# How to derive the cdf of a lognormal distribution from its pdf

I'm trying to understand how to derive the cumulative distribution function for a lognormal distribution from its probability density function.

I know that the pdf is:

$$f(x) =\frac{e^{-\frac{1}{2}\Bigl(\frac{\ln(x) - \mu}{\sigma}\Bigr)^2}}{x \sigma \sqrt{2\pi}},\ x \gt 0$$

and the cdf is:

$$\Phi(x) = \int_{-\infty}^x f(y) dy = \frac{1}{\sigma\sqrt{2\pi}}\int_0^x e^{-\frac{1}{2}\Bigl(\frac{\ln(y) - \mu}{\sigma}\Bigr)^2}\frac{dy}{y}.$$

Now, I don't know how to get the last formula. I tried the substitution $t=\ln(y)$ but I don't know how to deal with the $1/y$ term.

Am I on the wrong way, or did I make any mistake?

I'd like to know it for my personal knowledge, it's not for a class.

• Please add the self-study tag and read its tag wiki, modifying your question as needed to comply with the outline there. – Glen_b May 8 '15 at 11:00

If $Y$ ~ $\log N(\mu,\sigma^2)$ then $$f_Y(y) = \frac{1}{y\sqrt{2\pi \sigma^2}}\exp\left[-\frac{1}{2}\left(\frac{\log y-\mu}{\sigma}\right)^2\right]$$

Now,

$$F_Y(y)=\int_{-\infty}^{y_1}f_Y(y)dy = \int_{-\infty}^{y_1}\frac{1}{y\sqrt{2\pi \sigma^2}}\exp\left[-\frac{1}{2}\left(\frac{\log y-\mu}{\sigma}\right)^2\right]dy$$

$$\frac{\log y - \mu}{\sigma} = z \implies \frac{dy}{y\sigma} = dz, \text{ and } z_1 = \frac{\log y_1 - \mu}{\sigma}$$

$$F_Y(y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z_1}\exp(-z^2/2)dz$$

Oh boy, that's a standard normal distribution. I think it's easy from here.

• This appears to go a little beyond the 'hints and guidance' appropriate for self-study posts – Glen_b May 8 '15 at 12:27

I suppose you have the error function in mind.

You can always turn the CDF back into a normal CDF. Recall that $X$ is lognormal iff $\log X\sim N \left(\mu,\sigma^2 \right)$.

1) You need to take account of the Jacobian of the transformation

2) It might be easier to get the pdf from the cdf and then you might spot your error.