Correlation and Covariance

Quick question that would really help me when studying for probability exam.

A fair die is rolled $9$ times. with $S(k)$ denoting the total number of appearances of labeled $k$, where $k = 1,...,6$. It is clear that the $\mathbb{corr}(S(1), S(6))$ does not equal $0$ because the two events are not independent. But what is the correlation?

I know the formula for Covariance, but I am having trouble figuring out what $\mathbb{E}(S(1),S(6))$ is. Can anyone lead me in the right direction for finding this expectation?

• How are rolling different numbers on a die not independent events? It seems like the next roll will have no dependance on the previous roll. I would expect the correlation to be zero. Commented Jun 19, 2018 at 22:49
• OP is correct here. Although outcomes of the rolls are independent, the counts of outcomes over a fixed number of trials will be negatively correlated, since the occurrence of one outcome means the other did not occur.
– Ben
Commented Jun 20, 2018 at 1:17
• It's easy (almost trivial) to calculate the covariance for a single roll (from the expectation of the product and the product of the expectations). For there it's easy to do the case with more than one roll. Commented Jun 20, 2018 at 17:44

Instead of looking at the case where the die has been rolled 9 times, consider the case where the die has only been rolled once.

Let $X_{1,i}$ be an "indicator" for the die rolling a 1 on the $i$th roll. Put explicitly: $$X_{1,i} = \begin{cases} 1, \ \ \text{die rolls 1 on ith roll}\\ 0, \ \ \text{die does not roll 1 on ith roll}\\ \end{cases}$$ Similarly, let $X_{6,i}$ be the indicator variable for the die rolling a 6 on the $i$th roll: $$X_{6,i} = \begin{cases} 1, \ \ \text{die rolls 6 on ith roll}\\ 0, \ \ \text{die does not roll 6 on ith roll}\\ \end{cases}$$

What is the covariance between $X_{i,1}$ and $X_{6,i}$? We can use the decomposition formula: $$\text{Cov} (X_{1,i},X_{6,i}) = E(X_{1,i}*X_{6,i}) - E(X_{1,i})E(X_{6,i})$$ $E(X_{1,i})$ and $E(X_{6,i})$ are easy since $X_{1,i}$ are Bernoulli random variables with probability 1/6. Their expected variables will both be 1/6.

To calculate $E(X_{1,i}*X_{6,i})$, note that the random variable $Z = X_{1,i} *X_{6,i}$ equals 0 with probability 1.

If the $i$th roll is 1, $Z = 1*0 = 0$, if the $i$th roll is 6, $Z = 0*1 = 0$, and if the $i$th roll is neither 1 nor 6, then $Z = 0*0 = 0$. Therefore, the expected value of Z is 0. $$E(Z) = E(X_{1,i} X_{6,i}) = 0$$ Putting into the formula above, $$\text{Cov} (X_{1,i},X_{6,i}) = E(X_{1,i}*X_{6,i}) - E(X_{1,i})E(X_{6,i})$$ $$= 0 - (1/6)(1/6) = -1/36$$

Now, that is for just one roll, and I hope it's a good start to you solving this. To finish this problem, note:

(1) What is be the relationship between $X_{i,1}$, $X_{i,6}$, $S(1)$, and $S(6)$?

(2) Look into the summation formulas for Covariance

(3) The throws are independent, so Cov$(X_{1,i},X_{6,j}) = 0$ for $i \neq j$.

Once you have the covariance, the correlation can be calculated in a straightforward by noting that $S(1)$ and $S(6)$ are both Binomial random variables (information that can be used to get the variances, which are then used to scale the covariance).

• very helpful thank you. that got me in the right direction. Commented Jun 20, 2018 at 4:16