 𝑇 =

𝑆2

𝑛=

1

𝑛∑ ln(1 + 𝑋𝑖)𝑛𝑖=1 will suffice by the LST.

e) We previously found that the MLE for 𝜃 is 𝜃𝑀𝐿𝐸 =𝑛


. From Chapter 9, we

know that the MLE for some unknown parameter 𝜃 has an asymptotic normal

distribution with 𝜇 = 𝜃 and 𝜎2 = 𝐶𝑅𝐿𝐵; that is, 𝜃𝑀𝐿𝐸~𝑁(𝜃, 𝐶𝑅𝐿𝐵) for large 𝑛. We

must therefore find the Cramer-Rao Lower Bound, which can be easily done from the

work in part c) above with 𝜏(𝜃) = 𝜃, so that 𝐶𝑅𝐿𝐵 =𝜃2

𝑛. This means that we have

𝜃𝑀𝐿𝐸~𝑁 (𝜃,𝜃2

𝑛) for large 𝑛. We can similarly argue for the MLE of 𝜏(𝜃) =

1

𝜃, where we

see that �̃�𝑀𝐿𝐸 =1

�̂�𝑀𝐿𝐸=

1

𝑛∑ ln(1 + 𝑋𝑖)𝑛𝑖=1 by the Invariance Property of the Maximum

Likelihood Estimator. Then using the work done in part c) above for the Cramer-Rao

Lower Bound, we can conclude that �̃�𝑀𝐿𝐸~𝑁 (1

𝜃,

1

𝑛𝜃2) for large 𝑛.

f) We previously verified that this density is a member of the REC and that the statistic

𝑆2 = ∑ ln(1 + 𝑋𝑖)𝑛𝑖=1 is complete and sufficient for 𝜃 where 𝐸(𝑆2) =

𝑛

𝜃. As in the

previous question, we have that 𝐸 (1

𝑆2) = 𝐸 (

1


) =𝜃

𝑛−1 which implies that

𝑇 =𝑛−1

𝑆2=

𝑛−1


is unbiased and a UMVUE for the unknown parameter 𝜃.

Chapter #11 – Interval Estimation

Question #5: If 𝑋1, … , 𝑋𝑛 is a random sample from 𝑓𝑋(𝑥; 𝜂) = 𝑒𝜂−𝑥1{𝑥 > 𝜂} with 𝜂 unknown,

then a) show that 𝑄 = 𝑋1:𝑛 − 𝜂 is a pivotal quantity and find its distribution, and b) derive a

100𝛾% equal-tailed confidence interval for the unknown parameter 𝜂.

a) We first find the distribution of the smallest order statistic 𝑋1:𝑛 using the formula

𝑓1(𝑦; 𝜂) = 𝑛𝑓𝑋(𝑦; 𝜂)[1 − 𝐹𝑋(𝑦; 𝜂)]𝑛−1. We thus need the CDF of the population, which

is given by 𝐹𝑋(𝑥; 𝜂) = ∫ 𝑓𝑋(𝑡; 𝜂)𝑥

𝜂𝑑𝑡 = ∫ 𝑒𝜂−𝑡

𝑥

𝜂𝑑𝑡 = 𝑒𝜂 ∫ 𝑒−𝑡

𝑥

𝜂𝑑𝑡 = 𝑒𝜂[−𝑒−𝑡]𝜂

𝑥 =

𝑒𝜂(−𝑒−𝑥 + 𝑒−𝜂) = 1 − 𝑒𝜂−𝑥 whenever 𝑥 > 𝜂. We therefore have that 𝑓1(𝑦; 𝜂) =

𝑛𝑒𝜂−𝑦[1 − (1 − 𝑒𝜂−𝑦)]𝑛−1 = 𝑛𝑒𝜂−𝑦[𝑒𝜂−𝑦]𝑛−1 = 𝑛𝑒𝜂−𝑦𝑒𝑛𝜂−𝜂−𝑛𝑦+𝑦 = 𝑛𝑒𝑛(𝜂−𝑦) when

𝑥 > 𝜂. Now that we have the density of 𝑋1:𝑛, we can use the CDF technique to find the

