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?
Best Answer
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$$\frac{S}{S_0}\sim N\left(\frac{\mu_S}{S_0},\,\frac{\sigma_S^2}{S_0^2}\right).$$