Actuarial Statistics · Generating functions

MGFs of Standard Distributions: Derivations and Uses

Q: How do I identify a distribution from its MGF?

Rewrite the MGF so it matches a standard form, such as exp(λ(e^t − 1)) for Poisson. Read off the parameters. Then state that the MGF determines the distribution uniquely.

Q: What is the MGF of the exponential distribution?

For rate λ, M(t) = λ ÷ (λ − t) for t < λ. The mean is 1/λ and the variance is 1/λ². If the question uses mean θ, then M(t) = 1 ÷ (1 − θt).

Q: How do I derive the Poisson MGF?

Write M(t) = Σ e^(tx) e^(−λ) λ^x ÷ x! = e^(−λ) Σ (λe^t)^x ÷ x!. The sum is the exponential series for exp(λe^t). So M(t) = exp(λ(e^t − 1)).

Q: Does every distribution have an MGF?

No. The expectation E[e^(tX)] must be finite for t in an interval around 0. The lognormal, for example, has no MGF. In that case you use other tools, such as the characteristic function.

Updated 11 October 2026 · Fact-checked

The moment generating function (MGF) of X is M(t) = E[e^(tX)]. For standard distributions you derive it by summing or integrating e^(tx) against the pdf or pmf. You then use it to find moments, to find the MGF of sums of independent variables, and to identify an unknown distribution.

Understand MGFs of Standard Distributions

The moment generating function of a random variable X is M_X(t) = E[e^(tX)]. It exists if this expectation is finite for t in some interval around 0. For a discrete X you sum e^(tx) P(X = x). For a continuous X you integrate e^(tx) f(x) dx.

Why is it useful? Expand e^(tX) as a power series. Then M(t) = 1 + t E[X] + t² E[X²] ÷ 2! + ... So the k-th derivative of M at t = 0 gives E[X^k]. You get all moments from one function.

The second big use is uniqueness. If an MGF exists in an interval around 0, it determines the distribution uniquely. So if you recognise the form of an MGF, you know the distribution and its parameters. This is how you identify a distribution from its MGF.

The third use is sums. For independent X and Y, M_(X+Y)(t) = M_X(t) × M_Y(t). Multiply the MGFs, then recognise the result. This proves, for example, that a sum of independent Poissons is Poisson.

You must know the MGFs of the standard distributions and be able to derive the main ones. The derivations use one trick each: the binomial theorem, the exponential series, or a gamma integral that equals 1. For the normal, you complete the square.

Key rules to remember

Definition: M_X(t) = E[e^(tX)]; Valid for t in an interval around 0 where the expectation is finite.
Moments from MGF: E[X^k] = M_X^(k)(0); The k-th derivative at t = 0. So E[X] = M'(0) and Var(X) = M''(0) − (M'(0))².
Linear transformation: M_(aX+b)(t) = e^(bt) M_X(at); Use this to standardise or shift.
Independent sum: M_(X+Y)(t) = M_X(t) M_Y(t); Only for independent X and Y.
Binomial(n, p): M(t) = (1 − p + p e^t)^n; Valid for all t. Bernoulli is the case n = 1.
Poisson(λ): M(t) = exp(λ(e^t − 1)); Valid for all t.
Geometric (number of trials, support 1, 2, ...): M(t) = p e^t ÷ (1 − (1 − p) e^t); Valid for e^t (1 − p) < 1. Check which definition of the geometric the question uses.
Exponential(λ), mean 1/λ: M(t) = λ ÷ (λ − t); Valid for t < λ.
Gamma(α, λ), mean α/λ: M(t) = (λ ÷ (λ − t))^α; Valid for t < λ. The rate parameter λ is used here, so check how the question defines it.
Normal(μ, σ²): M(t) = exp(μt + σ²t² ÷ 2); Valid for all t.
Uniform(a, b): M(t) = (e^(tb) − e^(ta)) ÷ (t(b − a)), t ≠ 0; M(0) = 1.

How to solve MGFs of Standard Distributions questions

Use this method for any question on MGFs of standard distributions, whether you derive, use or identify.

1Write down the definition M(t) = E[e^(tX)] and the pmf or pdf, including its support.
2Combine e^(tx) with the pmf or pdf. Look for the pattern: a binomial sum, an exponential series, or a gamma-type integral.
3Rearrange so the remaining sum or integral is a full pmf or pdf (which equals 1) with a changed parameter. Read off M(t).
4State the range of t for which M(t) is valid, such as t < λ for the exponential.
5To find moments, differentiate and set t = 0. For the variance, use M''(0) − (M'(0))².
6To identify a distribution, match the MGF to a standard form and read off the parameters. Use the uniqueness property in your wording.
7For sums of independent variables, multiply the MGFs first, simplify, then identify the result.

Quickest way: Match and read off

When to use it: Use this for MCQs and short parts where you are given an MGF or asked for a sum of independent variables.

