Lognormal Distribution

Plain-Language Definition

A random variable is Lognormal if its logarithm is Normal. Take any Normal random variable $Y$ with mean $\mu$ and variance $\sigma^2$, and define $X = e^Y$. The distribution of $X$ is Lognormal. Equivalently, $X$ is Lognormal whenever $\ln X$ is Normal.

The Normal arises as the limit of sums of independent random variables (the central limit theorem). The Lognormal arises as the limit of products of independent positive random variables, because the logarithm of a product is a sum. Anywhere multiplicative compounding is more natural than additive (asset returns, particle size after repeated random splits, biological growth), the Lognormal is a more honest baseline than the Normal.

Definition

Definition

Lognormal Distribution$X\sim \mathrm{Lognormal}(\mu ,{\sigma }^2)$

A positive random variable $X$ has a Lognormal distribution with parameters $\mu \in \mathbb{R}$ and $\sigma^2 > 0$ when $\ln X \sim N(\mu, \sigma^2)$. Equivalently, the density of $X$ is

$f_X(x) = \frac{1}{x \sigma \sqrt{2\pi}} \exp\!\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right), \quad x > 0.$

The parameters $\mu$ and $\sigma^2$ are the mean and variance of $\ln X$, not of $X$ itself.

The density vanishes at zero, rises to a single mode, and decays super-polynomially in the right tail. The shape parameter is $\sigma$; the location of the mode and the spread of the upper tail both stretch as $\sigma$ grows. As $\sigma \to 0$ the distribution concentrates near $e^\mu$ and becomes approximately Normal in a local sense.

Why This Matters

The Lognormal is the natural multiplicative analogue of the Normal. Three places where this matters in practice.

1. Multiplicative processes. Asset returns compounded over $n$ periods are products of $n$ independent (or near-independent) gross returns. Under modest assumptions about the per-period log-return distribution, the central limit theorem applied to the log of the product gives an approximately Lognormal price after many periods. This is the formal basis of Black-Scholes option pricing and any geometric-Brownian-motion model.

2. Right-skewed positive data. Particle sizes, drug-trial responses, household incomes, file sizes, insurance claim severities. All are nonnegative and right-skewed, and many fit a Lognormal model better than a Normal at the cost of a single nonlinear transformation.

3. Survival and reliability work. The Lognormal hazard rises and then falls, which is a useful shape for failure-time data where early-life and end-of-life failures dominate but the middle of the life has comparatively few failures. Weibull is a more common reliability default, but the Lognormal is the standard alternative when the Weibull hazard shape is wrong.

The classical warning is that financial returns are heavier-tailed than Lognormal at long horizons. The empirical excess kurtosis of equity returns is decades of literature; a Lognormal model gets the bulk right and underestimates tail risk. For insurance severity, Pareto and Weibull are the standard alternatives once the data shows a heavier tail than Lognormal can support.

Moments

Theorem

Lognormal Mean, Variance, Median, Mode

Statement

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

Intuition

The median is the simplest of the four: $\ln X$ is Normal with median $\mu$, and the monotone exponential preserves quantiles, so $X$ has median $e^\mu$. The mean is strictly larger because of the convexity of the exponential and Jensen's inequality: $\mathbb{E}[e^Y] > e^{\mathbb{E}[Y]}$, and the correction is exactly $e^{\sigma^2/2}$. The mode is strictly smaller because the density is right-skewed.

Proof Sketch

Use the moment generating function of a Normal: $\mathbb{E}[e^{tY}] = \exp(\mu t + \sigma^2 t^2 / 2)$ for $Y \sim N(\mu, \sigma^2)$. Setting $t = 1$ gives $\mathbb{E}[X] = \mathbb{E}[e^Y] = \exp(\mu + \sigma^2/2)$. Setting $t = 2$ gives $\mathbb{E}[X^2] = \exp(2\mu + 2\sigma^2)$, so $\operatorname{Var}(X) = \exp(2\mu + 2\sigma^2) - \exp(2\mu + \sigma^2) = \exp(2\mu + \sigma^2)(\exp(\sigma^2) - 1)$. For the mode, differentiate the density, set to zero, and solve $\ln x = \mu - \sigma^2$.

Why It Matters

Forgetting the $\sigma^2/2$ correction in the mean is one of the most common errors in applied work. If you fit a Normal to log-returns and then exponentiate the estimated log-mean, you get the median of returns, not the mean. The two differ by the multiplicative factor $e^{\sigma^2/2}$, which for a typical equity sigma of $0.2$ is about $1.020$ over a year and grows quadratically with horizon.

