Let X be a random variable with m.g.f given by M ( t ) = ∑ j = 0 ∞ e ( t j − 1 ) j !

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Let X be a random variable with m.g.f given by  M  (  t  )  =      ∑          j      =      0              ∞                  e              (        t        j        −        1        )                    j      !

boomzwamhc · Accepted Answer

First, we simplify the MGF:  M  (  t  )  =      e          −      1            ∑          j      =      0        ∞              (              e        t                    )        j                    j      !        =      e          −      1        exp  ⁡  (      e    t    )  =  exp  ⁡  (      e    t    −  1  )which we identify as the MGF of a Poisson distribution with parameter   λ  =  1. Then,  P  (  X  =  2  )  =                    e                  −          λ                    ⋅              λ        2                    2      !        =      1          2      e

Brenden Tran · Accepted Answer

Here is another solution that works for any probability measure concentrated on             Z        +    =  {  0  }  ∪      N  Recall that for any r.v. taking values on             Z        +    M  (  t  )  =  E  [      e          t      X        ]  =      ∑    n    P  [  X  =  n  ]      e          n      t      Thus, for each n, the coefficient of       e          n      t       is   P  [  X  =  n  ] In your case, the coefficient of       e          2      t       is       e          −      1            1          2      !      , thus,   P  [  X  =  2  ]  =            e              −        1                    2      !

Let X be a random variable with m.g.f given by M ( t ) = <munderover> ∑<!--

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