Chapter 6B: Moment Generating Functions

MATH/STAT 394: Probability I

Arman Jahangiri

University of Washington Department of Mathematics

Summer 2026

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

1

Learning Goals

By the end of this lecture, students should be able to:

▶ deﬁne the moment generating function (MGF);

▶ compute MGFs of common distributions;

▶ extract moments from MGFs;

▶ understand how MGFs determine distributions;

▶ compute distributions of sums using MGFs;

▶ derive moments of Exponential and Normal distributions using MGFs.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

2

Deﬁnition of the MGF

Deﬁnition

The moment generating function (MGF) of X is

M_X(t) = E(e^tX

)

provided the expectation is ﬁnite in some open interval around 0.

Equivalently,

X







e^txp_X(x),

if X is discrete,





x

Z

M_X(t) =

∞



_−∞e^txf_X(x) dx, if X is continuous.

Quick check

Every valid MGF satisﬁes

M_X(0) = 1.

Chapter 6B: Moment Generating Functions

Arman Jahangiri

Summer 2026

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MGFs: Discrete Distributions I

X

M_X(t) = E(e^tX) =

e^txp_X(x).

x

Bernoulli

If X ∼ Bern(p), then

M_X(t) = e^t·1p + e^t·0(1 − p) = pe^t+ q.

Discrete Uniform

If X ∼ DUnif{1, . . . , m}, then

m

e^t(e^mt− 1)

X

ꢀ

ꢁ

1

M_X(t) =

e^tk

=

e^t+ e^2t+ · · · + e^mt

=

.

m

m(e^t− 1)

k=1

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

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MGFs: Discrete Distributions II

Binomial

If X ∼ Bin(n, p), then

n

ꢂ ꢃ

X

n

M_X(t) =

e^tk

p^kq^n−k

=

(pe^t)^kq^n−k= (q + pe^t)ⁿ.

k

k=0

Poisson

If X ∼ Pois(λ), then

∞

t

k

_−λλ^k

(λe )

X

t

M_X(t) =

e^tk

e

= e^−λ

= e^−λe^λe= e^λ(e

−1)

.

k!

k=0

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

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MGFs: Discrete Distributions III

Geometric

If X ∼ Geom(p), where X counts failures before the ﬁrst success, then

∞

X

p

M_X(t) =

e^tkq^kp = p

(qe^t)^k

=

,

qe^t< 1.

1 − qe^t

k=0

Negative Binomial

If X ∼ NegBin(r, p), where X counts failures before the r-th success, then

ꢄ

ꢅ

∞

r

ꢂ

ꢃ

ꢂ

ꢃ

X

k + r − 1

p

M_X(t) =

e^tk

p^rq^k= p^r

(qe^t)^k

=

.

k

1 − qe^t

k=0

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

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MGFs: Discrete Distributions IV

Hypergeometric

If

X ∼ HGeom(w, b, n),

then

ꢀ ꢁꢀ

ꢁ

w

k

ꢀ

b

n−k

P(X = k) =

,

ꢁ

w+b

n

where

max(0, n − b) ≤ k ≤ min(w, n).

Therefore,

ꢀ ꢁꢀ

ꢁ

min(w,n)

w

k

ꢀ

b

X

n−k

M_X(t) =

e^tk

.

ꢁ

w+b

n

k=max(0,n−b)

Note: Unlike Binomial or Poisson, this MGF usually does not simplify to a nice elementary closed form.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

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MGFs: Continuous Distributions I

Z

∞

tx

M_X(t) = E(e^tX) =

e

f_X(x) dx.

−∞

Uniform

If X ∼ Unif(a, b), then

ꢆ

ꢇ

Z

b

a

b

1

e^tx

t

e^tb− e^ta

M_X(t) =

e^tx

dx =

=

,

t

̸= 0.

b − a

t(b − a)

a

Also, M_X(0) = 1.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

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MGFs: Continuous Distributions II

Exponential

If X ∼ Exp(λ), then

Z

∞

λ

M_X(t) =

e^txλe^−λxdx = λ

e^−(λ−t)xdx =

,

t < λ.

