suppose that $N$ is a Poisson$(μ)$ random variable. Given $N=n$, random variables $X_1,X_2,X_3,\cdots,X_n$ are independent with uniform∼$(0,1)$ distribution. So there are a random number of $X$'s.
(a) Given $N=n$ what is the probability that all the $X$'s are less than $t$?
So I set up the problem as:
$P(X<t\mid N=n)=\frac{P(X<t,N=n)}{P(N=n)}$
How do I compute this if it's correct, or what do I do next?
(b) What is the (unconditional) probability that all the $X$'s are less than $t$?
No idea how to start this one.
Best Answer
If $0\le t \le 1$, and there are $n$ random variables $X_i$, the probability they are all $\lt t$ is, by independence, equal to $t^n$.
For the unconditional probability, note that $\Pr(N=n)=e^{-\mu} \frac{\mu^n}{n!}$. So the probability that all the $X_i$ are $\lt t$ is $$\sum_0^\infty e^{-\mu}\frac{\mu^n}{n!} t^n.$$ Since $\mu^n t^n=(\mu t)^n$, the above sum simplifies to $$e^{-\mu}e^{\mu t},$$ which can be "simplified" to $e^{-\mu(1-t)}$. For completeness, one should observe that the probability is $0$ if $t\lt 0$, and $1$ if $t\gt 1$.