Thread: Deriving MGF of Gaussian distribution
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Old December 30th, 2008, 04:37 PM
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Quote:
Originally Posted by Eight^squared View Post
Hi there,
I'm revising at the moment for a stats exam in about a week's time and I'm having trouble deriving the MGF of the Normal distribution.

I know that i should get:
\exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right)

but i just can't seem to get my integration to work.

Any help would be greatly appreciated!
The easiest method to derive the moment-generating function of a general normal distribution N(\mu, \sigma^2) is to find the moment for a standard normal N(0,1) and then use the formula for the linear transformation of a moment. Given (\Omega, F, P), we have a probability space.

Lemma: Let X: \Omega \longrightarrow R be an absolutely continuous random variable whose moment-generating function is M_x (s). Then if Y = aX +b, then M_y (s) = e^{sb} M_x (sa)

Proof: Let a second random variable Y = aX + b. Then the moment generating function for Y is M_y (s) = E(e^{sY})
= E(e^{s(aX+b)})
= E(e^{saX+sb})
= E(e^{saX} e^{sb})
= e^{sb} E(e^{saX})
= e^{sb} M_x (sa)

Therefore, if Y = aX +b, then M_y (s) = e^{sb} M_x (sa) - this completes the lemma.

Consider X, which is standard normally distributed with mean 0 and variance 1, so that f_x (x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}

The moment generating function for X is calculated by M_x (s) = E(e^{sx})
= \int_{-\infty}^{\infty} e^{sx} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}
= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{sx -x^2/2}
= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} ((x-s)^2 - s^2)}
= e^{s^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} (x-s)^2}
= e^{s^2/2}

Finally, consider Y, which is normally distributed with mean \mu and variance \sigma, so that Y = aX +b = \sigma X + \mu

Recalling that M_y (s) = e^{sb} M_x (sa) from our lemma, we have M_y (s) = e^{s \mu} M_x (s \sigma) = e^{s \mu} e^{s^2 \sigma^2/2} = e^{s \mu + s^2 \sigma^2/2} \qquad QED

Sorry, I know that you wanted the variables in terms of t instead of s but s is the convention that I am used to, haha...
Last edited by Last_Singularity; December 30th, 2008 at 05:03 PM.
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