BAU Journal - Science and Technology

Volume 1 Issue 1 ISSN: 2706-784X Article 7

December 2019

ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES

Therrar Kadri Faculty of Sciences, Lebanese International University, Lebanon, [email protected]

Khaled Smaili Faculty of Sciences, Lebanese University, Zahle, Lebanon, [email protected]

Seifedine Kadry Associate Professor, Faculty of Science, Beirut Arab University, Beirut, Lebanon, [email protected]

Ali El-Joubbeh Faculty of Sciences, Lebanese University, Zahle, Lebanon, [email protected]

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Recommended Citation Kadri, Therrar; Smaili, Khaled; Kadry, Seifedine; and El-Joubbeh, Ali (2019) "ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES," BAU Journal - Science and Technology: Vol. 1 : Iss. 1 , Article 7. Available at: https://digitalcommons.bau.edu.lb/stjournal/vol1/iss1/7

This Article is brought to you for free and open access by Digital Commons @ BAU. It has been accepted for inclusion in BAU Journal - Science and Technology by an authorized editor of Digital Commons @ BAU. For more information, please contact [email protected]. ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES

Abstract In this paper, we consider the product of two independent Hypoexponential distributions which has many applications in sciences and engineering. We find analytically the probability density function, the cumulative distribution function, moment generating function, the reliability function and hazard function, which was proved to be a linear combination of the K distribution. Finally, we will apply our results application in stochastic PERT Network.

Keywords Product Distribution, Hypoexponential Distribution; K Distribution; Probability Density Function; Reliability Function.

This article is available in BAU Journal - Science and Technology: https://digitalcommons.bau.edu.lb/stjournal/vol1/ iss1/7 Kadri et al.: ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES

1. INTRODUCTION The product distributions 푋푌 are widely used in many applied problems of engineering, physics, number theory, order statistics, economics, biology, genetics, medicine, hydrology, psychology, classification, and ranking and selection. Examples of 푋푌 include traditional portfolio selection models, relationship between attitudes and behavior, number of cancer cells in tumor biology and stream flow in hydrology. The product distribution have been studied by several authors especially when 푋 and 푌 are independent random variables and come from the same family. For a historical review, see the papers Sakamoto (Sakamoto, 1943) for uniform family, Harter (Harter, 1951) and Wallgren (Wallgren, 1980) for Students t family, Springer and Thompson (Springer, 1970) for normal family, Stuart (Stuart, 1962) and Podolski (Podolski, 1972) for gamma family, Steece (Steece, 1976), Bhargava and Khatri (Bhargava, 1981), and Tang and Gupta (Tang, 1984) for beta family, AbuSalih (Abu-Salih, 1983) for power function family, and Malik and Trudel (Malik, 1986) for Exponential family, Pham-Gia and Turkkan (Pham-Gia, 2002) for Generalized F family, Nadarajah (Nadarajah, 2007) for the Laplace family, Coelho and Mexia (Coelho, 2007) for the Generalized Gamma-Ratio family, see also Rathie and Rohrer (Rathie, 1987) for a comprehensive review of known results). However, there is no work done when 푋 and 푌 are two independent Hypoexponential random variables for their general case. The particular case of the Hypoexponential distribution, the Erlang distribution, was solved knowing that the Erlang distribution is a particular case of the gamma distribution and that latter was examined by Podolski (Podolski, 1972), Redding (Redding, 1999), and Withers and Nadarajah (Withers, 2013). They showed that the product of two independent Gamma distributions is the 퐾-distribution. In our paper, we consider the product of two independent Hypoexponential random variables. We examine the general case of this problem when the stages of the Hypoexponential distribution do not have to be distinct. We use the expression of the probability density function (PDF) for the general case of the Hypoexponential distribution given by Smaili et al. in (Smaili, 2013b), that has a form of a linear combination of the PDF of the Erlang distribution and also we use the expression of the PDF for particular case of the Hypoexponential distribution with different stages given by Smaili et al. (Smaili, 2013a). In this manner, we solve the general problem of 푋푌 and derive the exact expressions of the PDF, cumulative distribution function (CDF), reliability function, hazard function and the moment generating function (MGF). We showed that the PDF of 푋푌 is a linear combination of the 퐾-distribution. Next, we find a more closed and simple form of the expressions of PDF and CDF for the product distribution when 푋 and 푌 are independent Hypoexponential distribution of different stages. Eventually, we apply our theoretical results to stochastic activity network used in project management by determining the probability of the total execution time in stochastic PERT Network.

