Statistics and Probability Letters 78 (2008) 3152–3158

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Statistics and Probability Letters

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Stochastic inequalities for weighted sum of two random variablesindependently and identically distributed as exponentialMehmet Yilmaz a,∗, Birol Topçu ba Ankara University, Faculty of Science, Department of Statistics, Ankara, Turkeyb Afyon Kocatepe University, Faculty of Science and Literature, Department of Statistics, Afyonkarahisar, Turkey

a r t i c l e i n f o

Article history:Received 27 September 2007Received in revised form 27 May 2008Accepted 28 May 2008Available online 12 June 2008

a b s t r a c t

Let X and Y be two random variables which are independently and identically distributed(i.i.d.) as exponential. Given two nonnegative numbers a and b, it is of interest to establishbounds on the tail probability of aX + bY , i.e. P (aX + bY > t). The present work attemptsto provide first some basic inequalities for P (aX + bY > t) and then shows that thisprobability increases in (a, b) defined on the set D =

{(a, b) : a2 + b2 = 1, a > b

}for

some t . This result is further supported and enhanced by numerical computation.© 2008 Elsevier B.V. All rights reserved.

1. Introduction

Let X1, X2, . . . , Xn be i.i.d. random variables and consider the probability as follow

P

(n∑i=1

aiXi > t

). (1)

This probability in (1) is obviously useful in such applications areas as portfolio analyses and industrial productionprocess like glass manufacturing and iron–steel manufacturing. For example, civil engineers are interested in resistanceof steel construction. This resistance is related to amount of the raw materials contained in the construction. Engineershould obtain optimal mixture since the probability level of mixture resistance is important for him. Consequently, tailprobabilities such as (1) are used to determine the reliability of mixtured structures in industry, commerce, communicationsystems etc.It is known that if the common density of X1, X2, . . . , Xn is symmetric about zero and log-concave, then the probability

(1), which can be viewed as the function of (a1, . . . , an), increases in (a1, . . . , an) (c.f., e.g., Proschan (1965)). No doubt,exponential distributions are basic enquiry material for reliability analysis and the results obtained can easily be extendedto larger scale family of Gamma distributions. For the two-variable-case with nonsymmetrical density functions such asexponential distributions, on the other hand, Diaconis and Perlman (1987) provides the following results: If X and Y arei.i.d with Exp(1), then, for n = 2, (1) decreases in (a1, a2) for t ≤ a1 + a2 and increases for t ≥ 3

2 (a1 + a2), wherea1 + a2 = constant. These results are further improved and the proof is achieved by Merkle and Petrović (1994).

∗ Corresponding author.E-mail addresses: yilmazm@science.ankara.edu.tr (M. Yilmaz), btopcu@aku.edu.tr (B. Topçu).

0167-7152/$ – see front matter© 2008 Elsevier B.V. All rights reserved.doi:10.1016/j.spl.2008.05.038

M. Yilmaz, B. Topçu / Statistics and Probability Letters 78 (2008) 3152–3158 3153

To find a more general result, the well-known concept of majorization can be utilized, as it implies certain usefulinequalities for the tail probability in question (for definition, c.f., e.g., Hardy et al. (1934), p.45).

Definition 1. Let a = {ai}ni=1 and b = {bi}ni=1 stand for two sequences of real numbers. Suppose that (i)

∑ni=1 ai =

∑nk=1 bi,

(ii) a1 ≥ a2 ≥ · · · ≥ an and b1 ≥ b2 ≥ · · · ≥ bn (iii)∑ki=1 ai ≥

∑kk=1 bi for 1 ≤ k ≤ n, it is said that b is majorised by a and

this fact is denoted as b≺m a.

In addition, the notion of stochastic dominance will prove useful for further discussions (c.f., e.g., for definition Shakedand Shanthikumar (1994), p.3).

Definition 2. Let X and Y be two random variables such that P (Y > t) ≥ P (X > t) for all t ∈ R, then X is said to bestochastically smaller than Y and it is denoted as X ≺st Y .

We will prove in Lemma 1 that a1X + b1Y ≺st a2X + b2Y for all λt ∈ (0, 4.23] when(a22, b

22

)is majorised by

(a21, b

21

).

Lemma 1 shows that weighted exponential variates does not hold such an ordering given Definition 2 for all t . Although theanalytic techniques used in Lemma 1 can give exact bound for t , our attempts are to obtain larger interval than found inLemma 1 by using computational approach and to support Lemma 1’s result. Such a stochastic dominance which holds forall t is shown for the Rayleigh distribution by Hu and Lin (2000). It is of interest to check that it also holds for exponentialdistribution.

