PDF of a random variable depending on another random variable Let $X$ be a random variable with probability density function $f_X(x), x\in\mathbb R$ and let $Y=e^X$. What is the probability density function $f_Y(y), \ y>0$ of the random variable Y?
The solution is $f_Y(y)=\frac{f_X(log(y)}{y}$ but how do I actually compute it?
The reason I am asking is so I can understand how to calculate the PDF of a random variable that depends on another.
 A: You can use the fundamental tranformation theorem directly deriving $f_Y(y)$
$$f_Y(y)=f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|=f_X(log(y))\cdot \frac{1}{y}$$
Proof of the theorem:
As per definition
$$F_Y(y)=\mathbb{P}[Y \leq y]=\mathbb{P}[g(X) \leq y]=$$
Suppose g is monotone increasing (that is  your case) we have
$$F_Y(y)=\mathbb{P}[X \leq g^{-1}(y)]=F_X(g^{-1}(y))$$
to derive $f_Y(y)$ simply derivate F obtaining
$$f_Y(y)=f_X(g^{-1}(y))\frac{d}{dy}g^{-1}(y)$$
you can do the same brainstorming with g decreasing  and also with g non monotone getting similar results
A: $P(Y \leq y)=P(X \leq \ln y)=\int_{-\infty}^{\ln y} f_X(t)dt$ for $y>0$.
Put $s=e^{t}$ to write the integral as $\int_{-\infty}^{y} f_X(\ln s) \frac 1  s ds$. Differentiate this to get the density as $\frac 1 y f_X(\ln y)$.
A: By chain rule since $x=\log y$ and $\frac{dy}{dx}=e^x$  we have
$$f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|=\frac{f_X(\log(y))}{e^x}=\frac{f_X(\log(y))}{y}$$
For the proof, assuming wlog $g$ strictly increasing, note that for the CDF we have
$$F_Y(y)=P(Y\le y)=P(g(X)\le y)=P(X\le g^{-1}(y))=F_X(g^{-1}(y))=F_X(x)$$
therefore
$$f_Y(y)=\frac{d}{dy}F_X(x)=\frac{d}{dx}F_X(x)\cdot \frac{dx}{dy}=f_X(x)\frac{dx}{dy}$$