2. SOME PRELIMINARIES 2.1 The PDF and CDF of Product Distributions Determining the PDF and the CDF of the distribution of the product of two independent random variables are well known, see (Rohatgi, 1976) and (Grimmett, 2001). If we suppose that 푋 and 푌 be independent random variables having the respective PDF’s 푓푋(푥) and 푓푌(푦), then the CDF of 푍 = 푋푌 is given by

푡 0 ∞ ∞ 푡 푥 퐹푍(푡) = ∫−∞ [∫ 푓푌(푦)푑푦] 푓푋(푥)푑푥 + ∫0 [∫−∞ 푓푌(푦)푑푦] 푓푋(푥)푑푥 푥

and the PDF of 푍 is given as

+∞ 1 푡 푓 (푡) = ∫ 푓 ( )푓 (푥)푑푥. 푍 −∞ |푥| 푌 푥 푋

Moreover, when 푓푋(푥) = 0 for 푥 < 0, then

푡 ∞ 푡 퐹 (푡) = ∫ [∫푥 푓 (푦)푑푦] 푓 (푥)푑푥 = 퐸[퐹 ( )] (1) 푍 0 −∞ 푌 푋 푌 푥

and the PDF is given as

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+∞ 1 푡 1 푡 푓 (푡) = ∫ 푓 ( )푓 (푥)푑푥 = 퐸[ 푓 ( )], (2) 푍 0 푥 푌 푥 푋 푥 푌 푥

where 퐸[X] is the expectation of the random variable 푋. We note that the above expressions are symmetric in 푋 and 푌.

2.2 The Hypoexponential Distribution

The Hypoexponential distribution is the distribution of the sum of 푚 ≥ 2 independent Exponential random variables. The general case of the Hypoexponential distribution is when the 푚 exponential stages do not have to be distinct. This general case can be written as 푆푚 = 푛 푘푖 ∑푖=1 ∑푗=1 푋푖푗, where 푋푖푗 is an Exponential RV of parameter 훼푖, 푖 = 1,2,...,푛, and written ⃑⃑ ⃑⃑ as 푆푚 ∼ 퐻푦푝표푒푥푝(훼⃑, 푘) of parameters 훼⃑= (훼1, 훼2, . . . , 훼푛) and 푘 = (푘1, 푘2, . . . , 푘푛) with 푛 푚 = ∑푖=1 푘푖, see (Smaili, 2013b). Smaili et al. in (Smaili, 2013b) gave a modified expression for the PDF of the Hypoexponential random variable as

∑푛 ∑푘푖 푓푆푚 (푡) = 푖=1 푗=1 퐴푖푗푓퐸푖푗(푡) (3)

where (훼 푡)푗−1훼 푒−훼푖푡 푓 (푡) = 푖 푖 if 푡 > 0 (4) 퐸푖푗 (푗−1)!

( ) and 푓퐸푖푗 푡 = 0 if 푡 ≤ 0, is the PDF of the Erlang distribution 퐸푖푗 with 푗 the shape parameter and 훼푖 the rate parameters, written as 퐸푖푗 ∼ 퐸푟푙(푗, 훼푖) for 1 ≤ 푖 ≤ 푛, 1 ≤ 푗 ≤ 푘푖. They showed that

푛 푘푖 ∑푖=1 ∑푗=1 퐴푖푗 = 1 (5)

Moreover, they gave an expression of 퐴푖푗 in a simple iterated form as for 1 ≤ 푖 ≤ 푛, 1 ≤ 푗 ≤ 푘푖 as −푘푗 ∏푛 훼푖 퐴푖,푘푖 = 푗=1,푗≠푖 (1 − ) (6) 훼푗

and for 푗 = 푘푖 − 1, down to 1.

1 푘푖−푗 푛 훼푝 −푙 퐴푖푗 = ∑푙=1 [(∑푝=1,푝≠푖 푘푝(1 − ) ) 퐴푖,푗+푙]. (7) 푘푖−푗 훼푖

However, the mean of 푆푚 is given as

푛 푘푖 푛 푘푖 푗 퐸[푆푚] = ∑푖=1 = ∑푖=1 ∑푗=1 퐴푖푗 (8) 훼푖 훼푖

A particular case of this distribution is when the 푚 stages are different. This case is ⃑⃑ written as 푋 ∼ 퐻푦푝표푒푥푝(훼⃑), where 훼⃑= (훼1, 훼2, . . . , 훼푛), 푘 = (1,1, . . . ,1), see (Smaili, ∑푛 ∑푘푖 ∑푛 2013a). From Eq. (3) we have the PDF is 푓푆푚 (푡) = 푖=1 푗=1 퐴푖푗푓퐸푖푗(푡) = 푖=1 퐴푖1푓퐸푖1(푡). However, 퐸푖1 is the Exponential random variable with parameter 훼푖 and 퐴푖1 = 푛 훼푗 Δ ∏푗=1,푗≠푖 ( ) = 퐴푖, for 1 ≤ 푖 ≤ 푛, see (Smaili, 2013b). Then the PDF is 훼푗−훼푖

∑푛 −훼푖푡 푓푆푚 (푡) = 푖=1 퐴푖훼푖푒 for푡 > 0 (9)

https://digitalcommons.bau.edu.lb/stjournal/vol1/iss1/7 2 Kadri et al.: ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES

Another particular case is when the 푚 stages are identical and the distribution is the Erlang distribution. Taking in Eq (3) 훼⃑= (훼), 푘⃑⃑ = (푘) and 푛 = 1, we write 푋 ∼ 퐻푦푝표푒푥푝(훼, 푘) = 퐸푟푙(푘, 훼), see (Smaili, 2013b). Moreover, in (Smaili, 2013b), they showed that 퐴1푗 = 0 for 1 ≤ 푗 ≤ 푘 − 1 and 퐴1,푘 = 1.