2. Basic results

Consider two independent random pair such as X and Ywith the respective distribution functions FX and FY defined onan identical support. Needless to add that the distribution of the sum aX + bY can easily be obtained through convolutiontechnique,

P (aX + bY ≤ t) =

∫ t

b

0FX

(t − bya

)dFY (y)∫ t

a

0FY

(t − axb

)dFX (x).

(2)

Two simple inequalities in (2) can then be expressed as follows;∫ tb

0FX

(t − bya

)dFY (y) ≤

∫ tb

01dFY (y) = FY

(tb

)and ∫ t

a

0FY

(t − axb

)dFX (x) ≤

∫ ta

01dFX (x) = FX

(ta

).

Hence,

P (aX + bY ≤ t) ≤ min {P (aX ≤ t) , P (bY ≤ t)} . (3)

One can find narrower upper bound for (2), such that∫ tb

0FX

(t − bya

)dFY (y) ≤

∫ tb

0FX

(ta

)dFY (y) = FX

(ta

)FY

(tb

).

Hence,

P (aX + bY ≤ t) ≤ P (max {aX, bY } ≤ t) . (4)

On the basis of the geometry of aX + bY depicted in Fig. 1, it is clearly possible to write an expression for the tail probabilityof aX + bY such as

P (aX + bY > t) = P(X >

ta+ b

, Y >t

a+ b

)+ P

(aX + bY > t, Y ≤

ta+ b

)+ P

(aX + bY > t, X ≤

ta+ b

).

3154 M. Yilmaz, B. Topçu / Statistics and Probability Letters 78 (2008) 3152–3158

Fig. 1. Geometric representation of ax+ by.

This latter expression implies P (aX + bY > t) ≥ P(X > t

a+b , Y >ta+b

), i.e,

(a+ b)min {X, Y }≺st aX + bY . (5)

It can be simply concluded from Fig. 1,

P(aX + bY > t, Y ≤

ta+ b

)≤ P

(X >

ta+ b

, Y ≤t

a+ b

)and hence

P(aX + bY > t, X ≤

ta+ b

)≤ P

(X ≤

ta+ b

, Y >t

a+ b

)hold. Recall the basic fact that

P ((a+ b)max {X, Y } > t) = P(X >

ta+ b

, Y >t

a+ b

)+ P

(X >

ta+ b

, Y ≤t

a+ b

)+ P

(X ≤

ta+ b

, Y >t

a+ b

).

By combining these latter two facts, we have another simple result:

aX + bY ≺st (a+ b)max {X, Y }�. (6)

Now, assume that a ≥ b and X ≺st Y , then P(Y > t

a

)≥ P

(X > t

a

)≥ P

(X > t

b

). This implies bX ≺st aY .

3. Main result

Majorization allows us to compare tail probabilities of weighted sums of exponential variates, when the weightcoefficients are composed of interior and boundary points of a unit circle. This method can be applied to give our results forexponential distributions.

Lemma 1. Let X and Y be i.i.d. as Exp(λ). Consider nonnegative real numbers a1, b1, a2 and b2 such that(a22, b

22

)≺m

(a21, b

21

).

Then P (a1X + b1Y > t) ≤ P (a2X + b2Y > t) for all λt ∈ (0, 4.23], i.e P (aX + bY > t) is non decreasing in b and nonincreasing in a, where (a, b) is defined on D =

{(a, b) : a2 + b2 = 1, a > b

}.

Proof. According to Hu and Lin (2000), it suffices to prove that the function ht(α) = P (cosαX + sinαY > t), α ∈[0, π4

]satisfies ∂ht (α)

∂α≥ 0 for all λt ∈ (0, 4.23]. Using independence of X and Y we can get the following expression for ht(α)

ht(α) =∫∫

(cosα)x+(sinα)y>tfX (x)fY (y)dxdy =

∫∫(cosα)x+(sinα)y>t

λ2e−λ(x+y)dxdy. (7)

Using polar coordinates (x, y) = (r cos θ, r sin θ)we can express (7) as

ht(α) =∫ π

2

0

∫r> tcos(α−θ)

λ2e−λr(cos θ+sin θ)rdrdθ. (8)

M. Yilmaz, B. Topçu / Statistics and Probability Letters 78 (2008) 3152–3158 3155

If we take φt = φt(α, θ) =t

cos(α−θ) , then∂φt∂θ=−t sin(α−θ)cos2(α−θ)

= −∂φt∂α. Applying the Fundamental Theorem of Calculus, we

have

∂ht(α)∂α

=

∫ π2

0λ2e−λφt (cos(θ)+sin(θ))φt

(−∂φt

∂α

)dθ

=

∫ π2

0λ2e−λφt (cos(θ)+sin(θ))φtdφt . (9)

The trigonometric equalities of the above expressions can be re-arranged as:

cos θ + sin θ = cos (α − θ) [cosα + sinα]− sin (α − θ) [cosα − sinα]

and

φt (cos θ + sin θ) = t [(cosα + sinα)− tan(α − θ) (cosα − sinα)] .