density of 𝑄 = 𝑋1:𝑛 − 𝜂. Thus, we have 𝐹𝑄(𝑞) = 𝑃(𝑄 ≤ 𝑞) = 𝑃(𝑋1:𝑛 − 𝜂 ≤ 𝑞) =

𝑃(𝑋1:𝑛 ≤ 𝑞 + 𝜂) = 𝐹1(𝑞 + 𝜂) so 𝑓𝑄(𝑞) =𝑑

𝑑𝑞𝐹1(𝑞 + 𝜂) = 𝑓1(𝑞 + 𝜂) = 𝑛𝑒−𝑛𝑞 whenever

𝑞 + 𝜂 > 𝜂 → 𝑞 > 0. This reveals that 𝑄 = 𝑋1:𝑛 − 𝜂~𝐸𝑋𝑃 (1

𝑛), so it is clearly a pivotal

quantity since it is a function of 𝜂 but its distribution does not depend on 𝜂.

b) We have 𝑃(𝑥(1−𝛾)/2 < 𝑄 < 𝑥(1+𝛾)/2) = 𝛾 → 𝑃(𝑥(1−𝛾)/2 < 𝑋1:𝑛 − 𝜂 < 𝑥(1+𝛾)/2) = 𝛾, so

after solving for the unknown parameter we obtain the 100𝛾% equal tailed

confidence interval 𝑃(𝑋1:𝑛 − 𝑥(1+𝛾)/2 < 𝜂 < 𝑋1:𝑛 − 𝑥(1−𝛾)/2) = 𝛾. This can also be

expressed as the random interval (𝑋1:𝑛 − 𝑥(1+𝛾)/2, 𝑋1:𝑛 − 𝑥(1−𝛾)/2). Finally, we know

that the 𝐸𝑋𝑃 (1

𝑛) distribution has CDF 𝐹𝑄(𝑞) = 1 − 𝑒−𝑛𝑞 so that 𝐹𝑄(𝑥𝛼) = 𝛼 implies

1 − 𝑒−𝑛𝑥𝛼 = 𝛼. We solve this last equality for 𝑥𝛼 = −1

𝑛ln(1 − 𝛼). This means that the

confidence interval becomes (𝑋1:𝑛 +1

𝑛ln (

1−𝛾

2) , 𝑋1:𝑛 +

1

𝑛ln (

1+𝛾

2)), where each term is

found by substituting 𝛼 =1−𝛾

2 and 𝛼 =

1+𝛾

2 into the expression 𝑥𝛼 = −

1

𝑛ln(1 − 𝛼).

Question #7: If 𝑋1, … , 𝑋𝑛 is a random sample from 𝑓𝑋(𝑥; 𝜃) =2

𝜃2𝑥𝑒−𝑥

2/𝜃21{𝑥 > 0} with

unknown parameter 𝜃, a) show that 𝑄 =2∑ 𝑋𝑖

2𝑛𝑖=1

𝜃2~𝜒2(2𝑛), b) use 𝑄 =

2∑ 𝑋𝑖2𝑛

𝑖=1

𝜃2 to derive an

equal-tailed 100𝛾% confidence interval for 𝜃, c) find a lower 100𝛾% confidence limit for

𝑃(𝑋 > 𝑡) = 𝑒−𝑡2/𝜃2 , d) find an upper 100𝛾% confidence limit for the 𝑝𝑡ℎ percentile.

a) Since 𝑓𝑋(𝑥; 𝜃) =2

𝜃2𝑥𝑒−𝑥

2/𝜃21{𝑥 > 0}, we know that 𝑋~𝑊𝐸𝐼(𝜃, 2). The CDF technique

then reveals that 𝑋2~𝐸𝑋𝑃(𝜃2) so that ∑ 𝑋𝑖2𝑛

𝑖=1 ~𝐺𝐴𝑀𝑀𝐴(𝜃2, 𝑛). A final application of

the CDF technique shows that 2

𝜃2∑ 𝑋𝑖

2𝑛𝑖=1 ~𝜒2(2𝑛), proving the desired result. This

also shows that 𝑄 =2∑ 𝑋𝑖

2𝑛𝑖=1

𝜃2 is a pivotal quantity for the unknown parameter 𝜃.

