Moment generating function Pt 2.

Alright, not really needed, but I'm going to throw in a few things that may come up. Essentially nifty little fun-facts. Let's start with the scenario that you'd want a moment about the mean. To get that result we just start over with:

$$MGF\equiv\int_{-\infty}^{\infty}e^{tX}f(X)dx,~t\in\mathbb{R}$$

And replace $e^{tX}$ with $e^{t(X-\mu)}$. Now, the rest is pretty straight forward. We could sub in the expression for the expansion of $e$ and get:

 $$e^{t(X-\mu)}=1+\frac{t(X-\mu)}{1!}+\frac{t^{2}(X-\mu)^{2}}{2!}+\frac{t^{3}(X-\mu)^{3}}{3!}+...$$

Following the steps of last time would give you essentially the same idea. However, there's another way we could go about it. Since we have $e^{t(X-\mu)}$, we can transform it into $e^{tX}e^{-t\mu}$. You'll notice the former is just what we had earlier, and the latter is a new term. But let's write it out:

$$\int_{-\infty}^{\infty}e^{-t\mu}e^{tX}f(X)dx$$

Well, the integral is just with respect to $X$, so we can treat  $e^{-t\mu}$ as a constant and pull it out to get:

$$e^{-t\mu}\int_{-\infty}^{\infty}e^{tX}f(X)dx$$

Where the integral on the right is exactly the same as the MGF we had before. So this now becomes:

 $$e^{-t\mu}M_{x}(t)$$

Differentiating and setting $t$ equal to zero like last time will give us the moments we want. To see this, let's try the first ones. We know the variance is defined as $E[(x-\mu)^2]$, and can alternatively be defined as $E[x^2]-\mu^2$. Well, this is the second moment about the mean, so let's try that in our alternate and simpler MGF. We'd have to take the second derivative and set it equal to zero. So let's start by differentiating it.  Here:
$$ \frac{d(e^{-t\mu}M_{x}(t)}{dt}=M'_{x}(t)e^{-t\mu}+-\mu M_{x}(t)e^{-t\mu}$$
Now, this is the first moment about the mean. So it's essentially $E[x-\mu]=E[x]-E[\mu]=\mu-\mu=0$. Well, let's set $t=0$ to test that. We have:
$$M'_{x}(0)e^{0}+-\mu M_{0}(t)e^{0}=\mu(1)-\mu(1)(1)=\mu-\mu=0$$
Using the different moments of $x$ and the fact that $e^{0}=1$. The second derivative being:
$$\frac{d^{2}M_{x}(t)}{dt^2}=M''_{x}(t)e^{-t\mu}+-\mu M'_{x}(t)e^{-t\mu}+-\mu M'_{x}(t)e^{-t\mu}+\mu^{2} M_{x}(t)e^{-t\mu}$$
Setting $t=0$, and using the moments from the last post, we get:
$$M''_{x}(0)e^{0}+-2\mu M'_{x}(0)e^{0}+-\mu^{2} M'_{x}(0)e^{0}=E[x^2]+-2\mu^{2}+\mu^{2}=E[x^2]-\mu^{2}$$

Thus what we were trying to prove. Next post in this series will be about multiple variables. I may also mix the MGF series with the discrete distributions to get the MGFs of them.