Question

Find the probability density function of Y = eX when X is normally distributed with parameters μ and σ2. The random variable Y is said to have a lognormal distribution (since log Y has a normal distribution) with parameters μ and σ2.

Short answer

The result will be

$f_Y\left(y\right)=\frac{1}{\sigma \sqrt{2\pi }}\exp \left(-\frac{(\log y-\mu {)}^2}{2{\sigma }^2}\right)\cdot \frac{1}{y}$

Step-by-step solution

Step:1 Given Information

When X is normally distributed with parameters and 2, find the probability density function for Y = e X. Because log Y has a normal distribution, the random variable Y is said to have a lognormal distribution with parameters and.

Step:2 Definition

A probability density function (PDF) is used in probability theory to signify the random variable's likelihood of falling into a particular range of values as opposed to taking up a single value. The feature illustrates the normal distribution's probability density function and how mean and deviation are calculated.

Step:3 Explanation of the solution

We are given that $X~\mathcal{N}\left(\mu ,{\sigma }^2\right)$and $Y=e^X$, Observe that $Y>0$with probability 1 . So, for any $y>0$we have that

$F_Y\left(y\right)=P(Y\le y)=P\left(e^X\le y\right)=P(X\le \log (y\left)\right)=F_X\left(\log \right(y\left)\right)$

where $F_X$is CDF of X. By differentiating, we have that

$f_Y\left(y\right)=\frac{d}{dy}{F}_Y\left(y\right)=f_X\left(\log \right(y\left)\right)·\frac{1}{y}=\frac{1}{\sigma \sqrt{2\pi }}\exp \left(-\frac{(\log y-\mu {)}^2}{2{\sigma }^2}\right)·\frac{1}{y}$