Base-10 lognormal PDF integrated over log10(x)

Question

From what I understand, the lognormal probability density function in base-10 is mathematically defined thus:

$$ p(x; \mu, \sigma) = \frac{log_{10}(e)}{x \sigma \sqrt{2 \pi}} e^{-\frac{(log_{10}(x) - \mu)^2}{2 \sigma^2}} $$

As any PDF is bound to be, this function is normalised - that is, integrated over all $x$ (I integrate numerically, with $x$ distributed uniformly between the limits of integration) gives 1. In addition, this function, plotted against $x$, shows a unimodal distribution that is skewed/heavy-tailed.

Now, if my understanding is correct, a lognormal distribution is a distribution in which the random variable's logarithm is normally distributed. That is, if I plot the above equation with respect to $log_{10}(x)$, I get something that looks like a Gaussian.

Correct me if I am wrong, but what I am plotting is this:

$$ g(y=log_{10}(x); \mu, \sigma) = \frac{log_{10}(e)}{10^y \sigma \sqrt{2 \pi}} e^{-\frac{(y - \mu)^2}{2 \sigma^2}} $$

But if I integrate this function over all $y$ (I integrate numerically, with $y$ distributed uniformly between the limits of integration), I do not get 1 (due to the $log_{10}(e)/10^y$ factor). In other words, what I thought was an alternate representation of the same function, is not normalised.

So my question is, are the two forms of the same PDF not equivalent, as it seems to be in this case? If not, what is the reason? If yes, what am I doing wrong?

Please help!

Thank you.

Answer 1

A crucial term in the PDFs has been omitted: the measure $\mathrm{d}x$. Really we should think in terms of the probability element

$$p(x; \mu, \sigma)\ \mathrm{d}x = \frac{\log_{10}(e)}{x\,\sigma \sqrt{2 \pi}} e^{-\frac{(\log_{10}(x) - \mu)^2}{2 \sigma^2}}\ \mathrm{d}x,$$

whence, substituting $10^y=x$ (and equivalently $y=\log_{10}(x)$), we may rewrite the probability element in terms of $y$ as

$$\eqalign{ g(y; \mu, \sigma)\ \mathrm{d}y &= \frac{\log_{10}(e)}{10^y \sigma \sqrt{2 \pi}} e^{-\frac{(y - \mu)^2}{2 \sigma^2}}\ \mathrm{d}(10^y) \\ &=\frac{\log_{10}(e)}{10^y \sigma \sqrt{2 \pi}} e^{-\frac{(y - \mu)^2}{2 \sigma^2}}\ \log(10)10^y\ \mathrm{d}y \\ &= \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(y - \mu)^2}{2 \sigma^2}}\ \mathrm{d}y }$$

(because $\log_{10}(e)\log(10) = 1$).

This is the standard normal PDF. Normally, one begins at the end, substitutes for $y$, and works backwards to obtain the pdf for the lognormal distribution.