probablity

1

Module 2

Random Variable and Its Distribution

1. Random Variable

Let ��, ℱ, �� be a probability space. On many occasions we may not be directly interested in

the whole sample space�. Rather we may be interested in some numerical characteristic of the

sample space �, as the following example illustrates.

Example 1.1

Let three distinguishable dice be labeled as �, and . Consider the random experiment of

rolling these three dice. Then the sample space is � = �� , �, ��: , �, � ∈ �1, 2, … ,6��; here an

outcome � , �, �� ∈ � indicates that the dice �, , and show, respectively, , �and� number

of dots on their upper faces. Suppose that our primary interest is on the study of random

phenomenon of sum of number of dots on the upper faces of three dice. Here we are primarily

interested in the study of the function �:� → ℝ, defined by

�� , �, �� = + � + �, � , �, �� ∈ �. ▄

Moreover, generally, the sample space � is quite abstract and thus may be tedious to deal with.

In such situations it may be convenient to study the probability space ��, ℱ, �� through the

study of a real-valued function defined on �. Example 1.2

Consider the random experiment of tossing a fair coin twice. Here the sample space � =�HH,HT, TH, TT�, where H and T stand for head and tail respectively and in an outcome (e.g., HT) the first letter (e.g., H in HT) indicates the result of the first toss and the second letter (e.g., T in HT) indicates the result of the second toss. Since we are more comfortable in dealing with

real numbers it may be helpful to identify various outcomes in� with different real numbers

(e.g., identify HH,HT, TH and TT with 1, 2, 3 and 4 respectively). This amounts to defining a

function �:� → ℝ on the sample space (e.g., �:� → ℝ, defined as ��HH� = 1, ��HT� = 2,��TH� = 3, and ��TT� = 4�. ▄

The above discussion suggests the desirability of study of real valued functions �:� → ℝ

defined on the sample space �.

Consider a function �:� → ℝ defined on the sample space �.Since the outcomes �in�� of the

random experiment cannot be predicted in advance the values assumed by the function � are

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also unpredictable. It may be of interest to compute the probabilities of various events

concerning the values assumed by function�. Specifically, it may be of interest to compute the

probability that the random experiment results in a value of � in a given set ⊆ ℝ. This

amounts to assigning probabilities,

�)�� ≝ ��+ ∈ �:��+� ∈ ��, ⊆ ℝ, to various subsets of ℝ. Note that, for B ⊆ ℝ, �)�� = ��+ ∈ �:��+� ∈ �� is properly

defined only if �+ ∈ �:��+� ∈ � ∈ ℱ. This puts restrictions on kind of functions � and/or

kind of sets ⊆ ℝwe should be considering. An approach to deal with this issue is to

appropriately choose an event space (a sigma-field) ℬ of subsets of ℝand then put restriction(s)

on the function � so that �)�� = ��+ ∈ � ∶ ��+� ∈ �� is properly defined for each ∈ℬ, i. e. , �+ ∈ �:��+� ∈ � ∈ ℱ, ∀ ∈ ℬ.

Let 1�ℝ� and 1�� denote the power sets of ℝ and� , respectively. Define �23: 1�ℝ� →1�� by

�23�� = �+ ∈ �:��+� ∈ �, ∈ 1�ℝ�.

The following proposition, which follows directly from the definition of �23, will be useful for

further discussion (see Problem 2).

Lemma 1.1

Let �, ∈ 1�ℝ� and let �4 ∈ 1�ℝ�, α ∈ 6, where 6 ⊆ ℝ is an arbitrary index set. Then

(i) �23�� − � = �23�� − �23��. In particular �23�8� = ��23�� 9; (ii) �23�⋃ �<<∈= � = ⋃ �23<∈= ��<� and �23�⋂ �<<∈= � = ⋂ �23<∈= ��<�; (iii) � ∩ = @ ⇒�23�� ∩ �23�� = @. ▄

Let B denote the class of all open intervals in ℝ, i.e., B= ��C, D�: −∞ ≤ C < D ≤ ∞�. In the real

line ℝ an appropriate event space is the Borel sigma-field ℬ3 = H�B�, the smallest sigma-field

containing B. Now, for �)�� = ��+ ∈ ℝ:��+� ∈ �� to be properly defined for every Borel

set ∈ ℬ3, we must have

�23�� = �+ ∈ �:��+� ∈ � ∈ ℱ, ∀ ∈ ℬ3.

This leads to the introduction of the following definition.

Definition 1.1

Let ��, ℱ, �� be a probability space and let �:� → ℝ be a given function. We say that � is a

random variable (r.v.) if �23�� ∈ ℱ, ∀ ∈ ℬ3. ▄

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Note that if ℱ = 1�� then any function �:� → ℝ is a random variable. The following theorem

provides an easy to verify condition for checking whether or not a given function �:� → ℝ is a

random variable.

Theorem 1.1

Let ��, ℱ, �� be a probability space and let �:� → ℝ be a given function. Then � is a random

variable if, and only if, �23��−∞, C]� = �+ ∈ �: ��+� ≤ C� ∈ ℱ, ∀C ∈ ℝ. Proof. First suppose that � is a random variable. Then �23�� ∈ ℱ, ∀ ∈ ℬ3 and, in particular

�23��J, K�� ∈ ℱ, whenever −∞ ≤ J < K ≤ ∞ (since B ⊆ ℬ3).

Fix C ∈ ℝ. Then �−∞, C� = ⋃ L−M, C − 3NOPNQ3 and �C� = ⋂ LC − 3N , C + 3NOPNQ3 . Therefore

�−∞, C] = �−∞, C� ∪ �C�= STU−M,C − 1MV

PNQ3 W ∪ SXUC − 1M , C + 1MV

PNQ3 W .

Now using Lemma 1.1 (ii), it follows that

�23��−∞, C]� = �T �23�U−M, C − 1MV�YZZZZZ[ZZZZZ\�∈ℱ,∀N]3

PNQ3YZZZZZZZ[ZZZZZZZ\∈ℱ

∪ �X �23�UC − 1M , C + 1MV�YZZZZZZ[ZZZZZZ\�∈ℱ,∀N]3

PNQ3 YZZZZZZZZ[ZZZZZZZZ\∈ℱ

YZZZZZZZZZZZZZZZZZZZZZ[ZZZZZZZZZZZZZZZZZZZZZ\∈ℱ

i.e.,�23��−∞, C]� ∈ ℱ.

Conversely suppose that �23��−∞, C]� ∈ ℱ, ∀C ∈ ℝ . Then, for −∞ ≤ J < K ≤ ∞,�−∞, K� = ⋃ L−∞, K − 3NOPNQ3 , and

�23��J, K� = �23��−∞,K�� − ��−∞, J]� = �23��−∞, K�� − �23��−∞, J]��usingLemma1.1�i��

= �23 ST�−∞, K − 1M]P

NQ3 W − �23��−∞, J]�

= T �23��−∞, K − 1M]�YZZZZZ[ZZZZZ\∈ℱ,∀N]3P

NQ3 YZZZZZZZ[ZZZZZZZ\∈ℱ− �23��−∞, J]�YZZZZZ[ZZZZZ\∈ℱ

YZZZZZZZZZZZZZZZ[ZZZZZZZZZZZZZZZ\∈ℱ

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⇒ �23�c� ∈ ℱ,∀c ∈ B.�1.1� Define,

d = �� ⊆ ℝ:�23�� ∈ ℱ�. Using Lemma 1.1 it is easy to verify that d is a sigma-field of subsets of ℝ. Thus d = σ�d�.

Using (1.1) we have B ⊆ d = σ�d�, i.e., B ⊆ σ�d�. This implies that σ�B� ⊆ σ�d� = d, i.e., ℬ3 ⊆ d. Consequently �23�� ∈ ℱ, ∀ ∈ ℬ3, i.e., � is a random variable. ▄

The following theorem follows on using the arguments similar to the ones used in proving

Theorem 1.1.

Theorem 1.2

Let ��, ℱ, �� be a probability space and let �:� → ℝ be a given function. Then � is a random

variable if, an only if, one of the following equivalent conditions is satisfied.

