Correlation and Covariance

2018-06-19 23:55:37

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 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 E(S(1),S(6)) is. Can anyone lead me in the right direction for finding this expectation?

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{d

  • 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

    2018-06-20 00:39:31