# Poisson process condition

A certain toy store sells toys. During a particular hour, customers arrive at the store according to a Poisson Process with mean $$10$$ per hour. Let $$X$$ be the number of customers who arrive during this hour. For each customer, there is a $$\frac{1}{5}$$ chance that the customer will buy exactly one toy and a $$\frac{4}{5}$$ chance that the customer will not buy any toy, independent of other customers. Let $$Y$$ be the number of toys sold in the one hour period. By conditioning $$Y$$ on $$X$$, use the law of iterated expectation to determine $$\mathbb{E}(Y)$$ and then use the law of total variance to determine $$\mathrm{Var}(Y)$$.

I have been trying to solve this problem for hours but without success. Any assistance you can provide is much appreciated.

• $Y \sim \mathsf{Pois}(\lambda = 20(1/5) = 2)?$ Commented Mar 18, 2021 at 22:52

Since customer arrival is a Poisson process with an intensity of $$10$$, the Number of customers in an hour will have distribution Poisson$$(10)$$, and each customer buys a toy with probability $$1/5$$ therefore expected number of toys sold given $$x$$ many customers arrived ($$\mathbb{E}(Y|X=x)$$) is $$x/5$$. Thus

$$\mathbb{E}(Y) = \sum_{x=0}^{\infty}\mathbb{E}(Y|X=x)\mathbb{P}(X=x) = \sum_{x=0}^{\infty}\frac{x}{5}\frac{e^{-10}10^{x}}{x!} = \frac{1}{5}\sum_{x=0}^{\infty}\frac{x}{5}\frac{e^{-10}10^{x}}{x!} =2,$$ since $$\sum_{x=0}^{\infty}{x}\frac{e^{-10}10^{x}}{x!}$$ is the expectation of Poisson$$(10)$$, which is $$10$$.

You can write $$Var(Y) = Var(\mathbb{E}(Y|X)) +\mathbb{E}(Var(Y|X)).$$ you can use this formula and compute the two terms separately. Try it!

Hint 1: $$Y|X \sim$$ Binomial$$(X,\frac{1}{5})$$.

Hint 2: $$X=$$ the Number of customers in an hour. It has distribution Poisson$$(10)$$

• Oh I see. How about $Var(Y)$?
– user902292
Commented Mar 18, 2021 at 22:35
• For the Variance, if $Y|X$ is binomial then $E(Y|X) = \frac{X}{5}$ and $Var(Y|X) = \frac{4X}{25}$. So $Var(Y) = Var(\frac{X}{5}) + E(\frac{4X}{25}) = \frac{1}{25} Var(X) + \frac{4}{25} E(X)$. How would I then continue?
– user902292
Commented Mar 18, 2021 at 22:59
• @Turnipsavocados See another hint. Commented Mar 19, 2021 at 10:19