(i) �23��−∞, C�� ∈ ℱ,∀C ∈ ℝ; (ii) �23��C,∞�� ∈ ℱ,∀C ∈ ℝ; (iii) �23�[C,∞�� ∈ ℱ,∀C ∈ ℝ; (iv) �23��C, D]� ∈ ℱ, whenever −∞ ≤ C < D < ∞; (v) �23�[C, D�� ∈ ℱ, whenever −∞ < C < D ≤ ∞; (vi) �23��C, D�� ∈ ℱ, whenever −∞ ≤ C < D ≤ ∞. ▄

2. Induced Probability Measure

Let ��, ℱ, �� be a probability space and let �:� → ℝ be a random variable. Define the set

function �):ℬ3 → ℝ, by

�)�� = ��23�� = ��+ ∈ ℝ:��+� ∈ ��, ∈ ℬ3, where ℬ3 denotes the Borel sigma-field.

Since � is a r.v., �23�� ∈ ℱ, ∀ ∈ ℬ3 and, therefore, �) is well defined.

Theorem 2.1

�ℝ, ℬ3, �)� is a probability space.

Proof. Clearly,

�)�� = ��23�� ≥ 0, ∀ ∈ ℬ3.

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Let 3, i, ⋯ be a countable collection of mutually exclusive events �k ∩l = @,if ≠ � in ℬ3. Then �23�3�, �23�i�,⋯ is a countable collection of mutually exclusive events in ℱ

(Lemma 1.1 (iii)). Therefore

�) STkPkQ3 W = �o�23 STk

PkQ3 Wp

= �oT�23PkQ3 �k�p�usingLemma1.1�ii��

= q�PkQ3 L�−1� �O

= q�)PkQ3 � �,

i.e., �) is countable additive.

We also have

�)�ℝ� = ��23�ℝ� = �� = 1. It follows that �) is a probability measure on ℬ3, i.e., �ℝ, ℬ3, �)� is a probability space. ▄

Definition 2.1

Let��, ℱ, �� be a probability space and let �:� → ℝbe a r.v.. Let �):ℬ3 → ℝbe defined by �)�� = ��23�� , ∈ ℬ3. The probability space �ℝ, ℬ3, �)� is called the probability space

induced by � and �) is called the probability measure induced by �. ▄

Our primary interest now is in the induced probability space �ℝ, ℬ3, �)� rather than the original

probability space ��, ℱ, ��.

Example 2.1

(i) Suppose that a fair coin is independently flipped thrice. With usual interpretations of

the outcomes HHH,HHT,…, the sample space is

� = �HHH,HHT,HTH,HTT, THH, THT, TTH, TTT�.

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Since � is finite we shall take ℱ = 1��. The relevant probability measure �:ℱ → ℝ is

given by

�� = |�|8 ,� ∈ ℱ, where |�| denotes the number of elements in � .Suppose that we are primarily

interested in the number of times a head is observed in three flips, i.e., suppose that our

primary interest is on the function �:� → ℝ defined by

��+� = t0,if+ = TTT1,if+ ∈ �HTT, THT, TTH�2,if+ ∈ �HHT,HTH, THH�3,if+ = HHH .

Since ℱ = 1��, any function u: � → ℝ is a random variable. In particular the function �:� → ℝ defined above is a random variable. The probability space induced by r.v. � is �ℝ, ℬ3, �)�, where �)��0�� = �)��3�� = 3v , �)��1�� = �)��2�� = wv, and

�)�B� = q �)�� k∈�x,3,i,w�∩y , ∈ ℬ3. (ii) Consider the probability space �ℝ, ℬ3, ��, where

�� = z {2|I~��K�Px

= z {2|c�∩[x,P��K�P2P ,

and, for ⊆ ℝ, cy�∙� denotes the indicator function of �i. e., cy�� = 1, if� ∈ ,=0, if� ∉ �. It is easy to verify that � is a probability measure on ℬ3. Define �:ℝ → ℝ by

��+� = �√+,if+ > 00,if+ ≤ 0.

We have

�23��−∞, a]� = �@,ifC < 0�−∞, ai],ifC ≥ 0

∈ ℬ3.

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Thus � is a random variable. The probability space induced by � is �ℝ, ℬ3, �)�, where,

for ∈ ℬ3

�)�� = ��+ ∈ ℝ:��+� ∈ ��

= ��+ ∈ ℝ:+ > 0, √+ ∈ � + ��+ ∈ ℝ:+ ≤ 0,0 ∈ ��

= z {2|cy�√� K� + 0Px

= 2z �{2��cy��K�.Px ▄

3. Distribution Function and Its Properties

Let ��, ℱ, �� be a probability space and let �:� → ℝ be a r.v. so that �23��−∞, a]� =�+ ∈ ℝ:��+� ≤ C� ∈ ℱ, ∀C ∈ ℝ. Throughout we will use the following notation:

�astatement�sayS�about�� = �+ ∈ �: statementSholds�, e. g.,

�C < � ≤ D� ≝ �+ ∈ �:C < ��+� ≤ D� ≝ �23��a, b]�, −∞ ≤ C < D < ∞ �� = J� ≝ �+ ∈ �:��+� = J� ≝ �23��c��,c ∈ ℝ,�� ∈ � ≝ �+ ∈ �:��+� ∈ � ≝ �23��, ∈ ℬ3, �� ≤ c� ≝ �+ ∈ �:��+� ≤ J� ≝�23��−∞, c]�, c ∈ ℝ.

Definition 3.1

The function �):ℝ → ℝ, defined by,

�)�� = �� ≤ �� = �)��−∞, �]�,� ∈ ℝ, is called the distribution function (d.f.) of random variable �. ▄

Example 3.1

(i) Let us revisit Example 2.1 (i). The induced probability space is �ℝ, ℬ3, �)�, where �)��0�� = �)��3�� = 3v , �)��1�� = �)��2�� = wv and �)�� = �� ∈ ��

8

= q �)�� k∈�x,3,i,w�∩y , ∈ -3. Clearly, for � ∈ �,

�)�� F �� )��7∞, �I�

� q �� ∈�0,1,2,3�∩�2P,�I

��0,if� G 018 ,if0 F � G 112 ,if1 F � G 278 ,if2 F � G 31,if� g 3

Figure 3.1. Plot of distribution function �)�� Note that �)�� is non-decreasing, right continuous,�)�7∞� ≝ lim�→2P �)�� 0

and �)�∞� ≝ lim�→P �)�� 1. Moreover �)�� is a step function having

discontinuities at points 0, 1, 2and3.

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(ii) Consider Example 2.1 (ii). The probability space induced by r.v. � is (�, ℬ3, �)),

where, for ∈ ℬ3,

�)�� = 2z �Px {2��cy��K�.

Therefore,

�)�� = �� ≤ �� = �)��−∞, �]�

= 2z �∞0 {−�2c�2P,�]��K�, � ∈ ℝ.

Clearly, for � < 0, �)�� = 0. For � ≥ 0

�)�� = 2z ��x {2��K� = 1 − {2�� .

Thus,

�)�� = �0,if� < 01 − {2�� , if� ≥ 0.

Note that�)�� is non-decreasing, continuous, �)�−∞� = lim�→2P �)�� = 0 and �)�∞� = lim�→P �)�� = 1. ▄

Now we will derive various properties of a distribution function. The following lemma, whose

proof is immediate and can be found in any standard text book on calculus, will be useful in

studying the properties of a distribution function.

Lemma 3.1

Let −∞ ≤ C < D ≤ ∞ and let �: �C, D� → ℝ be a non-decreasing function�i. e. , �� ≤ ��,∀C < � < � < D�.

Then

(i) for all � ∈ �C, D] and � ∈ [C, D�, �� −�and�� +� exist;

(ii) for all � ∈ �C, D�, �� −� ≤ �� ≤ �� +�; (iii) for C < � < � < D, ��+� ≤ ��−�;

(iv) � has at most countable number of discontinuities;

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where ��J 7� and ��J +� denote, respectively, the left hand and right hand limits of the

function � at point J ∈ �C, D�. ▄

Theorem 3.1

Let �)be the distribution function of a random variable�. Then

(i) �) is non-decreasing;

(ii) �) is right continuous;

(iii) �)�−∞� ≝ lim�→2P �) �� = 0 and �)�∞� ≝ lim�→P �) �� = 1. Proof.