λ − t

0

Gamma

If X ∼ Gamma(α, λ), with shape α and rate λ, then

Z

∞

λ^α

M_X(t) =

e^tx

x^α−1e^−λxdx =

x^α−1e^−(λ−t)xdx.

Γ(α)

0

Let u = (λ − t)x. Then x = u/(λ − t), dx = du/(λ − t), and

ꢄ

ꢅ

Z

α

∞

λ^α

1

λ^α

(λ − t)^α

λ

M_X(t) =

u^α−1e^−udu =

=

,

t < λ.

Γ(α) (λ − t)^α

λ − t

0

|

{z

}

Γ(α)

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026

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MGF of N(0, 1)

Standard Normal

Let Z ∼ N(0, 1). Then

Z

∞

2

1

tz

M_Z(t) = E(e^tZ) = √

e

e^{−z /2}dz.

2π

−∞

Complete the square:

Therefore,

z²

1

(z − t)²

t²

ꢀ

ꢁ

tz −

= − (z²− 2tz) = − (z − t)²− t²= −

+

.

2

Z

∞

2

1

−(z−t)

/2

−(z−t)

e

^/2dz.

M_Z(t) = √

e

^/2e^{t /2}dz = e^t

√

2π

−∞

√

Since the last integral equals 2π,

2

M_Z(t) = e^t

.

/2

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Chapter 6B: Moment Generating Functions

Summer 2026 10

MGF of N(µ, σ²)

Normal

Let X ∼ N(µ, σ²). Write

X = µ + σZ,

Z ∼ N(0, 1).

Then

ꢂ

ꢃ

M_X(t) = E(e^tX) = E e^t(µ+σZ)= e^µtE(e^σtZ).

Using

2

M_Z(s) = e^s

,

/2

with s = σt, we get

2

1

2

= e^µt+

.

t

/2

σ

t

M_X(t) = e^µtM_Z(σt) = e^µte^σ

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Chapter 6B: Moment Generating Functions

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Moments from MGFs

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Chapter 6B: Moment Generating Functions

Summer 2026 12

MGFs Generate Moments

Theorem

The n^thpopulation moment is obtained by diﬀerentiating the MGF:

E(Xⁿ) = M_X⁽ⁿ⁾(0).

Examples:

E(X) = M_X^′(0),

E(X²) = M_X^′′(0).

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Chapter 6B: Moment Generating Functions

Summer 2026 13

Proof: MGFs Generate Moments

Recall that the moment generating function is

M_X(t) = E(e^tX).

Diﬀerentiate once:

ꢀ

ꢁ

d

M_X^′(t) =

dt

d

E(e^tX) = E

e^tX= E(Xe^tX).

dt

Now plugging in t = 0,

M_X^′(0) = E(Xe⁰) = E(X).

Diﬀerentiate another time noting that X is a constant w.r.t. t.

ꢀ

ꢁ

d

M_X^′′(t) =

E(Xe^tX) = E

Xe^tX= E(X²e^tX).

dt

Plugging in t = 0,

M_X^′′(0) = E(X²e⁰) = E(X²).

Similarly, diﬀerentiating n times gives

Therefore, at t = 0,

M_X⁽ⁿ⁾(t) = E(Xⁿe^tX).

M_X⁽ⁿ⁾(0) = E(Xⁿe⁰) = E(Xⁿ).

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026 14

Another way of seeing it...

The Taylor expansion of

is

e^tX

t²X²

e^tX= 1 + tX +

+ · · · .

2!

Taking expectations:

t²

M_X(t) = 1 + tE(X) +

E(X²) + · · · .

2!

The moments appear as Taylor coeﬃcients.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026 15

Example: Bernoulli Moments

For

M_X(t) = pe^t+ q,

M_X^′(t) = pe^t.

M_X^′(0) = p.

we compute:

Thus:

So

E(X) = p.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

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MGF Uniqueness Theorem

Theorem (Uniqueness of MGF)

If two random variables have the same MGF on an interval around 0, then they have the same distribution,

i.e., the moment generating function uniquely determines the distribution.