(9) can now be rewritten as ∂ht (α)∂α= −λ2t2e−λt(cosα+sinα)I1, where

I1 =∫ π

2

0

tan(α − θ)cos2(α − θ)

exp {λt tan(α − θ)(cosα − sinα)} dθ.

I1 can be easily evaluated using integration by parts as

(1+ V1)e−V1 − (1− V2)eV2

(λt)2(cosα − sinα)2(10)

here V1 = λt(cosα − sinα) cotα and V2 = λt(cosα − sinα) tanα, such that (9) is equal to ∂ht (α)∂α= At(α)gt(α) with

At(α) = e−λt(cosα+sinα)

(cosα−sinα)2and

gt(α) = (1− V2)eV2 − (1+ V1)e−V1 .

Thus the sign of ∂ht (α)∂αis the same as gt(α).

If ϑ(α) denotes cosα−sinαcosα sinα λt then let gt(α) = ϕ (α). Hence,

ϕ (α) =(1− ϑ(α) sin2 α

)eϑ(α) sin

2 α−(1+ ϑ(α) cos2 α

)e−ϑ(α) cos

2 α.

ϕ (α) is positive only in the case 1− ϑ(α) sin2 α > 0. ϕ (α) = 0 is equivalent to

log(1− ϑ(α) sin2 α

)− log

(1+ ϑ(α) cos2 α

)+ ϑ(α) = ψ (α) = 0

with ψ(π4

)= 0. We want to show that ψ (α) is positive on

[0, π4

]. Therefore we consider the first derivative of ψ (α) for

further discussions.

ψ ′ (α) = −ϑ ′(α) sin2 α + ϑ(α) sin 2α

1− ϑ(α) sin2 α−ϑ ′(α) cos2 α − ϑ(α) sin 2α

1+ ϑ(α) cos2 α+ ϑ ′(α).

After rearrangements on the expression above we have

ψ ′ (α) =ϑ(α)

[ϑ ′(α)

(cos2 α − sin2 α − ϑ(α) sin2 α cos2 α

)− ϑ(α) sin 2α

](1− ϑ(α) sin2 α

) (1+ ϑ(α) cos2 α

) . (11)

For the sake of brevity, let A = cosα− sinα, B = cosα+ sinα and C = sinα cosα. So, A′ = −B, B′ = A and C ′ = AB. Henceϑ(α) = A

C λt and

ϑ ′(α) = −BC2[C + A2

]λt = −

BC2[1− C] λt.

If above statements are substituted in (11), then

ψ ′ (α) =

A2C (λt)

2[−BC2 (1− C)

(B− 1

C λtC2)−1C 2C

](1− A

C λt sin2 α) (1+ A

C λt cos2 α)

is obtained. After some arrangements, we have

ψ ′ (α) =

A2

C2 (λt)2(

1− AC λt sin

2 α) (1+ A

C λt cos2 α) [− (1+ C)

C+ B (1− C) λt

]. (12)

3156 M. Yilmaz, B. Topçu / Statistics and Probability Letters 78 (2008) 3152–3158

We see that ψ ′(π4

)= 0, and the sign of ψ ′ (α) can be determined by second term in (12). Now, assume that λt ≤ 1+C

B(1−C)C ,then ψ (α) is a decreasing function on

(0, π4

)and considering ψ ′

(π4

)= ψ

(π4

)= 0, ψ (α) will not be negative in

[0, π4

].

However, this situation depends on λt . According to the assumption, upper bound of λt varies depending on the values of α.Let U(α) denote 1+C

C(1−C)B , then

U ′(α) =A[3C3 + 5C2 − C − 1

]B2(1− C)2C2

.

U ′(α) has only two roots on[0, π4

]which are α1 = π

4 and α2 = 0.641220597653598. Since U(0) = ∞ and it has aminimum point at α1 or α2, magnitude of λt is less than minimum value of U(α). Hence, λt ≤ min {U(α1),U(α2)} whereU(α1) = 3

√2. If we use a computer program for calculation of U(α2) then we get U(α2) = 4.235340877719861. We can

say that λt is bounded by U(α2) from above.Recall that the case 1 − ϑ(α) sin2 α > 0. Note that, This condition gives another upper bound for λt . After some calcu-

lation, We have easily obtain this bound as follows

λt < 4.43042464198544.