b) We find that the confidence interval is 𝑃 (𝜒1−𝛾2

2 (2𝑛) < 𝑄 < 𝜒1+𝛾2

2 (2𝑛)) = 𝛾 →

𝑃 (𝜒1−𝛾2

2 (2𝑛) <2

𝜃2∑ 𝑋𝑖

2𝑛𝑖=1 < 𝜒1+𝛾

2

2 (2𝑛)) = 𝛾 → 𝑃(2∑ 𝑋𝑖

2𝑛𝑖=1

𝜒1+𝛾2

2 (2𝑛)< 𝜃2 <

2∑ 𝑋𝑖2𝑛

𝑖=1

𝜒1−𝛾2

2 (2𝑛)) = 𝛾.

Taking square roots gives the desired random interval (√2∑ 𝑋𝑖

2𝑛𝑖=1

𝜒1+𝛾2

2 (2𝑛), √

2∑ 𝑋𝑖2𝑛

𝑖=1

𝜒1−𝛾2

2 (2𝑛)).

c) From the work done above, a lower confidence limit for 𝜃 is √2∑ 𝑋𝑖

2𝑛𝑖=1

𝜒𝛾2(2𝑛)

. Since the

quantity 𝜏(𝜃) = 𝑒−𝑡2/𝜃2 is a monotonically increasing function of 𝜃, we can simply

substitute √2∑ 𝑋𝑖

2𝑛𝑖=1

𝜒𝛾2(2𝑛)

for 𝜃 into the expression 𝜏(𝜃) = 𝑒−𝑡2/𝜃2 by Corollary 11.3.1.

d) We must solve the equation 𝑃(𝑋 > 𝑡𝑝) = 1 − 𝑝 for 𝑡𝑝. From the question above, we

are given that 𝑃(𝑋 > 𝑡) = 𝑒−𝑡2/𝜃2 so we must solve 𝑒−𝑡𝑝

2/𝜃2 = 1 − 𝑝 for 𝑡𝑝, which gives

𝑡𝑝 = 𝜃√ln(1 − 𝑝). By the same reasoning as above, we substitute √2∑ 𝑋𝑖

2𝑛𝑖=1

𝜒1−𝛾2 (2𝑛)

in for 𝜃

into the expression 𝑡𝑝 = 𝜏(𝜃) = 𝜃√ln(1 − 𝑝) to obtain 𝑡𝑝 = √2∑ 𝑋𝑖

2𝑛𝑖=1

𝜒1−𝛾2 (2𝑛)

ln(1 − 𝑝).

Question #8: If 𝑋1, … , 𝑋𝑛 is a random sample from 𝑋~𝑈𝑁𝐼𝐹(0, 𝜃) with 𝜃 > 0 unknown and

𝑋𝑛:𝑛 is the largest order statistic, then a) find the probability that the random interval given

by (𝑋𝑛:𝑛, 2𝑋𝑛:𝑛) contains 𝜃, and b) find the value of the constant 𝑐 such that the random

interval (𝑋𝑛:𝑛, 𝑐𝑋𝑛:𝑛) is a 100(1 − 𝛼)% confidence interval for the parameter 𝜃.

a) We have that 𝜃 ∈ (𝑋𝑛:𝑛, 2𝑋𝑛:𝑛) if and only if 𝜃 < 2𝑋𝑛:𝑛, since the inequality 𝜃 < 𝑋𝑛:𝑛

will always be true by the definition of the density. We must therefore compute

𝑃(2𝑋𝑛:𝑛 > 𝜃) = 𝑃 (𝑋𝑛:𝑛 >𝜃

2) = 1 − 𝑃 (𝑋𝑛:𝑛 ≤

𝜃

2) = 1 − [𝑃 (𝑋𝑖 ≤

𝜃

2)]

𝑛

= 1 − 2−𝑛.

b) As above, we have that 𝑃[𝜃 ∈ (𝑋𝑛:𝑛, 𝑐𝑋𝑛:𝑛)] = 1 − 𝑐−𝑛, so if we set this equal to 1 − 𝛼

and solve for the value of the constant, we obtain 1 − 𝑐−𝑛 = 1 − 𝛼 → 𝑐 = 𝛼−1/𝑛.

