Are there random variables that sum up to a Bernoulli random variable, analogous to the Poisson process?

Question

For $Y \sim \operatorname{Pois}(\mu)$, a Poisson random variable and $X_i \sim \operatorname{Pois}(\frac{\mu}{n})$, a sequence (in time) of independent and identically distributed RVs, it is well known that: $$Y \sim \sum_i^n X_i.$$

My question is whether there exist analogous random variables $X_i$ (I don't think they can be identically distributed nor independent), such that $$Y \sim \sum_i^n X_i,$$ when $Y \sim \operatorname{Bernoulli}(p)$.

I will try to use some eaxmples to demonstrate what I mean by analogous, apologising in advance for the vagueness, as I do not know what this random variable could possible look like.

1. Suppose $p$ is the probability that something happens in 10 minutes. I would like a random variable that distributes the probability mass as fairly as possible across each 2 minute period.
2. Suppose an urn contains one ball, and there is the probability $p$ of drawing this ball from the urn after some arbitrary process. I would like to simulate the process by attempting to draw the ball (using a random process) from the urn in $n$ attempts. When $n = 1$, it is clear that I can just flip a coin which is weighted such that heads turns up with probability $p$. What do I do when $n = 2, 3, \dots$? Note that the support of the random variable defined by this process has to be $\{0, 1\}$ as we cannot draw the ball from the urn when it has been drawn already. Finally, I would like the probability mass to be distributed as evenly across the $n$ draws. For example, the solution $X_1 \sim \operatorname{Bern}(p)$ and $X_i \sim \operatorname{Bern}(0)$, when $i >1$ is not acceptable.

Answer 1

So success is Bernoulli and time of success, conditioned on success occuring, is uniform?

The constraint of max 1 success would mean the $X_i$ cannot be independent. However they can easily be identically distributed.

One way to reproduce this is with $n$ urns and a coin:

• first roll an ($n$-sided) die to choose an urn $k$
• then toss a (possibly biased) coin, and place the ball in urn $k$ if it lands heads

The coin flip is a Bernoulli variable $b\sim\mathrm{Bern}(p)$, and the die roll is a discrete uniform variable $k\sim\mathrm{Unif}(n)$. The $X_i$ are then indicator variables $$X_i=\begin{cases}1 & \text{if ball in urn }i \\ 0 & \text{otherwise}\end{cases}$$ which can be expressed as $X_i=[k=i]b$, using Iverson bracket notation (i.e. $\vec{X}/b$ is a one-hot encoding of $k$).

So the $X_i\sim\mathrm{Bern}(p/n)$ are identically distributed, but not independent, as $Y=\sum_iX_i=b\sim\mathrm{Bern}(p)$. (In contrast, i.i.d. $X_i$ would give Binomial $Y\sim\mathrm{Binom}(n,p/n)$.)

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In the special case $p=\frac{n}{N}$, a simpler procedure is to always place the ball in one of $N$ urns, but only search the first $n$ urns.

That is, take $k\sim\mathrm{Unif}(N)$, then define $\vec{X}$ to be the one-hot encoding of $k$, and $\vec{Y}$ to be the prefix sum of $\vec{X}$, i.e. \begin{align} X_i &= [k=i] \\ Y_n &= \sum_{i=1}^nX_i \end{align} Then $X_i\sim\mathrm{Bern}(\frac{1}{N})$ and $Y_n\sim\mathrm{Bern}(\frac{n}{N})$.