1 problem found
A discrete random variable \(X\) takes only positive integer values. Define \(\E(X)\) for this case, and show that \[\E(X) =\sum^{\infty}_{n=1}\P\left(X\ge n \right).\] I am collecting toy penguins from cereal boxes. Each box contains either one daddy penguin or one mummy penguin. The probability that a given box contains a daddy penguin is \(p\) and the probability that a given box contains a mummy penguin is \(q\), where \(p\ne0\), \(q\ne0\) and \(p+q=1\,\). Let \(X\) be the number of boxes that I need to open to get at least one of each kind of penguin. Show that \(\P(X\ge 4)= p^{3}+q^{3}\), and that \[ \E(X)=\frac{1}{pq}-1.\, \] Hence show that \(\E(X)\ge 3\,\).
Solution: \[ \E[X] := \sum_{n=1}^{\infty} n \mathbb{P}(X=n) \] \begin{align*} && \sum^{\infty}_{n=1}\mathbb{P}\left(X\ge n \right) &= \sum^{\infty}_{n=1}\sum_{k=n}^\infty \mathbb{P}(X=k) \\ &&&= \sum_{k=1}^\infty k \cdot \mathbb{P}(X=k) \\ &&&= \E[X] \end{align*} \begin{align*} &&\mathbb{P}(X \geq 4) &= \mathbb{P}(\text{first 3 are daddies}) +\mathbb{P}(\text{first 3 are mummies}) \\ &&&= p^3 + q^3 \\ \Rightarrow && \E[X] &= \sum_{n=1}^{\infty} \mathbb{P}\left(X\ge n \right) \\ &&&= 1+\sum_{n=2}^{\infty} \left ( p^{n-1} + q^{n-1}\right) \\ &&&= 1+\frac{p}{1-p} + \frac{q}{1-q} \\ &&&= 1+\frac{p}q + \frac{q}p \\ &&&= 1+\frac{p^2+q^2}{pq} \\ &&&= 1+\frac{(p+q)^2-2pq}{pq} \\ &&&= \frac{1}{pq} -1 \\ &&& \underbrace{\geq}_{AM-GM} \frac{1}{4}-1 = 3 \end{align*}