Determining the pdf of a log-normal distribution?

Question

I am learning probability and have just gotten to transformations, which I do not really understand. I have the following problem:

Use the transformation method to derive the density function f of a log-normal distribution.

There seem to be multiple different methods that qualify as a "transformation method," and I don't know which one to use, nor really how any of them work. What is a transformation? What is it good for in probability? And how does it let me find the pdf of a distribution?

EDIT: This was marked as duplicate linking to another question which showed how to derive the density function for this distribution. I am not asking how to find it in general, I am asking specifically how to find it using the transformation method, including an explanation of what this method is, what it does, and how to use that, as this has all been unclear to me from my course and I am looking to understand it.

Answer 1

The idea here is the same as in the other question you asked recently.

A log normal random variable is one which satisfies $\ln(X)\sim\mathcal{N}(\mu,\sigma^{2})$. Equivalently, $X=e^{Z}$ where $Z\sim\mathcal{N}(\mu,\sigma^{2})$. Note that $$ \Phi_{X}(x)\equiv\mathbb{P}(X\leq x)=\mathbb{P}(e^{Z}\leq x)=\mathbb{P}(Z\leq\ln(x))=\Phi_{Z}(\ln(x)) $$ where $\Phi_{X}$ and $\Phi_{Z}$ are the CDFs of $X$ and $Z$. In particular, $\Phi_{Z}$ is the CDF of a normal distribution. To get the PDF $\varphi_{X}$ in terms of the PDF of the normal distribution $\varphi_{Z}$, just differentiate using the chain rule: $$ \varphi_{X}(x)\equiv\Phi_{X}^{\prime}(x)=\Phi_{Z}^{\prime}(\ln(x))\frac{1}{x}=\varphi_{Z}(\ln(x))\frac{1}{x}. $$

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Transformation just means function.

I would guess that the "transformation method" refers to the technique I highlight above to find the PDF of $X\equiv f(Z)$ where

• $Z$ is a continuous random variable and
• $f$ is a strictly increasing function with differentiable inverse $g\equiv f^{-1}$.

Indeed, in this case, $$\Phi_{X}(x)=\mathbb{P}(f(Z) \leq x)=\mathbb{P}(Z\leq g(x))=\Phi_{Z}(g(x))$$ and hence $\varphi_{X}(x)=\varphi_{Z}(g(x))g^{\prime}(x)$.

The above idea also works if $f$ is strictly decreasing, but in this case $$\Phi_{X}(x)=\mathbb{P}(f(Z)\leq x)=\mathbb{P}(Z\geq g(x))=1-\Phi_{Z}(g(x))$$ and hence $\varphi_{X}(x)=-\varphi_{Z}(g(x))g^{\prime}(x)$.

You can summarize both the strictly increasing and decreasing case by $$ \boxed{\varphi_{X}(x)=\varphi_Z(g(x))\left|g^{\prime}(x)\right|} $$