$X\sim N(0,1)$, 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!
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 $f_{Y}(y)=\frac{1}{y}\frac{1}{\sqrt{2\pi}}\exp(-\frac{\ln^{2}(y)}{2})\,\,,\,y>0$
With the substitution $x=g(y)$ we have $dx = g'(y) \, dy$, so $$f(x) \, dx = f(g(y)) g'(y) \, dy = f(\ln y) \frac{1}{y} \, dy = \frac{1}{y} \frac{1}{\sqrt{2 \pi}} e^{-(\ln y)^2/2} \, dy$$ which is the PDF of the standard log-normal distribution.
No it is not correct
$$f_Y(y)=\frac{1}{y\sqrt{2\pi}}e^{-\log^2(x)/2}$$
It's a LogNormal density
The cumulative probability function of $Y$ is :
$$P(Y\leq t) =P(e^X\leq t)=P(X\leq \ln(t))=\int_{-\infty}^{\ln(t)}f_X(u)du$$
since $f_Y (t) = \frac{d}{dt} P(Y\leq t)$, by the FTC and the chain rule you get:
$$f_Y(t)=\frac{f_X(\ln(t))}{t}$$
Using that $X\sim N(0,1)$
$$f_Y(t)=\frac{1}{t\sqrt{2\pi}}e^{-\frac{\ln(t)^2}{2}}$$
