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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?

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1 Answer 1

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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.

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  • $\begingroup$ Thanks! So I should calculate the integral and then take the differential of it in order to get rid of the $u$'s? $\endgroup$
    – user732574
    Commented Dec 7, 2019 at 6:32
  • $\begingroup$ 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). $\endgroup$ Commented Dec 7, 2019 at 6:33
  • $\begingroup$ 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? $\endgroup$
    – user732574
    Commented Dec 7, 2019 at 6:38
  • $\begingroup$ 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. $\endgroup$ Commented Dec 7, 2019 at 6:40
  • $\begingroup$ 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. $\endgroup$
    – user732574
    Commented Dec 7, 2019 at 7:10

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