Problem source

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:
\begin{questionparts}
\item$W + L + D$ has variance $0\,$;
\item $W + L$ has a  binomial distribution.
\end{questionparts}
Find the value of $\displaystyle {\mathrm{Cov}(W,L) \over \sqrt{\mathrm{Var}(W) \mathrm{Var}(L)}}\;$.

Solution source

\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*}

\begin{questionparts}
\item $W+L+D = m$ where $m$ is the number of games, which has variance $0$. Therefore $W+L+D$ has variance $0$.

\item 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.
\end{questionparts}

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*}