Poisson Distribution Moment Generating Function: Derivation, Mean and Variance

The moment generating function (MGF) of a Poisson random variable X with rate λ is M(t) = exp[λ(e^t − 1)], defined for every real t. Differentiating it at t = 0 gives the mean (λ) and the variance (λ) in two lines, and multiplying two such functions proves that independent Poisson counts add. This guide derives the result from the exponential series, checks it numerically for λ = 3, and shows the same check in Excel, R and Python.

Definition in one paragraph

The moment generating function of a random variable X is M(t) = E[e^tX], the expected value of e^tX viewed as a function of t. When it exists on an interval around t = 0, its derivatives at 0 are the moments of X: M′(0) = E[X], M′′(0) = E[X²], and so on. For the Poisson distribution the MGF exists for all real t. For the full distribution (pmf, assumptions, regression use) see the CASRAI guide to the Poisson distribution.

Formula: the moment generating function of the Poisson distribution

If X ~ Poisson(λ), then

M_X(t) = exp[λ(e^t − 1)] = e^{λ(et − 1)}

• X: the count of events (0, 1, 2, …).
• λ (lambda): the rate parameter, the average number of events in the interval; the NIST/SEMATECH e-Handbook describes it as the shape parameter that indicates the average number of events in the given time interval and gives the pmf p(x; λ) = e^−λλ^x/x!.
• t: a real number used only as a dummy argument; you evaluate derivatives at t = 0.
• e^t: the exponential function evaluated at t.

Wolfram MathWorld lists the same function, with the mean and variance both equal to its rate parameter and the cumulant-generating function K(h) = λ(e^h − 1).

Derivation from the exponential series

Start from the definition and the pmf:

1. M(t) = E[e^tX] = ∑_x=0^∞ e^tx · e^−λλ^x/x!
2. Pull out the constant: M(t) = e^−λ ∑ (λe^t)^x/x!
3. The sum is the series for e^z with z = λe^t, which converges for every real z, so the sum equals e^λet.
4. Therefore M(t) = e^−λ e^λet = exp[λ(e^t − 1)].

Siegrist’s Statistics LibreTexts chapter on generating functions reaches the same function in a second way: the probability generating function of a Poisson variable is P(t) = e^λ(t−1), and substituting e^t for t gives the MGF exp(λe^t − λ).

Worked example with real numbers: λ = 3

Moment generating function for the Poisson distribution: deriving the mean and variance

Let X ~ Poisson(3), so M(t) = exp[3(e^t − 1)].

• First derivative. By the chain rule, M′(t) = 3e^t · M(t). At t = 0: M(0) = exp(0) = 1, so M′(0) = 3 × 1 × 1 = 3. That is E[X].
• Second derivative. M′′(t) = (3e^t + 9e^2t) M(t). At t = 0: (3 + 9) × 1 = 12. That is E[X²].
• Variance. Var(X) = E[X²] − (E[X])² = 12 − 3² = 3, equal to λ, as expected.

In general M′(0) = λ, M′′(0) = λ + λ², and Var = λ.

Numerical check at t = 0.5. The formula gives M(0.5) = exp[3(e^0.5 − 1)] = exp[3 × 0.648721] = exp(1.946164) ≈ 7.0018. Summing e^0.5x × pmf(x) over x = 0 to 199 directly also gives 7.0018 (computed directly), so the closed form and the definition agree.

Interpretation

Quantity	How to read it from M(t)	Poisson(λ) result
M(0)	Must equal 1 for any MGF (total probability)	exp(0) = 1
M′(0)	Mean, E[X]	λ
M′′(0)	Second moment, E[X2]	λ + λ2
M′′(0) − M′(0)2	Variance	λ
ln M(t)	Cumulant-generating function	λ(et − 1)

Sums of independent Poisson variables

If X₁ ~ Poisson(λ₁) and X₂ ~ Poisson(λ₂) are independent, the MGF of the sum is the product of the MGFs: exp[λ₁(e^t − 1)] · exp[λ₂(e^t − 1)] = exp[(λ₁ + λ₂)(e^t − 1)], which is the MGF of Poisson(λ₁ + λ₂). Because an MGF that exists near 0 identifies the distribution, the sum is Poisson with the added rates. The same MGF-product idea is used for other families; compare the negative binomial distribution, which is often used when counts are more variable than a Poisson allows.