Question #13: Let 𝑋1, … , 𝑋𝑛 be a random sample from 𝑋~𝐺𝐴𝑀𝑀𝐴(𝜃, 𝜅) such that their

common distribution is 𝑓(𝑥; 𝜃, 𝜅) =1

𝜃𝜅Γ(𝜅)𝑥𝜅−1𝑒−𝑥/𝜃1{𝑥 > 0} with the parameter 𝜅 known

but 𝜃 unknown. Derive a 100(1 − 𝛼)% equail-tailed confidence interval for 𝜃 based on the

sufficient statistic for the unknown parameter 𝜃.

We begin by noting that the given density is a member of the Regular Exponential

Class (REC) since 𝑓(𝑥; 𝜃, 𝜅) = 𝜃−𝜅Γ(𝜅)−1𝑥𝜅−1𝑒−𝑥/𝜃 = 𝑐(𝜃)ℎ(𝑥) exp{𝑞1(𝜃)𝑡1(𝑥)}

where 𝑡1(𝑥) = 𝑥. Then we know that 𝑆 = ∑ 𝑡1(𝑋𝑖)𝑛𝑖=1 = ∑ 𝑋𝑖

𝑛𝑖=1 is complete sufficient

for the unknown parameter 𝜃. Next, we need to create a pivotal quantity from 𝑆; from

the distribution in question 7 which is similar, we guess that 𝑄 =2

𝜃𝑆 =

2

𝜃∑ 𝑋𝑖𝑛𝑖=1

might be appropriate. We now derive the distribution of 𝑄 and, by showing that it is

simultaneously a function of 𝜃 but its density does not depend on 𝜃, will verify that it

is a pivotal quantity. Since the 𝑋𝑖~𝐺𝐴𝑀𝑀𝐴(𝜃, 𝜅), we know that the random variable

𝐴 = ∑ 𝑋𝑖𝑛𝑖=1 ~𝐺𝐴𝑀𝑀𝐴(𝜃, 𝑛𝜅). Then 𝐹𝑄(𝑞) = 𝑃(𝑄 ≤ 𝑞) = 𝑃 (

2

𝜃∑ 𝑋𝑖𝑛𝑖=1 ≤ 𝑞) =

𝑃 (∑ 𝑋𝑖𝑛𝑖=1 ≤

𝑞𝜃

2) = 𝑃 (𝐴 ≤

𝑞𝜃

2) = 𝐹𝐴 (

𝑞𝜃

2) so that 𝑓𝑄(𝑞) =

𝑑

𝑑𝑞𝐹𝐴 (

𝑞𝜃

2) = 𝑓𝐴 (

𝑞𝜃

2)𝜃

2=

1

𝜃𝑛𝜅Γ(𝑛𝜅)(𝑞𝜃

2)𝑛𝜅−1

𝑒−(𝑞𝜃

2)/𝜃 𝜃

2= ⋯ =

1

2𝑛𝜅Γ(𝑛𝜅)𝑞𝑛𝜅−1𝑒−𝑞/2, which shows that the

transformed random variable 𝑄 =2

𝜃∑ 𝑋𝑖𝑛𝑖=1 ~𝐺𝐴𝑀𝑀𝐴(2, 𝑛𝜅) ≡ 𝜒2(2𝑛𝜅). This allows

to compute 𝑃[𝜒𝛼/22 (2𝑛𝜅) < 𝑄 < 𝜒1−𝛼/2

2 (2𝑛𝜅)] = 1 − 𝛼, so after substituting in for 𝑄

and solving for 𝜃, we have 𝑃 [2∑ 𝑋𝑖

𝑛𝑖=1

𝜒1−𝛼/22 (2𝑛𝜅)

< 𝜃 <2∑ 𝑋𝑖

𝑛𝑖=1

𝜒𝛼/22 (2𝑛𝜅)

] = 1 − 𝛼, which is the desired

100(1 − 𝛼)% equail-tailed confidence interval for 𝜃 based on the sufficient statistic

𝑆 = ∑ 𝑋𝑖𝑛𝑖=1 for the unknown parameter 𝜃.

chapter #6 functions of random variables question #2reeder/5080/5080solutions/elird haxhiu...

Documents