Using generating functions one can see that the n t h 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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Using generating functions one can see that the       n          t      h       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?

svirajueh · Accepted Answer

1.       B    n    =      ∑          k      =      0        n        {          n      k        }  , where       {          n      k        }   is a Stirling number of the second kind. (The number       {          n      k        }   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    =      ∑          k      =      0        n        x            k      _            {          n      k        }  , where       x            n      _        =  x  (  x  −  1  )  (  x  −  2  )  ⋯  (  x  −  n  +  1  ).3. The factorial moments of a Poisson  (  1  ) distribution are all   1; i.e.,   E  [      X            n      _        ]  =  1.  E  [      X    n    ]  =      ∑          k      =      0        n    E  [      X            k      _        ]      {          n      k        }    =      ∑          k      =      0        n        {          n      k        }    =      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            n      _        ]  =  E  [  X  (  X  −  1  )  (  X  −  2  )  ⋯  (  X  −  n  +  1  )  ]  =      ∑          x      =      0          ∞    x  (  x  −  1  )  ⋯  (  x  −  n  +  1  )            e              −        1                    x      !        =      ∑          x      =      n          ∞    x  (  x  −  1  )  ⋯  (  x  −  n  +  1  )            e              −        1                    x      !        =      ∑          x      =      n          ∞              x      !              (      x      −      n      )      !                  e              −        1                    x      !        =      ∑          y      =      0          ∞              e              −        1                    y      !        =  e  /  e  =  1.

Using generating functions one can see that the n t h </mrow>

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