Q:

Given X, a random variable with CDF (cumulative distribution function) F, how do you find the CDF (G) of Y which is a function of X?

Given $X$, a random variable with CDF (cumulative distribution function) $F$, how do you find the CDF $G$ of $Y$ which is a function of $X$?

I've read my course materials on CDFs but I'm not finding many resources on this kind of problem. For example if $Y = e^X$ then my intuition tells me that in the cases where $u \leq 0$, $G = 0$ since $Y$'s lower limit is $0$. And for cases where $u > 0$ then $G = e^F$? I'm not sure about the latter...
Another type of problem is $Y = mX+b$ (where $m$ is non-zero). Again my intuition is that there are 2 cases: 1 for which m>0 and 1 for m<0. One of the given hints for this type of problem is that the $m<0$ case might need to be expressed in terms of both the CDF and PMF of X, not just the CDF.
Any help in understanding how to break these types of problems down would be appreciated.

A:

If $Y=h(X)$ where $h$ is strictly increasing, then: $G(y)=P(h(X)<y)=P(X<h^{-1}(y))$ where $h^{-1}(y)=-\infty$ if $y<h(x)$ for all $x$ and $h^{-1}(y)=+\infty$ if $F(x)>y$ or all $x$.
So if $Y=e^X$, then if $y>0$ then $G(y)=P(Y<y)=P(X<\log y)=F(\log y)$.
If $h$ is strictly decreasing, then $P(Y<y)=P(X>h^{-1}(y))$. So:
$$P(Y<y)=1-P(X<h^{-1}(y)) - P(X=h^{-1}(y)) = 1-F(h^{-1}(y)) - P(X=h^{-1}(y))$$
If $X$ is a continuous random variable, then $P(X=h^{-1}(y))=0$ and you just have:
$$P(Y<y)=1-F(h^{-1}(y)) $$
It becomes more complicated when $h$ is not $1-1$, for example, if $h(x)=x^2$ or $h(x)=\sin x$.

