Question

Show that the sum of two stochastically independent poisson-distributed random quantities again is poisson distributed!
A computer system consists of three sub-devices. The manufacturers of the sub-devices indicated that the lifetimes were exponentially distributed and the mean lifetimes 2, 3 and 5
Years. What is the average lifespan of the overall system, i. H. the mean time until the first failure of a subsystem?

Answer 1

Show that the sum of two stochastically independent poisson-distributed random quantities again is poisson distributed!

Let X and Y be two stochastically independent random variables having Poisson distributions with parameters µ and ν, respectively. Then the sum X + Y has a Poisson distribution with parameter µ + ν.

Proof. By the law of total probability,

P(X + Y = n) = P(X = 1, Y = n – i) i=0

$=\sum_{i=0}^{n}P(X=i) P(Y=n-i)$

(Since X and Y are independent)

$=\sum_{i=0}^{n}\frac{\mu ^{i}e^{-\mu }}{i!} \frac{\nu^{n-i}e^{-\nu }}{(n-1)!}$

$=\frac{e^{-(\mu+\nu) }}{n!}\sum_{i=0}^{n}\frac{n!}{i!(n-i)!} \mu ^{i}*\nu^{n-i}$

The binomial expansion of $(\mu+\nu)^{n}$ is, of course,

$(\mu+\nu)^{n}=\sum_{i=0}^{n}\frac{n!}{i!(n-i)!} \mu ^{i}*\nu^{n-i}$

$P(X+Y=n)=\frac{e^{-(\mu+\nu) }(\mu+\nu)^{n}}{n!}$

n=0,1,2,......

therefore, The sum X + Y has a Poisson distribution with parameter µ + ν.

A computer system consists of three sub-devices. The manufacturers of the sub-devices indicated that the lifetimes were exponentially distributed and the mean lifetimes 2, 3 and 5
Years. What is the average lifespan of the overall system, i. H. the mean time until the first failure of a subsystem?

For the exponential distribution, the reliability R(t) =e - λt

$R(t)=e^{-\lambda t}$

$\lambda_{1}=2, \lambda_{2}=3, \lambda_{3}=5$

The average lifespan of the overall system is

$R_{s}(t)=e^{-\lambda_{1} t}e^{-\lambda_{2} t}e^{-\lambda_{3} t}=e^{-2t}e^{-3 t}e^{-5 t}=e^{-10t}$

The average lifespan of the overall system is $\lambda_{1}+ \lambda_{2}+ \lambda_{3}=2+3+5=10$

The mean time until the first failure of a subsystem

MTTF=$MTTF=\int_{0}^{\infty }R(t)dt=\int_{0}^{\infty }e^{-\lambda t}dt=\frac{1}{\lambda }=\frac{1}{10}=0.1$