# Transformation and log-normal distribution

Let $$X\sim N(\mu ,\sigma^2)$$ for $$\mu\in\mathbb{R}$$ and $$\sigma > 0$$. Let also $$Y=e^X$$. Find the PDF for $$Y$$.

I get that

\begin{align*} F_Y(t)&=P(Y\le t) \\ &=P(e^X \le t) \end{align*}

Since $$t>0$$ I now get that \begin{align*} P(X\le \ln(t))&=\int_{-\infty}^{\ln(t)} f_X(x)\mathrm{d}x \\ &=\int_{-\infty}^{\ln(t)} \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrm{d}x \\ \end{align*}

Let $$u=e^x$$, then $$\frac{du}{dx}=e^x=u$$ and $$\mathrm{d}x=\frac{\mathrm{d}u}{u}$$.

and then \begin{align*} \int_{-\infty}^{\ln(t)} \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrm{d}x &=\int_{0}^{t} \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\frac{1}{u}\mathrm{d}u \\ &=\int_{0}^{t} \frac{1}{\sqrt{2\pi}}e^{-\frac{(\ln(u)-\mu)^2}{2\sigma^2}}\frac{1}{u}\mathrm{d}u. \end{align*}

Is it correct and how do I proceed?

This is almost correct and you are nearly there! A couple of notes:

1. When writing the PDF of the normal distribution, you forgot a $$\sigma$$ in the denominator. The PDF of the normal should be $$\frac{1}{\sqrt{2\pi}\color{blue}{\sigma}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$.

2. To proceed, recall that $$f_Y(t) =\frac{d}{dt}(P(Y\le t))$$. So you just have to differentiate the integral to get the answer (after fixing up the point from above). A hint to do this: use the Fundamental Theorem of Calculus.

• Thanks! So I should calculate the integral and then take the differential of it in order to get rid of the $u$'s?
– user732574
Commented Dec 7, 2019 at 6:32
• You're welcome! And there's no need to calculate the integral -- you can differentiate without calculating the integral, thanks to the Fundamental Theorem of Calculus ($\frac{d}{dt}\int_a^t g(u)\, du = g(t)$ -- that is, just replace $u$ with $t$ in the integrand and that's the answer). Commented Dec 7, 2019 at 6:33
• So the answer is just: $\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(\ln(u)-\mu)^2}{2\sigma^2}}\frac{1}{u}$? As I understand it, the fundamental theorem of calculus means I can simply "remove" the integral by differentiating since the integral is the antiderivative. Am I missing something embarrassingly obvious here?
– user732574
Commented Dec 7, 2019 at 6:38
• Almost, but the $u$ needs to be replaced by $t$, and there must be a $\sigma$ in the denominator for the final answer because of what I said in point 1. of my answer. By the way, this distribution is called the log-normal distribution. Commented Dec 7, 2019 at 6:40
• Another question: Is it common to use $t$ instead of $y$ as the parameter in a problem like this? Wouldn't it make more sense to use $y$ instead of $t$? An alternative textbook I found used $t$ like this.
– user732574
Commented Dec 7, 2019 at 7:10