The Moment-Generating Function of Severity

Why Use a Moment-Generating Function?

Raw moments such as $E[X]$ and $E[X^2]$ describe important features of a claim severity model. The moment-generating function packages those moments into a function whose derivatives, when the required conditions hold, recover them. It can therefore make moment calculations systematic and can help compare models.

The qualification “when the required conditions hold” matters. A distribution can have a finite mean—or even several finite moments—while its moment-generating function is infinite for every positive argument. This often occurs with heavy-tailed severity models. This tutorial develops the definition, the differentiation result, and the domain checks needed to avoid that error. The treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For a loss $X$, its moment-generating function (MGF) is $M_X(t)=E[e^{tX}]$ at each real value of $t$ for which this expectation is finite. In the usual sense, the MGF exists if it is finite on some open interval containing $t=0$. For a nonnegative severity, $M_X(0)=1$, and $M_X(t)\leq 1$ for $t\leq 0$; finiteness for positive values near zero is the crucial additional condition.

For a continuous loss with density $f_X$, the expectation can be written as an integral. For a discrete loss, it is a probability-weighted sum. The parameter $t$ must be interpreted in units reciprocal to the units of $X$: if $X$ is measured in dollars, $t$ is measured per dollar, so that $tX$ is dimensionless.

M_X(t)=\int_{-\infty}^{\infty} e^{tx}f_X(x)\,dx \quad\text{or}\quad M_X(t)=\sum_x e^{tx}P(X=x).

Derivatives and Raw Moments

If the MGF is finite on an open interval containing zero, its derivatives at zero give the raw moments of $X$. A raw moment is $E[X^n]$, not a central moment such as the variance.

Theorem: If $M_X(t)$ is finite for all $t$ in some open interval containing zero, then for each nonnegative integer $n$, $M_X^{(n)}(0)=E[X^n]$, provided the moment is interpreted under the same probability model. In particular, $M_X(0)=1$, $M_X'(0)=E[X]$, and $M_X''(0)=E[X^2]$.

The reason is that differentiating $e^{tX}$ with respect to $t$ brings down a factor of $X$ each time. For example, the first two derivatives of the integrand are $Xe^{tX}$ and $X^2e^{tX}$. Finiteness in an interval around zero justifies moving those derivatives inside the expectation. More precisely, choose a positive $a$ strictly inside an interval where the MGF is finite. For $t$ sufficiently close to zero, the terms $X^n e^{tX}$ are bounded by a constant multiple of $e^{aX}$ when $X\geq0$. The latter has finite expectation, providing an integrable bound for differentiation. Evaluating the resulting derivative formula at zero gives $E[X^n e^{0X}]=E[X^n]$.

M_X^{(n)}(t)=E[X^n e^{tX}], \qquad M_X^{(n)}(0)=E[X^n].

The variance is obtained from the first two raw moments, rather than directly from the second derivative alone:

\operatorname{Var}(X)=M_X''(0)-\bigl(M_X'(0)\bigr)^2.

Worked Example: Recovering Moments for an Exponential Severity

A hypothetical equipment policy has claim severity $X$ with an exponential distribution of rate $\lambda=0.001$ per dollar. Find its MGF and use it to calculate the mean, second raw moment, and variance. State the MGF domain.

Method and domain. The exponential density is $f_X(x)=\lambda e^{-\lambda x}$ for $x\geq0$. Thus

M_X(t)=\int_0^\infty e^{tx}\lambda e^{-\lambda x}\,dx =\lambda\int_0^\infty e^{-(\lambda-t)x}\,dx.

This integral converges precisely when $\lambda-t>0$, or $t<0.001$ per dollar. Integrating gives

M_X(t)=\frac{\lambda}{\lambda-t} =\frac{0.001}{0.001-t} =\frac{1}{1-1000t}.

First derivative. Differentiating and evaluating at zero,

M_X'(t)=\frac{1000}{(1-1000t)^2}, \qquad E[X]=M_X'(0)=1000.

The mean severity is therefore $\$1{,}000$.

Second derivative.

M_X''(t)=\frac{2(1000)^2}{(1-1000t)^3}, \qquad E[X^2]=M_X''(0)=2{,}000{,}000.

The variance is the second raw moment minus the square of the mean:

\operatorname{Var}(X) =2{,}000{,}000-(1{,}000)^2 =1{,}000{,}000\text{ dollars}^2.

The standard deviation is $\$1{,}000$. As a check, the MGF gives $M_X(0)=1$, and the calculated variance is positive. This example uses the MGF derivatives to obtain raw moments, then applies the usual raw-moment formula for variance.

MGFs for Mixtures

A mixture severity first selects a component distribution according to specified probabilities and then generates a loss from that component. Because expectation is linear, the mixture MGF is the probability-weighted sum of the component MGFs, at values of $t$ where all terms are finite. This is useful when a portfolio contains distinct claim types or risk groups.

Worked Example: A Two-Component Severity Mixture

A hypothetical repair portfolio has two claim types. With probability $0.75$, severity is exponential with rate $0.001$ per dollar; with probability $0.25$, it is exponential with rate $0.002$ per dollar. Find the MGF domain, mean, second raw moment, and variance of the overall severity.

Method. For an exponential component with rate $\lambda$, the MGF is $\lambda/(\lambda-t)$ for $t<\lambda$. The first component is finite for $t<0.001$, and the second for $t<0.002$. Both terms are finite when $t<0.001$, which is the domain of the mixture MGF around zero. Weighting the component MGFs gives

M_X(t)=0.75\frac{0.001}{0.001-t} +0.25\frac{0.002}{0.002-t} =\frac{0.75}{1-1000t}+\frac{0.25}{1-500t}, \qquad t<0.001.

