Moment Generating Functions For Severity Models

Why Use a Moment Generating Function?

A severity distribution can be described by its cumulative distribution function, density, or probability mass function. A moment generating function (MGF) offers another description: when it exists near zero, its derivatives at zero give the distribution’s raw moments. This can turn a moment calculation that would require summing or integrating across the whole severity distribution into a short differentiation problem.

The key qualification is “when it exists near zero.” An MGF is not guaranteed to exist for every severity model, and its value at zero alone does not establish that its derivatives can be used to find moments. These conditions matter in applications and are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For a loss severity random variable $X$, its moment generating function is $$ M_X(t)=E[e^{tX}], $$ for each real number $t$ for which this expectation is finite. The set of such $t$ is the domain of the MGF. In particular, $M_X(0)=1$ whenever $X$ is a proper random variable, since $e^{0X}=1$.

For a nonnegative severity, $M_X(t)$ is finite for every $t\leq 0$, because $0<e^{tX}\leq 1$ there. It may or may not be finite for positive $t$. To use the standard derivative formulas for moments, the MGF must be finite on an open interval containing zero; it is not enough that it is defined only at zero or only on one side of zero.

Moments From Derivatives

Suppose $M_X(t)$ is finite on some open interval containing zero. Under this condition, differentiating the MGF gives the raw moments:

Theorem: If $M_X(t)$ exists on an open interval containing zero, then, for each positive integer $r$, $$ M_X^{(r)}(0)=E[X^r], $$ provided the derivative is of the indicated order. In particular, $$ E[X]=M_X'(0), \qquad E[X^2]=M_X''(0), $$ and $$ \operatorname{Var}(X)=M_X''(0)-\bigl(M_X'(0)\bigr)^2. $$

The second derivative gives the second raw moment, not the variance. The variance is the second raw moment minus the square of the mean. This distinction is essential: the raw second moment is at least as large as the square of the mean, so a negative variance signals a calculation error.

Why Differentiation Produces Moments

For a fixed value of $X$, the first two derivatives of $e^{tX}$ with respect to $t$ are $Xe^{tX}$ and $X^2e^{tX}$. If $M_X(t)$ is finite in a neighborhood of zero, the expectation and differentiation can be interchanged under the usual conditions for differentiating an expectation. Evaluating those differentiated expressions at zero gives the moments.

Start from the definition.
$M_X(t)=E[e^{tX}]$.

Differentiate inside the expectation.
In a neighborhood of zero where differentiation under the expectation is justified, $M_X'(t)=E[Xe^{tX}]$ and $M_X''(t)=E[X^2e^{tX}]$.

Set $t=0$.
Since $e^{0X}=1$, $M_X'(0)=E[X]$ and $M_X''(0)=E[X^2]$.

Convert raw moments to variance.
Use $\operatorname{Var}(X)=E[X^2]-(E[X])^2$, not $E[X^2]$ alone.

Worked Calculations

Worked Example: Mean and Variance From a Given MGF

A hypothetical nonnegative severity model has $M_X(t)=(1-t/4)^{-3}$ for $t<4$. Find $E[X]$ and $\operatorname{Var}(X)$. The model and its MGF are supplied; no distribution-family identification is needed.

Method. Differentiate the given MGF, evaluate the first two derivatives at zero, and then convert the raw moments to variance. The stated domain includes an open interval around zero, so the derivative formulas apply.

First derivative. By the chain rule,

M_X'(t)=\frac{3}{4}(1-t/4)^{-4}, \qquad M_X'(0)=\frac{3}{4}.

Thus $E[X]=0.75$.

Second derivative. Differentiating once more,

M_X''(t)=\frac{3}{4}(1-t/4)^{-5}, \qquad M_X''(0)=\frac{3}{4}.

Therefore $E[X^2]=0.75$, and

\operatorname{Var}(X) =E[X^2]-(E[X])^2 =\frac{3}{4}-\left(\frac{3}{4}\right)^2 =\frac{12}{16}-\frac{9}{16} =\frac{3}{16} =0.1875.

Check. The variance is positive, as required. The common error is to report $M_X''(0)=0.75$ as the variance, or to make an arithmetic error in subtracting the squared mean. The correct variance is $0.1875$.

Worked Example: Using a Logarithm to Simplify Differentiation

For another hypothetical nonnegative severity, suppose $M_X(t)=e^{2t}(1-t/3)^{-2}$ for $t<3$. Calculate the mean and variance. The expression is finite on an open interval containing zero.

Method. Use logarithmic differentiation. Let $L(t)=\log M_X(t)$. Then $M_X'(t)=M_X(t)L'(t)$, and differentiating again gives $M_X''(t)=M_X(t)\{L''(t)+(L'(t))^2\}$.

First,

L(t)=2t-2\log(1-t/3), \qquad L'(t)=2+\frac{2}{3}(1-t/3)^{-1}, \qquad L''(t)=\frac{2}{9}(1-t/3)^{-2}.

At $t=0$, $M_X(0)=1$, $L'(0)=2+2/3=8/3$, and $L''(0)=2/9$. Hence

E[X]=M_X'(0)=M_X(0)L'(0)=\frac{8}{3}.

