Consider n independent tosses of a k-sided fair die. Let $X_i$ be the number of tosses that result in i. What would be the covariance between $X_1$ and $X_2$ ?
I can't seem to get my head around this problem. I know that
$Cov(X_1,X_2) = E[X_1 X_2] - E[X_1]E[X_2]$.
I believe $P(X_1) = {{N}\choose{X_1}} * (1/k)^{X_1} * (k-1/k)^{n-X_1} $ and $E[X_1] = \sum_{i=0}^{n} X_1 * P(X_1)$
but I can't take it from here any further. Any help would be appreciated. Thanks
Are you familiar with the technique of using Indicator Random Variables to simplify covariance of counts of events among sequences of trials?
Begin like this:-
Let $Y_k$ be the indicator random variable that roll $k$ is a
1, and $Z_k$ be that for being a2.So $X_1=\sum_{k=1}^n Y_k\;, X_2=\sum_{j=1}^n Z_j$
By the Bilinarity of Covariance:
$$\mathsf {Cov}(X_1,X_2) = \sum_{k=1}^n \sum_{j=1}^n\mathsf {Cov}(Y_k,Z_j) $$
Because these are Indicator Random Variables,
$$\begin{align}\mathsf{Cov}(Y_k,Z_j) &= \mathsf P(Y_k=1,Z_j=1)-\mathsf P(Y_k=1)\mathsf P(Z_j=1) \\[1ex]&= \mathsf P(Y_k=1, Z_j=1)-\tfrac 1{36}\end{align}$$
Next, you must consider the cases where $k=j$ and where $k\neq j$
$\vdots$