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Actuarial Statistics · Generating functions

Cumulant Generating Function: Mean, Variance and Skewness

Updated 11 October 2026 · Fact-checked

The cumulant generating function (CGF) is the natural log of the moment generating function: K(t) = ln M(t). Differentiate it at t = 0. The first derivative gives the mean, the second gives the variance, and the third gives the third central moment, which you use for skewness.

Understand Cumulant Generating Function

The moment generating function (MGF) of X is M(t) = E[e^(tX)]. Its derivatives at t = 0 give the raw moments E[X], E[X²], E[X³] and so on. This works, but turning raw moments into variance and skewness takes extra algebra.

The cumulant generating function removes most of that work. You define K(t) = ln M(t), wherever M(t) exists in an interval around 0. Now the derivatives at t = 0 are called cumulants. The r-th cumulant is κ_r = K^(r)(0).

The first three cumulants have a direct meaning. κ1 is the mean. κ2 is the variance. κ3 is the third central moment, E[(X − μ)³]. From the fourth cumulant onwards, the cumulants are no longer equal to central moments. For example, κ4 = E[(X − μ)⁴] − 3σ⁴.

Why does the log help? If X and Y are independent, M_{X+Y}(t) = M_X(t) × M_Y(t). Taking logs turns the product into a sum: K_{X+Y}(t) = K_X(t) + K_Y(t). So cumulants of a sum of independent variables simply add. This is a big saving for sums of independent claims.

Skewness is the third central moment divided by σ³. In cumulant form, skewness = κ3 ÷ κ2^(3/2). You never need to compute E[X³] and then adjust it.

Key rules to remember

Definition of the CGF
K(t) = ln M(t) = ln E[e^(tX)]
Valid where the MGF exists in an interval around t = 0. Note K(0) = ln 1 = 0.
Cumulants
κ_r = K^(r)(0), the r-th derivative of K at t = 0
Equivalently, K(t) = κ1 t + κ2 t²/2! + κ3 t³/3! + ...
Mean
E[X] = κ1 = K'(0)
Same as M'(0), because M(0) = 1.
Variance
Var(X) = κ2 = K''(0)
Easier than M''(0) − (M'(0))².
Third central moment
E[(X − μ)³] = κ3 = K'''(0)
This equality holds for the third cumulant only, together with the first two.
Skewness
γ1 = κ3 ÷ κ2^(3/2) = K'''(0) ÷ (K''(0))^(3/2)
Positive means a longer right tail. Zero for symmetric distributions with finite third moment.
Independent sums
K_{X+Y}(t) = K_X(t) + K_Y(t)
Requires X and Y to be independent. Cumulants then add.
Linear change
K_{aX+b}(t) = bt + K_X(at)
Shifting by b changes only κ1. Higher cumulants scale by a^r.

How to solve Cumulant Generating Function questions

Use this method whenever a question gives you an MGF, a distribution, or asks for mean, variance or skewness through cumulants.

  1. 1Write down the MGF M(t). If it is not given, derive it from E[e^(tX)] or recall it for the standard distribution.
  2. 2Take the natural log to get K(t) = ln M(t). Simplify using log rules so the expression is as short as possible.
  3. 3Differentiate K(t) three times if skewness is needed. Simplify at each stage.
  4. 4Substitute t = 0 into each derivative to get κ1, κ2 and κ3.
  5. 5Read off: mean = κ1, variance = κ2, third central moment = κ3.
  6. 6Compute skewness as κ3 ÷ κ2^(3/2) if asked.
  7. 7For sums of independent variables, add the CGFs first, then differentiate. For aX + b, use the linear change rule.
  8. 8Check the answer: the variance must be positive, and the result should match any known standard value.

Quickest way: Expand K(t) as a power series

When to use it: Use this when the MGF is a standard form like an exponential of a polynomial, or when the log simplifies to a short sum of terms.

  1. Take ln M(t) and simplify to a short expression.
  2. Expand any logs or exponentials as a series in t, for example ln(1 − x) = −x − x²/2 − x³/3 − ...
  3. Match the coefficient of t^r to κ_r ÷ r!. So κ1 is the t coefficient, κ2 is 2 × the t² coefficient, κ3 is 6 × the t³ coefficient.
  4. Compute skewness from κ3 ÷ κ2^(3/2).
  5. This avoids repeated differentiation and the quotient rule.

Common mistakes in Cumulant Generating Function

  • Using E[X²] as the variance, or forgetting to subtract the mean squared when working from the MGF.

    Students carry over the MGF habit, where M''(0) gives E[X²], not the variance.

