How to prove the sum of n squared binomial probabilities does not increase as n increases

Question

Let $F\left( n \right) = \sum\limits_{k = 0}^n {{{\left( {C_n^k{p^k}{{\left( {1 - p} \right)}^{n - k}}} \right)}^2}} $, prove $F\left( n \right) \ge F\left( {n + 1} \right)$.

UPDATE: More general, denote $F\left( n \right) = \sum\limits_{k = 0}^n {C_n^kp_1^kq_1^{n - k}C_n^kp_2^kq_2^{n - k}}$, where ${q_1} = 1 - {p_1}$ and ${q_2} = 1 - {p_2}$, prove $F\left( n \right) \ge F\left( {n + 1} \right)$.

$\color{red}{^\bf{{\rm{New}}}}$ UPDATE: More general, denote $F\left( n \right) = \sum\limits_{k = 0}^n {C_n^kp_1^kq_1^{n - k}\sum\limits_{i = 0}^k {C_k^ip_2^iq_2^{k - i}C_{n - k}^{k - i}p_3^{k - i}q_3^{\left( {n - k} \right) - \left( {k - i} \right)}} }$, where $q_1=1-p_1$, $q_2=1-p_2$, and $q_3=1-p_3$, is it true that $F\left( n \right) \ge F\left( {n + 1} \right)$?

Answer 1

Denote $q=1-p$, write $[x^a]f(x)$ for a coefficient of $x^a$ in the polynomial (or Laurent polynomial) $f(x)$. We have $$F(n)=[1](px+q)^n(px^{-1}+q)^n=[1]((p^2+q^2)+qp(x+x^{-1}))^n=\\=(2\pi)^{-1}\int_{0}^{2\pi}((p^2+q^2)+qp(e^{it}+e^{-it}))^ndt.$$ The function $(p^2+q^2)+qp(e^{it}+e^{-it})=p^2+q^2+2pq\cos t$ is non-negative and its values do not exceed 1, thus the result.

I use the useful relation $[1]\varphi(x)=(2\pi)^{-1}\int_0^{2\pi} \varphi(e^{it})dt$ for any Laurent polynomial $\varphi$.

UPD. For the updated problem, use the same integral representation for $p=\sqrt{p_1p_2},q=\sqrt{q_1q_2} $. We need to check that $p+q\leqslant 1$, this follows from $p\leqslant (p_1+p_2)/2, $
$q\leqslant (q_1+q_2)/2. $

$$ \sum\limits_{i = 0}^k {C_k^ip_2^iq_2^{k - i}C_{n - k}^{k - i}p_3^{k - i}q_3^{\left( {n - k} \right) - \left( {k - i} \right)}}=[t^k](tp_2+q_2)^k(tp_3+q_3)^{n-k} $$

About updated function $F(n)$. Denoting $h(z)=z^{-1}p_1(zp_2+q_2)+q_1(zp_3+q_3)$ we have $F(n)=[1](h(z))^n$, thus $$ F(n)=\frac1{2\pi i}\int h(z)^ndz/z,$$ where integral is taken by any circuit around 0. It is natural to take the circuit with real value of $h(z)=p_1p_2+q_1q_3+q_1p_3 z+p_1q_2 z^{-1}$. The natural choice is the contour $z=\sqrt{\frac{p_1q_2}{q_1p_3}}e^{i\theta}$, we get $h(z)=p_1p_2+q_1q_3+2\sqrt{p_1q_1q_2p_3}\cos \theta$. We have $|h(z)|\leqslant p_1p_2+q_1q_3+p_1q_2+q_1p_3=p_1+q_1=1$, thus everything works nicely if we have $h(z)\geqslant 0$. This is not always the case. For example, if $p_2=q_3=0$ we get $h(z)=2\sqrt{p_1q_1}\cos \theta$ and the integral equals 0 for odd values of $n$ and is positive for even values of $n$. Needless to say, this could be seen from the very definition of $F(n)$, but for what it worth I leave here the general explanation.

Answer 2

$F(n)$ is the density maximum of the $n$th convolution power of a fixed (three point) probability density on $\mathbb Z$, and hence is decreasing in $n$.

To spell the first claim out, let $\ast$ indicate convolution products or powers of probability densities on $\mathbb Z$, let $\mathrm{b}_{n,p}$ denote the binomial density with the indicated parameters, $\mathrm{b}_p:= \mathrm{b}_{1,p}$ the Bernoulli density, and $q:=1-p$. Then $$ F(n) = \sum_{k=0}^n \mathrm{b}_{n,p}(k) \mathrm{b}_{n,q}(n-k) = \big(\mathrm{b}_{n,p}\ast \mathrm{b}_{n,q}\big)(n) = \big(\mathrm{b}_p \ast \mathrm{b}_q\big)^{\ast n}(n) = \left\| \big(\mathrm{b}_p \ast \mathrm{b}_q\big)^{\ast n} \right\|_\infty . $$ To check the final step above, one may use the fact that convolutions of symmetric lattice unimodal laws are again symmetric unimodal (Wintner's theorem).

Of course, this is not shorter than Fedor's proof.