Memorise the standard MGFs in the formula list.
Rewrite the given MGF to match one form. For example, (2 ÷ (2 − t))³ is Gamma(3, 2).
For sums of independent variables of the same family, add the parameters that add. Binomials with the same p add n. Poissons add λ. Gammas with the same λ add α. Normals add means and variances.
For the mean, use the known formulas rather than differentiating. For example, the mean of Gamma(α, λ) is α/λ and the variance is α/λ².
Use the log trick if differentiating: for ln M(t), the first derivative at 0 is the mean and the second derivative at 0 is the variance.

Common mistakes in MGFs of Standard Distributions

Forgetting the validity range, such as writing the exponential MGF without t < λ.
Students focus on the formula and forget that the integral diverges for large t.
Fix: Always state the range. It costs one line and earns a mark in derivations.
Mixing up rate and scale for the exponential and gamma.
Some sources use mean θ and others use rate λ. The MGF changes form: 1 ÷ (1 − θt) versus λ ÷ (λ − t).
Fix: Read the question's definition first. Write the pdf before the MGF. Check M'(0) gives the correct mean.
Adding MGFs for a sum of variables instead of multiplying.
Students confuse it with adding expectations.
Fix: Remember: sum of variables means product of MGFs, and only if independent.
Getting the variance as M''(0) without subtracting the square of the mean.
M''(0) is E[X²], not the variance.
Fix: Always compute Var(X) = M''(0) − (M'(0))².
Applying the MGF of a·X as M_X(t) × a or M_X(a) instead of M_X(at).
Students substitute loosely.
Fix: Use M_(aX+b)(t) = e^(bt) M_X(at). Check by testing a simple case such as aX with a = 2.
Declaring two sums of different families are the same distribution, for example adding Gammas with different rates.
Students apply the closure result without checking its conditions.
Fix: Gammas add α only if the rate λ is the same. Binomials add n only if p is the same.

Worked examples

Example 1

Derive the MGF of X ~ Exponential(λ), and use it to find E[X] and Var(X).

Show the solution

The pdf is f(x) = λ e^(−λx) for x > 0.
M(t) = ∫₀^∞ e^(tx) λ e^(−λx) dx = λ ∫₀^∞ e^(−(λ − t)x) dx.
For t < λ the integral converges and equals 1 ÷ (λ − t). So M(t) = λ ÷ (λ − t).
Differentiate: M'(t) = λ ÷ (λ − t)². So M'(0) = 1/λ.
M''(t) = 2λ ÷ (λ − t)³. So M''(0) = 2/λ².
Var(X) = 2/λ² − (1/λ)² = 1/λ².

Answer: M(t) = λ ÷ (λ − t) for t < λ. E[X] = 1/λ and Var(X) = 1/λ².

Example 2

X and Y are independent. X has MGF (1 − 3t)^(−2) and Y has MGF (1 − 3t)^(−5), both for t < 1/3. Identify the distribution of X + Y and find its mean and variance.

Show the solution

Because X and Y are independent, M_(X+Y)(t) = (1 − 3t)^(−2) × (1 − 3t)^(−5) = (1 − 3t)^(−7).
Write (1 − 3t)^(−7) = (λ ÷ (λ − t))^7 with λ = 1/3. Check: (1/3) ÷ (1/3 − t) = 1 ÷ (1 − 3t). This matches.
By uniqueness of the MGF, X + Y ~ Gamma(α = 7, λ = 1/3).
Mean = α/λ = 7 ÷ (1/3) = 21.
Variance = α/λ² = 7 ÷ (1/9) = 63.

Answer: X + Y ~ Gamma(7, 1/3), with mean 21 and variance 63.

Exam tips

Learn the standard MGFs cold. MCQs often give an MGF and ask for a mean, variance or probability, so recognition saves time.
In written derivations, show the key step: the remaining integral or sum is a full density or pmf, so it equals 1. State the validity range of t.
For sums, state independence explicitly before multiplying MGFs, and state uniqueness before naming the result.
Check rate versus scale in the question before using any exponential or gamma formula. Check your mean with M'(0).
In derivation of the normal MGF, show completing the square clearly. Examiners award marks for the method.

Practice questions from Generating functions

MGFs of Standard Distributions: frequently asked questions

How do I identify a distribution from its MGF?

Rewrite the MGF so it matches a standard form, such as exp(λ(e^t − 1)) for Poisson. Read off the parameters. Then state that the MGF determines the distribution uniquely.

What is the MGF of the exponential distribution?

For rate λ, M(t) = λ ÷ (λ − t) for t < λ. The mean is 1/λ and the variance is 1/λ². If the question uses mean θ, then M(t) = 1 ÷ (1 − θt).

How do I derive the Poisson MGF?

Write M(t) = Σ e^(tx) e^(−λ) λ^x ÷ x! = e^(−λ) Σ (λe^t)^x ÷ x!. The sum is the exponential series for exp(λe^t). So M(t) = exp(λ(e^t − 1)).

Does every distribution have an MGF?

No. The expectation E[e^(tX)] must be finite for t in an interval around 0. The lognormal, for example, has no MGF. In that case you use other tools, such as the characteristic function.