2.3 K distribution Let 푋 be a random variable that has a 퐾-Distribution with three distribution having the shape parameters 퐿 and 푣 and mean 휇, denoted as 푋 ∼ 퐾(퐿, 푣, 휇), The PDF of 푋 is given by

퐿+푣 2 퐿푣푡 1 퐿푣푡 푓(푡) = ( ) 2 퐾 (2√ ) , for 푡 > 0 (10) 푡 휇 Γ(퐿)Γ(푣) 푣−퐿 휇

where 퐾푝(⋅) is the modified Bessel function of second kind of order 푝, see (Redding, 1999). 퐾-distribution arises as the consequence of a statistical or probabilistic model used in Synthetic Aperture Radar (SAR) imagery. The 퐾-distribution is formed by compounding two separate probability distributions, one representing the radar cross-section, and the other representing speckle that is a characteristic of coherent imaging. In the following Lemma, we present a modified form of the 퐾-distribution as the product of two Erlang random variables. This form shall help us through out the paper to reach our results. Lemma 1: Let 푋 ∼ 퐸푟푙(푘, 훼) and 푌 ∼ 퐸푟푙(푙, 훽) be two independent random variables where 푘 and 푙 are the shape parameters and the values 훼 and 훽 are the rate parameters. Let 푍 = 푋푌. Then 푘푙 푍 ∼ 퐾(푘, 푙, ) and its PDF has the following expression 훼훽 푙+푘 2(훼훽푡) 2 푓 (푡) = 퐾 (2√훼훽푡), (11) 푍 (푙−1)!(푘−1)!푡 푙−푘

where 퐾푝(⋅) is the modified Bessel function of second kind of order 푝.

Proof. Let 푋 ∼ 퐸푟푙(푘, 훼) and 푌 ∼ 퐸푟푙(푙, 훽). Then the PDF of 푍 = 푋푌 is given from Eq (2) as

1 푡 푓 (푡) = 퐸[ 푓 ( )] (12) 푍 푥 푌 푥

훽푡 훽푡 푙−1 − 푡 푡 ( ) 훽푒 푥 But 푌 ∼ 퐸푟푙(푙, 훽) then 푓 ( ) is obtained from Eq (4) as 푓 ( ) = 푥 . Hence, we have 푌 푥 푌 푥 (푙−1)!

훽푡 훽푡 훽푡 푙−1 − 훽푡 푙−1 − 1 ( ) 훽푒 푥 ∞ 1 ( ) 훽푒 푥 (훼푥)푘−1훼푒−훼푥 푓 (푡) = 퐸[ 푥 ] = ∫ 푥 푑푥 푍 푥 (푙−1)! 0 푥 (푙−1)! (푘−1)! 훽푡 훽푙훼푘푡푙−1 ∞ − −훼푥 = ∫ 푥−푙+푘−1푒 푥 푑푥 (푙−1)!(푘−1)! 0

푐 1−푎 − −푑푥 +∞ 푒 푥 푐 However, we have ∫ 푑푥 = 2 ( ) 2 퐾 (2√푐푑). By setting 푐 = 훽푡, 푑 = 훼, 푎 = 푙 − 푘 + 0 푥푎 푑 푎−1 1, we obtain that

푘−푙 훽푙훼푘푡푙−1 훽푡 푓 (푡) = 2 ( ) 2 퐾 (2√훼훽푡) 푍 (푙−1)!(푘−1)! 훼 푙−푘 푙+푘 푙+푘 푘+푙 −1 2훽 2 훼 2 푡 2 = 퐾 (2√훼훽푡) (푙−1)!(푘−1)! 푙−푘 푙+푘 2(훼훽푡) 2 = 퐾 (2√훼훽푡) (푙−1)!(푘−1)!푡 푙−푘

푘 푙 푘푙 Now, the means of 푋 and 푌 are given by 휇 = and 휇 = , then 휇 = 휇 휇 = we can rewrite 푋 훼 푌 훽 푍 푋 푌 훼훽 the PDF of 푍 as

Published by Digital Commons @ BAU, 2019 3 BAU Journal - Science and Technology, Vol. 1, Iss. 1 [2019], Art. 7