(i) Let −∞ < � < � < ∞ . Then �−∞, �] ⊆ �−∞, �] and therefore, on using

monotonicity of probability measures, we get �)�� = �)��−∞, �] ≤ �)��−∞, �]� = �)��.

(ii) Fix � ∈ ℝ. Since �) is non-decreasing, it follows from Lemma 3.1 that �)�� +�

exists. Therefore �)�� +� = limN→P�) U� + 1MV = limN→P�) U�−∞, � + 1M]V.

Note that L−∞, � + 3N� ↓ and LimN→P L−∞, � + 3N� = ⋂ L−∞, � + 3N�PNQ3 = �−∞, �]. Now using continuity of probability measures (Theorem 4.1, Module 1) we have �)�� +� = limN→P�) U�−∞, � + 1M]V

= �) LLimN→P�−∞, � + 3N]O

= �)��−∞, �]�

= �)��. (iii) Using standard arguments of calculus it follows that �)�−∞� = limN→P�)�−M�

and�)�∞� = limN→P�)�M� , where limits are taken along the sequence �M:M =1, 2,⋯ � . Note that �−∞,−M] ↓, �−∞, M] ↑, LimN→P�−∞,−M] = ⋂ �−∞,−M]PNQ3 = @ and LimN→P�−∞, M] = ⋃ �−∞, M]P¡Q3 = ℝ. Again using the continuity of probability

measures, we get

�)�−∞� = limN→P�)�−M� = limN→P�)��−∞,−M] = �) LLimN→P�−∞,−M]O = �)�@� = 0, and

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�)�∞� = limN→P�)�M� = limN→P�)��−∞, M] = �) LLimN→P�−∞, M]O = �)�ℝ� = 1. ▄

Remark 3.1

(i) Using Lemma-3.1 (i)-(ii) and Theorem 3.1 (i) it follows that for a d.f. �), �)�� +� and �)�� −�exist for every � ∈ ℝ and �) is discontinuous at � ∈ ℝ if and only if �)�� −� < �)�� +� = �)��. Consequently a d.f. has only jump discontinuities (a

discontinuity point � ∈ ℝ of �)is called a jump discontinuity if �)�� +� and �)�� −�

exist but �)�� −� = �)�� +� = �)�� does not hold). Moreover the size of the jump

at a point � ∈ ℝ of discontinuity is ¢� = �)�� − �)�� −�.

(ii) Using Lemma 3.1 (iv) and Theorem 3.1 (i) it follows that any d.f. �) has atmost

countable number of discontinuities.

(iii) Let C ∈ ℝ . Since �−∞, a − 3¡] ↑ and LimN→P �−∞, a − 3¡] = ⋃ �−∞, a − 3¡]PNQ3 =�−∞, C�, the continuity of probability measures implies �� < C�� = �)��−∞,C�

= �) £LimN→P�−∞, a − 1n]¤

= limN→P�) £�−∞, a − 1n]¤

= limN→P�) UC − 1MV

= �)�C −�. Therefore, P�� < �� = �)�� −�,∀� ∈ ℝ. Also,

�)�� = �)�� +� ≤ �)�� −�, ∀ −∞ < � < � < ∞�usingLemma3.1�iii�� and

�� = �� = �� ≤ �� − �� < �� = �)�� − �)�� −�, ∀� ∈ ℝ.

Thus �) is continuous (discontinuous) at a point � ∈ ℝ if, and only if, �� = �� =0�� = �� > 0�. (iv) Let ¦) denote the set of discontinuity points (jump points) of d.f. �). Then ¦) is a

countable set and q [�)�� − �)�� −�]�∈§¨= q �� = ��∈§¨

= �)�¦)� ≤ 1,

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i.e., the sum of sizes of jumps of a d.f. does not exceed 1.

(v) Let 7∞ < C < D < ∞. Then

��C < � ≤ D�� = �� ≤ D�� − �� ≤ C�� = �)�D� − �)�C� ��C < � < D�� = �� < D�� − �� ≤ C�� = �)�D −� − �)�C� ��C ≤ � < D�� = �� < D�� − �� < C�� = �)�D −� − �)�C−� ��C ≤ � ≤ D�� = �� ≤ D�� − �� < C�� = �)�D� − �)�C −�, and, for −∞ < C < ∞,

�� ≥ C�� = 1 − �� < C�� = 1 − �)�C −�, and �� > C�� = 1 − �� ≤ C�� = 1 − �)�C�.▄

We state the following theorem without providing the proof. The theorem essentially states

that any function ©:ℝ → ℝ that is non-decreasing and right continuous with ©�−∞� =lim�→2P ©�� = 0 and ©�∞� = lim�→P ©�� = 1 can be regarded as d.f. of a random variable.

Theorem 3.2

Let ©:ℝ → ℝ be a non-decreasing and right continuous function for which ©�−∞� = 0 and ©�∞� = 1. Then there exists a random variable � defined on a probability space ��, ℱ, �� such

that the distribution function of � is ©. ▄

Example 3.2

(i) Consider a function©:ℝ → ℝ, defined by,

©�� = �0,if� < 01 − {2� ,if� ≥ 0.

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Figure 3.2. Plot of distribution function ©�� Clearly © is non-decreasing, continuous and satisfies ©�7∞� � 0 and ©�∞� � 1.

Therefore G can be treated as d.f. of some r.v., say �. Since© is continuous we have

�� ©�� 7 ©�� 7� � 0, ∀� ∈ �,

and, for 7∞ G C G D G ∞,

��C G � G D�� C F � G D�� P�C F � F D� � ��C G � F D�� ©�D� 7 G�a�. Moreover, for 7∞ G C G ∞, �� g C�� C�� 1 7 ©�C� and

�� G C�� F C�� ©�C�. In particular

��2 G � F 3�� ©�3� 7 ©�2� � {2i 7 {2w; ��72 G � F 3�� ©�3� 7 ©�72� � 1 7 {2w; ��1 F � G 4�� ©�4� 7 ©�1� � {23 7 {2«;

��5 F � G 8�� ©�8� 7 ©�5� � {2 7 {2v; �� g 2�� 1 7 ©�2� 7 {2i;

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and

�� 5�� 1 7 ©�5� � {2. Note that the sum of sizes of jumps of © is 0.

(ii) Let ®:� → � be given by

®�� 0,if� G 0�« ,if0 F � G 1�w if1 F � G 2w�v if2 F � G i1,if� g i

.

Figure 3.3. Plot of distribution function ®�� Clearly ® is non-decreasing, right continuous and satisfies ®�7∞� � 0 and ®�∞� � 1 . Therefore ® can be treated as d.f. of some r.v., say u . ® is continuous everywhere except at points 1, 2,and 5/2 where it has jump

discontinuities with jumps of sizes ��u � 1�� ®�1� 7 ®�1 7� � 1/12, ��u � 2�� ®�2� 7 ®�2 7� � 1/12 and ��u � 5/2�� ®�5/2� 7®�5/27� � 1/16 . Moreover for � ∈ � 7 �1, 2, 5/2�, ��u � �� 0 . We also

have

� U�1 G u F 52°V � ® U52V 7 ®�1� � 1 7 13 � 23 ;

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� U�1 G u G 52°V � ® U52 7V 7 ®�1� � 1516 7 13 � 2948 ; � L²1 F u G i³O � ® Li 7O 7®�1 7� � 33´ 7 3« � 333´;

��72 F u G 1�� ®�1 7� 7 ®�72 7� � 3«7 0 � 3«;

��u g 2�� 1 7 ®�2 7� � 1 7 23 � 13 ; and��u � 2�� 1 7 ®�2� � 1 7 w« � 3« ∙ Note that sum of sizes of jumps of ® is 11/48 ∈ �0, 1�.

