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If I have a binomial $X \sim B(n,p)$, and another binomial $X' \sim B(n,p)$ conditioned on $X'$ being of even parity. Is it true that there always exists a coupling for $(X,X')$ with $|X-X'| \le 1$? (i.e. for any $n$ and $p := p(n)$ possibly a function of $n$.)

It seems intuitively obvious; is there a clean proof?

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    $\begingroup$ The case $p=1/2$ is easy: first sample $X$ and if its even set $X'=X$ and if it is odd set $X' = X\pm 1$ with a minus sign chosen independently with probability $X/n$. $\endgroup$ Dec 16 '20 at 16:47
  • $\begingroup$ That's super nice, thanks! I'll work on trying something similar for other $p$. $\endgroup$
    – DJA
    Dec 16 '20 at 19:24
  • $\begingroup$ in other words, $X=\xi_1+\ldots+\xi_n$, $X'=\xi_1+\ldots+\xi_{n-1}+(\xi_1+\ldots+\xi_{n-1} \pmod 2)$ for i.i.d 1/2-Bernoulli $\xi_i$'s. $\endgroup$ Dec 16 '20 at 20:04
  • $\begingroup$ This is a super special symmetry, though. I'm finding the case of $p \neq 1/2$ much more challenging. $\endgroup$
    – DJA
    Dec 16 '20 at 20:19
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This is possible for all $n$ and $p$.

I start with a direct construction.

Obviously, if $X$ is even, then we should have $X'=X$. So we should construct the corresponding coupling between $Y$ and $X'$, where $Y$ is the $B(n,p)$ restricted to odd outcomes.

Choose $2n$ i.i.d. Bernoulli$(p)$ variables $\xi_1,\ldots,\xi_n;\eta_1,\ldots,\eta_n$ and condition to $$\sum (\xi_i+\eta_i) \quad \text{is odd}.$$

Denote by $\Omega$ the set of possible $2^{2n-1}$ outcomes and consider the map $\Phi:\Omega\to \Omega$: choose the minimal $i$ for which $\xi_i\ne \eta_i$ and switch $\xi_i$ and $\eta_i$. This is a measure-preserving involution. Note that $\Phi$ changes the parity of $S=\eta_1+\ldots+\eta_n$, so $S$ is even with probability $1/2$. Next, if we further condition to ($S$ is even), then $S$ becomes distributed as $X'$. Indeed, this clearly holds even we fix all $\xi_i$'s (with odd sum). Analogously, if $S$ is odd, it is distributed as $Y$.

Now our coupling: choose $\omega\in \Omega$ at random, set $\{X',Y\}=\{S(\omega), S(\Phi(\omega)\}$.


Well, now goes a boring explanation how to get this coupling using generating functions.

Let $c_0,c_2,\ldots$ be probabilities of outcomes $0,2,\ldots$ for $X'$, we have $c_0+c_2x^2+\ldots=\frac{(q+px)^n+(q-px)^n}{1+\delta^n}$, $\delta:=q-p$ (and $q=1-p$). Denote the probabilities for $Y$ by $c_1,c_3,\ldots$, then $c_1+c_3x^2+\ldots=\frac{(q+px)^n-(q-px)^n}{1-\delta^n}$.

How may our coupling between $Y$ and $X$ look like? There is no freedom: if $Y=1$, then $X'\in \{0,2\}$ with probabilities corr. $c_0$ and $c_1-c_0$ (these are not conditional probabilities, I mean, $c_1-c_0={\rm prob}(Y=1,X'=2)$ etc.) If $Y=3$, then $X'\in \{2,4\}$ with probabilities $c_2-c_1+c_0$ and $c_3-c_2+c_1-c_0$, etc. Thus what we need is that all alternating sums $c_k-c_{k-1}+c_{k-2}-\ldots$ must be non-negative, or: all coefficients of $$ F(x):=(c_0+c_1x+c_2x^2+\ldots)(1-x+x^2+\ldots) $$ must be non-negative. We have $$ F(x)=2\frac{(q+px)^n-\delta^n(q-px)^n}{(1+x)(1-\delta^{2n})}= 2\frac{((q+p)(q+px))^n-((q-p)(q-px))^n}{(1+x)(1-\delta^{2n})}=\\ 2\frac{((q^2+p^2x)+pq(1+x))^n-((q^2+p^2x)-pq(1+x))^n}{(1+x)(1-\delta^{2n})}, $$ and expanding $((q^2+p^2x)\pm pq(1+x))^n$ by Binomial we see that $F(x)$ is indeed a polynomial with non-negative coefficients.

