Jenna has two cell phones a and b. We know that the probability of receiving a message from a is p and the total number of received messages in an interval of t seconds follows the Poisson distribution with the rate of $\lambda$. If N(t) is the total number of received messages from mobile a, What is N(t) distribution?
my attempt:
if we take :
$a+b \sim {\rm Poisson}(\lambda )$
$\lambda=\lambda_{a}+\lambda_{b}$
$\lambda_{a}=N(t)p$
so N(t) has a poisson distribution.
but I'm not sure about my answer. any help would be appreciated!
Take $X(t)\sim \text{Poisson}(\lambda t)$ as the total number of calls received in $[0,t)$. Then $N(t)|X(t)\sim \text{Binomial}(X(t),p)$ and $$\begin{eqnarray*}\mathbb{P}(N(t)=x)&=&\sum_{n=x}^{\infty}\mathbb{P}(N(t)=x|X(t)=n)\mathbb{P}(X(t)=n) \\ &=& \sum_{n=x}^{\infty}{n \choose x}p^x(1-p)^{n-x}\cdot \frac{e^{-\lambda t}(\lambda t)^n}{n!}\end{eqnarray*}$$ After a little work you'll see that this sum evaluates to $$e^{-\lambda p t}\cdot \frac{(\lambda p t)^x}{x!}$$ In other words, $N(t)\sim \text{Possion}(\lambda p t)$