Thus MGFs characterize distributions, similarly to PDFs and CDFs.

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Chapter 6B: Moment Generating Functions

Summer 2026 17

MGFs of Sums

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Chapter 6B: Moment Generating Functions

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MGF of a Sum

Theorem

If X and Y are independent, then

M_X+Y(t) = M_X(t) × M_Y(t).

More generally, if X₁, . . . , X_nare independent, then

n

Y

M^P

(t) =

M_X(t)

n

i=1

X

i

i=1

This is one of the main reasons MGFs are useful. Since using the MGF Uniqueness theorem, the MGF characterize

the distribution in full.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

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Binomial MGF

A Binomial random variable can be written as:

X = I₁+ · · · + I_n,

where the I_jare independent Bernoulli(p) random variables.

Since each Bernoulli MGF is

pe^t+ q,

we get

M_X(t) = (pe^t+ q)ⁿ.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026 20

Negative Binomial MGF

If

X ∼ NBin(r, p),

then

X = G₁+ · · · + G_r,

where the G_jare independent Geometric(p) random variables.

Thus,

ꢀ

ꢁ

r

p

M_X(t) =

.

1 − qe^t

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Chapter 6B: Moment Generating Functions

Summer 2026 21

Sums via MGFs

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Chapter 6B: Moment Generating Functions

Summer 2026 22

Sum of Independent Poissons

If

X ∼ Pois(λ),

Y ∼ Pois(µ),

independently, then

t

−1)

M_X(t) = e^λ(e

,

and

t

−1)

M_Y(t) = e^µ(e

.

Therefore,

Hence:

t

−1)

M_X+Y(t) = e^(λ+µ)(e

.

X + Y ∼ Pois(λ + µ).

Remark: This can be generalized to n independent Poisson r.v.s using induction.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026 23

Sum of Independent Normals

If

X₁∼ N(µ₁, σ₁²),

X₂∼ N(µ₂, σ₂²),

independently, then

Thus,

2

1

2

+µ )t+ (σ +σ )t

2

₂(t) = e^(µ

.

1

2

M_X

+X

1

X₁+ X₂∼ N(µ₁+ µ₂, σ₁²+ σ₂²).

Remark: This can be generalized to n independent Normal r.v.s using induction.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026 24

Distributions Closed Under Addition Under Independence

Distribution

Assumption

Sum Distribution

n

X

Bernoulli

X_i∼ Bern(p)

X_i∼ Bin(n, p)

i=1

n

ꢂ

ꢃ

X

Binomial

Poisson

X_i∼ Bin(n_i, p)

X_i∼ Bin

n_i, p

i=1

n

ꢂ

ꢃ

X

X_i∼ Pois(λ_i)

X_i∼ Pois

λ_i

i=1

n

ꢂ

ꢃ

X

Negative Binomial X_i∼ NegBin(r_i, p)

X_i∼ NegBin

r_i, p

i=1

n

X

Geometric

Normal

X_i∼ Geom(p)

X_i∼ N(µ_i, σ_i²)

X_i∼ Gamma(α_i, λ)

X_i∼ Exp(λ)

X_i∼ NegBin(n, p)

i=1

n

ꢂ

X

X_i∼ N

µ_i,

σ_i²

i=1

n

ꢂ

ꢃ

X

Gamma

X_i∼ Gamma

α_i, λ

i=1

n

X

Exponential

X_i∼ Gamma(n, λ)

i=1

Arman Jahangiri

Chapter 6B: Moment Generating Functions

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Summary

▶ MGFs encode moments through derivatives.

▶ MGFs determine distributions.

▶ Independent sums correspond to products of MGFs.

▶ MGFs simplify moment calculations dramatically.

▶ The Normal and Exponential MGFs are especially important.

Next lecture

Joint distributions.

Arman Jahangiri

Chapter 6B: Moment Generating Functions

Summer 2026 26