Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we have a large number $n$ of bins and throw $\lambda n$ balls into them independently with uniform probability then we expect $X_r$, the number of bins that end up with exactly $r$ balls is $$E(X_r)=\frac{\lambda^r}{r!}e^{-\lambda}.$$

With help from André Nicolas and Robert Israel I can compute the covariance matrix for these counts as

$$\Sigma = n(D - p'p - q'q)$$

where $p = p_0,p_1,\dots$, and $q = q_0,q_1,\dots$ where $$p_i = \frac{\lambda^i}{i!}e^{-\lambda},$$ and $$q_i = \frac{i-\lambda}{\sqrt{\lambda}}\frac{\lambda^i}{i!}e^{-\lambda},$$ and $D$ is the diagonal matrix made of the elements of $p$.

Is there a simple explanation for the simple form of $\Sigma$? It looks vaguely related to the covariance of a multinomial (which I'd expect) with an extra perturbation by $q'q$.

share|cite|improve this question
In the title of this post you have "of counts" twice; I don't think you mean that, do you? (If you do, there is some subtlety in this question that I'm missing.) – Michael Lugo Dec 2 '11 at 18:04
@MichaelLugo I am looking at the count of the number of of cells that have a specified number of balls, so one count for the balls, and one for the cells. – deinst Dec 2 '11 at 18:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.