Failure Mode

The MGF $\mathbb{E}[e^{sX}]$ of a Lognormal $X$ is infinite for every $s > 0$. The Lognormal does not have an MGF in the usual sense, and identities that depend on MGF uniqueness do not apply. The Lognormal still has moments of every order, but the moment sequence does not uniquely determine the distribution; there is a Stieltjes moment-determinacy failure that produces non-identifiable density modifications.

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The Multiplicative Central Limit Theorem

Theorem

Product of Independent Positive r.v.s converges to Lognormal

Statement

Let $X_1, X_2, \dots$ be iid positive random variables with $\mu_L = \mathbb{E}[\ln X_1] \in \mathbb{R}$ and $\sigma_L^2 = \operatorname{Var}(\ln X_1) \in (0, \infty)$. Then for the product $P_n = X_1 X_2 \cdots X_n$,

$\frac{\ln P_n - n\mu_L}{\sigma_L \sqrt{n}} \xrightarrow{d} N(0, 1) \text{ as } n \to \infty.$

Equivalently, $P_n$ is approximately Lognormal with parameters $n\mu_L$ and $n\sigma_L^2$ for large $n$.

Intuition

This is the central limit theorem applied to $\ln P_n = \sum \ln X_i$. The product of many independent factors is Lognormal in the same way a sum of many independent terms is Normal. The Normal limit for the log is exact in the limit; the Lognormal claim for the product is its exponentiated counterpart and is approximate in the same sense.

Proof Sketch

Define $Y_i = \ln X_i$. The $Y_i$ are iid with finite mean and finite variance by assumption, so the classical central limit theorem applied to the $Y_i$ gives the displayed convergence in distribution. Continuous mapping under the exponential transforms convergence of the log-sum into convergence in distribution of the product to a Lognormal.

Why It Matters

The result formalizes the intuition that multiplicative compounding produces Lognormal aggregates. Geometric Brownian motion in finance is exactly this limit when the time step shrinks and the number of independent multiplicative shocks grows. The same logic gives Lognormal-shaped distributions for biological growth and for repeated-fragmentation particle sizes.

Failure Mode

The result requires finite variance of $\ln X_1$. If $X_i$ have a heavy enough left tail (e.g. $X_i$ can be very close to zero), $\ln X_i$ has infinite variance and the Lognormal limit fails. Power-law tails in $X_i$ also break finite-variance assumptions and can produce stable-law limits for the product instead of Lognormal.

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Worked Example: Median vs Mean of Equity Returns

A daily equity log-return has approximately $\mu = 0.0003$ and $\sigma = 0.012$ (about 19 percent annualized). Over $n = 252$ trading days, the cumulative log-return is approximately $N(252 \cdot 0.0003, 252 \cdot 0.012^2) = N(0.0756, 0.0363)$, with $\sigma_{\text{annual}} \approx 0.1904$.

The cumulative gross return $P_n = X_1 \cdots X_n$ is approximately Lognormal with parameters $0.0756$ and $0.0363$. Therefore:

• $\operatorname{Median}(P_n) = e^{0.0756} \approx 1.0785$, a typical annual gross return near $7.85$ percent.
• $\mathbb{E}[P_n] = e^{0.0756 + 0.0363/2} \approx e^{0.0937} \approx 1.0982$, an annual expected gross return near $9.82$ percent.
• $\operatorname{Var}(P_n) = (e^{0.0363} - 1) \cdot e^{2 \cdot 0.0756 + 0.0363} \approx 0.0440$, $\operatorname{SD}(P_n) \approx 0.2098$.

The gap between $7.85$ percent (median) and $9.82$ percent (mean) is the volatility drag. Quoting "average return" without specifying mean versus median misrepresents one or the other by close to $200$ basis points per year at this volatility level.

Common Misconceptions

Watch Out

The parameters mu and sigma are not the mean and variance of X

$\mu$ is the mean of $\ln X$, not of $X$. The mean of $X$ is $e^{\mu + \sigma^2/2}$. Mistaking $\mu$ for $\mathbb{E}[X]$ is the most common applied error with this distribution, and it propagates through every downstream calculation.

Watch Out

The Lognormal MGF is not finite at any positive value

$\mathbb{E}[e^{sX}] = \infty$ for every $s > 0$, because the density tail of $X$ decays sub-exponentially. Any inference procedure that relies on the existence of the MGF (Chernoff bounds, tilted measures, MGF uniqueness in the Stieltjes sense) breaks. The Lognormal still has all moments and a well-defined cumulant generating function on the imaginary axis, but the right-half-plane MGF is unavailable.

