Suppose we toss a fair dice $n$ times. For $i=1,\dots,6$, let $X_i$ denote the number of times the dice comes up with value $i$. Find the $\textrm{cov}(X1, X2)$.

By definition, $\textrm{cov}(X1, X2)=E[X_1 X_2] - E[X_1]E[X_2]$.

Since the dice is fair, we know that each $X_i$ follows a $\textrm{Binomial}(n, 1/6)$ distribution. So $E[X_i]=n/6$.

My question is how do we find $E[X_1 X_2]$? Thanks a lot for your help in advance.


1 Answer 1


Note that $X_i$ can be considered as the sum of independent Bernoulli random variables, i.e., $X_i=\sum_{j=1}^nY_i^j$, where $Y_i^j$ is $1$ when $i$ occurs at the $j$-th trial, and $0$ if otherwise. Then, we have $\mathrm{E}[X_1X_2]=\sum_{j=1}^n\mathrm{E}[Y_1^jY_2^j]+\sum_{i\neq j}\mathrm{E}[Y_1^iY_2^j]=\sum_{i\neq j}\mathrm{E}[Y_1^i]\mathrm{E}[Y_2^j]=(n^2-n)/36$. Therefore, it follows that $\operatorname{Cov}(X_1,X_2)=\mathrm{E}[X_1X_2]-\mathrm{E}[X_1]\mathrm{E}[X_2]=-n/36$.

A more general result is $Cov(X_i,X_j)=-np_ip_j$, where $p_i$ is the occurrance probability of the $i$-th outcome at a single trial.


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