# Derivation of the LogNormal CDF from PDF [duplicate]

I've been trying to derive the CDF of the lognormal distribution. I got this far but now I'm stuck.

$$F(x) = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x\frac{1}{z}e^{-t^2}dz$$

where $$t = \frac{\ln(z)-\mu}{\sigma\sqrt{2}}$$

so $$z = e^{\sigma t \sqrt{2} + \mu}$$

$$F(x) = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x\frac{1}{e^{\sigma t \sqrt{2}}}e^{-t^2}d(e^{\sigma t \sqrt{2}})$$

$$F(x) = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x e^{-\sigma t \sqrt{2}} e^{-t^2}d(e^{\sigma t \sqrt{2}})$$

I hope someone can point me in the right direction or link me a full derivation since I haven't been able to find one.

By definition, $$\ln X\sim\mathcal N(\mu, \sigma^2).$$

So, the cdf would be

$$F_X(x) =\Phi\left(\frac{\ln x-\mu}{\sigma}\right).\tag I\label a$$

The pdf is ($$x>0$$)

$$f_X(x) = \frac{1}{x\sigma\sqrt{2\pi}} e^{-\frac{(\ln x-\mu)^2}{2\sigma^2}}\tag 1\label 1$$

OP can proceed as taking the integral of $$\eqref 1$$ and substituting

$$\frac{\ln x-\mu}{\sigma}:= t.$$

This would be easier to yield $$\eqref a.$$