Dice throw, joint distribution.

Question

The two dice were thrown. $X$ is random variable of the amount of 5 in throw. $Y$ is a random variable of 6 in throw. Compute $\mathrm{cov}(X,Y)$.

I suppose, that for $E(X)$ and $E(Y)$ we should use the expected value formula for binomial coefficient, and that gives us $\frac{1}{3}$ (I am also not sure in this answer).

But how we calculate $E(XY)$?

Answer 1

Hint:

You should be able to work any of these statistics from a probability table:

X   Y   Probability
0   0   16/36
0   1    8/36
0   2    1/36
1   0    8/36
1   1    2/36 
2   0    1/36

For example $E[X]= 0 \times \frac{25}{36}+1 \times \frac{10}{36} +2 \times \frac{1}{36} = \frac13$, and similarly for $E[Y]$, as you found.

Answer 2

Let $X_i, Y_i$ be the indicator of having $5$ and $6$ respectively in the $i$-th dice. We can generalize to the $n$ dice case which $$ X = \sum_{i=1}^n X_i, Y = \sum_{j=1}^n Y_j $$ Note that when $i \neq j$, $X_i$ and $Y_j$ are independent and thus uncorrelated. Therefore $$ \begin{align*} Cov[X, Y] & = Cov\left[\sum_{i=1}^n X_i, \sum_{j=1}^n Y_j\right] \\ & = \sum_{i=1}^n Cov[X_i, Y_i] + \sum_{i=1}^n \sum_{j\neq i} Cov[X_i, Y_j]\\ & = n Cov[X_1, Y_1]\end{align*}$$

Note that $X_1, Y_1$ are in the same multinomial trials - the indicators of mutually exclusive events, therefore $$ \Pr\{X_1 = 1, Y_1 = 1\} = E[X_1Y_1] = 0$$ and thus $$ Cov[X_1, Y_1] = 0 - E[X_1]E[Y_1] = \frac {1} {36}$$ Hence you obtain the result $$ Cov[X, Y] = \frac {n} {36}$$