푙+푘 푘푙 2 2( 푡) 휇푍 푘푙푡 푓푍(푡) = 퐾푙−푘(2√ ) (13) (푙−1)!(푘−1)!푡 휇푍

푘푙 Then 푍 is the 퐾-distribution with parameters 푘 and 푙 and mean 휇 = . 푧 훼훽

In the next Lemma we review some characteristics of this distribution. These expressions can be found in (Jakeman, 1978), (Redding, 1999), and (Ward, 1981). 푘푙 Lemma 2: Let 푋 ∼ 퐾(푘, 푙, 휇), and 휇 = in the case 푋 is the product of 퐸푟푙(푘, 훼) and 퐸푟푙(푙, 훽). 훼훽 Then

1. The Moment of 푋 of order 푚 is given by

Γ(푘+푚)Γ(푙+푚) Γ(푘+푚)Γ(푙+푚) 퐸[푋푟] = 휇푚 = . 푘푚푙푚Γ(푘)Γ(푙) 훼푚훽푚Γ(푘)Γ(푙) 푘푙 2. 퐸[푋] = 휇 = . 훼훽

푘+푙+1 푘푙(푘+푙+1) 3. 푉푎푟[푋] = 휇2 ( ) = . 푘푙 (훼훽)2

4. The moment generating function of 푋 is, 푘+푙−1 푘+푙−1 푘푙 훼훽 푘푙 2 푘푙 훼훽 2 훼훽 Φ (푡) = ( ) 푒2푡휇푊 푘+푙−1 푘−푙 ( ) = ( ) 푒 2푡 푊 푘+푙−1 푘−푙 ( ), 푋 휇푡 − , 휇푡 푡 − , 푡 2 2 2 2

where 푊푎,푏(⋅) where is the Whittaker function, see (Abramowitz, 1965).

3. PRODUCT OF TWO INDEPENDENT HYPOEXPONENTIAL DISTRIBUTIONS In this section, we give the exact closed expression of the PDF of the product of two independent Hypoexponential distributions. The expression is given in the case of the general case of any two independent Hypoexponential distributions.

3.1 General Case We suppose that 푋 and 푌 follow the general case of the Hypoexponential distribution that are ⃑⃑ ⃑⃑ independent. We take 푋 ∼ 퐻푦푝표푒푥푝(훼⃑, 푘), 훼⃑= (훼1, 훼2, . . . , 훼푛), 푘 = (푘1, 푘2, . . . , 푘푛) and 푌 ∼ ⃑ ⃑ ⃑ ⃑ 퐻푦푝표푒푥푝(훽, 푙), 훽 = (훽1, 훽2, . . . , 훽푟), 푙 = (푙1, 푙2, . . . , 푙푟). From Eq. (3) the PDFs of 푋 and 푌 are

푛 푘푖 푟 푙푝 푓푋(푡) = ∑ ∑ 퐴푖푗푓퐸 (푡)and푓푌(푡) = ∑푝=1 ∑ 퐵푝푞푓퐸 (푡) (14) 푖=1 푗=1 훼푖,푗 푞=1 훽푝,푞

respectively, where 퐸휔,훿 ∼ 퐸푟푙(훿, 휔) and 퐴푖푗 and 퐵푝푞 are given in (6) and (7) for 1 ≤ 푖 ≤ 푛, 1 ≤ 푗 ≤ 푘푖, 1 ≤ 푝 ≤ 푟, and 1 ≤ 푞 ≤ 푙푝.

⃑⃑ ⃑⃑ Theorem 1: Let 푋 ∼ 퐻푦푝표exp(훼⃑, 푘), 훼⃑= (훼1, 훼2, . . . , 훼푛), 푘 = (푘1, 푘2, . . . , 푘푛) ⃑ ⃑ ⃑ ⃑⃑ and 푌 ∼ 퐻푦푝표exp(훽, 푙), 훽 = (훽1, 훽2, . . . , 훽푟), 푙 = (푙1, 푙2, . . . , 푙푟). Then the PDF of 푍 = 푋푌 is given by 푛 푘푖 푟 푙푝 푓푍(푡) = ∑푖=1 ∑푗=1 ∑푝=1 ∑푞=1 퐴푖푗퐵푝푞푓퐾(푡) (15)

푞푗 where 푓퐾(푡) is the PDF of the 퐾-distribution with parameters 푞, 푗 and mean 휇퐾 = , and 퐴푖푗 훼푖훽푝 and 퐵푝푞 are obtained from in (6) and (7).

