Problems

Solution:

1. 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 \((1-p)\lambda\)
2. 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*}
3. 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)