The moment generating function can be quite obscure for students. We have a definition here. Quite obscure, isn’t it! In this blog, I will show a simple example that will convince you that moment generating functions are very practicalâ€¦ to generate moments.

The example is based on the very pedagogical YouTube video of Lawrence Leemis.

In the example, you have to use the moment generating function to find *E*[*X*], *E*[*X ^{2}*], and

*E*[

*X*] for the continuous random variable

^{3}*X*with the following probability density function:

f(x)=e^{-x} \quad x>0

The moment generating function of *X* is by definition over the support *x > 0*:

M(t) =E\left[e^{t X}\right]

By the definition of the expected value of a random variable with density, we have:

=\int_0^{\infty} e^{t x} e^{-x} d x

From the exponent laws, we have:

=\int_0^{\infty} e^{(t-1) x} d x

For example, see *e ^{-3 x}* versus

*e*on Wolfram Alpha with the following prompt command

^{+3 x}`Plot[{E^(-3 x), E^(3 x)}, {x, -1., 1.}]`

:We can see from above that *(t-1)* needs to be inferior to 1 in order to have a convergent integral in blue. So let’s put that *(t-1) < 0*.

For the integrand *e ^{(t-1) x}*, substitute

*u =(t-1) x*and

*d u = (t-1) d x*. Check this with Wolfram Alpha :

=\frac{1}{t-1} \int e^u d u

Come back to the integral over the support *x > 0*:

\frac{1}{t-1}\left[e^{(t-1) x}\right]_0^{\infty}

You can easily see on Wolfram Alpha with the following prompt `Integrate[E^((-1 + t) x), {x, 0, Infinity}]`

that the moment generating function is

M(t)=\frac{1}{1-t} \quad t-1<0

The moment generation function is defined for values of t inferior to 1. Indeed, for the value of *t = 0*, it exists some values that satisfy *-h < t < h*. For example, *h* could be equal to 1/3 or 1/2 or some other values. That suffices to have a valid moment generating function.

I can now generate the three first moments quite easily by taking the first three derivatives of the moment generating function:

\begin{aligned} M^{\prime}(t)&=(1-t)^{-2} \\ \quad M^{\prime \prime}(t)&=2(1-t)^{-3} \\ M^{\prime \prime \prime}(t)&=6(1-t)^{-4} \end{aligned}

for *t < 1*. Using *t = 0* ( for the value of *t = 0*, it exists some values that satisfy *-h < t < h*) as an argument yields

\small{E[X]=1 \quad E\left[X^2\right]=2 \quad E\left[X^3\right]=6}

This is easier than

E\left[X^3\right]=\int_0^{\infty} x^3 e^{-x} d x

where you have to use integration by parts.