Let $X$ be a random variable. If $f : \mathbb{R} → \mathbb{R}$ is right-continuous and nondecreasing. Is the following claim true: $$F_{f(X)}(f(\hat{x}))=F_X(\hat{x})$$

I guess that it is correct since, given that $f$ is non-decreasing $x'>x \Leftrightarrow f(x')>f(x)$, but I have not found any way to prove it or any proof.


1 Answer 1


If $f$ is not strictly increasing then this is usually not true, though it has the right idea, which is to write the event $f(X) \leq f(x_0)$ in the form $X \leq z(x_0)$ for some suitable $z(x_0)$.


$$z(x_0)=\inf \{ x : f(x) \geq f(x_0) \}.$$

Then $\{ X \leq z(x_0) \}$ and $\{ f(X) \leq f(x_0) \}$ are just the same event.

For a counterexample to your version, consider $f(x)=\begin{cases} 0 & x<1/2 \\ 1 & x \geq 1/2 \end{cases}$, $X \sim U(0,1)$ and $x_0=1/4$. Then $F_{f(X)}(f(1/4))=F_{f(X)}(0)=1/2 \neq F_X(1/4)=1/4$.

As far as I know there is no general term relating $f(x)$ to $g(y)=\inf \{ x : f(x) \geq y \}$ when $f$ is right-continuous and nondecreasing. But when $f$ is a CDF, $g$ is called the quantile function; cf. "Inverse" of nondecreasing, right-continuous function?


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