TITLE: How to find the PDF of one random variable when the PDF of another random variable and the relationship between the two random variables are known?
QUESTION [8 upvotes]: The probability density function of random variable X is given as  $ f_x(x) = \lambda e^ {-\lambda x} , x \ge 0. $ A new random variable $ Y = e^ {-\lambda X}$ is formed. Find the PDF of Y.

REPLY [4 votes]: There is a nice "change of variables" formula for doing this.  If $g$ is a monotonic function, then 
$$
f_Y(y)=\left\vert\frac{d}{dy}(g^{-1}(y))\right\vert f_X(g^{-1}(y))
$$
So, let $g(x)=e^{-\lambda x}$; this is a monotonic function, so the formula applies.  $g^{-1}(y)=-\ln(y)/\lambda$, and so 
$$
\left\vert\frac{d}{dy}g^{-1}(y)\right\vert=\frac{1}{\lambda y}
$$
putting the pieces together, we get: 
$$
f_Y(y)=\frac{1}{\lambda y}\lambda \exp[-\lambda g^{-1}(y)]=\frac{1}{y}\exp[\ln(y)]=1
$$
Notice that since $x\geq 0$, $0<y\leq 1$ (assuming $\lambda>0$) so this distribution makes sense.