Bell numbers and also minutes of the Poisson circulation

Using creating features one can see that the $n^{th}$ Bell number, i.e., the variety of all feasible dividings of a set of $n$ components, amounts to $E(X^n)$ where $X$ is a Poisson arbitrary variable with mean 1. Exists a means to clarify this link with ease?

2019-05-18 22:57:21
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Answers: 1

One means might be to make use of these realities. You can determine if this is instinctive sufficient or otherwise.

  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 variety of means to dividing a set of $n$ components right into $k$ collections.)

  2. Stirling varieties of the 2nd kind are made use of to transform average powers to dropping powers using $x^n = \sum_{k=0}^n x^{\underline{k}} \left\{n \atop k \right\}$, where $x^{\underline n} = x(x-1)(x-2) \cdots (x-n+1)$.

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

Placing them with each other generates

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

Facts 1 and also 2 are well - well-known buildings of the Bell and also Stirling numbers. Below is a fast evidence of # 3. The 2nd action is the definition of anticipated value, making use of the Poisson probability mass function. The 2nd - to - last action is the Maclaurin collection development for $e^x$ reviewed at $1$.

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

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

2019-05-21 07:17:02