Question

# Find the probability density function of Y=e^{X}

Upper level probability

Find the probability density function of $$Y=e^{X}$$, when X is normally distributed with parameters $$\mu\ \text{and}\ \sigma^{2}$$. The random variable Y is said to have a lognormal distribution (since log Y has a normal distribution) with parameters $$\mu\ \text{and}\ \sigma^{2}$$

2021-02-02

We are given that

$$X\sim N(\mu, \ \sigma^{2}) \ \text{and}$$ $$Y-e^{X}$$.

Observe that $$Y>0$$ with porbability 1.

So that for ant $$y>0$$ we have that

$${F}_{{Y}}{\left({y}\right)}={P}{\left({Y}\le{y}\right)}={P}{\left({e}^{x}\le{y}\right)}={P}{\left({X}\le \log{{\left({y}\right)}}\right)}={F}_{{X}}{\left( \log{{\left({y}\right)}}\right)}$$

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

$${{f}_{{Y}}{\left({y}\right)}}=\frac{d}{{\left.{d}{y}\right.}}{F}_{{y}}{\left({y}\right)}={{f}_{{X}}{\left( \log{{\left({y}\right)}}\right)}}$$. $$\frac{1}{{y}}=\frac{1}{{\sigma\sqrt{{2}}\pi}} \exp{{\left(-\frac{{{\left( \log{{u}}-\mu\right)}^{2}}}{{{2}\sigma^{2}}}\right)}}\cdot\frac{1}{{y}}$$