We are given that $(X,Y)$ are in a trinomial distribution. Find the conditional probability mass function of $X$ given $Y = y$ and the $E(X | Y = y)$.
Attempt at a solution:
First note that the trinomial distribution is going to have a probability mass function of $\frac{n!}{x!y!(n-x-y)!} p^xq^y(1-p-q)^{n-x-y}$ and thus we can easily find that the marginal probability mass function for Y is given by $P(Y = y) = \frac{n!}{y!(n-y)!}q^y(1-q)^{n-y}$. Then combining these two we find that the conditional probability mass function of $X$ gien that $Y = y$ will be determined by:
$\frac{(n-y)!}{x!(n-x-y)!} p^xq^{y-n}(1-p-q)^{n-x-y}$ However, this doesn't seem to follow along a normal distribution as I expected it to. Would the conditional probability be something different in this case then?
As for the expected value, how can you simplify the sum of $x$ times that from $y = 0$ to $n-x$.
Thanks