Suppose $X$ and $Y$ are independent. $X$ is binomial based on $m$ trials and success probability $\theta$. $Y$ is binomial based on n trials and success probability
(i) What is the conditional probability mass function of $X$ given that $X+Y =s$?
Confusion in Calculating Conditional Probability mass function
How do they calculate that total probability in the denom? I have this as the numerator:
\begin{equation} \left[nCx \text{ }\theta^x (1-\theta^{n-x})\right] \left[ mC(s-x) \text{ } \theta^{s-x} (1-\theta^{m - s-x}\right] \end{equation}
Should I follow suit and break apart the denominator in a similar way using i and (s - i)? When I do that I can't tell if something is supposed to cancel or what?
Any hints for whats to come below are appreciated
(ii) What is $E(X|X + Y )$?
(iii) What is $Var(X|X + Y)$?
$\endgroup$ 21 Answer
$\begingroup$A way to simplify the denominator is to recognize that $X+Y \sim \text{Binomial}(m+n, \theta)$. So the denominator can be written as $\binom{n+m}{s} \theta^x (1-\theta)^{n+m-s}$. In the linked question they write as a sum by iterating over the events $\{X=i, Y=s-i\}$, which is equivalent but hard to simplify.
In the end you should get $$P(X=x \mid X+Y=s) = \frac{\binom{m}{x} \binom{n}{s-x}}{\binom{n+m}{s}},$$i.e. the conditional distribution is a hypergeometric distribution. You can then cite/refer to derivations of the mean/variance for that distribution.
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