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.

  • $\begingroup$ Hi David. Please add the self-study tag to your post. $\endgroup$ Dec 20, 2022 at 14:56

1 Answer 1


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. $


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