# Independent sampling of dependent random variables

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

• 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. – Ron P Aug 17 '20 at 11:32
• @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 $f_i$. – Mathman Aug 17 '20 at 13:00

## 1 Answer

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.

• This is excellent Ron. Thank you! – Mathman Aug 17 '20 at 15:08