1. ## Limiting Poisson distribution

Starting at some fixed time, $\displaystyle 0,$ satellites are launched at times of a Poisson process with rate $\displaystyle \lambda.$ After an independent amount of time having distribution $\displaystyle F$ and mean $\displaystyle \mu$, the satellite stops working. Let $\displaystyle X(t)$ be the number of satellites working at time $\displaystyle t.$

(a) Find the distribution of $\displaystyle X(t).$
(b) Let $\displaystyle t \rightarrow \infty$ in (a) to show the limiting distribution is $\displaystyle Poisson(\lambda\mu).$

2. Originally Posted by Anonymous1
Starting at some fixed time, $\displaystyle 0,$ satellites are launched at times of a Poisson process with rate $\displaystyle \lambda.$ After an independent amount of time having distribution $\displaystyle F$ and mean $\displaystyle \mu$, the satellite stops working. Let $\displaystyle X(t)$ be the number of satellites working at time $\displaystyle t.$

(a) Find the distribution of $\displaystyle X(t).$
(b) Let $\displaystyle t \rightarrow \infty$ in (a) to show the limiting distribution is $\displaystyle Poisson(\lambda\mu).$
Hi,
the number $\displaystyle N$ of satellites launched before $\displaystyle t$ is Poisson with parameter $\displaystyle \lambda t$ and, given $\displaystyle N$, the times of the launches are (without order) i.i.d. random variables $\displaystyle X_1,\ldots,X_N$ uniformly distributed on $\displaystyle [0,t]$, ok?
Let $\displaystyle \tau_1,\ldots,\tau_N$ be their lifespan: i.i.d. with distribution $\displaystyle F$.
If you prove that $\displaystyle P(X_1+\tau_1>t)=\frac{1}{t}E[\min(\tau_1,t)]$, then you see that the number of satellites that are still working at time $\displaystyle t$ is obtained by picking each of the launched satellites with this same probability, hence $\displaystyle X(t)$ is Poisson with parameter $\displaystyle (\lambda t)(\frac{1}{t}E[\min(\tau_1,t)])=\lambda E[\min(\tau_1,t)]$. (It's the usual "picking each point of a rate $\displaystyle \lambda$ Poisson process with probability $\displaystyle p$ yields a rate $\displaystyle \lambda p$ Poisson process"). You can also check this conclusion by computing $\displaystyle P(X(t)=n)$ explicitly: $\displaystyle P(X(t)=n)=E[P(X(t)=n|N)]=E[{N\choose n}P(X_1+\tau_1>t)^nP(X_1+\tau_1<t)^{N-n}]$, etc. I let you fill in the dots... If you need further help, please tell us extendedly what you tried and so on.

3. Not many dots left to fill in . So for part (b)...

$\displaystyle \lambda E[\min(\tau_1,t)] \rightarrow \lambda E[\tau_1]$ as $\displaystyle n \rightarrow \infty.$

Also, $\displaystyle E[\tau_1]=\mu.$

Thanks!

4. $\displaystyle E[{N\choose n} P(X_1+\tau_1>t)^n P(X_1+\tau_1)^{N-n}] = N \times P(X_1+\tau_1>t)$ Since we have a $\displaystyle Bin(N,P(X_1+\tau_1>t))$

Therefore, $\displaystyle X(t)$ ~ $\displaystyle Poi(\lambda\times E[\min(\tau_1, t)]).$