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Let $X$ and $Y$ be iid with distribution $\text{Bin}(n,p)$. I would like to show $P(X=Y)$ is minimzed when $p=1/2$.

$$P(X=Y)=\sum_{i=0}^n P(X=i)P(Y=i)=(1-p)^{2 n} \, _2F_1\left(-n,-n;1;\frac{p^2}{(p-1)^2}\right)$$ ($_2F_1$ is the Hypergeometric Function)

I'm wondering how to show that such a complicated function is minimzed at $p=1/2$ or if there is an all-around simpler approach for this problem.

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  • $\begingroup$ Very nice question. If we let $f(p)=P(X=Y)$, it is clear that $f(p)=f(1-p)$, and the problem would be solved if one could show $f$ was convex. $\endgroup$ Mar 22, 2022 at 1:42

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The characteristic function (CF) of $X-Y$ has a tractable form: $$\begin{aligned}\varphi_{n,p}(a)&=E[e^{ia(X-Y)}]=\\ &=E[e^{iaX}]E[e^{-iaY}]=\\ &=[(q+pe^{ia})(q+pe^{-ia})]^n=\\ &=(q^2+2qp\cos(a)+p^2)^n=\\ &=(\varphi_{1,p}(a))^n\end{aligned}$$ where $q=1-p$. We want to minimize (wrt $p$) the integral $$P_{n,p}(X=Y)=\frac{1}{2\pi}\int_{(-\pi,\pi]}\varphi_{n,p}(a)da$$ It remains to argue that $\varphi_{n,1/2}\leq \varphi_{n,p}$. Indeed: $$\varphi_{n,1/2}(a)=(\varphi_{1,1/2}(a))^n\leq (\varphi_{1,p}(a))^n=\varphi_{n,p}(a),\,a \in (-\pi,\pi]$$

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  • $\begingroup$ Thanks for this! Why is the integral from $(-\pi, \pi]$ and not over $\mathbb{R}$ $\endgroup$ Mar 22, 2022 at 16:04
  • $\begingroup$ @user2757771 You're welcome. In this case, the CF is a DTFT $\endgroup$
    – Snoop
    Mar 22, 2022 at 16:08
  • $\begingroup$ Right, understood $\endgroup$ Mar 22, 2022 at 17:56

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