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Using generating functions one can see that the $n^{th}$ Bell number, i.e., the number of all possible partitions of a set of $n$ elements, is equal to $E(X^n)$ where $X$ is a Poisson random variable with mean 1. Is there a way to explain this connection intuitively?

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One way may be to use these facts. You can decide if this is intuitive enough or not. :)

  1. $B_n = \sum_{k=0}^n \left\{n \atop k \right\}$, where $\left\{n \atop k \right\}$ is a Stirling number of the second kind. (The number $\left\{ n \atop k \right\}$ counts the number of ways to partition a set of $n$ elements into $k$ sets.)

  2. Stirling numbers of the second kind are used to convert ordinary powers to falling powers via $x^n = \sum_{k=0}^n x^{\underline{k}} \left\{n \atop k \right\}$, where $x^{\underline n} = x(x-1)(x-2) \cdots (x-n+1)$.

  3. The factorial moments of a Poisson$(1)$ distribution are all $1$; i.e., $E[X^{\underline{n}}] = 1$.

Putting them together yields

$$E[X^n] = \sum_{k=0}^n E[X^{\underline{k}}] \left\{n \atop k \right\} = \sum_{k=0}^n \left\{n \atop k \right\} = B_n.$$

Facts 1 and 2 are well-known properties of the Bell and Stirling numbers. Here is a quick proof of #3. The second step is the definition of expected value, using the Poisson probability mass function. The second-to-last step is the Maclaurin series expansion for $e^x$ evaluated at $1$.

$$E[X^{\underline{n}}] = E[X(X-1)(X-2) \cdots (X-n+1)] = \sum_{x=0}^{\infty} x(x-1) \cdots (x-n+1) \frac{e^{-1}}{x!}$$

$$= \sum_{x=n}^{\infty} x(x-1) \cdots (x-n+1) \frac{e^{-1}}{x!} = \sum_{x=n}^{\infty} \frac{x!}{(x-n)!} \frac{e^{-1}}{x!} = \sum_{y=0}^{\infty} \frac{e^{-1}}{y!} = e/e = 1.$$

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  • $\begingroup$ Very interesting argument . $\endgroup$ – Arin Chaudhuri Oct 8 '10 at 21:33

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