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 for each $i \in \{1, \ldots, n\}$ let $y_i$ be a Bernoulli random variable that is 1 independently with probability $f_i$ and 0 otherwise.

Is the following conjecture correct or is there a distribution on $x_i$'s refuting it?

Conjecture: There is a fixed $\epsilon > 0$ (i.e. $\epsilon$ being independent of $n$) such that with probability at least $\epsilon$, there is exactly one index $i$ where $y_i = 1$.

Related question: Bounds on variance of sum of dependent random variables

  • $\begingroup$ A clarifying question. Is the following description correct: first the $f_i$s are sampled, and then the $y_i$s are independent conditioned on the $f_i$s. $\endgroup$
    – Ron P
    Commented Aug 17, 2020 at 11:32
  • $\begingroup$ @RonP Correct. The $f_i$'s are first drawn according to the distribution of $x_i$'s and are not independent. Then each $y_i$ is drawn independently according to probability $f_i$. $\endgroup$
    – Mathman
    Commented Aug 17, 2020 at 13:00

1 Answer 1


The answer is "no" (if i understand the question correctly).

Consider the following exchangeble joint distribution of the $x_i$s. In event $A$, which occur with probabiluty $1/\sqrt n$, all the $x_i$s are 1, exept for one 2. In the complement event $B$, all the $x_i$s are 0 exept for one 2.

Under this distribution, $f_i$ is either 0 or $1/\sqrt n$. Let $Y=\sum y_i$. Since $E[ Y|A]=\sqrt n$, and $E[Y|B]=1/\sqrt n$, in either event it is too far from 1; therefore the probability that there is exaxtly one positive $b_i$ is vanishing.

  • $\begingroup$ This is excellent Ron. Thank you! $\endgroup$
    – Mathman
    Commented Aug 17, 2020 at 15:08

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