Problem source

The number of customers arriving at a builders' merchants each day follows a Poisson distribution with mean $\lambda$. Each customer is offered some free sand. The probability of any given customer taking the free sand is $p$.
\begin{questionparts}
\item Show that the number of customers each day who take sand follows a Poisson distribution with mean $p\lambda$.
\item The merchant has a mass $S$ of sand at the beginning of the day. Each customer who takes the free sand gets a proportion $k$ of the remaining sand, where $0 \leq k < 1$. Show that by the end of the day the expected mass of sand taken is
$$\left(1 - e^{-kp\lambda}\right)S.$$
\item At the beginning of the day, the merchant's bag of sand contains a large number of grains, exactly one of which is made from solid gold. At the end of the day, the merchant's assistant takes a proportion $k$ of the remaining sand. Find the probability that the assistant takes the golden grain. Comment on the case $k = 0$ and on the limit $k \to 1$.
In the case $p\lambda > 1$ find the value of $k$ which maximises the probability that the assistant takes the golden grain.
\end{questionparts}

Solution source

\begin{questionparts}

\item Let $X$ be the number of people arriving on a given day, and $Y$ be the number taking sand, then \begin{align*}
&& \mathbb{P}(Y = k) &= \sum_{x=k}^{\infty} \mathbb{P}(x \text{ arrive and }k\text{ of them take sand}) \\
&&&= \sum_{x=k}^{\infty} \mathbb{P}(X=x)\mathbb{P}(k \text{ out of }x\text{ of them take sand})\\
&&&=  \sum_{x=k}^{\infty} e^{-\lambda} \frac{\lambda^x}{x!}\binom{x}{k}p^k(1-p)^{x-k}\\
&&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \sum_{x=k}^{\infty} \frac{((1-p)\lambda)^x}{k!(x-k)!} \\
&&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \frac{((1-p)\lambda)^k}{k!} \sum_{x=0}^{\infty} \frac{((1-p)\lambda)^x}{x!} \\
&&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \frac{((1-p)\lambda)^k}{k!}e^{(1-p)\lambda)} \\
&&&= e^{-p\lambda} \frac{(p\lambda)^k}{k!}
\end{align*}

which is precisely a Poisson with parameter $p\lambda$.

Alternatively, $Y = B_1 + B_2 + \cdots + B_X$ where $B_i \sim Bernoulli(p)$ so $G_Y(t) = G_X(G_B(t)) = G_X(1-p+pt) = e^{-\lambda(1-(1-p+pt))} = e^{-p\lambda(1-t)}$ so $Y \sim Po(\lambda)$


Alternatively, alternatively, let $Z$ be the number of people not taking sand, so \begin{align*}
&& \mathbb{P}(Y = y, Z= z) &= \mathbb{P}(X=y+z) \cdot \binom{y+z}{y} p^y(1-p)^z \\
&&&= e^{-\lambda} \frac{\lambda^{y+z}}{(y+z)!} \frac{(y+z)!}{y!z!} p^y(1-p)^z \\
&&&=\left ( e^{-p\lambda} \frac{(p\lambda)^y}{y!} \right) \cdot \left ( e^{-(1-p)\lambda} \frac{((1-p)\lambda)^z}{z!}\right)
\end{align*}

So clearly $Y$ and $Z$ are both (independent!) Poisson with parameters $p\lambda $ and $(1-p)\lambda$

\item The amount taken is $Sk + S(1-k)k  + \cdots +Sk(1-k)^{Y-1} = Sk\cdot \frac{1-(1-k)^Y}{k} = S(1-(1-k)^Y)$ so

\begin{align*}
\E[\text{taken sand}] &= \E \left [ S(1-(1-k)^Y)\right] \\
&= S-S\E\left [(1-k)^Y \right] \\
&= S - SG_Y(1-k)\\
&=S - Se^{-p\lambda(1-(1-k))} \tag{pgf for Poisson} \\
&= S\left (1-e^{-kp\lambda} \right)
\end{align*}

\item The fraction of grains the assistant takes home is:

$(1-k)^Yk$, which has expected value $ke^{-kp\lambda}$. This the the probability he takes home the golden grain.

When $k = 0$ the probability is $0$ which makes sense (no-one takes home any sand, including the merchant's assistant).

As $k \to 1$ we get $e^{-p\lambda}$ which is the probability that no-one gets any sand other than him.

\begin{align*}
&& \frac{\d }{\d k} \left ( ke^{-kp\lambda} \right) &= e^{-kp\lambda} - (p\lambda)ke^{-kp\lambda} \\
&&&= e^{-kp\lambda}(1 - (p\lambda)k) 
\end{align*}

Therefore maximised at $k = \frac{1}{p\lambda}$. (Clearly this is a maximum just by sketching the function)

\end{questionparts}