Distribution of a product of Multinomials Consider the following: $(X_1, X_2, X_3, X_4) \sim \mathrm{Multinomial} (n,\mathbf{p})$ where $\mathbf{p} = (p_1,p_2,p_3,p_4)$.  I would like to find the distribution of $X_1 X_4$, or at least know some bounds on the variance of $X_1 X_4$.  I know that $E(X_1X_4) = n^2 p_1 p_4 - np_1 p_4 = n(np_1p_4 - p_1p_4)$.  Is $X_1X_4$ in any way 
 A: $$Var(X_1X_4)=\mathbb{E}[X_1X_4]^2-\left(\mathbb{E}[X_1X_4]\right)^2$$
and
$$\mathbb{E}[X_1X_4]=p_1p_4\times n(n-1)$$
$$\mathbb{E}[X_1X_4]^2=p_1p_4\times n\left[(n-1)+(p_1+p_4)(n^2-3n+2)+p_1p_4(n^3-6n^2+11n-6)\right]$$
A: $\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}\newcommand{cov}{\operatorname{cov}}$The multinomial with parameter $n$ is the sum of $n$ independent copies of the multinomial with parameter $n$.  Let's look first at $n=1$.  You have
$$
(Y_1,Y_2,Y_3,Y_3) = \begin{cases} 
(1,0,0,0) & \text{with probability }p_1, \\
(0,1,0,0) & \text{w.p. }p_2, \\
(0,0,1,0) & \text{w.p. }p_3, \\
(0,0,0,1) & \text{w.p. }p_4.
\end{cases}
$$
Then
$$
\cov(Y_3, Y_4) = \E(Y_3 Y_4) - (\E Y_3)(\E Y_4) = 0 - p_3 p_4.
$$
Therefore
$$
\cov(X_3,X_4) = -np_3 p_4.
$$
But
$$
\cov(X_3,X_4) = \E(X_3 X_4) - (\E X_2)(\E X_4)
$$
so
$$
-n p_3 p_4 = \E(X_3 X_4) - (np_3)(np_4),
$$
and so
$$
\E(X_3 X_4) = n^2 p_3 p_4 - np_3 p_4 = n(n-1)p_3 p_4.
$$
I'm not sure yet about bounds on the variance.  I'll be back${}\,\ldots$