(iii) Let �:� → � be given by

�� 0,if� G 03v ,if0 F � G 23« if2 F � G 33i if3 F � G 6« ,if6 F � G 12µv ,if12 F � G 151,if� g 15

.

Figure 3.4. Plot of distribution function ��

16

As � is non-decreasing and right continuous with ��7∞� = 0 and ��∞� = 1, it can

be regarded as d.f. of some r.v., say ¶. Clearly, except at points 0, 2, 3, 6, 12and15,� is continuous at all other points and at discontinuity points 0, 2, 3, 6,12and15 it

has jump discontinuities with jumps of sizes

��¶ = 0�� = ��0� − ��0 −� = 18, ��¶ = 2�� = ��2� − ��2 −� = 18, ��¶ = 3�� = ��3� − ��3 −� = 14, ��¶ = 6�� = ��6� − ��6 −� = 310, ��¶ = 12�� = ��12� − ��12 −� = 340,

and ��¶ = 15�� = ��15� − ��15 −� = 18. Moreover��¶ = �� = �� = �� −� = 0, ∀� ∈ ℝ − �0, 2, 3, 6, 12, 15�.Note

that in this case sum of sizes of jumps of � is 1. ▄

Remark 3.2

Let � be a r.v. defined on a probability space ��, ℱ, �� and let �ℝ, 3, �)� be the probability

space induced by �. In advanced courses on probability theory it is shown that the d.f. �)

uniquely determines the induced probability measure �) and vice-versa. Thus to study the

induced probability space �ℝ, 3, �)� it suffices to study the d.f. �). ▄

4. Types of Random Variables: Discrete, Continuous and Absolutely Continuous

Let � be a r.v. defined on a probability space ��, ℱ, �� and let �ℝ, 3, �)�be the probability

space induced by�. Let �) be the d.f. of �. Then �) will either be continuous everywhere or it

will have countable number of discontinuities. Moreover the sum of sizes of jumps at the point

of discontinuities of �) will be either 1 or less than 1. These properties can be used to classify a

r.v. into three broad categories.

Definition 4.1

17

A random variable �is said to be of discrete type if there exists a non-empty and countable set ·) such that �� )�� 7 �)�� 7� � 0, ∀� ∈ ·) and �)�·)� = ∑ ��∈¹¨ �� =�� = ∑ [�)�� − �)�� −�]�∈¹¨ = 1. The set ·) is called the support of the discrete random

variable �. ▄

Remark 4.1

If a r.v. � is of discrete type then �)�·)8� = 1 − �)�·)� = 0 and, consequently �� = �� =0, ∀� ∈ ·)8 , i. e. , �)�� − �)�� −� = 0, ∀� ∈ ·)8 and �) is continuous at every point of ·)8.Moreover, �)�� − �)�� −� = �� = �� > 0, ∀� ∈ Sº. It follows that the support ·)

of a discrete type r.v. � is nothing but the set of discontinuity points of the d.f. �). Moreover

the sum of sizes of jumps at the point of discontinuities is

∑ [�)�� − �)�� −�]�∈¹» = ∑ �� = ��∈¹» = �)�·)� = 1. ▄

Thus we have the following theorem.

Theorem 4.1

Let � be a random variable with distribution function �) and let ¦) be the set of discontinuity

points of �). Then � is of discrete type if, and only if, �� ∈ ¦)�� = 1. ▄

Definition 4.2

Let � be a discrete type random variable with support ·). The function �): ℝ → ℝ, defined by,

�)�� = �� = ��,if� ∈ ·)0,otherwise is called the probability mass function (p.m.f.) of �.

Example 4.1

Let us consider a r.v. ¶ having the d.f. � considered in Example 3.2 (iii). The set of discontinuity

points of � is ¦¾ = �0, 2, 3, 6, 12, 15�and��¶ ∈ ¦¾�� = ∑ [�� − �� −�]�∈§¿ = 1. Therefore the r.v. ¶ is of discrete type with support ·¾ = ¦¾ = �0, 2, 3, 6, 12, 15� and p.m.f.

�¾�� = �[�� − �� −�],if� ∈ ·¾0,otherwise

18

��18 ,if� ∈ �0, 2, 15�14 ,if� � 3310 ,if� � 6340 ,if� � 120,otherwise

.

Figure 4.1. Plot of p.m.f. �¾��

Note that the p.m.f. �) of a discrete type r.v. �, having support ·), satisfies the following

properties:

(i) �)�� 0, ∀� ∈ ·) and �)�� 0, ∀� ∉ ·), (4.1)

(ii) ∑ �)�� ∈¹¨ ∑ �� ∈¹¨ � 1. (4.2)

Moreover, for ∈ -3,

�)�� )� ∩ ·)� + �)� ∩ ·)8� � �)� ∩ ·)� (since ∩ ·)8 ⊆ ·)8 and�)�·)8� � 0)

� q �)��∈y∩¹¨ . This suggest that we can study probability space ��, -3, �)�, induced by a discrete type r.v. �,

through the study of its p.m.f. �) . Also

�)�� q �)À∈�2P,�I∩¹¨ ��,� ∈ �

and

19

�)�� )�� 7 �)�� 7�,� ∈ ℝ. Thus, given a p.m.f. of a discrete type of r.v., we can get its d.f. and vice-versa. In other words,

there is one-one correspondence between p.m.f.s and distribution functions of discrete type

random variables.

The following theorem establishes that any function Á:ℝ → ℝ satisfying (4.1) and (4.2) is p.m.f.

of some discrete type random variable.

Theorem 4.2

Suppose that there exists a non-empty and countable set · ⊆ ℝ and a function Á:ℝ → ℝ

satisfying: (i) Á�� > 0, ∀� ∈ ·; (ii)Á�� = 0, ∀� ∉ · , and (iii) ∑ Á��∈¹ = 1. Then there

exists a discrete type random variable on some probability space �ℝ, ℬ3, �� such that the p.m.f.

of � is Á.

Proof. Define the set function �:ℬ3 → ℝ by

�� = q Á��∈y∩¹ , ∈ ℬ3. It is easy to verify that � is a probability measure onℬ3, i.e., �ℝ, ℬ3, ��is a probability space.

Define �:ℝ → ℝby ��+� = +,+ ∈ ℝ. Clearly � is a r.v. on the probability space �ℝ, ℬ3, �� and

it induces the same probability space �ℝ, ℬ3, ��. Clearly �� = �� = Á��, � ∈ ℝ, and ∑ Á��∈¹ = 1. Therefore the r.v. � is of discrete type with support · and p.m.f. Á. ▄

Example 4.2

Consider a coin that, in any flip, ends up in head with probability 3« and in tail with probability

w«.

The coin is tossed repeatedly and independently until a total of two heads have been observed.

Let � denote the number of flips required to achieve this. Then �� = �� = 0, if� ∉�2, 3, 4,⋯ �. For ∈ �2, ,3,4… � �� = �� = oL − 11 O 14 U34Vk2ip14

= − 116 U34Vk2i. Moreover, ∑ �� = ��PkQi = 1 . It follows that � is a discrete type r.v. with support ·) = �2, 3, 4, … � and p.m.f.

20

�)�� Â�233´ Lw«O�2i ,if� ∈ �2, 3, 4,⋯ �0,otherwise .

Figure 4.2. Plot of p.m.f. �)�� The d.f. of � is

�)�� F �� t0,if� G 2116q�� 7 1� U34Vl2i

klQi , if F � G + 1, � 2, 3, 4,⋯

� Ã0,if� G 21 7 + 34 U34Vk23 ,if F � G + 1, � 2, 3, 4,⋯

▄

Example 4.3

A r.v. � has the d.f.

