Proof: Cumulative distribution function of the log-normal distribution
Index:
The Book of Statistical Proofs ▷
Probability Distributions ▷
Univariate continuous distributions ▷
Log-normal distribution ▷
Cumulative distribution function
Metadata: ID: P325 | shortcut: lognorm-cdf | author: majapavlo | date: 2022-06-29, 22:20.
Theorem: Let $X$ be a random variable following a log-normal distribution:
\[\label{eq:norm} X \sim \ln \mathcal{N}(\mu, \sigma^2) \; .\]Then, the cumulative distribution function of $X$ is
\[\label{eq:lognorm-cdf} F_X(x) = \frac{1}{2} \left[ 1 + \mathrm{erf}\left( \frac{\ln x-\mu}{\sqrt{2} \sigma} \right) \right]\]where $\mathrm{erf}(x)$ is the error function defined as
\[\label{eq:erf} \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} \exp(-t^2) \, \mathrm{d}t \; .\]Proof: The probability density function of the log-normal distribution is:
\[\label{eq:lognorm-pdf} f_X(x) = \frac{1}{x \sigma \sqrt{2 \pi}} \cdot \exp \left[ - \left( \frac{\ln x-\mu}{\sqrt{2} \sigma} \right)^2 \right] \; .\]Thus, the cumulative distribution function is:
\[\label{eq:lognorm-cdf-s1} \begin{split} F_X(x) &= \int_{-\infty}^{x} \mathcal{\ln N}(z; \mu, \sigma^2) \, \mathrm{d}z \\ &= \int_{-\infty}^{x} \frac{1}{z\sigma \sqrt{2 \pi}} \cdot \exp \left[ -\left( \frac{\ln z-\mu}{\sqrt{2} \sigma} \right)^2 \right] \, \mathrm{d}z \\ &= \frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{x} \frac{1}{z} \cdot \exp \left[ -\left( \frac{\ln z-\mu}{\sqrt{2} \sigma} \right)^2 \right] \, \mathrm{d}z \; . \end{split}\]From this point forward, the proof is similar to the derivation of the cumulative distribution function for the normal distribution. Substituting $t = (\ln z-\mu)/(\sqrt{2} \sigma)$, i.e. $\ln z = \sqrt{2} \sigma t + \mu$, $z = \exp( \sqrt{2} \sigma t + \mu) $ this becomes:
\[\label{eq:lognorm-cdf-s2} \begin{split} F_X(x) &= \frac{1}{\sigma \sqrt{2 \pi}} \int_{(-\infty-\mu)/(\sqrt{2} \sigma)}^{(\ln x-\mu)/(\sqrt{2} \sigma)} \frac{1}{\exp( \sqrt{2} \sigma t + \mu)} \cdot \exp \left(-t^2 \right) \, \mathrm{d} \left[ \exp \left( \sqrt{2} \sigma t + \mu \right) \right] \\ &=\frac{\sqrt{2} \sigma}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{\frac{\ln x-\mu}{\sqrt{2} \sigma}} \frac{1}{\exp( \sqrt{2} \sigma t + \mu)} \cdot \exp(-t^2) \cdot \exp \left( \sqrt{2} \sigma t + \mu \right) \, \mathrm{d}t \\ &= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\frac{\ln x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \\ &= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{0} \exp(-t^2) \, \mathrm{d}t + \frac{1}{\sqrt{\pi}} \int_{0}^{\frac{\ln x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \\ &= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \exp(-t^2) \, \mathrm{d}t + \frac{1}{\sqrt{\pi}} \int_{0}^{\frac{\ln x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \; . \end{split}\]Applying \eqref{eq:erf} to \eqref{eq:lognorm-cdf-s2}, we have:
\[\label{eq:lognorm-cdf-s3} \begin{split} F_X(x) &= \frac{1}{2} \lim_{x \to \infty} \mathrm{erf}(x) + \frac{1}{2} \, \mathrm{erf}\left( \frac{\ln x-\mu}{\sqrt{2} \sigma} \right) \\ &= \frac{1}{2} + \frac{1}{2} \, \mathrm{erf}\left( \frac{\ln x-\mu}{\sqrt{2} \sigma} \right) \\ &= \frac{1}{2} \left[ 1 + \mathrm{erf}\left( \frac{\ln x-\mu}{\sqrt{2} \sigma} \right) \right] \; . \end{split}\]∎
Sources: - skdhfgeq2134 (2015): "How to derive the cdf of a lognormal distribution from its pdf"; in: StackExchange, retrieved on 2022-06-29; URL: https://stats.stackexchange.com/questions/151398/how-to-derive-the-cdf-of-a-lognormal-distribution-from-its-pdf/151404#151404.
Metadata: ID: P325 | shortcut: lognorm-cdf | author: majapavlo | date: 2022-06-29, 22:20.