    Fix: With the CGF, K''(0) is already the variance. Do not subtract anything.

  • Assuming κ4 equals the fourth central moment.

    The first three cumulants match the mean and central moments, so students assume the pattern continues.

    Fix: Remember κ4 = μ4 − 3σ⁴. Only κ1, κ2 and κ3 follow the simple pattern.

  • Writing K(t) = ln t or differentiating ln M(t) without the chain rule.

    Rushing the derivative. K'(t) = M'(t) ÷ M(t), not M'(t).

    Fix: Write K'(t) = M'(t) ÷ M(t) explicitly, or simplify ln M(t) first so the chain rule is easy.

  • Adding CGFs for dependent variables.

    Students remember that cumulants add and forget the independence condition.

    Fix: State independence before adding. Without it, K_{X+Y} ≠ K_X + K_Y in general.

  • Computing skewness as κ3 ÷ κ2 or κ3 ÷ κ2².

    Mixing up the power of the standard deviation.

    Fix: Skewness divides by σ³, which is κ2^(3/2). Write σ = √κ2 first, then cube it.

  • Forgetting that scaling by a multiplies κ_r by a^r, and shifting affects only κ1.

    Students apply the mean-shift logic to every cumulant.

    Fix: Use K_{aX+b}(t) = bt + K_X(at). Then the t-derivatives give the pattern directly.

Worked examples

Example 1

X has MGF M(t) = exp(λ(e^t − 1)), for λ > 0. Find the CGF and use it to find the mean, variance and skewness of X.

Show the solution
  1. K(t) = ln M(t) = λ(e^t − 1).
  2. K'(t) = λe^t, so κ1 = K'(0) = λ.
  3. K''(t) = λe^t, so κ2 = K''(0) = λ.
  4. K'''(t) = λe^t, so κ3 = K'''(0) = λ.
  5. Skewness = κ3 ÷ κ2^(3/2) = λ ÷ λ^(3/2) = 1 ÷ √λ.

Answer: Mean = λ, variance = λ, skewness = 1/√λ. This is the Poisson distribution.

Example 2

X ~ Gamma(α, λ) has MGF M(t) = (1 − t/λ)^(−α) for t < λ. Use the CGF to find the mean, variance and skewness. Then find the skewness when α = 4.

Show the solution
  1. K(t) = −α ln(1 − t/λ).
  2. K'(t) = −α × (−1/λ) ÷ (1 − t/λ) = (α/λ) ÷ (1 − t/λ). At t = 0, κ1 = α/λ.
  3. K''(t) = (α/λ²) ÷ (1 − t/λ)². At t = 0, κ2 = α/λ².
  4. K'''(t) = 2(α/λ³) ÷ (1 − t/λ)³. At t = 0, κ3 = 2α/λ³.
  5. Skewness = κ3 ÷ κ2^(3/2) = (2α/λ³) ÷ (α^(3/2)/λ³) = 2 ÷ √α.
  6. With α = 4, skewness = 2 ÷ 2 = 1.

Answer: Mean = α/λ, variance = α/λ², skewness = 2/√α. For α = 4 the skewness is 1.

Exam tips

  • In written questions, state K(t) = ln M(t) first and show each derivative. Marks go for method even if arithmetic slips.
  • If the question says sum of independent variables, add CGFs. This is usually the intended shortcut.
  • Simplify ln M(t) fully before differentiating. Many MGFs become a simple expression after taking logs.
  • Check your skewness sign. A right-skewed claims distribution such as the gamma should give a positive value.
  • In MCQs, mean = κ1, variance = κ2 and third central moment = κ3 are the most tested links. Do not assume anything about κ4 without computing it.

Practice questions from Generating functions

Cumulant Generating Function: frequently asked questions

What is the difference between the MGF and the CGF?

The MGF is E[e^(tX)] and its derivatives give raw moments. The CGF is the natural log of the MGF and its derivatives give cumulants. The CGF makes variance and skewness easier and turns products of MGFs into sums.

How do I find skewness using the CGF?

Find K'''(0) and K''(0). Skewness is K'''(0) divided by (K''(0)) raised to the power 3/2. This is the third central moment over the cube of the standard deviation.

Why do cumulants add for independent variables?

For independent X and Y, the MGF of the sum is the product of the MGFs. Taking logs turns that product into a sum. So the CGF, and hence every cumulant, of the sum is the sum of the individual ones.

Is the fourth cumulant the same as the fourth central moment?

No. The fourth cumulant is the fourth central moment minus 3σ⁴. Only the first three cumulants equal the mean, variance and third central moment.