21

�)�� 0,if� < 2iw ,if2 ≤ � < 5µ2´Ä´ ,if5 ≤ � < 9wÄ�2´ÄÅµ´ ,if9 ≤ � < 143´Ä�23´ÄÅ3Æ3´ ,if14 ≤ � ≤ 201,if� > 20

,

where � ∈ ℝ. (i) Find the value of constant �; (ii) Show that the r.v. � is of discrete type and find its support;

(iii) Find the p.m.f. of �. Solution. (i) Since �) is right continuous, we have

�)�20� = �)�20+�

⇒ 16�i − 16� + 3 = 0 ⇒ � = 14 or� = 34.�4.3�

Also �) is non-decreasing. Therefore

�)�5 −� ≤ �)�5�

⇒ � ≤ 12.�4.4�

On combining (4.3) and (4.4) we get � = 1/4. Therefore

�)�� =��0,if� < 2iw ,if2 ≤ � < 5333i ,if5 ≤ � < 9Æ3Æ´ ,if9 ≤ � < 141,if� ≥ 14

.

(ii) The set of discontinuity points of �) is ¦) = �2, 5, 9, 14�. Moreover �� = 2�� = �)�2�–�)�2 −� = iw, �� = 5�� = �)�5�–�)�5 −� = 3«, �� = 9�� = �)�9�–�)�9 −� = 3wi,

22

�� 14�� = �)�14�–�)�14 −� = Æ´,and�� ∈ ¦)�� = �� = 2�� + �� = 5�� + �� = 9�� + �� = 14�� = 1. Therefore the r.v. � is of discrete type with support ·) = �2, 5, 9, 14�.

(iii) Clearly the p.m.f. of � is given by

�)�� = �� = �� = ��

iw ,if� = 23« ,if� = 53wi ,if� = 9Æ´ ,if� = 140,otherwise.

▄

Example 4.4

A r.v. � has the p.m.f.

�)�� = Â J�2� − 1��2� + 1� , if� ∈ �1, 2, 3,⋯ �0,otherwise , where J ∈ ℝ.

(i) Find the value of constant J;

(ii) For positive integers È and M, such that È < M, evaluate �� < È + 1��, �� ≥È��, ��È ≤ � < M�� and ��È < � ≤ M��;

(iii) Determine the d.f. of �.

Solution.

(i) Let ·) be the support of � so that ·) = �� ∈ ℝ:�)�� > 0� and ∑ �)�� = 1�∈¹¨ .

Clearly, ·) = �1, 2, 3, … � and

q c�2 − 1��2 + 1� = 1PkQ3

⇒ limM→∞q J�2 − 1��2 + 1� = 1NkQ3

⇒ J2 limM→∞qÉ 12 − 1 − 12 + 1ÊN

kQ3 = 1

23

⇒ J2 limM→∞ Ëq 12 − 1 − q 12 + 1N

kQ3N

kQ3 ÌN

= 1

⇒ J2 limM→∞ É1 − 12M + 1Ê = 1

⇒ c = 2. (ii) We have �� < È + 1�� = �� ≤ È��

= q 2�2 − 1��2 + 1�ÍkQ3

= qÉ 12 − 1 − 12 + 1ÊÍkQ3

= 1 − 12È + 1

= 2È2È + 1, �� ≥ È�� = 1 − �� < È�� = 1 − 2�È − 1�2�È − 1� + 1

= 12È − 1, ��È ≤ � < M�� = �� < M�� − �� < È��

= 2�M − 1�2M − 1 − 2�È − 1�2È − 1

= 2�M − È��2M − 1��2È − 1�, and��È < � ≤ M�� = ��È + 1 ≤ � < M + 1��

= 2�M − È��2M + 1��2È + 1�. (iii) Clearly, for � < 1, �)�� = 0. For ≤ � < + 1, = 1, 2, 3,⋯

24

�)�� G + 1�� = 2 2 + 1.�using�ii�� Therefore,

�)�� = Â0,if� < 12 2 + 1 ,if ≤ � < + 1, = 1, 2, 3,⋯ .▄ Definition 4.3

(i) A random variable � is said to be of continuous type if its distribution function �) is

continuous everywhere.

(ii) A random variable � with distribution function �) is said to be of absolutely

continuous type if there exists an integrable function �):ℝ → ℝ such that �)�� ≥ 0, ∀� ∈ ℝ, and

�)�� = z��t�K�,�2P � ∈ ℝ.

The function �) is called the probability density function (p.d.f.) of random variable �

and the set ·) = �� ∈ ℝ:�)�� > 0� is called the support of random variable � (or

of p.d.f. �) ). ▄

Note that if �) is p.d.f. of an absolutely continuous type r.v. � then �)�� ≥ 0, ∀� ∈ℝand Î �)�t�K�P2P = �)�∞� = 1, where �)�∞� = lim�→∞�)��.

Example 4.5

Let � be a r.v. having the d.f.

�)�� = �0,if� < 01 − {2� ,if� ≥ 0. Clearly �) is continuous at every � ∈ ℝ and therefore � is of continuous type. Also

�)�� = Î �)�2P �t�K�,� ∈ ℝ,

where �):ℝ → [0,∞� is given by

�)�� = �0,if� < 0{2|,if� ≥ 0. �4.5�

It follows that � is also if absolutely continuous type with p.d.f. given by (4.5).

25

Figure 4.3. Plot of p.d.f. �)�� ▄

Remark 4.2

(i) Suppose that � is a r.v. of continuous type. Then �)�� )�� 7�)�� 7� � 0, ∀� ∈ �. In general, for any countable set , �� ∈ �� )�� )�⋃ ��∈8 � � ∑ �)��∈8 � 0. (ii) Since the p.d.f. �) of an absolutely continuous type r.v. � determines its d.f., using

Remark 3.2, it follows that we may study the induced probability space ��, -3, �)� through the study of p.d.f. �).

(iii) Suppose that � is a r.v. of absolutely continuous type. Then, for � ∈ �andÏ � 0, |�)�� 7 Ï� 7 �)��| � �)�� 7 �)�� 7 Ï� � z�)�t�K��

2P 7 z �)�t�K��2Ð2P

� z �)�t�K��2Ð

→ 0,asÏ ↓ 0,

26

i.e., �) is also left continuous on �. It follows that if � is an absolutely continuous

type r.v. then its d.f. �) is continuous everywhere on � and hence � is of continuous

type.

(iv) Let � be a r.v. of absolutely continuous type. Then � is also of continuous type (see

(iii) above) and therefore �� 0, ∀� ∈ ℝ. Consequently,

�� < �� = �� ≤ �� = �)�� = z�)��K��2P , � ∈ ℝ

�� ≥ �� = 1 − �� < �� = z �)��K�P� , � ∈ ℝ,�since z �)��K�P

2P = 1�

and, for −∞ < C < D < ∞,

��C < � ≤ D�� = ��C < � < D�� = ��C ≤ � < D�� = ��C ≤ � ≤ D��

= �)�D� − �)�C�

= z�)��K�Ñ2P − z�)��K�Ò

2P

= z�)��K�ÑÒ

= z �)��c�Ò,Ñ��K�P2P ,

where, for a set � ⊆ ℝ, c� denotes its indicator function, i.e.,

c�� = �1,if� ∈ �0,otherwise. In general, for any ∈ ℬ3, it can be shown that

�� ∈ �� = z �)�t�P2P cy�t�K�.

27

Figure 4.4. Probability of a region

(v) We know that if Ï:� → � is a non-negative integrable function then, for any

countable set ¦�n @� in �, and for 7∞ F C G D F ∞,

z Ï��ÑÒ c§�t�K� � zÏ�t�Ñ

Ò c⋃ ��»∈Ó ��K� � zÏ��Ñ

Ò Sq c��t��∈§ WK� � qzÏ�t�Ô

Ò c��t�K��∈§

� 0,�4.6� since Î Ï��c��t�ÑÒ K� � 0, ∀� ∈ �. Now let � be a r.v. of absolutely continuous type with p.d.f. �) and d.f. �) so that

�)�� z ��t�K�,�7∞ � ∈ �.

