Transformation and log-normal distribution

Question

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?

Answer 1

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.