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: $M_{X} (t) = E [e^{tX}]$
Moments: $M_{X}^{(k)} (0) = E [X^{k}]$
Sum of independents: $M_{X + Y} (t) = M_{X} (t) M_{Y} (t)$
Linear transform: $M_{a X + b} (t) = e^{b t} M_{X} (a t)$
Uniqueness: $M_{X} = M_{Y} \Rightarrow X = d Y$

Where it comes from

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

Worked example

A random variable has MGF $M(t) = \left(0.3 + 0.7e^{t}\right)^{8}$. Find its variance.

This is the MGF of a binomial with $n = 8$ and $p = 0.7$ , by inspection of $(1 - p + p e^{t})^{n}$ .
Uniqueness of the MGF makes that identification exact — no differentiation needed.
$Var (X) = n p (1 - p) = 8 (0.7) (0.3)$ .
$Var (X) = 1.68$ .

Answer: 1.68

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.

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

Moment generating functions

The formulas

Where it comes from

Worked example

Memory hooks

Traps

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