Consider $N$ balls randomly distributed in $M$ boxes with $N\gg M$ and probability $ p_i$ of each individual ball going into box $i$, where each box can contain any number of balls. Let $X_i$ be the occupancy of box $i$. It is clear that $E(X_i) = N p_i$. I thought it was reasonable to assume that $X_i$ are independent and follow a Poisson distribution, so $\sigma^2(X_i) = E(X_i) = N p_i$. However if I define the new random variable $Y_i = \sum_{j=1}^i X_j$ then $\sigma^2(Y_M) = \sum_{j=1}^M \sigma^2(X_j) = \sum_{j=1}^M E(X_j) = N\sum_{j=1}^M p_j$ which is wrong because $Y_M$ is always exactly equal to $N$ and its variance should be zero. So, what are the correct distributions for $X_i$ and $Y_i$?


If the location of each ball is independent of the other balls then each $X_i$ has a binomial distribution $\operatorname{Bin}\left(N,p_i\right)$ and so mean $Np_i$ and variance $Np_i(1-p_i)$

But $X_i$ is not independent of $X_j$: for example they cannot both take the value $N$ when $i \not = j$

If $\displaystyle Y_i = \sum_{j=1}^i X_j$ and letting $\displaystyle q_i=\sum_{j=1}^i p_j$, then each $Y_i$ has a binomial distribution $\displaystyle \operatorname{Bin}\left(N,q_i\right)$ and so mean $Nq_i$ and variance $Nq_i(1-q_i)$

In particular $\displaystyle q_M^{\,} = \sum_{j=1}^M p_j = 1$ so $Y_i$ has mean $N\times 1 =N$ and variance $N\times 1 \times 0=0$ as expected

  • $\begingroup$ Thank you Henry, I actually was searching for the distribution of $Y_i$ but now I'm curious about the distribution of $X_i$ too, is it harder to find? $\endgroup$ – Manuel Aug 26 '17 at 23:52
  • $\begingroup$ @Manuel: My answer says that each $X_i$ and $Y_i$ is binomially distributed $\endgroup$ – Henry Aug 27 '17 at 10:12
  • $\begingroup$ Right, I was confused about the independence of the location of balls. One more thing, what would be the covariance between $Y_i$ and $Y_j$? $\endgroup$ – Manuel Aug 27 '17 at 20:55
  • $\begingroup$ Assuming $i \le j$, it would be $Nq_i(1-q_j)$ $\endgroup$ – Henry Aug 27 '17 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.