Let Õ be any countable set and let Á:� → f0,∞� be any non-negative function such

that Á�� )��, ∀� ∈ Õ8 � �7 Õ and Á�� n �)��, ∀� ∈ Õ. Then, for � ∈ �,

�)�� z�)�t�K��2P

28

� z�)�t�Öc×�t� +c×Ø�t�ÙK��2P

= z�)��c×Ø��K��using�4.6��2P

= zÁ��c×Ø��K��since�)��2P c×Ø�� = Á��c×Ø��

= zÁ��c×Ø�t�K��2P + zÁ��c×�t�K��

2P �using�4.6�� = zÁ��K��

2P ,

i.e., g is also a p.d.f. of r.v. � . Note thatÁ�� = �)��, ∀� ∈ Õ8 andÁ�� ≠�)��, ∀� ∈ Õ. It follows that the p.d.f. of a r.v. of absolutely continuous type is not

unique. In fact if the values of the p.d.f. �) of a r.v. � of absolutely continuous type

are changed at a finite number of points with some other non-negative values then

the resulting function is again a p.d.f. of �. In other words a r.v. of absolutely

continuous type has different versions of p.d.f.s. Consequently the support of an

absolutely continuous type r.v. is also not unique and it depends upon the version of

p.d.f. chosen. However it is worth mentioning here that the d.f. of any r.v. is unique.

(vi) Suppose that the d.f. �) of a r.v. � is differentiable at every � ∈ ℝ. Then

�)�� = z�)Ú�2P ��K�, � ∈ ℝ.

It follows that if the d.f. �) is differentiable everywhere then the r.v. � is of

absolutely continuous type and one may take its p.d.f. to be �)�� = ��′ ��, � ∈ ℝ.

(vii) Suppose that the d.f. of a r.v. � is differentiable everywhere except on countable set ¦. Further suppose that z �)Ú ��c§Ü��K� = 1.P2P

Then, using a standard result in advanced calculus, it follows that the random

variable � is of absolutely continuous type with a p.d.f.

29

�)�� Ý��′ ��,if� ∉ C�,if� ∈ , where C� , � ∈ are arbitrary nonnegative constants. Here, note that z �)Ú ��c§Ü��K� = z �)��K� = 1P

2PP

2P

and �)�� = z �)Ú��c§Ü��K� = z �)��K�,� ∈ ℝ.�2P

�2P

(viii) There are random variables that are neither of discrete type nor of continuous type

(and hence also nor of absolutely continuous type). To see this let us consider a r.v. � having the d.f. �) (see Example 3.2 (ii)) given by

�)�� =��0,if� < 0�« ,if0 ≤ � < 1�w ,if1 ≤ � < 2w�v ,if2 ≤ � < i1,if� ≥ i

.

The set of discontinuity points of �) is ¦) = ²1, 2, i³. Since ¦) ≠ ∅ the r.v. is not of

continuous type. Moreover

�� ∈ ¦)�� = �� = 1�� + �� = 2�� + � L²� = i³O

= [�)�1� − �)�1 −�] + [�)�2� − �)�2 −�] + É�) U52V − �) U52 −VÊ = 33«v < 1,

implying that the r.v. � is also not of discrete type.

(ix) There are random variables which are of continuous type but not of absolutely

continuous type. These random variables are normally difficult to study. ▄

Example 4.6

Consider a r.v. � having the d.f. �) (see Example 4.5) given by

�)�� = �0,if� < 01 − {2� ,if� ≥ 0 ∙ Clearly �) is differentiable everywhere except at � ∈ ¦ = �0�. Also

z �)Ú ��c§Ü��K� = z {2|K� = 1.Px

P2P

30

Using Remark 4.2 (vii) it follows that the r.v. � is of absolutely continuous type and one may

take

�)�� Â 0,if� < 0C,if� = 0,{2� ,if� > 0

as a p.d.f. of �; here C is an arbitrary non-negative constant. In particular one may take

�)�� = �0,if� ≤ 0{2� ,if� > 0

as a p.d.f. of �. ▄

Note that the p.d.f. �) of a r.v. � of absolutely continuous type satisfies the following two

properties:

(i) �)�� ≥ 0, ∀� ∈ ℝ; (ii) Î �)�t�K�P2P = lim�→∞�)�� = 1.

The following theorem establishes that any function Á:ℝ → [0,∞] satisfying the above two

properties is a p.d.f. of some r.v. of absolutely continuous type.

Theorem 4.3

Suppose that there exists a non-negative function Á:ℝ → ℝ satisfying:

(i) Á�� ≥ 0, ∀� ∈ ℝ; (ii) Î Á��K� = 1P2P .

Then there exists an absolutely continuous type random variable � on some probability space ��, ℬ3, �) such that the p.d.f. � is Á.

Proof. Define the set function �:ℬ3 → ℝ by

�� = z Á��cy��K�,P2P ∈ ℬ3 ∙

It is easy to verify that � is a probability measure on ℬ3, i.e.,�ℝ, ℬ3, �� is a probability space.

Define �:ℝ → ℝ by ��+� = +, + ∈ ℝ. Clearly � is a random variable on the probability

space �ℝ, ℬ3, ��. The space (ℝ,ℬ3, �� is also the probability space induced by �. Clearly, for � ∈ ℝ,

�)�� = �)��−∞, �]�

31

� ��7∞, �]�

= z Á�t�P2P c�2P,�]�t�K�

= zÁ��K��2P .

It follows that � is of absolutely continuous type and Á is the p.d.f. of �. ▄

Example 4.7

Let � be r.v. with the d.f.

�)�� =��0,if� < 0��i ,if0 ≤ � < 1�i ,if1 ≤ � < 21,if� ≥ 2

.

Show that the r.v. � is of absolutely continuous type and find the p.d.f. of �.

Solution. Clearly �) is differentiable everywhere except at points 1 and 2. Let ¦ = �1, 2�, so

that

z �)Ú ��c§Ü��K� = z �K� + z 12i3 K� = 1.3

xP

2P

Using Remark 4.2 (vii) it follows that the r.v. � is of absolutely continuous type with a p.d.f.

�)�� =��,if0 ≤ � < 1C,if� = 13i ,if1 < � < 2D,if� = 20,otherwise

,

where C and D are arbitrary nonnegative constants. In particular a p.d.f. of � is

�)�� = Ã�,if0 < � < 13i ,if1 < � < 20,otherwise .

▄

32

Example 4.8

Let � be an absolutely continuous type r.v. with p.d.f.

�)�� Ý� 7 |�|,if|�| < 3i0,otherwise ,

where � ∈ ℝ.

(i) Find the value of constant k;

(ii) Evaluate: �� < 0��, �� ≤ 0��, � L²0 < � ≤ 3«³O , � L²0 ≤ � < 3«³O and� L²− 3v ≤ � ≤3«³O;

(iii) Find the d.f. of �.

Solution.

(i) Since �) is a p.d.f.

z �)��P2P K� = 1

⇒ z �� − |�|�3 iß23 iß

K� = 1

⇒ � = 54 ∙ Also, for � = « , �)�� ≥ 0, ∀� ∈ ℝ.

(ii) Since the r.v. �is of absolutely continuous type, �� = �� = 0, ∀� ∈ ℝ (see

Remark 4.2 (iv)).Therefore �� < 0�� = �� ≤ 0�� = Î �)x2P ��K� = Î L« + �OK� = 3ix23 iß ,

� L²0 < � ≤ 3«³O = � L²0 ≤ � < 3«³O = Î �)3 «ßx ��K� = Î L« − �O K� = Æwi3 «ßx ,

and

� U− 18 ≤ � ≤ 14V = z �)3 «ß

23 vß��K�

33

� z U54 + �VK� +x23 vß

z U54 − �V3 «ßx K�

= 55128 ∙ (iii) Clearly, for � < − 3i , �)�� = 0and, for� ≥ 3i , �)�� = 1. For − 3i ≤ � < 0,

�)�� = z�)�t��2P K�

�)�� = z U54 + tV�23 iß

K�

= ��i + « � + 3i ,

and, for 0 ≤ � < 3i,

�)�� = z�)�t��2P K�

= zU54 + tV K� + zU54 − tV�x K�x

2à�

= −�i2 + 54� + 12 ∙ Therefore the d.f. of � is

�)�� =��−0if� < − 3i�|�|i + « � + 3i ,if − 3i ≤ � < 3i1,if� ≥ 3i

.

▄

Theorem 4.4

Let �) be the distribution function of a random variable �. Then �) can be decomposed as �)(�) = á�â(�) + (1 − á)�9(�), � ∈ ℝ, whereá ∈ f0,1I, �â is a distribution function of

some random variable of discrete type and �9 is a distribution function of some random

variables of continuous type.