For the second raw moment,

E[X^2]=M_X''(0) =M_X(0)\{L''(0)+(L'(0))^2\} =\frac{2}{9}+\left(\frac{8}{3}\right)^2 =\frac{2}{9}+\frac{64}{9} =\frac{66}{9} =\frac{22}{3}.

Therefore,

\operatorname{Var}(X) =\frac{22}{3}-\left(\frac{8}{3}\right)^2 =\frac{66}{9}-\frac{64}{9} =\frac{2}{9}.

Check. The raw second moment is $22/3$, while the variance is $2/9$. Keeping these quantities distinct avoids the common mistake of treating $M_X''(0)$ as the variance.

When the MGF Does Not Exist Near Zero

A model can have $M_X(0)=1$ and still have no MGF on any open interval containing zero. For a nonnegative severity, negative arguments cause no problem, but positive arguments weight large losses by the rapidly growing factor $e^{tX}$. If the tail is sufficiently heavy, the expectation can be infinite for every $t>0$.

Worked Example: A Severity With No MGF Near Zero

A hypothetical loss $X$ has density $f_X(x)=2/(1+x)^3$ for $x\geq 0$, and zero otherwise. Determine where its MGF is finite and whether its first two moments can be obtained by the usual MGF derivative rule.

Check the density. Its total probability is

\int_0^\infty \frac{2}{(1+x)^3}\,dx =2\left[\frac{-(1+x)^{-2}}{2}\right]_0^\infty =1.

For $t>0$, the MGF integral is $\int_0^\infty e^{tx}2/(1+x)^3\,dx$. The exponential factor grows faster than the density decreases, so this integral diverges. For $t\leq0$, $e^{tX}\leq1$, and the expectation is finite. Thus the MGF is finite for $t\leq0$ but not on any open interval containing zero.

The first moment is finite and can be checked directly:

E[X] =2\int_0^\infty \frac{x}{(1+x)^3}\,dx =2\int_1^\infty (u^{-2}-u^{-3})\,du =2\left(1-\frac{1}{2}\right) =1,

where $u=1+x$. But the second moment diverges: its integrand $2x^2/(1+x)^3$ behaves like $2/x$ for large $x$, and $\int_1^\infty 1/x\,dx$ diverges. The usual theorem requiring an MGF near zero cannot be invoked. In particular, there is no finite second raw moment or finite variance here. A one-sided derivative at zero, if considered, is not a substitute for the neighborhood condition in the standard theorem.

Check. The MGF exists at zero and on the nonpositive side, but not near zero in the required two-sided sense. This example shows why checking the domain is part of using the method, not an optional detail.

Common Traps and a Reliable Procedure

In an exam calculation, the algebra is often short; most errors come from applying the right calculation to the wrong quantity or without checking its assumptions. Use this sequence whenever an MGF is supplied.

Check the domain. Confirm that the MGF is finite on an open interval containing zero before using the standard derivative formulas.
Differentiate carefully. Apply the chain rule to every nested expression and retain all factors.
Evaluate at zero. Substitute $t=0$ only after differentiating.
Label each result. The first derivative gives the mean; the second gives the second raw moment.
Calculate variance last. Subtract the square of the mean from the second raw moment, and check that the result is nonnegative.

A common wrong calculation is $\operatorname{Var}(X)=M_X''(0)$. For the first worked example, this would give $0.75$, even though the mean is also $0.75$. The correction is $\operatorname{Var}(X)=0.75-(0.75)^2=0.1875$. Another trap is assuming that because $M_X(0)=1$, the MGF exists in a neighborhood of zero. The third worked example shows that this implication is false.

Key takeaway: When an MGF exists near zero, its derivatives at zero give raw moments: $M_X'(0)=E[X]$ and $M_X''(0)=E[X^2]$. Obtain variance by subtracting the squared mean, and check the MGF’s domain before applying the formulas.

Check Your Understanding

For each problem, check the domain condition before using MGF derivatives.

A hypothetical severity has $M_X(t)=(1-t/5)^{-2}$ for $t<5$. Find its mean and variance. (Answer: $E[X]=0.4000$; $\operatorname{Var}(X)=0.0800$.)
A severity has $M_X(t)=e^t(1-t/4)^{-1}$ for $t<4$. Use logarithmic differentiation to find its mean and variance. (Answer: $E[X]=1.2500$; $\operatorname{Var}(X)=0.0625$.)
A severity has density $f_X(x)=3/(1+x)^4$ for $x\geq0$. Is its MGF finite for any $t>0$? Find $E[X]$, $E[X^2]$, and its variance by direct integration or an equivalent justified method. (Answer: The MGF is infinite for every $t>0$; $E[X]=0.5000$, $E[X^2]=1.0000$, and $\operatorname{Var}(X)=0.7500$.)
A model has $M_X(t)=(1-t/2)^{-1}$ for $t<2$. Find the first three raw moments. (Answer: $E[X]=0.5000$, $E[X^2]=0.5000$, and $E[X^3]=0.7500$.)
A supplied MGF is finite only for $t\leq0$, although $M_X(0)=1$. Can the standard two-sided MGF derivative theorem automatically be used to obtain moments? Explain the relevant condition. (Answer: No. The standard theorem requires the MGF to be finite on an open interval containing zero; finiteness only for $t\leq0$ does not meet that condition.)

What You'll Learn