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2002 Paper 3 Q14
D: 1700.0 B: 1500.0
probability variance covariance binomial distribution discrete random variables bivariate data proof correlation coefficient chess

Prove that, for any two discrete random variables \(X\) and \(Y\), \[ \mathrm{Var} \left(X + Y \right) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2 \, \mathrm{Cov}(X,Y), \] where \(\mathrm{Var}(X)\) is the variance of \(X\) and \(\mathrm{Cov}(X,Y)\) is the covariance of \(X\) and \(Y\). When a Grandmaster plays a sequence of \(m\) games of chess, she is, independently, equally likely to win, lose or draw each game. If the values of the random variables \(W\), \(L\) and \(D\) are the numbers of her wins, losses and draws respectively, justify briefly the following claims:

  1. \(W + L + D\) has variance \(0\,\);
  2. \(W + L\) has a binomial distribution.
Find the value of \(\displaystyle {\mathrm{Cov}(W,L) \over \sqrt{\mathrm{Var}(W) \mathrm{Var}(L)}}\;\).

Solution: \begin{align*} && \var[X+Y] &= \E\left [(X+Y-\E[X+Y])^2 \right] \\ &&&= \E \left [ (X - \E[X] + Y - \E[Y])^2 \right] \\ &&&= \E \left [(X - \E[X])^2 + (Y-\E[Y])^2 + 2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \E \left [(X - \E[X])^2 \right]+\E \left [(Y-\E[Y])^2 \right]+\E \left [2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \var[X] + \var[Y] + 2 \mathrm{Cov}(X,Y) \end{align*}

  1. \(W+L+D = m\) where \(m\) is the number of games, which has variance \(0\). Therefore \(W+L+D\) has variance \(0\).
  2. The probability of a decisive game is \(\frac23\) and \(W+L\) is the number of decisive games. Each game is independent so this meets the criteria for a binomial distribution.
Notice \(W+L \sim B(m, \tfrac23)\) and \(W, L, D \sim B(m, \tfrac13)\), in particular \(\var[W+L] = m \tfrac23 \tfrac13 = \tfrac29m\) and \(\var[W] = \var[D] = \var[D] = m \tfrac13 \tfrac13 = \tfrac29m\) \begin{align*} && \var[W+L] &= \var[W] + \var[L] + 2\mathrm{Cov}(W,L) \\ \Rightarrow && \mathrm{Cov}(W,L) &= -\tfrac19m \\ \Rightarrow && \frac{\mathrm{Cov}(W,L) }{\sqrt{\var[W]\var[L]}} &= -\frac12 \end{align*}

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