$X\sim N(0,1)$, find probability density function of $Y=e^X$. [duplicate]

$X\sim N(0,1)$, find probability density function of $Y=e^X$.

Define $\psi:=e^X$, since $\psi$ is a monotonic continues function then $\frac{f_X(X)}{|\psi'(X)|}=f_{\psi(X)}(X)=f_Y(Y)$.

$\frac{\frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2}}{ln(y)y}=f_Y $

I am not sure that i can use this theorem.

Is it correct ?

Thanks!

Solution 1:

Yes you can use the Jacobian Transformation

$f_{Y}(y)=|\frac{dx}{dy}|f_{X}(x)$

You have $y=e^{x}\implies \ln(y)=x$ and $\frac{dy}{dx}=e^{x}=y$.

And $f_{X}(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^{2}}{2})$

So $f_{Y}(y)=\frac{1}{y}\frac{1}{\sqrt{2\pi}}\exp(-\frac{\ln^{2}(y)}{2})\,\,,\,y>0$

Solution 2:

No it is not correct

$$f_Y(y)=\frac{1}{y\sqrt{2\pi}}e^{-\log^2(x)/2}$$

It's a LogNormal density