Proof. The PDF of 푍 = 푋푌 is given from Eq. (2) as

https://digitalcommons.bau.edu.lb/stjournal/vol1/iss1/7 4 Kadri et al.: ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES

1 푡 ∞ 1 푡 푓 (푡) = 퐸[ 푓 ( )] = ∫ 푓 ( )푓 (푥)푑푥 푍 푥 푌 푥 0 푥 푌 푥 푋

Next, using the PDF of 푋 and 푌 in (2) we get

∞ 1 푟 푙푝 푡 푛 푘푖 푓푍(푡) = ∫ (∑푝=1 ∑ 퐵푝푞푓퐸 ( ))(∑ ∑ 퐴푖푗푓퐸 (푥)) 푑푥 0 푥 푞=1 훽푝,푞 푥 푖=1 푗=1 훼푖,푗 푟 푙푝 푛 푘푖 ∞ 1 푡 = ∑푝=1 ∑ ∑ ∑ 퐵푝푞퐴푖푗 ∫ (푓퐸 ( )푓퐸 (푥)) 푑푥 푞=1 푖=1 푗=1 0 푥 훽푝,푞 푥 훼푖,푗

but 푡 푡 푞−1 −훽 ( ) (훽 ) 훽 푒 푝 푥 푗−1 −훼 푥 푡 푝푥 푝 (훼푖푥) 훼푖푒 푖 푓퐸 ( )푓퐸 (푥) = 훽푝,푞 푥 훼푖,푗 (푞−1)! (푗−1)! 푞 푗 푞−1 훽푝푡 훽푝훼푖 푡 − +훼 푥 = 푥−푞+푗푒 푥 푖 (푞−1)!(푗−1)! then we have 훽푞훼푗푡푞−1 훽푝푡 푟 푙푝 푛 푘푖 푝 푖 ∞ − +훼 푥 푓 (푡) = ∑ ∑ ∑ ∑ 퐵 퐴 ∫ 푥−푞+푗−1푒 푥 푖 푑푥. (16) 푍 푝=1 푞=1 푖=1 푗=1 푝푞 푖푗 (푞−1)!(푗−1)! 0

Now, by using the following integral expression 푐 1−푎 − +푑.푥 +∞ 푒 푥 푐 ∫ 푑푥 = 2 ( ) 2 퐾 (2√푐푑) 0 푥푎 푑 푎−1

we can evaluate the integral in (6) by setting 푎 = 푞 − 푗 + 1, 푐 = 훽푝푡, 푑 = 훼푖. Thus we may write 푗−푞 훽푞훼푗푡푞−1 푟 푙푝 푛 푘푖 푝 푖 훽푝푡 2 푓푍(푡) = ∑푝=1 ∑ ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗 2 ( ) 퐾푞−푗(2√훼푖훽푝푡) 푞=1 (푞−1)!(푗−1)! 훼푖 푞+푗 푞+푗 푞+푗 −1 푙 2훽 2 훼 2 푡 2 = ∑푟 ∑ 푝 ∑푛 ∑푘푖 퐵 퐴 푝 푖 퐾 (2 훼 훽 푡) 푝=1 푞=1 푖=1 푗=1 푝푞 푖푗 (푞−1)!(푗−1)! 푞−푗 √ 푖 푝 푞+푗 푙 2(훽 훼 푡) 2 = ∑푟 ∑ 푝 ∑푛 ∑푘푖 퐵 퐴 푝 푖 퐾 (2 훼 훽 푡) 푝=1 푞=1 푖=1 푗=1 푝푞 푖푗 (푞−1)!(푗−1)!푡 푞−푗 √ 푖 푝

By comparing the expression of PDF of the 퐾-distribution defined in Lemma 1 , we can conclude that 푞+푗 2(훽 훼 푡) 2 푝 푖 퐾 (2 훼 훽 푡) (푞−1)!(푗−1)!푡 푞−푗 √ 푖 푝

푞푗 is the PDF of a 퐾-distribution with parameter 푞 and 푗 and mean 휇퐾 = , or we write 퐾 ∼ 훼푖훽푝 푞푗 푟 푙푝 푛 푘푖 퐾(푞, 푗, ). Therefore, the PDF of 푍 can be written as 푓푍(푡) = ∑푝=1 ∑ ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗푓퐾(푡) 훼푖훽푝 푞=1 푞푗 where 푓퐾(푡) is the PDF of 퐾(푞, 푗, ). 훼푖훽푝

The PDF of the product of two independent Hypoexponential distributions, given in Theorem 1, is proven to be a linear combination of the PDF of the 퐾-distribution. However, some articles concentrated on finding the PDF of unknown distribution in this format as a linear combination of a PDF of more simple and known distribution see (Smaili, 2013a), (Smaili, 2013b), (Kadri, 2015a) and (Kadri, 2015b). Thus we can directly conclude the CDF, Reliability function, MGF, Moment of order 푚 and the Expectation of this unknown distribution as a linear combination of CDF, Reliability function, MGF, Moment of order 푚 and the Expectation of the known distribution respectively. In our paper this known distribution is the 퐾-distribution and we presented it characteristics previously in Section 2. By applying the above idea we proved that the PDF of 푍 is 푟 푙푝 푛 푘푖 푓푍(푡) = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗푓퐾(푡) thus we conclude that CDF of 푍 is 푟 푙푝 푛 푘푖 퐹푍(푡) = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗퐹퐾(푡)

