Suppose $$X$$ is a normal r.v. with mean $$\mu$$ and variance $$\sigma^2$$. Let $$Y = e^X$$ so that $$Y$$ is lognormally distributed. The pdf of a lognormal distribution in terms of the value $$x$$ is

$$F_Y(x) = \frac{1}{x\sqrt{2\pi\sigma^2}}\exp\bigg\{\frac{-(\log x-\mu)^2}{2\sigma^2}\bigg\}$$

Thus, given values from $$X$$, one can compute the probability density of the lognormal variable $$Y$$. However, what if you want to express the PDF in terms of $$y$$, the values $$Y$$ can take? Basically, I want to have a pdf $$F_Y(y)$$ in terms of the values $$y$$. Would it just be

$$F_Y(y) = \frac{1}{\log y\sqrt{2\pi\sigma^2}}\exp\bigg\{\frac{-(\log(\log y)-\mu)^2}{2\sigma^2}\bigg\}$$

I am asking because I am working on a problem related to finance where I have the pdf of a normally distributed random variable $$X$$, call it $$Q(X)$$. I want the PDF of $$S/S_0$$, call it $$P(S)$$, where $$X = \log(S/S_0)$$. So, would $$P(S)$$ be given by the second equation?

• $x$ is just a variable. $F(x)=x$ $\iff$ $F(y)=y$ $\iff$ $F(z)=z$ denotes the same function. – NCh Feb 6 '20 at 7:07
• I understand that. But $\log(\log(y))$ doesn't make much sense to me. Is it correct? – Josh Pilipovsky Feb 6 '20 at 7:15
• No. It is not well-defined function since $\log(y)$ can be negative and $\log(\log(y))$ is not defined at all. – NCh Feb 6 '20 at 7:17

If $$X\sim N(\mu,\,\sigma^2)$$ then $$P(X\le x)=\Phi\left(\frac{x-\mu}{\sigma}\right)$$, where$$\Phi(z):=\int_{-\infty}^z\phi(z^\prime)dz^\prime$$is the $$N(0,\,1)$$ CDF, which has PDF being$$\phi(z):=\frac{1}{\sqrt{2\pi}}\exp\frac{-z^2}{2}.$$If $$Y=e^X$$,$$P(Y\le y)=P(X\le\ln y)=\Phi\left(\frac{\ln y-\mu}{\sigma}\right).$$We can differentiate the CDFs of $$X$$ and $$Y$$, with respect to $$x$$ and $$y$$, to get these variables' PDFs. For $$X$$, the result is$$\color{blue}{\frac{1}{\sigma}}\phi\left(\frac{x-\mu}{\sigma}\right),$$the blue factor being$$\frac{d}{dx}\frac{x-\mu}{\sigma}$$due to the chain rule. For $$Y$$, the same strategy obtains a PDF of$$\frac{1}{\sigma y}\phi\left(\frac{\ln y-\mu}{\sigma}\right),$$ which is what you started with. So $$X$$ in this example is Normally distributed, with the famous PDF of that distribution. In your financial example, $$S$$ is normal, say $$S\sim N(\mu_S,\,\sigma_S^2)$$. Then$$\frac{S}{S_0}\sim N\left(\frac{\mu_S}{S_0},\,\frac{\sigma_S^2}{S_0^2}\right).$$
You seem to have confused yourself with an unfortunate choice of variable names. What you call “the pdf of a lognormal distribution in terms of the value $$x$$” is already the density function for $$Y$$ (as indicated by the index $$Y$$). Its independent variable would more conveniently be called $$y$$, since it represents values of $$Y$$, not of $$X$$. So there’s no issue of taking a double logarithm here; the one logarithm is fine; substituting $$\log y$$ for the misnamed variable $$x$$ that stands for values of $$Y$$ would make no sense.