Deriving Lognormal Moments From The Normal Model

Why derive moments on the log scale?

A lognormal loss is positive and often described by parameters for its logarithm. Those parameters are not the loss's mean and variance. The normal model for the logarithm provides a direct route to raw moments of the loss, but only if the variable in the moment-generating function is identified correctly. This distinction matters in calculations and when reasoning about whether a moment exists.

Throughout, take $X$ to be a positive loss and let $c$ denote a fixed positive unit of measurement, such as one dollar or one thousand dollars. The dimensionless quantity $X/c$ has a logarithm. Assume that logarithm is normal, with positive standard deviation $\sigma$.

Definition: $X$ has a lognormal distribution with log-scale parameters $\mu$ and $\sigma^2$, where $\sigma>0$, if $Z=\ln(X/c)\sim N(\mu,\sigma^2)$. Equivalently, $X=c e^Z$. The $k$th raw moment is $E[X^k]$, not a moment of $Z$.

Changing the measurement unit changes $c$ and the numerical value of $X$, but not the log-scale parameters if $c$ is changed consistently. The factor $c^k$ in the raw-moment formula accounts for this scaling.

Deriving the raw-moment formula

The moment-generating function of the normal variable $Z$ is $M_Z(t)=E[e^{tZ}]$. For $Z\sim N(\mu,\sigma^2)$, it is finite for every real $t$ and equals $\exp(\mu t+\sigma^2t^2/2)$. Since $X=c e^Z$, the $k$th raw moment of $X$ is obtained by evaluating this function at $t=k$, then applying the scale factor.

Express the loss in terms of the normal variable.
By definition, $X=c e^Z$, so $X^k=c^k e^{kZ}$.

Take expectations.
Thus $E[X^k]=c^kE[e^{kZ}]=c^kM_Z(k)$.

Substitute the normal moment-generating function.
Evaluating it at $k$ gives $E[X^k]=c^k\exp(k\mu+k^2\sigma^2/2)$.

E[X^k]=c^k\exp\left(k\mu+\frac{k^2\sigma^2}{2}\right).

This calculation applies for every real $k$, not only integers: the normal moment-generating function exists at every real argument. In particular, for each positive integer $k$, the expression is a finite positive number. Therefore every positive integer raw moment of a lognormal loss is finite.

The first two raw moments give useful checks on calculations. Setting $k=1$ and $k=2$ yields the mean and second raw moment. The variance follows from the definition $\operatorname{Var}(X)=E[X^2]-(E[X])^2$.

E[X]=c\exp\left(\mu+\frac{\sigma^2}{2}\right), \qquad E[X^2]=c^2\exp\left(2\mu+2\sigma^2\right),

\operatorname{Var}(X) =c^2\exp(2\mu+\sigma^2)\left(\exp(\sigma^2)-1\right).

For the variance identity, square the mean to obtain $c^2\exp(2\mu+\sigma^2)$, and subtract this from the second raw moment $c^2\exp(2\mu+2\sigma^2)$. Factoring out $c^2\exp(2\mu+\sigma^2)$ gives the stated result. The variance is nonnegative because $\exp(\sigma^2)-1>0$ when $\sigma>0$.

Worked Example: Mean and Variance With a Thousand-Dollar Unit

A hypothetical insurer models a claim amount $X$ in dollars by $Z=\ln(X/\$1{,}000)\sim N(0,\ln 2)$. Find $E[X]$, $E[X^2]$, and $\operatorname{Var}(X)$. Report the mean in dollars to the cent and the variance in dollars squared to the nearest dollar squared.

Method and parameters. This is a lognormal model with $c=\$1{,}000$, $\mu=0$, and $\sigma^2=\ln 2$. Apply the raw-moment formula separately for $k=1$ and $k=2$.

For the mean, $E[X]=1{,}000\exp(0+\ln 2/2)=1{,}000\sqrt{2}=\$1{,}414.21356\ldots$, so $E[X]=\$1{,}414.21$.

For the second raw moment, $E[X^2]=1{,}000^2\exp(0+2\ln 2)=1{,}000{,}000\times 4=4{,}000{,}000$ dollars squared.

The squared mean is $(1{,}000\sqrt{2})^2=2{,}000{,}000$ dollars squared. Therefore, $\operatorname{Var}(X)=4{,}000{,}000-2{,}000{,}000=2{,}000{,}000$ dollars squared.

Check. The second raw moment exceeds the squared mean by exactly $2{,}000{,}000$, as required for a positive variance. This calculation avoids the common mistake of treating the log-scale variance $\ln 2$ as the variance of the claim amount.

Higher moments and the moment-generating-function trap

The raw-moment formula also works directly for higher powers. The exponent contains $k^2\sigma^2/2$, not $k\sigma^2/2$. That square is a consequence of evaluating the normal moment-generating function at $k$; it makes higher moments grow rapidly when log-scale variability is substantial.

There is an important distinction between the moment-generating function of $Z=\ln(X/c)$ and that of $X$. The normal moment-generating function $M_Z(t)$ is finite for all real $t$. This does not mean that the loss moment-generating function $M_X(t)=E[e^{tX}]$ is finite for positive $t$. In fact, for a nondegenerate lognormal loss, $M_X(t)=\infty$ for every $t>0$.

