Cumulative distribution function of log-normal distribution

Question

In a lot of sources, such as the Wikipedia page for log-normal distributions, the cumulative distribution function for log-normal distributions is denoted as

$$F_X(x)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right)$$

where $\Phi$ is the cumulative distribution function of a normal distribution. That is, if $Z$ is normally distributed, its CDF is

$$\Phi(z)=\frac{1}{\sqrt{2\pi}}\int^x_{-\infty}\exp \left(-\frac{1}{2}\cdot z^2\right)dz.$$

This might be a stupid questions but I am not sure what the CDF for the log-normal distribution actually means. Is it

$$F_X(x)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right) \\ =\frac{1}{\sqrt{2\pi}}\int^x_{-\infty}\exp \left(-\frac{1}{2}\cdot\left(\frac{\ln x-\mu}{\sigma}\right)^2\right)dz$$ and is there a way to derive the CDF for log-normal distribution without using its probability density function? Is the CDF of a log-normal distribution the logarithm of the CDF for normal distribution?

Answer 1

You write: $$\Phi(z)=\frac{1}{\sqrt{2\pi}}\int^x_{-\infty}\exp \left(-\frac{1}{2} z^2\right)dz$$which is wrong.

It should be:$$\Phi(z)=\frac{1}{\sqrt{2\pi}}\int^z_{-\infty}\exp \left(-\frac{1}{2} x^2\right)dx$$or - switching the variables $z,x$:$$\Phi(x)=\frac{1}{\sqrt{2\pi}}\int^x_{-\infty}\exp \left(-\frac{1}{2} z^2\right)dz$$

This leads to $$F_X(x)=\frac{1}{\sqrt{2\pi}}\int^{\frac{\ln x-\mu}{\sigma}}_{-\infty}\exp \left(-\frac{1}{2} z^2\right)dz=\int^{\frac{\ln x-\mu}{\sigma}}_{-\infty}\phi(z)dz$$where $\phi$ denotes the PDF of standard normal distribution.

PDF $f_X$ can be found by differentiating the CDF which leads to:$$f_X=\frac1{x\sigma}\phi\left(\frac{\ln x-\mu}{\sigma}\right)=\frac1{x\sigma\sqrt{2\pi}}\exp\left(-\frac{(\ln x-\mu)^2}{2\sigma^2}\right)$$