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where 퐹퐾(푡) is the CDF of 퐾, and the Reliability function of 푍 is 푟 푙푝 푛 푘푖 푅푍(푡) = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗푅퐾(푡)

where 푅퐾(푡) is the reliability function of 퐾. Then we simply can find the hazard function from 푓푍(푡) ℎ푍(푡) = 푅푍(푡) that is computed directly from the 퐾-distribution. Moreover, the MGF of 푍 is also a linear combination of the MGF of the 퐾-distribution as 푟 푙푝 푛 푘푖 Φ푍(푡) = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗Φ퐾(푡) (17)

푡ℎ and as a consequence the moment of order 푚 of 푍, which the 푚 derivative of Φ푍(푡) at 푡 = 0, shall have the following linear combination form 푚 푟 푙푝 푛 푘푖 푚 퐸[푍 ] = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗퐸[퐾 ] (18)

푚 푚 and 퐸[퐾 ] is the moment of order 푚 of 퐾. Now with the expression of Φ퐾(푡) and 퐸[퐾 ] from 푞푗 푞푗 Lemma 1 with 퐾-distribution of parameters 푞, 푗 and mean 휇퐾 = , where 퐾 ∼ 퐾 (푞, 푗, ), we 훼푖훽푝 훼푖훽푝 get 푞+푗−1 훼푖훽푝 훼푖훽푝 2 훼푖훽푝 Φ (푡) = ( ) 푒 2푡 푊 푞+푗−1 푗−푞 ( ), 퐾 푡 − , 푡 2 2 and 푚 Γ(푞+푚)Γ(푗+푚) 퐸[퐾 ] = 푚 푚 . 훼푖 훽푗 Γ(푞)Γ(푗)

Therefore, we obtain by substituting these expressions into (17) and (18) that 푞+푗−1 훼푖훽푝 푟 푙푝 푛 푘푖 훼푖훽푝 2 훼푖훽푝 Φ (푡) = ∑ ∑ ∑ ∑ 퐵 퐴 ( ) 푒 2푡 푊 푞+푗−1 푗−푞 ( ) 푍 푝=1 푞=1 푖=1 푗=1 푝푞 푖푗 푡 − , 푡 2 2 and Γ Γ 푚 푟 푙푝 푛 푘푖 퐵푝푞퐴푖푗 (푞+푚) (푗+푚) 퐸[푍 ] = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 푚 푚 훼푖 훽푗 Γ(푞)Γ(푗)

On the other hand, by observing the mean of 푍 = 푋푌, that has the form 퐸[푍] = 퐸[푋]퐸[푌]. But the means of 푋 ∼ 퐻푦푝표푒푥푝(훼⃑, 푘⃑⃑) and 푌 ∼ 퐻푦푝표푒푥푝(훽⃑, 푙⃑) are obtained from Eq. (8) as 퐸[푋] = 푛 푘푖 푛 푘푖 푗 푟 푙푝 푟 푙푝 푞 ∑푖=1 = ∑푖=1 ∑푗=1 퐴푖푗 and 퐸[푌] = ∑푝=1 = ∑푝=1 ∑ 퐵푝푞 . Then 훼푖 훼푖 훽푝 푞=1 훽푝 푛 푘푖 푟 푙푝 퐸[푍] = ∑푖=1 ∑푝=1 훼푖 훽푝 푛 푘푖 푗 푟 푙푝 푞 = ∑푖=1 ∑푗=1 퐴푖푗 ∑푝=1 ∑ 퐵푝푞 훼푖 푞=1 훽푝 푛 푘푖 푟 푙푝 푞푗 = ∑푖=1 ∑푗=1 ∑푝=1 ∑ 퐴푖푗퐵푝푞 푞=1 훽푝훼푖

this expression verifies our expression of the moment of 푍 of order 푚 = 1 in Eq. (18) 푟 푙푝 푛 푘푖 퐸[푍] = ∑푝=1 ∑푞=1 ∑푖=1 ∑푗=1 퐵푝푞퐴푖푗퐸[퐾]

푞푗 however the mean of 퐾 is stated in Theorem 1 as 퐸[퐾] = 휇퐾 = . 훼푖훽푝

3.2 Particular cases In this part, we consider the particular case of the Hypoexponential distribution when the Exponential stages are different. Let 푋 and 푌 be two independent Hypoexponential random variables each distribution of different stages. Thus, 푋 ∼ 퐻푦푝표푒푥푝(훼⃑), 훼⃑= (훼1, 훼2, . . . , 훼푛) and 푌 ∼ ⃑ ⃑ 퐻푦푝표푒푥푝(훽), 훽 = (훽1, 훽2, . . . , 훽푟), and then from Eq. (9) the PDF of 푋 and 푌 are 푛 −훼푖푡 푟 −훽푝푡 푓푋(푡) = ∑푖=1 퐴푖훼푖푒 and푓푌(푡) = ∑푝=1 퐵푝훽푝푒 (19)

