The problem:
Toss a die 10 times. Let $X$ be the number of ones and let $Y$ be the number of sixes.
Find $P\left(X+Y=2\right)$ and then find the variance of $W=5X+2Y$.
I found $P\left(X+Y=2\right)$ using the binomial distribution.
I am attempting to find $W=5X+2Y$ using $E[aX+bY]\ =v^{T}Mv$, where
$v=
\begin{bmatrix}
5\\2\\
\end{bmatrix}$, $M=\begin{bmatrix}
Var(X)&Cov(X,Y)\\Cov(X,Y)&Var(Y)\end{bmatrix}$
So now I have,
$\begin{bmatrix}
5&2
\end{bmatrix}$ $\begin{bmatrix}
Var(X)&Cov(X,Y)\\Cov(X,Y)&Var(Y)\end{bmatrix}$
$\begin{bmatrix}
5\\2\\
\end{bmatrix}$
The problem I am having is that I can't figure out how to find the covariance of $Cov(X,Y)$.
From the binomial distribution, $\mu=n\theta$ and $\sigma^2=n\theta\left(1-\theta\right)$, but I don't know if either of those relationships will help me find $Cov(X,Y)$. Any feedback would be greatly appreciated.
Best Answer
A slick way to get the covariance:
Let $X_1, X_2, \ldots, X_6$ be the number of ones, twos, etc. respectively. So $X=X_1$ and $Y=X_6$.
Note that $X_1 + \cdots + X_6 = 10$, and that the sequence $X_1, \ldots, X_6$ is exchangeable since no face of the die is intrinsically different from any other face. Then $\text{Var}(X_i)= \text{Var}(X_1)$ for any $i$, and $\text{Cov}(X_i, X_j) = \text{Cov}(X_1, X_2)$ for any $i \ne j$. Thus, $$0 = \text{Var}(X_1 + \cdots + X_6) = \sum_{i=1}^6 \text{Var}(X_i) + 2 \sum_{1 \le i < j \le 6}\text{Cov}(X_i, X_j) = 6 \text{Var}(X_1) + 6 \cdot 5 \cdot \text{Cov}(X_1, X_2).$$ Thus $\text{Cov}(X_1, X_2) = - \frac{1}{5} \text{Var}(X_1) = - \frac{1}{5} \cdot 10 \cdot \frac{1}{6} \cdot \frac{5}{6} = -\frac{5}{18}$.