Assumptions and common mistakes

• Writing e^λt instead of e^λ(et−1). The argument of the inner exponential is e^t, not t. Check that M(0) = 1.
• Forgetting the constant e^−λ. Without it M(0) = e^λ, not 1.
• Reading the second derivative as the variance. M′′(0) is E[X²]; subtract the squared mean.
• Adding MGFs of independent variables. You multiply the MGFs; you add the variables.
• Applying the product rule to dependent counts. The sum result needs independence.
• Confusing the MGF with the probability generating function. They are related by t → e^t; the PGF is P(t) = e^λ(t−1).

How to report it (APA example)

An MGF is usually stated in a methods or appendix derivation rather than reported as a result. A sentence in APA 7th style might read:

Event counts were modelled as X ~ Poisson(λ), whose moment generating function is M(t) = exp[λ(e^t − 1)] (Siegrist, n.d.); the mean and variance are both λ.

Reference entry: Siegrist, K. (n.d.). Generating functions. In Probability, mathematical statistics, and stochastic processes. LibreTexts Statistics. https://stats.libretexts.org/Bookshelves/Probability_Theory/Probability_Mathematical_Statistics_and_Stochastic_Processes_(Siegrist)/04%3A_Expected_Value/4.06%3A_Generating_Functions

How to calculate it in Excel, R or Python

Each snippet evaluates the closed form at λ = 3, t = 0.5 and compares it with the defining sum. Expected output: about 7.0018.

Excel

Closed form:   =EXP(3*(EXP(0.5)-1))
Definition:    put 0..100 in A1:A101, then
               =SUMPRODUCT(EXP(0.5*A1:A101)*POISSON.DIST(A1:A101,3,FALSE))

R

lambda <- 3; t <- 0.5
exp(lambda * (exp(t) - 1))
x <- 0:100
sum(exp(t * x) * dpois(x, lambda))

Python

import numpy as np
from scipy.stats import poisson
lam, t = 3, 0.5
print(np.exp(lam * (np.exp(t) - 1)))
x = np.arange(0, 101)
print(np.sum(np.exp(t * x) * poisson.pmf(x, lam)))

FAQ

What is the moment generating function of the Poisson distribution?

M(t) = exp[λ(e^t − 1)] for all real t, where λ is the average number of events per interval.

How do you get the mean and variance from the Poisson MGF?

The first derivative at 0 gives the mean, λ. The second derivative at 0 gives λ + λ²; subtracting λ² leaves a variance of λ.

Does the Poisson MGF exist for every t?

Yes. The series in the derivation is the exponential series, which converges for every real argument.

How is the MGF related to the cumulative distribution function?

Both determine the distribution, but the MGF packages the moments while the CDF gives probabilities directly; see the guide to the cumulative distribution function. For a parallel expected-value derivation, see binomial distribution expected value. When the variance exceeds the mean, see negative binomial regression and overdispersion.

Sources

• Wolfram MathWorld, Poisson Distribution, https://mathworld.wolfram.com/PoissonDistribution.html (checked 2026-10-29). Lists Grimmett & Stirzaker, Probability and Random Processes (2nd ed., Oxford University Press, 1992) and Papoulis, Probability, Random Variables, and Stochastic Processes (2nd ed., McGraw-Hill, 1984) among its references.
• Siegrist, K., Generating Functions, Statistics LibreTexts (University of Alabama in Huntsville), https://stats.libretexts.org/Bookshelves/Probability_Theory/Probability_Mathematical_Statistics_and_Stochastic_Processes_(Siegrist)/04%3A_Expected_Value/4.06%3A_Generating_Functions (checked 2026-10-29).
• NIST/SEMATECH e-Handbook of Statistical Methods, Poisson Distribution, https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm (checked 2026-10-29), for the pmf, mean and standard deviation.