Exam PRandom variablesFree to read

Moment generating functions

M(t) = E[e^{tX}] encodes every moment in its derivatives at zero, and turns sums of independent variables into products.

The formulas

Definition

$$M_X(t) = E\!\left[e^{tX}\right]$$

Moments

$$M^{(k)}_X(0) = E[X^{k}]$$

Sum of independents

$$M_{X+Y}(t) = M_X(t)M_Y(t)$$

Linear transform

$$M_{aX+b}(t) = e^{bt}M_X(at)$$

Uniqueness

$$M_X = M_Y \Rightarrow X \stackrel{d}{=} Y$$

Where it comes from

1. Expand $e^{tX} = \sum_k \frac{t^k X^k}{k!}$ and take expectations term by term: the coefficient of $t^k/k!$ is $E[X^k]$.
2. Differentiating $k$ times and setting $t=0$ therefore isolates the $k$th moment.
3. For independent $X$ and $Y$, $E[e^{t(X+Y)}] = E[e^{tX}]E[e^{tY}]$ - the factorisation that makes MGFs worth using at all.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• Recognise the shape before you differentiate. Most MGF questions are really 'name this distribution'.
• M(0) = 1 always. If your MGF fails that, you have made an algebra slip.

Traps

• Forgetting that M'(0) is E[X], not the variance.
• Multiplying MGFs of variables that are not independent.

Related

• Expectation, variance and moments
• Binomial distribution
• Poisson distribution
• Normal distribution

Drill this: the Exam P question bank has original questions on this topic, and today’s free round is open to everyone.