Suppose that in a store, customers arrive as a Poisson process with rate $1/\mathrm{min}$ between time $0$ and $10$ minutes. Suppose there are ten kinds of items in the store (each kind has infinite supply) and each customer independently chooses one item uniformly at random and buy it. Let $M$ be the number of distinct kinds of items that are purchased up to time $10$. What is the expectation and variance of $M$?

n.b.: This is not homework; just review.

For the expectation, I let $N$ (which follows $\mathrm{Pois}(10)$) be the number of customers arriving in the first ten minutes, and I found that $\mathbb{E}(M|N=k)=10(1-(9/10)^k)$ so then $\mathbb{E}M=\sum_k \frac{e^{-10}10^k}{k!}(10(1-(9/10)^k)$. But I'm not sure that this yields the correct answer.

Any help would be appreciated. Thanks.

  • $\begingroup$ So you know the waiting time between consecutive events has an exponential distribution. If you know that and $\gamma$, I think that gives you everything else. $\endgroup$
    – user99680
    Jun 6, 2014 at 1:16

1 Answer 1


Let $N_j$ be the number of demands for type $j$. Since there is a probability 0.1 that a demand is type $j$, the total demand rate is split into 10 $independent$ Poisson flows of rate 1 each. Let
$$M=I_1+...+I_{10}$$ where $I_j:=1$ if $N_j>0$ and $:=0$ otherwise. From the Poisson, $P(N_j=0)=e^{-1}$. Due to independence, M is Binomial with $n=10$ and $p=1-e^{-1}$. The mean and variance of $M$ are $np$ and $np(1-p).$

  • $\begingroup$ Wow, thank you. I actually solved the problem myself earlier in a different way, but this is a really clever solution. Thanks again. $\endgroup$
    – N4v
    Jun 7, 2014 at 12:39

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