Watch Out

The Lognormal is not heavy-tailed in the regularly varying sense

The right tail of the Lognormal decays faster than every polynomial. It is heavier than Normal but lighter than any power law. For practical purposes, the Lognormal tail behaves like a slowly-decaying exponential in the log scale, and tail probabilities $\mathbb{P}(X > x)$ can be computed accurately from the Normal CDF on $\ln x$. Confusing Lognormal with Pareto leads to dramatic underestimates of extreme-quantile risk for genuinely power-law data.

Comparison: Normal vs Lognormal

The two are intimately related, and the natural way to choose between them is to think about whether shocks are additive or multiplicative on the variable of interest.

• Normal. Supported on the whole real line. Mean and variance are the parameters. Symmetric. Arises from additive aggregation. Right tail and left tail are equal.
• Lognormal. Supported on positive reals. Log-mean and log-variance are the parameters. Right-skewed. Arises from multiplicative aggregation. The mean is strictly greater than the median.

A useful test: simulate a thousand independent draws, take log, and check whether the resulting sample is symmetric. If yes, Lognormal is reasonable. If the log-sample is still skewed, neither Normal nor Lognormal fits well, and a heavier-tailed alternative (Pareto, Weibull, or a finite-mixture) is needed.

For a side-by-side severity-modeling treatment, see ActuaryPath's Lognormal page at https://www.actuarypath.com/concepts/lognormal-distribution/ , which approaches the same distribution from the loss-modeling angle (ASTAM and FAM-S, fitted by MLE, used as a parametric severity component in compound distributions).

Exercises

ExerciseCore

Problem

Let $X \sim \operatorname{Lognormal}(\mu = 0, \sigma^2 = 1)$. Compute $\mathbb{E}[X]$, $\operatorname{Var}(X)$, the median, the mode, and $\mathbb{P}(X > 1)$.

Hint

Reveal Solution

ExerciseCore

Problem

A particle's diameter $D$ is the product of $n = 20$ independent random shrinkage factors, each iid Lognormal with $\mu = 0$ and $\sigma = 0.1$ (a shrinkage of about 10 percent per step in log scale). Find the approximate distribution of $D / D_0$ where $D_0$ is the initial size.

Hint

Reveal Solution

ExerciseCore

Problem

An insurance loss is modeled as $X \sim \operatorname{Lognormal}(\mu = 9, \sigma^2 = 1.5)$ in dollars. Compute the expected loss, the standard deviation of the loss, the median, and the 95th percentile.

Hint

Reveal Solution

ExerciseAdvanced

Problem

Show that the MLE of $(\mu, \sigma^2)$ from an iid sample $X_1, \dots, X_n$ from a Lognormal distribution reduces to the MLE of the Normal mean and variance applied to $\ln X_1, \dots, \ln X_n$.

Hint

Reveal Solution

ExerciseAdvanced

Problem

Compute $\mathbb{E}[X \mid X > t]$ for $X \sim \operatorname{Lognormal}(\mu, \sigma^2)$ in closed form, and interpret the result for $\mu = 0, \sigma = 1, t = e$ (one standard log-deviation above the median).

Hint

Reveal Solution

Beyond Lognormal: When the Tail Is Heavier

The Lognormal tail decays faster than every polynomial but slower than any exponential. Real-world data sometimes has a tail heavier than Lognormal can support. Three standard moves when that happens:

• Fit a Pareto tail above a threshold and a Lognormal body below. This is the peaks-over-threshold approach from extreme-value theory.
• Fit a Weibull with shape parameter below 1, which gives a tail heavier than Lognormal but lighter than Pareto.
• Fit a finite mixture of Lognormals or a Lognormal-Pareto composite, accepting the loss of analytical tractability for a better tail fit.

Diagnostics: a log-log survival-function plot. Pareto-like data shows a straight line on a log-log plot. Lognormal data shows a curve that bends down. Weibull data with shape less than 1 sits between the two.

References

• Casella, G., and Berger, R. L. (2002). Statistical Inference, 2nd ed., Duxbury. Section 3.3 covers the Lognormal as a transformation of the Normal and lists the moment formulas.
• Blitzstein, J. K., and Hwang, J. (2019). Introduction to Probability, 2nd ed., Chapman and Hall / CRC. Chapter 5 includes the Lognormal in the catalog of continuous distributions with worked examples on financial returns.
• For the loss-modeling and severity-fitting perspective, see ActuaryPath's Lognormal page at https://www.actuarypath.com/concepts/lognormal-distribution/ and Klugman, Panjer, Willmot (2019), Loss Models: From Data to Decisions, 5th ed., Wiley, Chapter 5.