But this upper bound provides positivity of 1− ϑ(α) sin2 α. Hence this can be seen as necessary condition for positivity ofgt(α). On the other hand, the upper bound U(α2) can only be considered necessary and sufficient condition for positivity ofgt(α) on

[0, π4

].

Needless to add that, if λt ≥ 1+CB(1−C)C had been assumed, thenψ

′ (α)would have been positive in(0, π4

). This implies that

ψ (α) is an increasing function with ψ(π4

)= 0. Hence, ψ (α) could be a negative. But this assumption leads to some diffi-

culties as noted follows: Since U(0) goes to infinity, a lower bound for λt can not be obtained. Therefore, a proper intervalwhere gt(α) is negative can not be determined. Also, this result can be obtained by assuming 1−λt(cosα−sinα) tanα ≤ 0,since (cosα − sinα) tanα has two roots as α = 0 and α = π

4 , λt can not be bounded from below.For the purpose of obtaining more sensitive interval than (0, 4.23], we shall introduce a numeric approach in the next

section. This approach also supports our results. �

4. Computational enhancement of the boundary

For further improvement of the upper bound of λt , a computer programming is suggested and a corresponding graphicalrepresentation is provided below. This improvement supports to our results. Throughout Section 4, we consider unitexponential function (i.e., λ = 1).Since the positivity of gt(α) depends on α, and t also depends on values of α, we will attempt to find value of t approx-

imately. We first need graphical representations of gt(α) for some values of t = 2.5, 2√2, 3, 3.5, 4, 4.1, 4.2, 4.3 and 4.7

(Figs. 2–5). There are four graphs of gt(α) which are plotted by using Mathcad 2000 Program. As it can be seen from theFig. 5, gt(α) has negative values for t ≥ 4.3.

Fig. 2. Graphs of gt (α) for some t .

Roughly, graphics give value of t yields to negativity of gt(α). We then need to findmore sensitive value for upper boundof t . As it can be seen below, a computer program gives more sensitive value for t than graphical representation. Here ddenotes the number of partitions of the interval

(0, π4

)and DELTA denotes as increment for t . It is purposed that value of t

starts 0 with DELTA increment and computed values of gt(α) are counted for d partitions if its value is negative. When totalcount is zero, DELTA increment is added to t , otherwise algorithm is stopped. Tabulated results can be seen in Table 1 forsome partitions and some increments of t .As it can be seen in Table 1, the upper bound can be taken approximately 4.236, since the value of t still exceeds 4.236

with the largest partition (d) and smallest increment (DELTA).

M. Yilmaz, B. Topçu / Statistics and Probability Letters 78 (2008) 3152–3158 3157

Fig. 3. Graphs of gt (α) for some t (continued).

Fig. 4. Graphs of gt (α) for some t (continued).

Fig. 5. Graphs of gt (α) for some t (continued).

Table 1Computed value of upper bound for t with DELTA increment

d DELTA = 0.00001 DELTA = 0.0001 DELTA = 0.001Value of t∗ Value of t∗ Value of t∗

5 4.236300000000001 4.236000000000000 4.23600000000000050 4.236300000000000 4.236000000000001 4.236000000000000200 4.236300000000001 4.236300000000001 4.236000000000000

10000 4.236300000000001 4.236300000000001 4.236000000000000100000 4.236300000000001 4.236300000000001 4.236000000000000200000 4.236300000000001 4.236300000000001 4.236000000000000

3158 M. Yilmaz, B. Topçu / Statistics and Probability Letters 78 (2008) 3152–3158

We use a MATLAB program to get result. This program can be given as below:

clear alld = 10000;DELTA=.001;

for t=0:DELTA:100sayac=0;

for i=0.0000001 : 1/d: pi/4-0.0001v1 = t * (cos(i) - sin(i)) * tan(i);v2 = t * (cos(i) - sin(i)) *cot(i);%g = (1 - v1) * exp(v1) - (1 + v2) * exp(-v2);g = (1 - v1) * exp(v1+v2) - (1 + v2);

if g<0sayac=sayac+1;ig

endend

if sayac > 0break

endend

t-DELTA

Acknowledgements

We wish to thank the anonymous referees for careful reading of the manuscript and several helpful comments whichenhanced the presentation.

References

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