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Why is X2 | X1 binomial? I tried to explicitly calculate X2 | X1 but was unable to figure out the summation for the marginal of x1. Any help would be appreciated. Thanks.

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This question is not clear to me. I guess a lot of context is missing. For starters, what are $X_1$ and $X_2$ ? – Joel Cohen Oct 10 '11 at 0:26
@icobes You need to fill in a lot of the missing details before we can start to help you. Is this from a textbook? If so, it may help if you copy out the section that you don't understand. – Chris Taylor Oct 10 '11 at 0:35
I only know the joint pmf and what x1 and x2 vary from. x1 and x2 are simply the marginals and can be calculated from the joint pmf. – icobes Oct 10 '11 at 0:36
Actually I think there may be enough information here. – Michael Hardy Oct 10 '11 at 0:46
The question posed has x1 as an exponent. The answer is that X2 | X1 ~ binomial (x1, 0.5) – icobes Oct 10 '11 at 0:54
up vote 2 down vote accepted

$$ \Pr(X_1=x_1 \text{ and } X_2=x_2) = \binom{x_1}{x_2} 0.5^{x_1} \left(\frac{x_1}{15}\right) \text{ for }x_1\in\{1,2,3,4,5\},\ x_2\in\{1,\ldots,x_1\}. $$ (It's easy to see that the sum of these probabilities is 1.)

Then $$ \Pr(X_2=x_2 \mid X_1=x_1) = \frac{\Pr(X_1=x_1 \text{ and } X_2=x_2)}{\Pr(X_1=x_1)} = \frac{\binom{x_1}{x_2}0.5^{x_1}\left(\frac{x_1}{15}\right)}{C} = \frac{\binom{x_1}{x_2}0.5^{x_1} \cdot B}{C} $$ where $B$ and $C$ do not depend on $x_2$. It wouldn't be too hard at all to show that $B$ is actually equal to $C$, but we don't need that. The rules of probability imply that $B/C$ must be whatever constant it takes to make the sum over all values of $x_2$ equal to $1$. "Constant" in this case means: NOT DEPENDING ON $x_2$.

So there you have a binomial distribution with parameters $x_1$ and $0.5$.

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