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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?

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  • $\begingroup$ 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. $\endgroup$ – kilojoules Jun 19 '18 at 22:49
  • $\begingroup$ 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. $\endgroup$ – Reinstate Monica Jun 20 '18 at 1:17
  • $\begingroup$ 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. $\endgroup$ – Glen_b -Reinstate Monica Jun 20 '18 at 17:44
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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: $$ X_{1,i} = \begin{cases} 1, \ \ \text{die rolls 1 on $i$th roll}\\ 0, \ \ \text{die does not roll 1 on $i$th roll}\\ \end{cases} $$ Similarly, let $X_{6,i}$ be the indicator variable for the die rolling a 6 on the $i$th roll: $$ X_{6,i} = \begin{cases} 1, \ \ \text{die rolls 6 on $i$th roll}\\ 0, \ \ \text{die does not roll 6 on $i$th roll}\\ \end{cases} $$

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).

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  • $\begingroup$ very helpful thank you. that got me in the right direction. $\endgroup$ – Grant Egnatz Jun 20 '18 at 4:16

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