If $X=\dbinom m Y \left(\dfrac 1 2\right)^Y \left(\dfrac 1 2\right)^{m-Y}$, and $Y$ has a Binomial distribution with parameters $m$ and $p$, What can we say about the distribution of $X$?
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1$\begingroup$ Please modify either your post or the title so that they address the same question: presently, they appear to have almost nothing in common. $\endgroup$– whuber ♦Commented Feb 27, 2017 at 19:55
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$\begingroup$ I took the liberty of changing "$Y$ is a binomial distribution" to "$Y$ has a binomial distribution." A random variable has a distribution. $\endgroup$– Michael HardyCommented Feb 28, 2017 at 2:18
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2$\begingroup$ Very closely related: Distribution of $\binom{n}{X}$. Indeed, since $\left(\dfrac 1 2\right)^Y \left(\dfrac 1 2\right)^{m-Y} = \left(\dfrac 1 2\right)^m$ is a constant, the answers there are simply a scaled version of the answer sought here. $\endgroup$– Dilip SarwateCommented Feb 28, 2017 at 5:37
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1$\begingroup$ I agree with Dilip - a scaling constant aside this is just a duplicate of an earlier question of yours. $\endgroup$– Glen_bCommented Feb 28, 2017 at 17:04
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1 Answer
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If $Y \sim \text{Binomial}(m,p)$, then its pmf is $f_Y(y;p) = {m \choose y} p^y(1-p)^{m-y}$. You are transforming $Y$ by applying this pmf to it, and you're using the value $p=1/2$. So $$ X = f_Y(Y;.5), $$ which means it has support or takes on values in the range of all possible probabilities for $Y$'s pmf when using $p=.5$. You could easily find this range for any specific $m$ by evaluating the pmf of $Y$ on the values $0,1,\ldots,m$.
To find the probabilities of each one of these support values $x$, notice that $$ P(X=x) = P(f(Y;.5) = x) = P(Y = f^{-1}(x;.5)) = {m \choose f^{-1}(x;.5)}p^{f^{-1}(x;.5)}(1-p)^{f^{-1}(x;.5)} $$ where $f^{-1}$ is the pre-image (the inverse might not always exist).
Edit: Thanks to @DilipSarwate and @jbownman for pointing the following out: Since we're using the parameter $.5$ to transform $Y$ into $X$ in this case, the transformation $$ f(y;.5) = {m \choose y} .5^y.5^{m-y} = {m \choose y} .5^m = {m \choose m-y} .5^m = f(m-y;.5) $$ is symmetric for $0 \le y \le m$.
This means \begin{align*} P(X=x) &= P(f(Y;.5) = x) \\ &= P(Y = k) + P(Y = m-k) \end{align*} as long as $k < m-k$ is such that $f(k;.5) = x$. This only doesn't happen when $m$ is even and $k=m/2$. In that case, there is only one $k$ such that $f(k;.5) = x$, which means \begin{align*} P(X=x) &= P(f(Y;.5) = x) \\ &= P(Y = k). \end{align*} This is the case that @jbownman mentioned: the case where the inverse exists.
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$\begingroup$ You could simplify this a little by noting that if the inverse exists, $P(X=x) = x$, and if it doesn't, $P(X=x) = x * (\text{# of Y for which }f(Y) = x)$. $\endgroup$– jbowmanCommented Feb 28, 2017 at 4:08
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$\begingroup$ This answer is incorrect because it does not take into consideration the relationship between $n$ and $m$. For example, if $n < m$, then $X=0$ if $Y \in \{0, n+1, n+2, \ldots , m\}$. $\endgroup$ Commented Feb 28, 2017 at 12:44
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$\begingroup$ @DilipSarwate the $n$ wasn't there when I answered this. Perhaps it's a mistake $\endgroup$– TaylorCommented Feb 28, 2017 at 14:48
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$\begingroup$ You are correct; the change from $m$ to $n$ in $\binom{n}{Y}$ occurred in Michael Hardy's edit. I have changed it back. Nonetheless, your answer still needs fixing a little (or more clarification) because $\binom{m}{k} = \binom{m}{m-k}$ and so the event $\{X=x\}$ is, in general, the union of two disjoint events $\{Y=k\}$ and $\{Y=m-k\}$, with a special case when $m$ is even and $k=\frac m2$. $\endgroup$ Commented Feb 28, 2017 at 15:11