To see why, write $X=c e^Z$. The expectation defining $M_X(t)$ includes the factor $\exp(tc e^z)$ integrated against the normal density of $Z$. As $z$ tends to infinity, $tc e^z$ grows faster than the quadratic term in the normal density's exponent. Consequently the integrand does not decay sufficiently for the integral to converge. Thus the positive loss MGF does not exist as a finite value, even though the raw-moment formula proves that each fixed positive integer moment is finite.

Worked Example: Computing Third and Fourth Raw Moments

A hypothetical equipment insurer measures losses in thousands of dollars and assumes $\ln(X/(\$1{,}000))\sim N(\ln 2,0.25)$. Calculate the third and fourth raw moments in powers of thousands of dollars, and find the variance in those units squared. Round numerical results to four decimal places.

Method and parameters. Let $U=X/(\$1{,}000)$. Then $\ln U\sim N(\mu,\sigma^2)$ with $\mu=\ln 2$ and $\sigma^2=0.25$. Apply the formula with scale one to $U$; its moments are expressed in powers of thousands of dollars.

For $k=3$, $E[U^3]=\exp(3\ln 2+(9/2)(0.25))=\exp(\ln 8+1.125)=8e^{1.125}=24.6417$ (thousands of dollars)$^3$.

For $k=4$, $E[U^4]=\exp(4\ln 2+(16/2)(0.25))=\exp(\ln 16+2)=16e^2=118.2249$ (thousands of dollars)$^4$.

To obtain the variance, first calculate $E[U]=2e^{0.125}=2.2663$ and $E[U^2]=4e^{0.5}=6.5949$. Using unrounded values, $(E[U])^2=4e^{0.25}=5.1361$, so $\operatorname{Var}(U)=4e^{0.5}-4e^{0.25}=1.4588$ (thousands of dollars)$^2$.

Check. The variance is positive, and the reported third and fourth moments follow from the same substitution rule with $k=3$ and $k=4$. The common error avoided here is using $k\sigma^2/2$ in place of $k^2\sigma^2/2$.

Worked Example: A Fifth Moment and General Moment Existence

A hypothetical specialty claim is measured in units of \$10,000. Let $U=X/(\$10{,}000)$, with $\ln U\sim N(-\ln 2,\ln 2)$. Calculate $E[U^5]$ to four decimal places and explain whether the sixth raw moment exists.

Method and parameters. Here the scale for $U$ is one, $\mu=-\ln 2$, and $\sigma^2=\ln 2$. For the fifth moment, substitute $k=5$: $E[U^5]=\exp(-5\ln 2+(25/2)\ln 2)=\exp((15/2)\ln 2)=2^{15/2}=128\sqrt{2}=181.0193$.

Thus the fifth raw moment is $181.0193$ in units of (\$10,000)$^5$. For the sixth moment, the same formula gives $E[U^6]=\exp(-6\ln 2+18\ln 2)=2^{12}=4{,}096$, which is finite.

Check. The fifth-moment exponent is $-5\ln 2+25\ln 2/2=15\ln 2/2$, and the sixth-moment exponent is $-6\ln 2+36\ln 2/2=12\ln 2$. The negative log-scale mean does not prevent either moment from being finite.

Common exam trap

A tempting but incorrect step is to write $M_X(t)=\exp(\mu t+\sigma^2t^2/2)$. That expression is $M_Z(t)$, the MGF of the log loss, not the MGF of $X$. The correct calculation for a raw moment uses $X=c e^Z$, so $E[X^k]=c^kM_Z(k)$. Do not infer from the finiteness of all integer moments that $M_X(t)$ is finite for positive $t$; the lognormal loss MGF is infinite there.

Key takeaway: If $\ln(X/c)\sim N(\mu,\sigma^2)$, then $E[X^k]=c^k\exp(k\mu+k^2\sigma^2/2)$. This establishes finite moments for every positive integer $k$, while leaving the positive MGF of $X$ nonexistent.

Check Your Understanding

For each problem, use the normal model for the log loss and keep the stated units throughout.

A hypothetical loss satisfies $\ln(X/\$1{,}000)\sim N(0.2,0.09)$. Find $E[X^2]$ in dollars squared, to the nearest dollar squared. (Answer: $1{,}786{,}038$ dollars squared.)
Let $U=X/(\$100)$, where a hypothetical claim model has $\ln U\sim N(\ln 3,0.16)$. Find $E[X]$ to the nearest cent. (Answer: $\$324.99$.)
A hypothetical severity has $\ln(X/\$1{,}000)\sim N(-0.1,0.25)$. Find $E[X^3]$ in dollars cubed, to three significant figures. (Answer: $2.28\times 10^9$ dollars cubed.)
Suppose $\ln U\sim N(0,0.04)$, where $U=X/c$ for a fixed measurement unit $c$. Find $E[U^{10}]$ to four decimal places. (Answer: $7.3891$.)
A nondegenerate loss satisfies $\ln(X/c)\sim N(1,0.25)$. Find $E[X^2]/c^2$, and state whether $M_X(1)$ is finite. (Answer: $e^{2.5}\approx12.1825$; $M_X(1)=\infty$.)

What You'll Learn