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Original problem: A point X is randomly chosen from the interval (0,1). Suppose X=x is observed. Then a coin with P(Heads) = x is tossed independently n times. Let Y be the number of heads in n tosses. Find the unconditional distribution of Y.

I've setup the problem below using total probability and a uniform pdf of x (a confirmation others agree my formulation is correct would be nice), but not sure how to integrate this:

$\int_{0}^{1}$$\begin{pmatrix}n\\y\end{pmatrix}$$x^y(1-x)^{n-y}\,dx$

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2 Answers 2

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To approach this, I would apply the binomial theorem, which holds for non-negative integer $c$:

$$ (a+b)^c = \sum_{i=0}^c {c\choose i} a^ib^{c-i} $$

When you apply this identity to $(1-x)^{n-y}$, the integrand becomes a standard polynomial in $x$:

\begin{align*} Pr(Y=y) &= {n\choose y}\int_0^1 \sum_{i=0}^{n-y} {n-y\choose i}(-1)^ix^{i+y}~dx \\ &= {n\choose y}\sum_{i=0}^{n-y}{n-y\choose i}(-1)^i\frac{x^{i+y+1}}{i+y+1}\bigg|_0^1 \\ &= {n\choose y}\sum_{i=0}^{n-y}\frac{{n-y\choose i}(-1)^i}{i+y+1} \end{align*}

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  • $\begingroup$ wow nice....much appreciated -- i learned something new here! $\endgroup$
    – JPJ
    Jun 12, 2016 at 18:01
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    $\begingroup$ That also simplifies to $ \dfrac{1}{n+1}$ $\endgroup$
    – Henry
    Jun 12, 2016 at 23:59
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I just realized that the integral is an archetype of a beta distribution and thus can be manipulated as such...beta distribution

$\int_{0}^{1}\begin{pmatrix}n\\y\end{pmatrix}x^y(1-x)^{n-y}\,dx$ ~ $\frac {\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1} $$\Rightarrow$ which integrates to 1 over the support (0,1).

Thus, if $\alpha-1=y$ and $\beta-1=n-y$ $\implies$ $\begin{pmatrix}n\\y\end{pmatrix}$$\frac{\Gamma(y+1)\Gamma(n-y+1)}{\Gamma(n+2)} $

I'm sloppy so hope I didn't make an algebra mistake. It should (I hope!) come out to the same thing @josliber wrote, and he still gets the credit because he managed more creativity IMO.

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    $\begingroup$ $\displaystyle {n \choose y}\frac{\Gamma(y+1)\Gamma(n-y+1)}{\Gamma(n+2)}$ can be simplified to $\dfrac{1}{n+1}$ for integers $\endgroup$
    – Henry
    Jun 12, 2016 at 22:14

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