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푛 훼푗 푟 훽푗 respectively, with 퐴푖 = ∏푗=1,푗≠푖 ( ) and 퐵푝 = ∏푗=1,푗≠푝 ( ). 훼푗−훼푖 훽푗−훽푝

Theorem 2: Let 푋 and 푌 be two independent random variables with densities given in (19). Then the PDF of 푍 = 푋푌 is given by 푛 푟 푓푍(푡) = ∑푖=1 ∑푝=1 2퐴푖퐵푝훼푖훽푝퐾0(2√훼푖훽푝푡).

and the CDF is 푟 푛 퐹푍(푡) = 1 − 2 ∑푝=1 ∑푖=1 퐵푝퐴푖√훼푖훽푝푡퐾1(2√훼푖훽푝푡)

Proof. From Theorem 1, we take the particular cases of the two Hypoexponential distributions when their stages are different. Then 푘⃑⃑ = (1,1, . . . ,1) and 푙⃑ = (1,1, . . . ,1), this gives in the summation in Eq. (11) the only values are 푗 = 푞 = 1. Then the 퐾-distribution is 퐾 ∼ 퐾(1,1, 휇푖,푝) where 휇푖,푝 = 1 . Thus, we have from Eq. (10) 훼푖훽푝 2 푡 푓 (푡) = 퐾 (2√ ) = 2훼 훽 퐾 (2 훼 훽 푡) 퐾 휇 0 휇 푖 푝 0 √ 푖 푝

Moreover, knowing that in this particular case 퐵푝푞 = 퐵푝 and 퐴푖푗 = 퐴푖, we obtain the given form of PDF of 푍 as 푛 푟 푓푍(푡) = ∑푖=1 ∑푝=1 2퐴푖퐵푝훼푖훽푝퐾0(2√훼푖훽푝푡). On the other hand, the expression of the CDF in this manner is complicated because of the unknown primitive of 퐾0(2√훼푖훽푝푡). So we shall use the other approach in finding the CDF from the its definition from the product distribution formula in (1) as 푡 퐹 (푡) = 퐸[퐹 ( )] 푍 푌 푥 ⃑ 푟 −훽푝푥 However, the CDF of 푌 ∼ 퐻푦푝표푒푥푝(훽) is 퐹푌(푥) = 1 − ∑푝=1 퐵푝푒 . Thus 푡 푟 −훽푝 퐹푍(푡) = 퐸[1 − ∑푝=1 퐵푝푒 푥] 푡 ∞ 푟 −훽 ∑ 푝푥 ( ) = 1 − ∫0 ( 푝=1 퐵푝푒 ) 푓푋 푥 푑푥 푡 ∞ 푟 −훽 푛 ∑ 푝푥 (∑ −훼푖푥) = 1 − ∫0 ( 푝=1 퐵푝푒 ) 푖=1 퐴푖훼푖푒 푑푥 훽푝푡 ∞ 푟 푛 − −훼 푥 ∑ ∑ 푥 푖 = 1 − ∫0 ( 푝=1 푖=1 퐵푝퐴푖훼푖푒 ) 푑푥 훽푝푡 푟 푛 ∞ − −훼 푥 ∑ ∑ 푥 푖 = 1 − 푝=1 푖=1 퐵푝퐴푖훼푖 ∫0 (푒 ) 푑푥

However, the integral in the summation can be obtained from 푚 +∞ − −푛푥 푚 ∫ 푒 푥 푑푥 = 2√ 퐾 (2√푚푛) 0 푛 1

Therefore, we get 푟 푛 훽푝푡 퐹푍(푡) = 1 − ∑푝=1 ∑푖=1 퐵푝퐴푖훼푖2√ 퐾1(2√훼푖훽푝푡) 훼푖 푟 푛 = 1 − 2 ∑푝=1 ∑푖=1 퐵푝퐴푖√훼푖훽푝푡퐾1(2√훼푖훽푝푡)

We also can conclude for this case of 푋푌, the reliability function is 푟 푛 푅푍(푡) = 2 ∑푝=1 ∑푖=1 퐵푝퐴푖√훼푖훽푝푡퐾1(2√훼푖훽푝푡)

and thus the hazard function is 푛 푟 ∑ ∑푝=1 퐴푖퐵푝훼푖훽푝퐾0(2√훼푖훽푝푡). ℎ (푡) = 푖=1 . 푍 푛 푟 ∑푖=1 ∑푝=1 퐴푖퐵푝√훼푖훽푝푡퐾1(2√훼푖훽푝푡)

