The Lognormal Distribution

Options, Futures and Derivative Securities

Lorenzo Naranjo

Spring 2025

The Normal Distribution

We say that a random variable (RV) X is normally distributed with mean \mu and standard deviation \sigma if its probability density function is: f(x) = \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(x - \mu)^{2}}{2 \sigma^{2}}} and we usually write X \sim \mathcal{N}(\mu, \sigma^{2}).

Standard Normal Cumulative Probability

If Z \sim \mathcal{N}(0, 1), we have that: \operatorname{P}(Z \leq z) = \int_{-\infty}^{z} \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} \, dx = \operatorname{\Phi}(z)

Since the integral cannot be solved in closed-form, the probability must then be obtained from a table or using a computer.

Computing Normal Probabilities

In order to compute cumulative probabilities for X \sim \mathcal{N}(\mu, \sigma^{2}), we usually normalize the random variable X: \operatorname{P}(X \leq x) = \operatorname{P}\left( \tfrac{X - \mu}{\sigma} \leq \tfrac{x - \mu}{\sigma} \right) = \operatorname{P}\left( Z \leq \tfrac{x - \mu}{\sigma} \right) = \operatorname{\Phi}\left(\tfrac{x - \mu}{\sigma}\right) where Z = \frac{X - \mu}{\sigma} \sim \mathcal{N}(0, 1) is called a Z-score.

Example 1 Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. What is the probability that X \leq 0? \operatorname{P}(X \leq 0) = \operatorname{P}\left( Z \leq \tfrac{0 - 10}{25} \right) = \operatorname{\Phi}(-0.40) = 0.3446.

Computing a Right-Tail Probability

Right-Tail Probability: \operatorname{P}(X > x) = 1 - \operatorname{P}(X \leq x)

Example 2 Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. What is the probability that X > 12? \operatorname{P}(X \leq 12) = \operatorname{P}\left( Z \leq \tfrac{12 - 10}{25} \right) = \operatorname{\Phi}(0.08) = 0.5319. Therefore, \operatorname{P}(X > 12) = 1 - 0.5319 = 0.4681.

Computing an Interval Probability

Interval Probability: \operatorname{P}(x_{1} < X \leq x_{2}) = \operatorname{P}(X \leq x_{2}) - \operatorname{P}(X \leq x_{1})

Example 3

Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. What is the probability that 2 < X \leq 14? \begin{aligned} \operatorname{P}(X \leq 14) & = \operatorname{P}\left( Z \leq \tfrac{14 - 10}{25} \right) = \operatorname{\Phi}(0.16) = 0.5636, \\ \operatorname{P}(X \leq 2) & = \operatorname{P}\left( Z \leq \tfrac{2 - 10}{25} \right) = \operatorname{\Phi}(-0.32) = 0.3745. \end{aligned}

Therefore, \operatorname{P}(2 < X \leq 14) = 0.5636 - 0.3745 = 0.1891.

Computing a Standard Normal Right-Tail Percentile

For a standard normal variable Z, a right-tail percentile is the value z_{\alpha} above which we obtain a certain probability \alpha.

Mathematically, this means finding z_{\alpha} such that: \operatorname{P}(Z > z_{\alpha}) = \alpha \Leftrightarrow \operatorname{P}(Z \leq z_{\alpha}) = 1 - \alpha
The following table shows common values for z_{\alpha}:

\alpha	0.05	0.025	0.01	0.005
z_{\alpha}	1.64	1.96	2.33	2.58

Confidence Interval for a Standard Normal RV

A 1 - \alpha confidence interval (CI) defines left and right percentiles such that the probability on each side is \alpha/2.

Example 4 Since z_{2.5\%} = 1.96, the 95% confidence interval of Z is [-1.96, 1.96]. This means that if we randomly sample this variable 100,000 times, approximately 95,000 observations will fall inside this interval.

Confidence Interval for a Normal RV

If X \sim \mathcal{N}(\mu, \sigma^{2}), its confidence interval is determined by \xi and \zeta such that: \begin{aligned} \operatorname{P}(X \leq \xi) = \alpha / 2 & \Rightarrow \operatorname{P}(Z \leq \tfrac{\xi - \mu}{\sigma}) = \alpha/2, \\ \operatorname{P}(X > \zeta) = \alpha / 2 & \Rightarrow \operatorname{P}(Z > \tfrac{\zeta - \mu}{\sigma}) = \alpha/2, \end{aligned} which implies that -z_{\alpha/2} = \tfrac{\xi - \mu}{\sigma} and z_{\alpha/2} = \tfrac{\zeta - \mu}{\sigma}.
The 1 - \alpha confidence interval for X is then [\mu - z_{\alpha/2}\sigma, \mu + z_{\alpha/2}\sigma].

