1 problem found
A team of \(m\) players, numbered from \(1\) to \(m\), puts on a set of a \(m\) shirts, similarly numbered from \(1\) to \(m\). The players change in a hurry, so that the shirts are assigned to them randomly, one to each player. Let \(C_i\) be the random variable that takes the value \(1\) if player \(i\) is wearing shirt \(i\), and 0 otherwise. Show that \(\mathrm{E}\left(C_1\right)={1 \over m}\) and find \(\var \left(C_1\right)\) and \(\mathrm{Cov}\left(C_1 \, , \; C_2 \right) \,\). Let \(\, N = C_1 + C_2 + \cdots + C_m \,\) be the random variable whose value is the number of players who are wearing the correct shirt. Show that \(\mathrm{E}\left(N\right)= \var \left(N\right) = 1 \,\). Explain why a Normal approximation to \(N\) is not likely to be appropriate for any \(m\), but that a Poisson approximation might be reasonable. In the case \(m = 4\), find, by listing equally likely possibilities or otherwise, the probability that no player is wearing the correct shirt and verify that an appropriate Poisson approximation to \(N\) gives this probability with a relative error of about \(2\%\). [Use \(\e \approx 2\frac{72}{100} \,\).]
Solution: There are \(m!\) different ways of assigning the shirts, and in \((m-1)!\) of them player \(1\) gets their own shirt, ie \(\mathbb{E}(C_1) = \mathbb{P}(\text{player }1\text{ gets own shirt}) = \frac{(m-1)!}{m!} = \frac{1}{m}\). \(\var(C_1) = \mathbb{E}(C_1^2) - [\mathbb{E}(C_1)]^2 = \frac{1}{m} - \frac{1}{m^2} = \frac{m-1}{m^2}\). If we have two players, there are \((m-2)!\) ways they both get their own shirts, therefore \(\textrm{Cov}(C_1,C_2) = \mathbb{E}(C_1C_2) - \mathbb{E}(C_1)\mathbb{E}(C_2) = \frac{(m-2)!}{m!} - \frac{1}{m^2} = \frac{1}{m(m-1)} - \frac{1}{m^2} = \frac{m-m+1}{m^2(m-1)} = \frac{1}{m^2(m-1)}\). \begin{align*} \mathbb{E}(N) &= \mathbb{E}(C_1 + C_2 + \cdots + C_m) \\ &= \mathbb{E}(C_1) + \mathbb{E}(C_2) + \cdots + \mathbb{E}(C_m) \\ &= \frac{1}{m} + \frac{1}{m} +\cdots+ \frac1m \\ &= 1 \\ \\ \var(N) &= \sum_{r=1}^m \var(C_r) + 2\sum_{r=1}^{m-1} \sum_{s=2}^{m} \textrm{Cov}(C_r,C_s) \\ &= m \frac{m-1}{m^2} + 2 \frac{m(m-1)}{2}\frac{1}{m^2(m-1)} \\ &=\frac{m-1}{m} + \frac{1}{m} \\ &= 1 \end{align*} If we were to take a normal approximation, we would want to take \(N(1,1)\), but this would say things like \(-1\) is as likely as \(3\) shirts being correct, which is clearly a bad model. A Poisson is much more likely to be a sensible model as they have the same mean and variance as the parameter, and if \(m\) is large, the covariance between shirts is going to be very small, so it will appear similar to random events occurring. We can have \begin{align*} BADC \\ BCDA \\ BDAC \\ CADB \\ CDAB\\ CDBA \\ DABC\\ DCAB \\ DCBA \end{align*} Ie \(\frac{9}{24}\) ways to have no player wearing their own shirt with \(4\) players. \(Po(1)\) would say this probability is \(e^{-1}\), giving a relative error of: \begin{align*} \frac{e^{-1}-\frac{9}{24}}{\frac9{24}} &\approx \frac{\frac{100}{272} - \frac{9}{24}}{\frac9{24}} \\ &= -\frac{1}{51} \\ &\approx -2\% \end{align*}