34

Proof. Let ¦) denote the set of discontinuity points of �). We will prove the result for the case

when ¦) is finite. The idea of the proof for the case when ¦) is countably infinite is similar but

slightly involved. First suppose that ¦) � @. In this case the result follows trivially by taking á � 0 and �8 ≡ �). Now suppose that ¦) = �C3, Ci, … , CN� for some M ∈ ℕ.Without loss of

generality let −∞ < C3 < Ci < ⋯ < CN < ∞.

Define

¢k = �(�� = Ck�) = ��(Ck) − �)(Ck −), = 1, 2, … , M, so that ¢k > 0, = 1,… , M.

Let á = ∑ ¢kNkQ3 so that á ∈ (0, 1I. Define �â: ℝ → ℝ by

�â(�) = ��0,if� < 0∑ ¢lklQ3á ifCk ≤ � < CkÅ3, = 1, … , M − 1.1,if� ≥ CN

Clearly �â is non-decreasing, right continuous �â(−∞) = 0and�â(∞) = 1 . The set of

discontinuity points of �â is �C3, … , CN� and

qf�â(Ck) − �â(Ck −)INkQ3 = qÝ∑ ¢åklQ3á − ∑ ¢lk23lQ3á æN

kQ3

= 1áq¢kN

kQ3

= 1.

It follows that �â is a d.f. of some r.v. of discrete type. If á = 1 then the result follows on taking �â ≡ �). Now suppose that á ∈ (0, 1).

Define �9: ℝ → ℝ by

�8(�) = �)(�) − á�â(�)1 − á ,� ∈ ℝ. For � ⊆ ℝ,let ·(�) = � ∈ �1, … , M�:Ck ∈ ��. Then, for −∞ < � < � < ∞,

�â(�) − �â(�) = q ¢kák∈¹((2P,ÀI) − q ¢kák∈¹((2P,�I)

35

� q ¢ká ,k∈¹((�,ÀI)

�)(�) − �)(�) = �(�� < � ≤ ��)

≥ q ¢kk∈¹((�,ÀI)

= á��â(�) − �â(�) , where, for � ⊆ ℝ, ∑ ¢k = 0k∈¹(�) , if ·(�) = @.

Therefore, for −∞ < � < � < ∞,

�9(�) − �ç(�) = �)(�) − �)(�) − á��â(�) − �â(�) 1 − á

≥ 0, i.e., �ç is non-decreasing. Note that �)(Ck) − �)(Ck −) = á��â(Ck) − �â(Ck −) = ¢k, =1, … , M and �)(�) − �)(� −) = 0, if � ∉ {C3, … , CN}. It follows that

�9(�) − �ç(� −) = �)(�) − �)(� −) − á��â(�) − �â(� −) 1 − á

= 0, ∀� ∈ ℝ, i.e.,�ç is continuous everywhere. Since �)(−∞) = �â(−∞) = 0and�)(∞) = �â(∞) = 1 we

also have �ç(−∞) = 0and�9(∞) = 1. Therefore �8 is a d.f. of some r.v. of continuous type.

Hence the result follows. ▄

Example 4.9

Let � be a r.v. having the d.f. �) (see Example 3.2 (iii)) given by

�)(�) =��0,if� < 0�« ,if0 ≤ � < 1�w ,if1 ≤ � < 2w�v ,if2 ≤ � < i1,if� ≥ i

.

Decompose �) as�)(�) = á®â(�) + (1 − á)®9(�), � ∈ ℝ, where á ∈ f0,1I, ®â is a d.f. of

some r.v. �â of discrete type and ®9 is a d.f. of some r.v. �9 of continuous type.

36

Solution. The set of discontinuity points �) is ¦) � �1, 2, 5/2� with ¢3 = �� = 1� = �)(1) − �)(1 −) = 112, ¢i = �� = 2� = �)(2) − �)(2 −) = 112,

and

¢w = � U�� = 52°V = �) U52V − �) U52 −V = 116. Thus,

á = ¢3 + ¢i + ¢w = 1148, �(��â = 1�) = ¢3á = 411 , �(��â = 2�) = ¢iá = 411 , � U��â = 52°V = ¢wá = 311,

®â(�) =��0,if� < 1411 ,if1 ≤ � < 2811 ,if2 ≤ � < 521,if� ≥ 52

and

®8(�) = ®(�) − á®â(�)1 − á

=��0,if� < 03iwµ �,if0 ≤ � < 1«(«�23)wµ ,if1 ≤ � < 2i(Æ�2«)wµ ,if2 ≤ � < i1,if� ≥ i

.

37

Figure 4.5. Plot of distribution function ®â��

Figure 4.6. Plot of distribution function ®9�� ▄

Problems

1. Let -3 denote the Borel sigma-field of subsets of � and let 7∞ G � G � G ∞. Define -f�,ÀI � �f�, �I ∩ : ∈ -3�. Show that:

(i) �C� ∈ -3, ∀C ∈ �;

(ii) If is a countable subset of �, then ∈ -3;

38

(iii) -f�,ÀI is a sigma-field of subsets of f�, �I.

2. Let � be a given set and let �: � → ℝ be a given function. Define �23: 1(ℝ) → 1(�)

by �23() = �+ ∈ �: �(+) ∈ �, where, for a set ·, 1(·) denotes the power set of ·.

Let �, ∈ 1(ℝ) and let �< ∈ 1(ℝ), á ∈ 6, where 6 ⊆ ℝ is an arbitrary index set.

Show that:

(i) �23(� − ) = �23(�) − �23(); (ii) �23(9) = ��23() 9; (iii) �23(⋃ �<<∈= ) = ⋃ �23<∈= (�<); (iv) �23(⋂ �<<∈= ) = ⋂ �23<∈= (�<); (v) � ∩ = @ ⇒ �23(�) ∩ �23() = @.

3. Let (�, ℱ, �) be a probability space and let �:� → ℝ be a function. In each of the

following cases, verify whether or not � is a r.v..

(i) � = �−2,−1, 0, 1, 2, 3�, ℱ = �ϕ, Ω, �0�, �−1, 1�, �−2,−1, 1, 2, 3�, �−2, 0, 2, 3�, �−1, 0, 1�, �−2, 2, 3�� and �(+) = +i, + ∈ �; (ii) � = �+3, +i, +w, +«�, ℱ = �@, �, �+3�, �+i�, �+i, +w, +«�, �+3, +w, +«�, �+3, +i�, �+w, +«��, �(+3) = 0, �(+i) = �(+w) = 1 and �(+«) = 2; (iii) � = f0,1I, ℱ = ℬfx,3I, where ℬfx,3I is as defined in Problem 1, and

�(+) t +,if+ ∈ É0, 12Ê+ − 12 ,if+ ∈ U12 , 1Ê. 4. Let (�, ℱ, �) be a probability space and let �:� → ℝ be a r.v.. In each of the following

cases determine whether or not u: � → ℝ is a r.v.:

(i) u(+) = |��+�|, + ∈ �;

(ii) u(+) = ��(+) i, + ∈ �;

(iii) u(+) = ê�(+), + ∈ �, where �23�(−∞, 0) = @;

(iv) u(+) = max(�(+), 0) , + ∈ �;

(v) u(+) = max(−�(+), 0) , + ∈ �.

5. Consider a random experiment of two independent tosses of a coin so that the sample

space is � = �HH,HT, TH, TT� with obvious interpretations of outcomes HH,HT, THandTT. Let ℱ = 1(�) (the power set of �) and let �(∙) be a probability

measure defined on ℱ such that �(�HH�) = ¢i, �(�HT�) = �(�TH�) = ¢(1 − ¢) and �(�TT�) = (1 − ¢)i , where ¢ ∈ (0,1). Define the function �:� → ℝ by �(�HH�) =2, �(�HT�) = �(�ì®�) = 1 and �(�TT�) = 0 , i.e., �(+) denotes the number of Hs

(heads) in +. Show that � a r.v. and find the probability space (ℝ, ℬ3, �)) induced by �.

