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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.
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    $\begingroup$ Any random variable $X$ that takes on two values $a$ and $b$ can be expressed as $X = (a-b)Z + b$ where $Z$ is a Bernoulli random variable with parameter $p = P(X=a)$. The only solution for the problem you consider is the solution that you explicitly reject in the last sentence of your question (or its variant where $X_i$ takes on value $x_i$ wtht probability $1$.) $\endgroup$ – Dilip Sarwate Jul 18 '17 at 2:03
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    $\begingroup$ 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. $\endgroup$ – GeoMatt22 Jul 18 '17 at 2:13
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    $\begingroup$ If you take a collection of $n$ balls; $n-1$ labelled "0" and one labelled "Bernoulli(p)" (which will automatically generate a 1 with probability $p$ or a 0 with probability $1-p$, once it's drawn) and you draw the balls one-by-one without replacement, then the sum of the numbers on the balls will be Bernoulli($p$). Equivalently, roll an $n$ sided die (labelled 1 to $n$), call the result $J$. Then let $X_J$ be Bernoulli($p$) and all other $X$'s are $0$. $\endgroup$ – Glen_b Jul 18 '17 at 9:23
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    $\begingroup$ @Glen_b that is what I was thinking (my answer is less clear than your example though!). Another case would be if $p=n/N$, so you could place a ball into one of $N$ urns, but only search the first $n$, and no explicit Bernoulli draw is needed. $\endgroup$ – GeoMatt22 Jul 18 '17 at 22:28
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    $\begingroup$ In the second example the die roll is just an oracle which tells you (before you start drawing) which draw from the first example is to be the Bernoulli one. Note that before you draw any balls, the first, second, third ... draw all have the same chance to be the one that's Bernoulli($p$). So you can replace the sequence of draws with a die roll that tells you which draw that happens on. $\endgroup$ – Glen_b Jul 19 '17 at 0:02
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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 $Y\sim\mathrm{Binom}(n,p/n)$.)


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 $Y_n\sim\mathrm{Bern}(\frac{n}{N})$.

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  • $\begingroup$ Thank you for the answer, but I don't understand what $k = i$ means in terms of this uniform index $k$ $\endgroup$ – Alex Jul 18 '17 at 3:14
  • $\begingroup$ Does the one-hot explanation help? $\endgroup$ – GeoMatt22 Jul 18 '17 at 3:26
  • $\begingroup$ I am afraid not, would it be possible to flesh out an example in the case $n = 2$, or calculate the expected values of $X_1, X_2$ in this case? Does $k$ somehow encode which draw number from the urn, we are up to? $\endgroup$ – Alex Jul 18 '17 at 3:29
  • $\begingroup$ $k$ is the draw that succeeds, if one does. In Matlab I would do k=ceil(n*rand), b=(rand<p), i=1:n, X=b*(i==k), Y=sum(X), if that helps? $\endgroup$ – GeoMatt22 Jul 18 '17 at 3:44
  • $\begingroup$ ok, this is starting to make a bit more sense, especially the bit where each $X_i$ are now identically distributed. let me think about it. $\endgroup$ – Alex Jul 18 '17 at 3:45

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