4. Moment Generating Function (MGF)

Concept

Similar to PGF, but works for both discrete and continuous variables. It uses the exponential function $e^{tX}$.

Definition

$$ M_X(t) = E[e^{tX}] $$

• Discrete: $M_X(t) = \sum e^{tx} P(x)$
• Continuous: $M_X(t) = \int e^{tx} f(x) dx$

Why it works (The Taylor Series Connection)

The expansion of $e^{tX}$ is: $$ e^{tX} = 1 + tX + \frac{t^2X^2}{2!} + \frac{t^3X^3}{3!} + \dots $$ Taking expectation on both sides: $$ M_X(t) = 1 + tE[X] + \frac{t^2E[X^2]}{2!} + \frac{t^3E[X^3]}{3!} + \dots $$ Notice the coefficients contain the moments $E[X], E[X^2], \dots$

How to find Moments

1. Mean ($E[X]$): Differentiate once, set $t=0$.

$$ M'_X(0) = E[X] $$

2. Second Moment ($E[X^2]$): Differentiate twice, set $t=0$.

$$ M''_X(0) = E[X^2] $$

Detailed Example

Problem: A continuous variable $X$ has PDF $f(x) = 2x$ for $0 \le x \le 1$. Find Mean and Variance using MGF.

Solution:

1. Find MGF:

$$ M_X(t) = \int_{0}^{1} e^{tx} (2x) dx $$ Using integration by parts ($\int u dv$), let $u=2x, dv=e^{tx}dx$: $$ M_X(t) = \left[ \frac{2x e^{tx}}{t} \right]_0^1 - \int_{0}^{1} \frac{2 e^{tx}}{t} dx $$ $$ M_X(t) = \frac{2e^t}{t} - \left[ \frac{2e^{tx}}{t^2} \right]_0^1 = \frac{2e^t}{t} - \frac{2e^t}{t^2} + \frac{2}{t^2} $$

1. Find Mean ($E[X]$):

Differentiate $M_X(t)$ with respect to $t$ and find limit as $t \to 0$. Using L'Hopital's rule on the derived MGF terms: $M'_X(0) = 2/3$. Mean = 2/3.

1. Variance:

Find $E[X^2]$ using $M''_X(0)$. $E[X^2] = 1/2$. $Var(X) = E[X^2] - (E[X])^2 = 1/2 - (2/3)^2 = 1/2 - 4/9 = 1/18$.