39

6. A card is drawn at random from a deck of 52 playing cards so that the sample space

consists of names of 52 cards (e.g., jack of heart, ace of spade, etc,). Let ℱ � 1�� (the

power set of �). Define �: � → ℝ by

�(+) =��5,if+isanace4,if+isaking3,if+isaqueen2,if+isajack1,otherwise

.

Show that � is a r.v. and find the probability space (ℝ, ℬ3, �)) induced by �.

7. Let �3, �iand�w be three random variables with respective distribution functions �3, �iand�w, where

�3(�) = t0,if� < −1� + 24 ,if − 1 ≤ � < 11,if� ≥ 1 ; �i(�) =��0,if� < −1� + 14 ,if − 1 ≤ � < 0� + 34 ,if0 ≤ � < 11,if� ≥ 1

; and

�w(�) =��0,if� < −213 ,if − 2 ≤ � < 012 ,if0 ≤ � < 512 + (� − 5)i2 ,if5 ≤ � < 61,if� ≥ 6

.

(i) Sketch the graph of �3(�) and compute � L²− 3i < �3 ≤ 3i³O , �(��3 = 0�), �(��3 =1�), �(�−1 ≤ �3 < 1�) and �(�−1 < �3 < 1�);

(ii) Compute � L²3« ≤ �i ≤ w«³O , � L²�i ≥ 3i³O , �(��i ≥ 0�) and � L²0 < �i ≤ 3i³O;

(iii) Compute �(�−2 ≤ �w < 5�), � L²0 < �w < 33i ³O and the conditional probability � L²wi < �w ≤ 33i ³ |��w � 2�O.

8. Do the following functions define distribution functions?

40

�i)�3(�) = t0,if� < 0�,if0 ≤ � ≤ 3i ;1,if� > 3i (ii) �i(�) = �0,if� < 01 − {2�, if� ≥ 0;

and

(iii) �w(�) = 3i + ðñ¡òà(�)ó , −∞ < � < ∞.

9. Let �:ℝ → ℝ be defined by

�(�) = Â0,if� < 01 − 23 {2»� − 13 {2ô»��,if� ≥ 0,

where , for � ∈ ℝ, f�I denotes the largest integer ≤ �. Show that � is a d.f. of some r.v. �. Compute �(�� > 4�), �(�� = 4�), �(�� ≥ 4�), �(�� = 3�) and �(�3 ≤ � < 6�).

10. Let �(∙) and ©(∙) be two distribution functions. Verify whether or not the following

functions are distribution functions:

(i)®(�) = �(�) + ©(�); (ii) ®(�) = max��(�), ©(�) ; (iii) ®(�) = min��(�), ©(�) .

11. (i) Let �3(∙),⋯ , �N(∙) be distribution functions and let C3, ⋯ , CN be positive real

numbers satisfying ∑ CkNkQ3 = 1. Show that ©(�) = ∑ CkNkQ3 �k(�) is also a d.f.;

(ii) If �(∙) is a d.f. and á is a positive real constant, then show that ©(�) = ��(�) <

and ®(�) = 1 − �1 − �(�) < are also distribution functions.

12. Do there exist real numbers á, õ, ö and ÷ such that the following functions become

distribution functions?

(i)�(�) =��0,if� < 0��i ,if0 ≤ � < 03i + á(� − 1)i,if1 ≤ � ≤ 2õ + (�2i)øµ ,if2 < � ≤ 31,if� > 3

; (ii) ©(�) = Ý0,if� ≤ 0ö + ÷{2»�� ,if� > 0.

13. Do the following functions define probability mass functions of some random variables

of discrete type?

41

�i)�3(�) = Ý�i ,if� ∈ �−1, 0, 1, 2�0,otherwise ; (ii) �i(�) = Ýùòà�! ,if� ∈ �0, 1, 2,⋯ �0,otherwise ; (iii) �w(�) = ÂL50� O L3«O� Lw«Ox2� ,if� ∈ �1, 2,⋯ , 50�0,otherwise .

14. For each of the following, find the value of constant J so that �(∙) is a p.m.f. of some

discrete type r.v. (say�). Also, for each of the following, find �(�� > 2�), �(�� < 4�),

and �(�1 < � < 2�):

(i)�(�) = �J(1 − ¢)� ,if� ∈ �1, 2, 3,⋯ �0,otherwise ; (ii) �(�) = Ý9û»�! ,if� ∈ �1, 2,⋯ �0,otherwise ; here ¢ ∈ (0,1) and ü > 0 are fixed constants.

15. Do the following functions define probability density functions of some random

variables of absolutely continuous type?

(i)�3(�) = ÝÆÅ�3vx ,if − 10 < � < 100,otherwise ; (ii) �i(�) = Ý ��Å3 ùò»i , if� > 00,otherwise; (iii)�w(�) = Â2 + cos �2ý , if0 < � < ý0,otherwise .

16. In each of the following, find the value of constant J so that �(∙) is a p.d.f. of some r.v. (say�) of absolutely continuous type. Also, for each of the following, find �(�� >3�), �(�� ≤ 3�), and �(�3 < � < 4�):

(i)�(�) = �J�{2�� , if� > 00,otherwise; (ii) �(�) = �J�{2(�2i), if� > 20,otherwise. 17. (i) Let � be a discrete type r.v. with support ·) = �0, 1, 2, 3, 4�, �(�� = 0�) =�(�� = 1�) = 33x , �(�� = 2�) = �(�� = 3�) = �(�� = 4�) = «3 . Find the d.f. � and

sketch its graph.

(ii) Let the r.v. � have the p.m.f.

42

�)�� Â �5050 , if� ∈ �1, 2,⋯ ,100�0,otherwise . Show that the d.f. of � is

�)(�) = t0,if� < 1f�I(f�I + 1)10100 , if1 ≤ � < 1001,if� ≥ 100 . Also compute �(�3 < � < 50�).

18. For each of the following p.d.f.s of some r.v. (say�) of absolutely continuous type, find

the d.f. and sketch its graph. Also compute �(�|�| G 1�) and �(��i < 9�).

(i)�(�) = Ý��3v ,if − 3 < � < 30,otherwise ; (ii) �(�) = Ý�Åi3v ,if − 2 < � < 40,otherwise . (iii) �(�) = Ý 3i�� , if|�| � 10,otherwise.

19. (i) Let � be a r.v. of absolutely continuous type with p.d.f. �(�) = �J�i, if − 1 < � < 10,otherwise . Compute the values of J, �(�� = 0�), �(�� > 0�), �(�� > 1/2�), �(�|�| �1/2�), �(�1/2 < � < 3/4�), �(�1/2 < � < 2�) and the conditional probability �(�� < 3/4�|�� 1/2�);

(ii) Let � be a r.v. of absolutely continuous type with p.d.f. �(�) = �J(� + 1){2û� , if� > 00,otherwise, where ü > 0 is a given constant. Compute the values of J, �(�� = 2�), �(�� >2�), �(�� > 1�), �(�1 < � < 3�), �(�|� 7 2| � 1�) and the conditional probability �(�� < 3�|�� 1�).

20. Let � be a r.v. with d.f. �)(∙). In each of the following cases determine whether � is of

discrete type or of absolutely continuous type. Also find the p.d.f./p.m.f. of �:

43

�i)�)(�) =��0,if� < −23w ,if − 2 ≤ � < 03i ,if0 ≤ � < 5w« ,if5 ≤ � < 61,if� ≥ 6

; (ii) �)(�) = �0,if� < 01 − {2� , if� ≥ 0.

21. Let the r.v. � have the d.f.

�)(�) =��0,if� < 0�3 ,if0 ≤ � < 123 ,if1 ≤ � < 21,if� ≥ 2

. Show that � is neither of discrete type nor of absolutely continuous type.

22. For the three d.f.s considered in Problems 20 and 21, find the decomposition �)(�) = á�â(�) + (1 − á)�9(�), � ∈ ℝ, where á ∈ f0,1I, �â is a d.f. of some r.v. of

discrete type and �9 is a continuous d.f..

probablity

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