# Expected value of expression involving multinomial random variables and Heaviside step function of said variables

Question 1

I have the following expression:

$H/S$,

where $H=\begin{cases}0,& \text{ if } rm\leq dj \\ 1, & \text{ if } rm>dj \end{cases}$ and $S=n-j$, and both $m$ and $j$ are drawn from the same multinomial distribution with number of trials $n$. The support of the multinomial distribution includes three discrete random variables: $m$, $d=S-m$, and $j$, which occur respectively with probabilities $x$, $y$, and $z=1-x-y$. I want to take the expected value of this expression. I know that $E[tu]\neq E[t]E[u]$ in general.

So my first question is, how do I compute the expected value of $H/S$, i.e., $E[H/S]$?

Question 2

In a related question, what is the expected value of $H$? Is it $E[H]=\begin{cases}0,& \text{ if } rE[m]\leq dE[j] \\ 1, & \text{ if } rE[m]>dE[j] \end{cases}$?

Thank you, CV.

(Also, that is the most $\LaTeX$ I've written in a single document so far. Go me!)

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 Did you mean $H$ to be $E[H]$ in the very last occurrence of it in your post? Or did you intend something else? – cardinal Dec 7 '11 at 1:20 @cardinal You mean when I say "...is it..." etc? If so, yes, thanks! – Brash Equilibrium Dec 7 '11 at 1:34 @cardinal BTW, I edited it. – Brash Equilibrium Dec 7 '11 at 1:46 I doubt that the standard sum (as per the definition of expected value) can be simplified into something short like $np$ for a binomial random variable. Of course, if you have numerical values for $n$ etc, then various statistical programming packages will make short work of it, but I suspect you want something short and sweet instead of $E[H/S] = \text{humonguous sum}$. – Dilip Sarwate Dec 7 '11 at 3:19 @DilipSarwate I'm happy to take the answer as a humongous sum. I am passing these expected values for $H/S$ and $H$ as part of a payoff function in an evolutionary game, which I will simulate. I am extending a game for which there is a simple(ish) solution (not short and sweet, but not a humongous sum). Only in that case, there weren't any discontinuous functions like those in $H$. – Brash Equilibrium Dec 7 '11 at 5:11

If the random variable $H = H(X,Y,Z)$ is a function of three integer-valued random variables $X, Y$, and $Z$ that take on values in $[0, n]$ and whose joint probability mass function is $$p_{X,Y,Z}(n_1,n_2,n_3) = P\{X = n_1, Y = n_2, Z = n_3\}, ~0 \leq n_1,n_2,n_3 \leq n,$$ then $$E[H] = \sum_{n_1=0}^n\,\sum_{n_2=0}^n\,\sum_{n_3=0}^nH(n_1,n_2,n_3)p_{X,Y,Z}(n_1,n_2,n_3).$$ It is not necessary that $H$ be expressible as a "nice" formula such as $X+Y+Z$ in order to use the above formula.

Here, $$p_{X,Y,Z}(n_1,n_2,n_3) = \begin{cases}\dfrac{n!}{n_1!n_2!n_3!}x^{n_1}y^{n_2}(1-x-y)^{n_3}, & \text{if}~ n_1 + n_2 + n_3 = n,\\ \quad\\ 0, & \text{otherwise},\end{cases}$$ is a $(n+1)\times(n+1)\times(n+1)$ array with lots of zeroes in it, as is $H(n_1,n_2,n_3)$ which is an array of zeroes and ones. Thus, $E[H]$ is actually the probability of an event.

Similarly, $H(n_1,n_2,n_3)/S(n_1,n_2,n_3)$ is an array with lots of zeroes in it but the nonzero entries are $1/S(n_1,n_2,n_3)$ and $$E\left [\frac{H}{S}\right] = \sum_{n_1=0}^n\,\sum_{n_2=0}^n\,\sum_{n_3=0}^n \frac{H(n_1,n_2,n_3)}{S(n_1,n_2,n_3)}p_{X,Y,Z}(n_1,n_2,n_3).$$

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 I'll review this soon. Thanks. Blasted iPhone cuts off the margins of SO web pages. – Brash Equilibrium Dec 7 '11 at 15:49 Thanks! Very useful so far as I can tell. I suspected Otto be the case that $H$ was the probability of an event. Is it the probability that $rm > dj$? And as for the expected $H/S$, it reduces to the sum of $1/S(...)$ over those cases when $rm>dj$ times the probability of that case? – Brash Equilibrium Dec 7 '11 at 17:30