Derive the moment generating function of the poisson distribution?

Answer 1

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