Lognormal Moments and Shape

Introduction

A lognormal variable is positive, but its logarithm has a normal distribution. This link makes its moments accessible through the normal moment-generating function, while also producing a characteristic right-skewed shape. The parameters describe the distribution of the logarithm, so translating them into the mean, variance, skewness, and mode of the original variable requires care.

Let \(Y\sim N(\mu,\sigma^2)\), where \(\mu\in\mathbb{R}\) and \(\sigma>0\), and set \(X=e^Y\). Thus \(\log X=Y\), and \(X\) has a lognormal distribution. The parameter \(\mu\) is the mean of \(\log X\), not generally the mean of \(X\); likewise, \(\sigma^2\) is the variance of \(\log X\), not the variance of \(X\).

Definition: A random variable \(X\) is lognormal with log-scale parameters \(\mu\) and \(\sigma^2\), written \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\), if \(X=e^Y\) for \(Y\sim N(\mu,\sigma^2)\), with \(\sigma>0\). Its support is \(x>0\), and its density is

f(x)=\frac{1}{x\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(\log x-\mu)^2}{2\sigma^2}\right),\qquad x>0.

Raw Moments from the Normal MGF

The moment-generating function of \(Y\sim N(\mu,\sigma^2)\) is \(M_Y(t)=E[e^{tY}]=\exp(\mu t+\sigma^2t^2/2)\), finite for every real \(t\). Because \(X^r=e^{rY}\), this MGF directly gives moments of \(X\). In particular, every positive integer raw moment exists; in fact, the same formula holds for any real power \(r\).

Theorem: If \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\), then for every real \(r\),

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

Proof: By the definition \(X=e^Y\), \(X^r=e^{rY}\). The expectation exists because the normal MGF is finite at every real argument. Substituting \(t=r\) into that MGF gives

E[X^r]=E[e^{rY}]=M_Y(r) =\exp\!\left(r\mu+\frac{r^2\sigma^2}{2}\right).

This also establishes finiteness for negative \(r\), since the formula applies to every real \(r\). \(\square\)

For integer \(k\geq 0\), the raw moment of order \(k\) is therefore \(E[X^k]=\exp(k\mu+k^2\sigma^2/2)\). The first three are particularly useful:

E[X]=e^{\mu+\sigma^2/2},\qquad E[X^2]=e^{2\mu+2\sigma^2},\qquad E[X^3]=e^{3\mu+9\sigma^2/2}.

Worked Example: First Three Raw Moments

Suppose \(X\sim\operatorname{Lognormal}(\log 3,\log 2)\). Here \(\mu=\log 3\) and \(\sigma^2=\log 2\). Apply the moment formula at \(r=1,2,3\):

First moment. \(E[X]=\exp(\log 3+\tfrac12\log 2)=3\sqrt{2}\).

Second moment. \(E[X^2]=\exp(2\log 3+2\log 2)=3^2 2^2=36\).

Third moment. \(E[X^3]=\exp(3\log 3+\tfrac92\log 2)=27\cdot 2^{9/2}=432\sqrt{2}\).

These are raw moments, not central moments: for example, \(E[X^2]=36\) is not the variance. The variance requires subtracting the square of the mean.

Mean and Variance

The mean is the first raw moment. For the variance, use \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\). Substituting the raw moments yields

E[X]=e^{\mu+\sigma^2/2},\qquad \operatorname{Var}(X) =e^{2\mu+2\sigma^2}-e^{2\mu+\sigma^2} =e^{2\mu+\sigma^2}(e^{\sigma^2}-1).

The factor \(e^{2\mu+\sigma^2}\) sets the squared scale, while \(e^{\sigma^2}-1\) determines the relative spread. For example, the squared coefficient of variation, \(\operatorname{Var}(X)/(E[X])^2\), equals \(e^{\sigma^2}-1\) and does not depend on \(\mu\).

Worked Example: Mean and Variance with Simple Values

Let \(X\sim\operatorname{Lognormal}(\log 3,\log 2)\), as above. Its mean is \(3\sqrt{2}\), so the squared mean is \(18\). Its variance can be found from either the raw moments or the simplified formula:

Using raw moments: \(\operatorname{Var}(X)=36-(3\sqrt{2})^2=36-18=18\).

Using the variance formula: \(e^{2\log 3+\log 2}(e^{\log 2}-1)=18(2-1)=18\). The two calculations agree. The standard deviation is \(\sqrt{18}=3\sqrt{2}\), equal to the mean in this particular example.

Skewness from the Third Central Moment

Skewness is the standardized third central moment. For a random variable with finite, positive variance, it is

\gamma_1=\frac{E[(X-E[X])^3]}{\operatorname{Var}(X)^{3/2}}.

The lognormal skewness can be derived entirely from the first three raw moments. Write \(q=e^{\sigma^2}\), so \(q>1\). Expanding the cube gives \(E[(X-E[X])^3]=E[X^3]-3E[X]E[X^2]+2E[X]^3\). Substitution and factorization give the following result.

Result: For \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\),

E[(X-E[X])^3]=e^{3\mu+3\sigma^2/2}(e^{\sigma^2}-1)^2(e^{\sigma^2}+2), \qquad \gamma_1=(e^{\sigma^2}+2)\sqrt{e^{\sigma^2}-1}.

Derivation: Set \(q=e^{\sigma^2}\). Factoring \(e^{3\mu+3\sigma^2/2}\) from the expanded third central moment leaves

E[(X-E[X])^3] =e^{3\mu+3\sigma^2/2}(q^3-3q+2) =e^{3\mu+3\sigma^2/2}(q-1)^2(q+2).

