Let $x_1, \ldots, x_n$ be possibly dependent random variables, each taking values $x_i \in \{0, 1, 2\}$. Suppose further that in every outcome the number of random variables that equal 2 is exactly 1. Now for each $i \in \{1, \ldots, n\}$ define $$ f_i = \begin{cases} \Pr[x_i = 2 \mid x_i \geq 1] & \text{if } x_i \geq 1\\ 0 & \text{if } x_i =0 \end{cases}, $$ and let $ f = \sum_i f_i. $

My question is how large can the variance of $f$ be? My conjecture is that we should be able to bound it by $O(1)$ but don't know how to prove this.

Note: In case it helps, it is easy to prove that $E[f] = 1$: $$ E[f] = \sum_i E[f_i] = \sum_i \Pr[x_i \geq 1] \times \Pr[x_i = 2 \mid x_i \geq 1] = \sum_i \Pr[x_i = 2] = 1, $$ where the last equality comes from our initial assumption that in all outcomes exactly one of the $x_i$'s equals 2.


$Var\,f$ can be on the order of $n$ (but not more than that).

Indeed, let $U$ and $N$ be independent random variables such that $P(U=1)=:p=1-P(U=0)=:q$ and $P(N=i)=1/n$ for all $i\in[n]:=\{1,\dots,n\}$. Let $$x_i:=1(U=1,N\ne i)+2\times1(N=i). $$ Then with $p=1/n$ $$Var\,f\sim n/4\tag{1}$$ (as $n\to\infty$).

On the other hand, $$Var\,f\le Ef^2=\sum_{i,j\in[n]}Ef_if_j\le\sum_{i,j\in[n]}Ef_i =n\sum_{i\in[n]}Ef_i=n.$$

Details on (1): We have $$Ef^2=\sum_{i,j\in[n]}Ef_if_j \\ =\sum_{i,j\in[n]}P(x_i=2|x_i\ge1)P(x_j=2|x_j\ge1) P(x_i\ge1,x_j\ge1),\tag{2}$$ $$P(x_i\ge1)=1-P(x_i=0)=1-P(U=0)P(N\ne i)=1-q(1-1/n)=p+q/n,$$ $$P(x_i=2)=P(N=i)=1/n,$$ $$P(x_i=2|x_i\ge1)=\frac{P(x_i=2)}{P(x_i\ge1)}=\frac{1/n}{p+q/n},$$ and $$P(x_i\ge1,x_j\ge1)=1-P(x_i=0\text{ or }x_j=0)=1-P(x_i=0)-P(x_j=0)+P(x_i=0,x_j=0) =1-2q(1-1/n)+q(1-2/n)=1-q=p$$ for $i\ne j$. Choosing now $p=1/n$, we have
$$Ef^2\sim n/4.$$ Since $Ef=1$, (1) now follows.

Looking back at (2), now the idea behind the construction should become transparent: We want to make $P(x_i\ge1,x_j\ge1)$ for $i\ne j$ much greater than $P(x_i\ge1)P(x_j\ge1)$ and at the same time not to make $P(x_i\ge1,x_j\ge1)$ too small. The choice $p=1/n$ is nearly optimal in this regard.

  • $\begingroup$ Thank you for the prompt response and the excellent demonstration of the distribution Iosif! Now I understand why I wasn't able to prove my conjecture lol. $\endgroup$ – Mathman Aug 17 '20 at 1:36
  • $\begingroup$ Btw, I posted a related question (which was the main thing I was trying to prove via bounding the variance of $f$, which you refuted here). It would be great if you take a look at that one too if you have time. mathoverflow.net/questions/369355/… $\endgroup$ – Mathman Aug 17 '20 at 1:56

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