Let $X$ be a random variable and $Y=g(X)$
Define $$\tag{1} \chi = \{x: f_X(x)>0\}\quad \text{and}\quad \mathcal{Y} = \{y:y=g(x) \text{ for some } x \in \chi\} $$
Define $g^{-1}(y) = \{x\in \chi:g(x) = y\}$
Define: A random variable $X$ is continuous if $F_X(x)$ is a continuous function of $x$.
My question is: how come, in the theorem below, the statement in (b) requires X to be a continuous random variable but the statement in (a) does not
The relevant theorem is (Theorem 2.1.3 in Casella and Berger 2nd Edition)
Let $X$ have cdf $F_X(x)$, let $Y=g(X)$, and let $\chi$ and $\mathcal{Y}$ be defined as in (1)
(a) If $g$ is an increasing function on $\chi$, $F_Y(y) = F_X(g^{-1}(y))$ for $y\in \mathcal{Y}$
(b) If $g$ is a decreasing function on $\chi$ and $X$ is a continuous random variable, $F_Y(y) = 1-F_X(g^{-1}(y))$ for $y\in\mathcal{Y}$
Another way of stating what I am asking is that, prior to stating this theorem, Casella and Berger state
if $g(x)$ is an increasing function, then using the fact that $F_Y(y) = \int_{x\in\chi : g(x)\leq y} f_X(x)dx$, we can write $$ F_Y(y) = \int_{x\in\chi : g(x)\leq y} f_X(x) \, dx = \int_{-\infty}^{g^{-1}(y)} f_X(x) \, dx = F_X(g^{-1}(y)) $$
If $g(x)$ is decreasing, then we have
$$ F_Y(y) = \int_{g^{-1}(y)}^\infty f_X(x) \, dx = 1-F_X(g^{-1}(y)) $$ "The continuity of $X$ is used to obtain the second equality
My question(restated) is in yellow box below:
My question (restated) is: How come, when $g(x)$ is an increasing function we do not need to use continuity of $X$, but we do for the case when $g(x)$ is decreasing?
- (A side question, I will accept answer so long as answers the above question): this is continuity of the random variable, but the integral uses the PDF. what is the relation between continuity of $X$ and it's pdf? (specifically, I think there may be some strangeness if $F_X$, the CDF of $X$ is continuous but not differentiable)?
What came to my mind was Fundamental theorem of calculus maybe, but there is a version of it that doesn't require continuity of $f$ I think? Plus, here we have $X$ is continuous, if that matters -- I'm not sure.