Example 5 Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. Since z_{2.5\%} = 1.96, the 95% confidence interval of X is: [10-1.96(25), 10+1.96(25)] = [-39, 59].

The Lognormal Distribution

If X \sim N(\mu, \sigma^{2}) then Y = e^{X} is said to be log-normally distributed with the same parameters.
The distribution function of the lognormal distribution can be obtained as follows: \begin{aligned} \operatorname{P}(Y \leq y) & = \operatorname{P}(X \leq \ln(y)) \\ & = \int_{-\infty}^{\ln(y)} \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(x - \mu)^{2}}{2 \sigma^{2}}} \, dx \end{aligned}
We can then define z = e^{x} \Leftrightarrow x = \ln(z) \Rightarrow dx = (1 / z) dz and \operatorname{P}(Y \leq y) = \int_{-\infty}^{y} \frac{1}{z \sqrt{2 \pi \sigma^{2}}} e^{-\frac{(\ln(z) - \mu)^{2}}{2 \sigma^{2}}} \, dz

Probability Density Function Comparison

Unlike the normal density, the lognormal density function is not symmetric around its mean.
Normally distributed variables can take values in (-\infty, \infty), whereas lognormally distributed variables are always positive.

Computing Probabilities for Lognormal RVs

We can use the fact that the logarithm of a lognormal random variable is normally distributed to compute cumulative probabilities.

Example 6 Let Y = e^{4 + 1.5 Z} where Z \sim \mathcal{N}(0, 1). What is the probability that Y \leq 100? \begin{aligned} \operatorname{P}(Y \leq 100) & = \operatorname{P}(e^{X} \leq 100) \\ & = \operatorname{P}(X \leq \ln(100)) \\ & = \operatorname{P}\left(Z \leq \tfrac{\ln(100) - 4}{1.5}\right) \\ & = \operatorname{\Phi}(0.4034) \\ & = 0.6567 \end{aligned}

Confidence Interval for a Lognormal RV

Let Y = e^{\mu + \sigma Z} where Z \sim \mathcal{N}(0, 1). We have that: \begin{aligned} -z_{\alpha/2} < Z \leq z_{\alpha/2} & \Rightarrow \mu - \sigma z_{\alpha/2} < \mu + \sigma Z \leq \mu + \sigma z_{\alpha/2} \\ & \Rightarrow e^{\mu - \sigma z_{\alpha/2}} < e^{\mu + \sigma Z} \leq e^{\mu + \sigma z_{\alpha/2}} \end{aligned}
The 1 - \alpha confidence interval for Y is [e^{\mu - \sigma z_{\alpha/2}}, e^{\mu + \sigma z_{\alpha/2}}].

Example 7 Let Y = e^{4 + 1.5 Z} where Z \sim \mathcal{N}(0, 1). The 95% confidence interval for Y is: [e^{4 - 1.96(1.5)}, e^{4 + 1.96(1.5)}] = [2.89, 1032.71].

Moments of a Lognormal Random Variable

Let Y = e^{\mu + \sigma Z} where Z \sim \mathcal{N}(0, 1). We have that: \begin{aligned} \operatorname{E}(Y) & = e^{\mu + 0.5 \sigma^{2}} \\ \operatorname{Var}(Y) & = e^{2\mu + \sigma^{2}} (e^{\sigma^{2}} - 1) \\ \operatorname{SD}(Y) & = \operatorname{E}(Y) \sqrt{e^{\sigma^{2}} - 1} \end{aligned}

Example 8 Let Y = e^{4 + 1.5 Z} where Z \sim \mathcal{N}(0, 1). The expectation and standard deviation of Y are: \begin{aligned} \operatorname{E}(Y) & = e^{4 + 0.5(1.5^{2})} = 168.17 \\ \operatorname{SD}(Y) & = 168.17 \sqrt{e^{1.5^{2}} - 1} = 489.95 \end{aligned}

Partial Expectations

When pricing a call option, the payoff is positive if the option is in-the-money and zero otherwise.
We usually use an indicator function to quantify this behavior: \large\mathbb{1}_{\{Y > K\}} = \begin{cases} 0 & \text{if $Y \leq K$} \\ 1 & \text{if $Y > K$} \end{cases}
Let Y = e^{X} where X \sim \mathcal{N}(\mu, \sigma^{2}). Then we have that: \begin{aligned} \operatorname{E}\left(Y \large\mathbb{1}_{\{Y > K\}}\right) & = e^{\mu + \frac{1}{2}\sigma^{2}} \operatorname{\Phi}\left(\frac{\mu + \sigma^{2} - \ln(K)}{\sigma}\right) \\ \operatorname{E}\left(K \large\mathbb{1}_{\{Y > K\}}\right) & = K \operatorname{\Phi}\left(\frac{\mu - \ln(K)}{\sigma}\right) \end{aligned}