Compute $\operatorname{Cov}(X,Y)$ while $X$ is the sum of the two rolls and $Y$ is the number of $2$s in the two rolls

Question

I think I should start by defining $Y_i$ by: $Y_i=1$ if at the $i$-th roll number $2$ occurs an $0$ otherwise. And define $X_i$ as $R_1+R_2$.

But I do not know how to proceed. How would I use the formula $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$ ?

Answer 1

The expected number of $2$s each time you throw the die is $1/6.$ Therefore the expected number in two tosses is $1/3,$ i.e. $\operatorname E(Y) = 1/3.$

The expected outcome when a die is thrown is $(1+2+3+4+5+6)/6 = 7/2 = 3.5;$ therefore the expected sum in two trials is $7,$ i.e. $\operatorname E(X) =7.$

So what is $\operatorname E(XY)\text{ ?}$

One way is just brute force: First look at the list of $36$ outcomes: $$ \begin{array}{cccccc} 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \end{array} $$ What is the number of $2$s? Here it is: $$ \begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{array} $$ This of course will tell you that $Y = \begin{cases} 0 & \text{with probability } 25/36, \\ 1 & \text{with probability } 10/36, \\ 2 & \text{with probabiltiy } \phantom{0}1/36. \end{cases} \qquad$ So this is another way of seeing that $\operatorname E(Y) = 1/3.$

But now multiply the numbers above by the sum: $$ \begin{array}{cccccc} 0 & 3 & 0 & 0 & 0 & 0 \\ 3 & 8 & 5 & 6 & 7 & 8 \\ 0 & 5 & 0 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 & 0 & 0 \\ 0 & 7 & 0 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 & 0 & 0 \end{array} $$ The sum of these is $66,$ so their average is $\dfrac{66}{36}= \dfrac{11}{12}.$

Thus $$ \operatorname{cov}(X,Y) = \operatorname E(XY) - \operatorname E(X)\operatorname E(Y) = \frac{11}{12} - \left( 7 \times \frac 1 3\right) = \frac{11}{12} - \frac{21}{12} = \frac{-10}{12} = \frac {-5} 6. $$