Consider the following: $$ T= \binom{Y} {n/2} \left(\frac{1}{2}\right)^{Y}\left(\frac{1}{2}\right)^{Y-\frac{n}{2}},$$ where $Y$ is a Binomial random variable with sample size $n$ and probability $p$. I noticed that this looks like the Binomial pmf, is it right to consider $T$ as the pmf of $X \mid Y$ where $X$ follows a $B\left(Y,\frac{1}{2}\right)$?
Conditional Binomial variable
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probability
binomial-distribution
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0No, if $y$ is known, so is $T$,and there is no distribution. And the unconditional distribution of $X$ has little to do with a Binomial. – 2017-02-19
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0@YvesDaoust, I edited my question, what about now? – 2017-02-19
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0No significant change. – 2017-02-20
1 Answers
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If $X\mid Y~\sim~\mathcal{Bin}(Y, \tfrac 12)$, then $$\mathsf P(X{=}n/2\mid Y) ~=~ \dbinom {Y}{n/2}\,\left(\dfrac 12\right)^Y ~\mathbf 1_{n\in 2\Bbb N_y}~\mathbf 1_{Y\in \Bbb N}$$