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  • $\begingroup$ thanks so much! $\endgroup$
    – DJA
    Dec 16 '20 at 22:06
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Your conjecture is true. Indeed, \begin{equation*} P_k:=P(X=k),\quad P'_{2j}:=P(X'=2j)=P_{2j}/Q,\tag{0} \end{equation*} where $k$ and $j$ are integers, and \begin{equation*} Q:=\sum_j P_{2j}. \end{equation*} Here we need to assume that $n$ is even or $p<1$, in order to have $Q>0$.

It suffices to show that for all integers $j$ there are numbers $m_j\in[0,P_{2j-1}]$ such that \begin{equation*} P'_{2j}=P_{2j}+m_j+(P_{2j+1}-m_{j+1}) \tag{1} \end{equation*} for all $j$; here, $m_j$ is the probability mass to be transported from $2j-1$ forward to $2j$, with the remaining probability mass $P_{2j-1}-m_j$ to be transported from $2j-1$ backward to $2j-2$.

If one wants to avoid mass transportation language, here is an explicit description of the desired joint distribution of $X$ and $X'$: $$P(X'=2j,X=k)= \begin{cases} P_{2j}&\text{ if }k=2j, \\ m_j&\text{ if }k=2j-1, \\ P_{2j+1}-m_{j+1}&\text{ if }k=2j+1, \\ 0&\text{ otherwise. } \end{cases}$$ for all integers $j,k$. This and (1) will indeed imply that $P(X'=2j)=P'_{2j}$ and $P(X=k)=P_k$ for all integers $j$ and $k$, as well as $P(|X'-X|\le1)=1$, as desired.

In view of (0), rewrite (1) as \begin{equation*} m_{j+1}-m_j=P_{2j}(1-1/Q)+P_{2j+1}, \end{equation*} with the initial condition $m_{-\infty}:=\lim_{j\to-\infty}m_j=0$. So, (1) can be rewritten as \begin{equation*} m_j=U_j:=\sum_{i=-\infty}^{j-1}u_i,\quad\text{where}\quad u_i:=P_{2i}(1-1/Q)+P_{2i+1}. \end{equation*} Also, $P_{2j-1}=\sum_{i=-\infty}^{j-1}(P_{2i+1}-P_{2i-1})$. So, the conditions $m_j\in[0,P_{2j-1}]$ for all integers $j$ can be rewritten as \begin{equation*} U_j\ge0\ge L_j:=\sum_{i=-\infty}^{j-1} l_i,\quad\text{where}\quad l_i:=P_{2i}(1-1/Q)+P_{2i-1}. \tag{2} \end{equation*}

Next, $U_{-\infty}=L_{-\infty}=U_{\infty}=L_{\infty}=0$. Also, the sequence $(P_k)$ is log-concave: $P_k^2\ge P_{k-1}P_{k+1}$ for all integers $k$.

So, the $u_i$'s can change the sign at most once as $i$ increases, and only from $+$ to $-$. That is, the $U_j$'s can change only from increase to decrease as $j$ increases. Since $U_{-\infty}=0=U_{\infty}$, it follows that indeed $U_j\ge0$ for all integers $j$. Similarly, $L_j\le0$ for all integers $j$. So, the inequalities (2) are proved. $\Box$

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  • $\begingroup$ thank you! this is great. $\endgroup$
    – DJA
    Dec 16 '20 at 22:06
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    $\begingroup$ It is now made clear that the reasoning holds for any distribution on $\mathbb Z$ with a log-concave density (with respect to the counting measure), provided that the probability mass of $2\mathbb Z$ is nonzero. $\endgroup$ Dec 17 '20 at 14:38

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