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Lognormal distribution

A positive, right-skewed quantity whose logarithm is normal - the default model for claim severity and asset prices.

Parameters and support

$\mu$

mean of the UNDERLYING normal - $\mu \in \mathbb{R}$

$\sigma$

standard deviation of the UNDERLYING normal - $\sigma > 0$

Support

$x > 0$

The formulas

f(x)

$$f(x) = \dfrac{1}{x\sigma\sqrt{2\pi}}\exp\!\left(-\dfrac{(\ln x-\mu)^{2}}{2\sigma^{2}}\right)$$

F(x)

$$F(x) = \Phi\!\left(\dfrac{\ln x-\mu}{\sigma}\right)$$

Mean

$$E[X] = e^{\mu + \sigma^{2}/2}$$

Variance

$$\operatorname{Var}(X) = \left(e^{\sigma^{2}}-1\right)e^{2\mu+\sigma^{2}}$$

MGF

Not examinable for this distribution.

The MGF does not exist for any t > 0 - the tail is too heavy. Use the moment formula E[Xᵏ] = exp(kμ + k²σ²/2) instead.

Where the moments come from

1. $X = e^{Y}$ with $Y \sim N(\mu, \sigma^2)$, so $P(X \le x) = P(Y \le \ln x) = \Phi\!\left(\frac{\ln x - \mu}{\sigma}\right)$.
2. $E[X^k] = E[e^{kY}] = M_Y(k) = e^{k\mu + k^2\sigma^2/2}$ - the normal MGF evaluated at $k$.
3. $k = 1$ gives $E[X] = e^{\mu+\sigma^2/2}$; $k = 2$ gives $E[X^2] = e^{2\mu+2\sigma^2}$.
4. $\operatorname{Var}(X) = e^{2\mu+2\sigma^2} - e^{2\mu+\sigma^2} = (e^{\sigma^2}-1)e^{2\mu+\sigma^2}$.
5. The median is $e^{\mu}$ because $\Phi(0) = 0.5$ at $\ln x = \mu$.

Worked example

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run - summed over the support for a discrete distribution, integrated by quadrature for a continuous one - and compared with the closed forms printed here. A typo on this page fails that test run.

Traps

• Reporting e^μ as the mean - that is the MEDIAN.
• Using σ² where σ belongs inside the Φ argument.
• Looking for an MGF; it does not exist. Use the k-th moment formula.

Related

• Normal distribution
• Gamma distribution
• Pareto distribution

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.