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Derive the moment generating function of the poisson distribution?
Wiki User
∙ 11y ago
The probability mass function (pmf, you should know this) of the Poisson distribution is
p(x)=((e-λ)*λx)/(x!), where x= 0, 1, ........
Then you take the expected value of exp(tx), you should always keep in mind to find the moment generating function (mgf) you must always do
(etx)*p(x), where t is a random variable
Therefore,
(etx)*((e-λ*λx)/(x!))
(e-λ)*sum[(e-λ*λx)/(x!)]
Thee-λ is only a constant; thus, it can be pulled out of the sums.
Continuing,
(e-λ)*sum[(λ*et)x)/x!]
Let y=λ*et
(e-λ)*sum[(y)x/x!]
By Macalurins series, the sum[(yx)/x! ]= ey
Soonwards
(ey)*(e-λ)
Lets return the y by λ*et
Wiki User
∙ 11y ago
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