Poisson process, mean is random variable What is the probability that group of $5$ items will last in warehouse for more than $1$ day, if sale of this item can be described by the Poisson process:
a) with mean of $5$ items per day
b) with mean being geometrically distributed random variable with parameter $\frac{1}{2}$
a) $P(N(1)<5)=\sum\limits_{k=0}^4 P(N(1)=k)=\sum\limits_{k=0}^4 e^{-4} \frac{4^k}{k!}$
Is a) correct? How to do b)?
 A: a)
if you sell 5 items per day, that means your group of 5 items has an average duration of one day. Their random survival is an $Exp(1)$ density thus the answer is
$$P(X>1)=e^{-1}$$
b)
Y is a geometric distribution with support $y \in\{0,1,2,3..\}$ thus
$p(Y=y)=\frac{1}{2^{y+1}}$
Considering the group of 5 items, $X|Y$ is an exponential distribution with mean $\frac{5}{y}$
(if you sell i.e. 10 items per day, the expected duration of your 5 items' group is half day...)
Thus
$$f_{X|Y}(x|y)=\frac{y}{5}e^{-\frac{y}{5}x}$$
The requested probability is thus
$$\mathbb{P}[X>1]=\sum_{y=0}^{\infty}\frac{1}{2^{y+1}}\int_1^{\infty}\frac{y}{5}e^{-\frac{y}{5}x} dx=\sum_{y=0}^{\infty}\frac{e^{-\frac{y}{5}}}{2^{y+1}}$$
Without doing a lot of math reasoning, the result is $\approx 0.847$; it is enough to calculate the first 9-10 addends....the rest are all $\approx 0$
A: When mean $\lambda$ follows geometric distribution itself,$$\begin{align*}P(N(1)<5)&=\sum_{k=0}^\infty P(N(1)<5\wedge\lambda=k)\\&=\sum_{k=0}^\infty \color{red}{P(N(1)<5|\lambda=k)}\color{blue}{P(\lambda=k)}\\&=\sum_{k=0}^\infty\left[\color{red}{\sum\limits_{x=0}^4 e^{-k}\frac{k^x}{x!}}\right]\color{blue}{(1/2)^{k+1}}\end{align*}$$
