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Actuarial Statistics · Generalised linear models

Exponential Family of Distributions for GLMs

Updated 11 October 2026 · Fact-checked

A distribution is in the exponential family if its density can be written as f(y; θ, φ) = exp{[yθ − b(θ)] ÷ a(φ) + c(y, φ)}. To solve questions, take logs, match terms to find θ, b(θ) and a(φ), then use mean = b′(θ) and variance = a(φ)·b″(θ).

Understand Exponential Family of Distributions

GLMs need one common form for the response distribution. That form is the exponential family. Normal, Poisson, binomial and gamma all fit it. Once you write a distribution in this form, you get its mean and variance from one function, and you can fit all these models with the same method.

The form is f(y; θ, φ) = exp{[yθ − b(θ)] ÷ a(φ) + c(y, φ)}. Here θ is the canonical parameter. It is the parameter that multiplies y inside the exponent. φ is the dispersion parameter (scale). The function b(θ) is the cumulant function. The term c(y, φ) holds everything that depends on y and φ but not on θ. Often a(φ) = φ, or φ divided by a known weight.

Why is b(θ) so useful? The density must integrate (or sum) to 1. Differentiate that identity with respect to θ and you get E(Y) = b′(θ). Differentiate again and you get Var(Y) = a(φ)·b″(θ). So you never need to integrate y or y² directly. This works under the usual regularity conditions, which hold for the standard distributions in the syllabus.

The mean μ depends on θ through b′. So b″(θ) can be written as a function of μ. This is the variance function V(μ). Then Var(Y) = a(φ)·V(μ). For normal V(μ) = 1, for Poisson V(μ) = μ, for binomial counts V(μ) = μ(1 − μ/n), for gamma V(μ) = μ².

To put a distribution in the form, write the log of its density. Group the terms that contain both y and the parameter. That group gives yθ ÷ a(φ). The remaining parameter-only terms give −b(θ) ÷ a(φ). Whatever is left depends only on y and φ.

Key rules to remember

Exponential family form
f(y; θ, φ) = exp{ [yθ − b(θ)] ÷ a(φ) + c(y, φ) }
θ is the canonical parameter, φ the dispersion parameter, b(θ) the cumulant function.
Mean
E(Y) = μ = b′(θ)
Comes from differentiating the total probability once with respect to θ.
Variance
Var(Y) = a(φ) · b″(θ)
Do not forget the factor a(φ). Write b″ in terms of μ to get the variance function V(μ).
Normal(μ, σ²)
θ = μ; b(θ) = θ² ÷ 2; φ = σ²; a(φ) = φ
Mean = θ, variance = σ². V(μ) = 1.
Poisson(μ)
θ = ln μ; b(θ) = e^θ; a(φ) = 1
Mean = variance = e^θ = μ. V(μ) = μ.
Binomial count, Y ~ Bin(n, p)
θ = ln[p ÷ (1 − p)]; b(θ) = n ln(1 + e^θ); a(φ) = 1
Mean = n e^θ ÷ (1 + e^θ) = np. Variance = n e^θ ÷ (1 + e^θ)² = np(1 − p).
Gamma with mean μ and shape α
θ = −1 ÷ μ; b(θ) = −ln(−θ); φ = 1 ÷ α; a(φ) = φ
Mean = −1 ÷ θ = μ. Variance = φμ² = μ² ÷ α. V(μ) = μ².

How to solve Exponential Family of Distributions questions

Use this method to show a distribution belongs to the exponential family and to find its mean and variance.

  1. 1Write down the density or probability function and state the range of y.
  2. 2Take the natural log, or write the density as exp{ln f}. Expand all logs and powers.
  3. 3Collect the terms that contain both y and the parameter. Match them to yθ ÷ a(φ). This identifies θ and a(φ).
  4. 4Collect the terms that contain the parameter but not y. Match them to −b(θ) ÷ a(φ). Express the parameter in terms of θ to get b(θ).
  5. 5Put everything left over, which depends on y and φ only, into c(y, φ). Check that the range of y does not depend on the parameter.
  6. 6Differentiate: mean = b′(θ), variance = a(φ)·b″(θ). Substitute θ back to express these in terms of the original parameters.
  7. 7Check against the known mean and variance of the distribution. Write the variance function V(μ) if asked.

Quickest way: Match-the-pattern shortcut

When to use it: Use it when the question names a standard distribution and asks for θ, b(θ), φ and the moments.

  1. Recall the canonical θ: normal θ = μ, Poisson θ = ln μ, binomial θ = ln[p ÷ (1 − p)], gamma θ = −1 ÷ μ.
  2. Recall b(θ): θ² ÷ 2, e^θ, n ln(1 + e^θ), −ln(−θ).
  3. Differentiate b twice and multiply b″ by a(φ). This is quick and shows the working the examiner wants.
  4. Still show one line of log-density expansion to justify θ. Marks are for the derivation, not only the result.
  5. Check the answer: it must equal the known mean and variance.

Common mistakes in Exponential Family of Distributions

  • Taking θ to be the mean for every distribution.

    For the normal, θ = μ, so students assume it always holds.

    Fix: Find θ by matching the coefficient of y in the log-density. For Poisson it is ln μ. For gamma it is −1 ÷ μ.

  • Forgetting a(φ) in the variance and writing Var(Y) = b″(θ).

    For Poisson and binomial a(φ) = 1, so the factor never shows up.

    Fix: Always write Var(Y) = a(φ)·b″(θ). Check with the normal, where b″ = 1 and the variance is σ² = a(φ).

