In this question, the accepted answer claims that

for a random variable $X\geq 0$, the event $A=[X\leq x]$ is the same as $ [u(X)\geq u(x)]$, for any decreasing bounded invertible function $u$ (e.g., $u(X)=\frac{1}{1+tX}$ or $u(X)= e^{-tX}$, $t>0$).

I really don't see how this is true. Could anyone give me a hint here?


1 Answer 1


For any decreasing function $u$ we have

$$y \leq x \implies u(y) \geq u(x).$$

If we assume additionally that $u$ has an inverse $u^{-1}$, then $u^{-1}$ is decreasing, and therefore

$$w := u(y) \geq u(x)=:v \implies y = u^{-1}(w) \leq u^{-1}(v)=x.$$

Consequently, we have

$$y \leq x \Leftrightarrow u(y) \leq u(x) \tag{1}$$

for any bounded invertible function $u$. Hence,

$$\begin{align*} \omega \in [X \leq x] &\Leftrightarrow X(\omega) \leq x \\ &\stackrel{(1)}{\Leftrightarrow} u(X(\omega)) \geq u(x) \\ &\Leftrightarrow \omega \in [u(X) \geq u(x)] \end{align*}$$ which shows that $$[X \leq x] = [u(X) \geq u(x)]$$ for any bounded invertible function $u$.


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