The variance is \(e^{2\mu+\sigma^2}(q-1)\), so its \(3/2\) power is \(e^{3\mu+3\sigma^2/2}(q-1)^{3/2}\). Dividing the central moment by this quantity gives \((q+2)\sqrt{q-1}\), as claimed. Since \(q>1\), this skewness is positive. \(\square\)

Worked Example: Calculating Skewness

For \(X\sim\operatorname{Lognormal}(\log 3,\log 2)\), \(e^{\sigma^2}=2\). Thus

\(\gamma_1=(2+2)\sqrt{2-1}=4\).

This is a standardized measure, so it has no units and does not change when \(\mu\) changes. The value \(4\) indicates positive asymmetry: the distribution has a long right tail. It is not a statement that most observations lie above the mean.

Mode and the Ordering of Typical Values

The mode is the value of \(x\) at which the density is maximized. For a lognormal distribution, maximizing the density in \(x\) is not the same as maximizing the normal density of \(\log X\): the density transformation includes the factor \(1/x\). Differentiate the lognormal log-density to locate its maximum:

\frac{d}{dx}\log f(x) =-\frac{1}{x}-\frac{\log x-\mu}{\sigma^2x} =-\frac{\sigma^2+\log x-\mu}{\sigma^2x}.

The denominator is positive for \(x>0\). The derivative is positive when \(x<e^{\mu-\sigma^2}\) and negative when \(x>e^{\mu-\sigma^2}\); hence the density increases up to that point and decreases afterward. The unique mode is

\operatorname{mode}(X)=e^{\mu-\sigma^2}.

The median is \(e^\mu\), since \(\log X=Y\) is symmetric about \(\mu\) and \(P(Y\leq\mu)=1/2\). For \(\sigma>0\), the mode, median, and mean therefore satisfy

e^{\mu-\sigma^2}<e^\mu<e^{\mu+\sigma^2/2}.

This ordering is one concise description of the distribution’s right-skewed shape. Increasing \(\mu\) multiplies all three values by the same positive factor, while increasing \(\sigma^2\) moves the mode downward relative to the median and pushes the mean upward relative to it.

Worked Example: Mean, Median, and Mode

Take \(X\sim\operatorname{Lognormal}(0,1)\). Its mean, median, and mode are, respectively,

Mean: \(e^{0+1/2}=\sqrt{e}\), approximately \(1.649\).

Median: \(e^0=1\).

Mode: \(e^{0-1}=e^{-1}\), approximately \(0.368\).

Thus \(0.368<1<1.649\), in agreement with the general ordering. The mean is pulled above the median by the relatively long right tail.

Recovering Log-Scale Parameters from Moments

The moment formulas can also be rearranged. Suppose a lognormal model has mean \(m>0\) and variance \(v>0\). Since \(v/m^2=e^{\sigma^2}-1\), its log-scale variance must be \(\sigma^2=\log(1+v/m^2)\). Then \(m=e^{\mu+\sigma^2/2}\) determines \(\mu=\log m-\sigma^2/2\). These expressions are useful when a model is specified in terms of its original-scale mean and variance.

Worked Example: Finding Parameters from a Mean and Variance

Suppose a lognormal model is required to have mean \(10\) and variance \(300\). Its squared coefficient of variation is \(300/10^2=3\), so

\(e^{\sigma^2}-1=3\), giving \(e^{\sigma^2}=4\) and \(\sigma^2=\log 4\).

The mean equation gives \(10=e^{\mu+(\log 4)/2}=2e^\mu\), so \(e^\mu=5\) and \(\mu=\log 5\). The resulting model has mode \(e^{\log 5-\log 4}=5/4\), and skewness \((4+2)\sqrt{4-1}=6\sqrt{3}\).

As a check, its mean is \(e^{\log 5+(\log 4)/2}=10\), and its variance is \(e^{2\log 5+\log 4}(4-1)=100\cdot3=300\), as required.

Common Mistake and Practical Interpretation

Common mistake: “The normal variable \(Y=\log X\) has mode \(\mu\), so \(X\) has mode \(e^\mu\).” This reasoning treats the mode as if it were preserved by a change of variable. It is not: density values are affected by the Jacobian factor \(1/x\). Maximizing the lognormal density gives \(e^{\mu-\sigma^2}\), whereas \(e^\mu\) is the median.

A second useful distinction is between log-scale and original-scale variability. The quantity \(\sigma^2\) describes variation in \(\log X\); the original-scale variance is \(e^{2\mu+\sigma^2}(e^{\sigma^2}-1)\). It depends on both parameters and can be much larger than \(\sigma^2\). For a lognormal model, reporting the parameters together with the desired original-scale summaries prevents this scale confusion.

Key takeaway: The normal MGF gives \(E[X^r]=e^{r\mu+r^2\sigma^2/2}\). From it follow the mean \(e^{\mu+\sigma^2/2}\), variance \(e^{2\mu+\sigma^2}(e^{\sigma^2}-1)\), and skewness \((e^{\sigma^2}+2)\sqrt{e^{\sigma^2}-1}\); maximizing the density gives mode \(e^{\mu-\sigma^2}\).

Check Your Understanding

Try these questions using the log-scale parameters and results developed above.

For \(X\sim\operatorname{Lognormal}(1,\log 3)\), calculate \(E[X]\) and \(\operatorname{Var}(X)\).
Show from the general raw-moment formula that every positive integer moment of a lognormal variable is finite.
For \(X\sim\operatorname{Lognormal}(0,\log 4)\), find its median, mode, and skewness, and put the three locations in order.
Why does exponentiating the mode of \(\log X\) fail to give the mode of \(X\)? State the density feature responsible.
A lognormal variable has mean \(m\) and variance \(v\). Write formulas for its log-scale variance and log-scale mean.

What You'll Learn