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4. APPLICATION: PROBABILITY DISTRIBUTION ON THE TOTAL EXECUTION TIME IN STOCHASTIC PERT NETWORK. In this real life application, we will apply our theoretical result to stochastic activity network used in project management. In a project management a well known management tool called PERT is used to schedule, organize, and coordinate tasks within a project. PERT stands for Program Evaluation Review Technique, a methodology developed by the U.S. Navy in the 1950s to manage the Polaris submarine missile program. A PERT chart is a graphic representation of a project’s schedule, showing the sequence of tasks, which tasks can be performed simultaneously, and the critical path of tasks that must be completed on time in order for the project to meet its completion deadline. The chart can be constructed with a variety of attributes, such as earliest and latest start dates for each task, earliest and latest finish dates for each task, and slack time between tasks.

Example of project using PERT technique:

Task Predecessors Realistic time estimate (day) S None None A S 5 B A 2 C G 1 D C 3 E B and D 5 and 11 F E 3 G S 15

Fig.1: The PERT chart representing a projects schedule showing the sequence of tasks that can be run in parallel and in series

From the chart we can see that some tasks can be run in parallel other in series and some of them can’t be run before the termination of others. For certain task durations, a simple algorithm gives the total time required to finish the project. For uncertain task durations, which are the real life situation, we consider the task durations as random variables. The normal assumption is that these durations are described by independent distributions. In this case, a critical interest in the PERT analysis is the distribution of the random variable describing the project’s duration. For all tasks appearing in series, the total activity duration for that series is the sum of the random variables corresponding to each project task. Let us assume the durations of each task are independent exponential variables with different parameters, 푋 = 푋푆 + 푋퐴 + 푋퐵 + 푋퐸, and 푌 = 푌푆 + 푌퐺 + 푌퐶 + 푌퐷 + 푌퐸. In this application, we are interested to evaluate analytically the probability distribution of the total time of execution required to finish the project, which can be achieved by considering the random variable 푋푌. Numerical Examples: 푋 is a sum of independent exponential variables, 푋푆, 푋퐴, 푋퐵, and 푋퐸 with different parameter 5,2 and 3 respectively. Then 푋 follows a Hypoexponential distribution with different parameters. That is, 푋 ∼ 퐻푦푝표푒푥푝(훼⃑, 푘⃑⃑), 훼⃑= (5,2,3), 푘⃑⃑ = (2,1,1). and the PDF of 푋 is 3 푘푖 푓푋(푡) = ∑ ∑ 퐴푖푗푓퐸 (푡) and using the formulas (6) and (7) gives that 퐴11 = 25/6, 퐴21 = 푖=1 푗=1 훼푖,푗 1, 퐴21 = 25/3, 퐴31 = −25/2. Also 푌 is a sum of independent exponential variables, 푌푆, 푌퐺, 푌퐶, 푌퐷 https://digitalcommons.bau.edu.lb/stjournal/vol1/iss1/7 8 Kadri et al.: ON THE PRODUCT OF TWO HYPOEXPONENTIAL RANDOM VARIABLES

and 푌퐸 with different parameter 15, 1, 3, 11,and 3 respectively. Then 푌 follows a Hypoexponential with different parameters. That is, 푌 ∼ 퐻푦푝표푒푥푝(훽⃑, 푙⃑), 훽⃑ = (15,1,3,11), 푙⃑ = (1,1,2,1) and the PDF 푙 of 푌 is 푓 (푡) = ∑4 ∑ 푝 퐵 푓 (푡). Also the formulas (6) and (7) gives that, 퐵 = 11/896, 푌 푝=1 푞=1 푝푞 퐸훽푝,푞 11 퐵21 = 297/112, 퐵31 = −385/512, 퐵32 = −55/64 퐵41 = −27/512.

Next, we shall determine the PDF of 푋푌 using Theorem 1, 푛=3 푘푖 푟=4 푙푝 푓푍(푡) = ∑푖=1 ∑푗=1 ∑푝=1 ∑푞=1 퐴푖푗퐵푝푞푓퐾(푡)

where 푞+푗 2(훽 훼 푡) 2 푓 (푡) = 푝 푖 퐾 (2 훼 훽 푡) 퐾 (푞−1)!(푗−1)!푡 푞−푗 √ 푖 푝

The function is implemented into MATHEMATICA, and the graph is shown in Fig. 2.

Fig.2: The probability distribution of the total time of execution required to finish the project

5. CONCLUSIONS The exact expressions of the probability density function of the product of two independent general case of the Hypoexponential distribution and particular two independent Hypoexponential distribution of different stages are obtained. The expressions are given as a linear combination of the 퐾-distribution. As a consequence, the CDF, reliability function, hazard function, MGF and moment of order 푚 are established. Eventually, a theoretical result is applied to stochastic activity network used in project management by determining the total execution time in stochastic PERT Network.

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