Mean. Differentiating each term,

M_X'(t)=\frac{0.75(1000)}{(1-1000t)^2} +\frac{0.25(500)}{(1-500t)^2}.

Therefore,

E[X]=M_X'(0) =0.75(1000)+0.25(500) =750+125 =\$875.

Second raw moment. A second differentiation gives

M_X''(0)=0.75\bigl(2(1000)^2\bigr) +0.25\bigl(2(500)^2\bigr) =1{,}500{,}000+125{,}000 =1{,}625{,}000\text{ dollars}^2.

Variance.

\operatorname{Var}(X) =1{,}625{,}000-(875)^2 =1{,}625{,}000-765{,}625 =859{,}375\text{ dollars}^2.

The result is consistent with the mixture: the higher-mean component has substantial weight, so the overall mean is closer to $\$1{,}000$ than to $\$500$. Also, $M_X(0)=0.75+0.25=1$, as required.

Finite Moments Do Not Guarantee an MGF

For a nonnegative loss, negative values of $t$ cause $e^{tX}$ to decrease as the loss increases, so the expectation is bounded by one. Positive values of $t$, however, exponentially increase the contribution from large losses. For a sufficiently heavy tail, that increase makes the expectation infinite.

This is a different question from whether particular moments exist. A distribution may have $E[X]$ and $E[X^2]$ finite while $E[e^{tX}]$ is infinite for every $t>0$. In that case the usual MGF does not exist, even though the mean and variance can still be computed by other methods.

Worked Example: Finite Pareto Moments but No MGF

A hypothetical large-loss severity follows a Pareto Type I model with minimum severity $x_m=\$1{,}000$ and shape parameter $\alpha=3$. Its density is $f_X(x)=\alpha x_m^\alpha x^{-\alpha-1}$ for $x\geq x_m$. Determine whether the MGF exists in an interval around zero, and calculate the mean and variance.

Check positive $t$. For $t>0$, the proposed MGF is

M_X(t)=3(1000)^3\int_{1000}^{\infty} e^{tx}x^{-4}\,dx.

The exponential factor eventually grows faster than the power $x^4$, so the integrand does not have a finite integral over the unbounded upper tail. Thus $M_X(t)=\infty$ for every $t>0$. For $t\leq0$, $0<e^{tX}\leq1$, so the expectation is finite. Since there is no interval around zero that includes positive values where the MGF is finite, this severity has no MGF in the usual sense.

Calculate the mean. For this Pareto model, $E[X^k]=\alpha x_m^k/(\alpha-k)$ when $k<\alpha$. The mean exists because $1<3$:

E[X]=\frac{3(1000)}{3-1} =\$1{,}500.

Calculate the second raw moment and variance. The second moment exists because $2<3$:

E[X^2]=\frac{3(1000)^2}{3-2} =3{,}000{,}000\text{ dollars}^2, $$ $$ \operatorname{Var}(X) =3{,}000{,}000-(1{,}500)^2 =3{,}000{,}000-2{,}250{,}000 =750{,}000\text{ dollars}^2.

The mean and variance are finite, but the MGF still does not exist around zero. This is why one must check the MGF domain rather than assume that existing moments imply an MGF.

Common Exam Trap: Confusing a Raw Moment with Variance

The second derivative at zero is $E[X^2]$, not $\operatorname{Var}(X)$. For the exponential severity in the first example, using $M_X''(0)=2{,}000{,}000$ as the variance would be wrong. The correction is to subtract the square of the mean:

\underbrace{E[X^2]}_{2{,}000{,}000} \ne \operatorname{Var}(X), \qquad \operatorname{Var}(X) =E[X^2]-\bigl(E[X]\bigr)^2 =2{,}000{,}000-1{,}000{,}000 =1{,}000{,}000.

A related trap is differentiating a formula at zero without checking that it represents an MGF on a neighborhood of zero. If a heavy-tailed model has no MGF there, formal algebra with a nonexistent positive-$t$ expectation does not establish that its moments are given by MGF derivatives.

Key takeaway: First establish the domain of $M_X(t)=E[e^{tX}]$. If it is finite around zero, then $M_X^{(n)}(0)=E[X^n]$. Use the first two raw moments to calculate variance, and do not infer MGF existence merely from the finiteness of one or more moments.

Check Your Understanding

For each problem, check both the calculation and any stated MGF domain.

A hypothetical severity is exponential with mean $\$2{,}500$. Write its MGF and state its domain in terms of $t$, with $X$ measured in dollars. (Answer: $M_X(t)=1/(1-2500t)$, $t<0.0004$ per dollar.)
A loss is a mixture: with probability $0.6$ it is exponential with mean $\$400$, and with probability $0.4$ it is exponential with mean $\$100$. Find the overall mean and variance. (Answer: mean $\$280$; variance $121{,}600\text{ dollars}^2$.)
A severity is uniform on $[0,600]$ dollars. Its MGF is $M_X(t)=(e^{600t}-1)/(600t)$ for $t\ne0$, with $M_X(0)=1$. Find its mean and variance using raw moments. (Answer: mean $\$300$; variance $30{,}000\text{ dollars}^2$.)
A Pareto Type I severity has $x_m=\$500$ and $\alpha=4$. Does its MGF exist around zero? Find $E[X^2]$. (Answer: no MGF around zero; $E[X^2]=\$500{,}000\text{ dollars}^2$.)
A portfolio severity is exponential with probability $1/3$ and mean $\$300$, and exponential with probability $2/3$ and mean $\$900$. Find the MGF domain, mean, and variance. (Answer: $t<1/900$ per dollar; mean $\$700$; variance $650{,}000\text{ dollars}^2$.)

What You'll Learn