Question

Q. 2 (Gamma and exponential, 30 pts). A parallel system consists of two components with independent lifetimes. The lifetime Ly of the first component is memoryless: it has the exponential distribution with parameter 1. On the other hand, based on statistical analyses, it is found out that the lifetime L2 of the second component has two independent phases each of which has the same characteristics as L). Therefore, it is assumed that L2 has the gamma distribution with shape index k = 2 and scale parameter 1. What is the expected lifetime of this system?

Answer 1

Solution:-

Given that

A parallel system consists of two components with independent lifetimes

. The lifetime of L1 of the first component is memoryless; it has the exponential distribution with parameter $\lambda$ .

Based on statistical analysis, it is found out that the lifetime L2 of second component has two independent phases each of which has the same characterisitics as L1. Therefore, it is assumed that L2 has gamma distribution with shape index K = 2 and scale parameter $\lambda$ .

What is the expected lifetime of this system?

pdf for life time :

L1

  

f(0, X) = le-dz I > 0 otherwise 0

pdf for life time $L_2$ , which is gamma distribution with shape index k = 2 and scale parameter $\lambda$ .

1 far, ke, = ਨਾਲ-*-le-1/4 10. otherwise 0

where  $\Gamma(k)=\int_{0}^{\infty}x^{k-1}e^{-x}dx$

at k = 2

$\Gamma(2)=\int_{0}^{\infty}xe^{-x}dx$

  $=[-e^{-x}-xe^{-x}]^\infty_0$

$\Gamma(2)=1$

therefore pdf is

f(0, 2, 1) = -22-12-1/ 121(2) 2 > 0

$f(x,2,\lambda)=\left\{\begin{matrix} \frac{1}{\lambda^2}xe^{-x/\lambda} &x>0 \\ 0& otherwise \end{matrix}\right.$

As
L1
and $L_2$ are independent then the sum of
L1
and $L_2$ is also independent and their pdf is also.

pdf of the system is

$f(x,\lambda)=\left\{\begin{matrix} \lambda e^{-\lambda x}+\frac{x}{\lambda^2}e^{-x/\lambda} &x>0 \\ 0 &otherwise \end{matrix}\right.$

Expected life time for system

$E(x)=\int_{0}^{\infty}xf(x)dx$

$=\int_{0}^{\infty}x[\lambda e^{-\lambda x}+\frac{x}{\lambda^2}e^{-x/\lambda}]dx$

$=\int_{0}^{\infty}\lambda x e^{-\lambda x}dx+\int_{0}^{\infty}\frac{x}{\lambda^2}e^{-x/\lambda}dx$

After solving

$\lambda x e^{-\lambda x}dx\ and\ \int_{0}^{\infty}\frac{x}{\lambda^2}e^{-x/\lambda}dx$ we get

$\lambda x e^{-\lambda x}dx=\frac{1}{\lambda}\ and\ \int_{0}^{\infty}\frac{x}{\lambda^2}e^{-x/\lambda}dx=2\lambda$

Therefore,

Expected lifetime for system,

$E(x)=\frac{1}{\lambda}+2\lambda$

$E(x)=\frac{1+2\lambda^2}{\lambda}$

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