What is the moment generating function of a Gaussian distribution?

1 Answer
Bill K.
Dec 7, 2015

If #X# is Normal (Gaussian) with mean #mu# and standard deviation #sigma#, its moment generating function is: #m_{X}(t)=e^{mu t+(sigma^[2} t^{2})/2#.

Explanation:

Deriving this formula requires a calculation of the integral:

#m_{X}(t)=E[e^{tX}]=1/(sigma sqrt(2pi})\int_{-infty}^{infty} e^{tx}e^{-1/2(x-mu)^2/sigma^{2}}\ dt #.

This can be done by expanding #-1/2(x-mu)^2/sigma^{2}#, completing the square, and using properties of exponents, but the details are not very pleasant. It can also be done by a substitution #z=(x-mu)/sigma#. The details are found at the pdf file found at the following website: derivation of mgf of Normal

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