  • Mixing up the gamma parameters, so the variance comes out wrong.

    The gamma is written with rate and shape in the syllabus, but the GLM form uses mean μ and shape α.

    Fix: Reparametrise with rate = α ÷ μ. Then φ = 1 ÷ α and Var = φμ². Check: shape α, rate λ gives variance α ÷ λ² = μ² ÷ α.

  • Putting terms that mix y and the parameter into c(y, φ).

    Students rush to tidy the log-density without separating the terms.

    Fix: c(y, φ) may depend only on y and φ. Any term containing both y and the parameter belongs in yθ ÷ a(φ).

  • Applying the binomial form to the proportion without adjusting a(φ) and b(θ).

    Some texts use Y = X ÷ n, which changes the scale.

    Fix: State clearly whether Y is the count or the proportion. For the count use b(θ) = n ln(1 + e^θ) and a(φ) = 1. For the proportion use b(θ) = ln(1 + e^θ) and a(φ) = 1 ÷ n.

  • Claiming a distribution is in the family without checking the range of y.

    Students focus on the algebra only.

    Fix: The set of possible y values must not depend on the parameters. A uniform on (0, θ) fails this test.

Worked examples

Example 1

Show that the exponential distribution with density f(y) = λe^(−λy), y > 0, belongs to the exponential family. Identify θ, b(θ), a(φ) and c(y, φ), and use b(θ) to find the mean and variance.

Show the solution
  1. Take logs: ln f = ln λ − λy.
  2. Match the term in y: yθ ÷ a(φ) = −λy. Take a(φ) = φ = 1, so θ = −λ.
  3. The remaining term is ln λ. Since λ = −θ, ln λ = ln(−θ). This must equal −b(θ) ÷ a(φ), so b(θ) = −ln(−θ).
  4. Nothing is left, so c(y, φ) = 0. The range y > 0 does not depend on λ. So f = exp{yθ − b(θ)} with θ < 0.
  5. Mean = b′(θ) = −1 ÷ θ = 1 ÷ λ.
  6. Variance = a(φ)·b″(θ) = 1 × 1 ÷ θ² = 1 ÷ λ².

Answer: θ = −λ, b(θ) = −ln(−θ), a(φ) = 1, c(y, φ) = 0. Mean = 1 ÷ λ and variance = 1 ÷ λ². This is the gamma case with α = 1.

Example 2

Let Y ~ Binomial(10, 0.3), the number of successes. Write the probability function in exponential family form. Find θ and b(θ), and use b(θ) to find the mean and variance.

Show the solution
  1. P(Y = y) = C(10, y) p^y (1 − p)^(10 − y). Take logs: ln C(10, y) + y ln p + (10 − y) ln(1 − p).
  2. Group terms in y: y[ln p − ln(1 − p)] + 10 ln(1 − p) + ln C(10, y).
  3. So θ = ln[p ÷ (1 − p)] and a(φ) = 1. Then p = e^θ ÷ (1 + e^θ) and 1 − p = 1 ÷ (1 + e^θ).
  4. The parameter-only term is 10 ln(1 − p) = −10 ln(1 + e^θ). So b(θ) = 10 ln(1 + e^θ), and c(y) = ln C(10, y).
  5. Here θ = ln(0.3 ÷ 0.7) = ln(3 ÷ 7) ≈ −0.8473, so e^θ = 3 ÷ 7.
  6. b′(θ) = 10e^θ ÷ (1 + e^θ) = 10 × (3 ÷ 7) ÷ (10 ÷ 7) = 3.
  7. b″(θ) = 10e^θ ÷ (1 + e^θ)² = 10 × (3 ÷ 7) × (49 ÷ 100) = 2.1. Since a(φ) = 1, variance = 2.1.
  8. Check: np = 3 and np(1 − p) = 10 × 0.3 × 0.7 = 2.1.

Answer: θ = ln(3 ÷ 7) ≈ −0.8473, b(θ) = 10 ln(1 + e^θ), a(φ) = 1. Mean = 3 and variance = 2.1.

Exam tips

  • Always show the log-density and the matching of terms. Examiners award method marks for identifying θ, b(θ), a(φ) and c(y, φ) separately.
  • Memorise θ and b(θ) for the normal, Poisson, binomial and gamma. These four are the usual exam cases.
  • State whether the binomial response is a count or a proportion before you write b(θ) and a(φ).
  • In Paper B (computer-based), check an R or Excel result against b′(θ) and a(φ)·b″(θ) for the same parameters.
  • Finish by writing the variance function V(μ). It links directly to the next topic, link functions and fitting.

Practice questions from Generalised linear models

Exponential Family of Distributions: frequently asked questions

How do I show a distribution belongs to the exponential family?

Write the log of the density and separate the terms. Match the term in y and the parameter to yθ ÷ a(φ), the parameter-only term to −b(θ) ÷ a(φ), and put the rest in c(y, φ). Also check that the range of y does not depend on the parameter.

What is the canonical parameter?

It is θ, the parameter that multiplies y in the exponent of the density. For normal it is μ, for Poisson ln μ, for binomial the log-odds ln[p ÷ (1 − p)], and for gamma −1 ÷ μ. The canonical link function is the function that makes the linear predictor equal θ.

How do I get the mean and variance from the cumulant function?

Differentiate b(θ). The mean is b′(θ) and the variance is a(φ)·b″(θ). Then substitute θ in terms of the original parameters.

What is the dispersion parameter?

It is φ, which scales the variance through a(φ). For Poisson and binomial it is fixed at 1. For the normal it equals σ², and for the gamma it equals 1 ÷ α.