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
Best Answer
Your first two expressions are correct. Then
\begin{align} \textsf{Pr}(X=x\mid Y=y) &=\frac{\textsf{Pr}(X=x\land Y=y)}{\textsf{Pr}(Y=y)}\\ &=\frac{\frac{n!}{x!y!(n-x-y)!} p^xq^y(1-p-q)^{n-x-y}}{\frac{n!}{y!(n-y)!}q^y(1-q)^{n-y}}\\ &=\frac{(n-y)!}{x!(n-x-y)!}\frac{p^x(1-p-q)^{n-x-y}}{(1-q)^{n-y}}\\ &=\binom{n-y}x\left(\frac p{1-q}\right)^x\left(\frac{1-p-q}{1-q}\right)^{n-y-x}\;, \end{align}
which is the binomial distribution for $n-y$ events known not to be events of type $Y$.