Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?

If $$f(x)$$ is a monotonic increasing function, then does $$\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$$? My intuition says it's true but I cannot prove the case nor find the name of the theorem.

• By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) \leq f(b)$)? – Artem Mavrin Dec 17 '18 at 5:15

Consider the set of $$x$$, call it $$S$$, where $$x. You seek for the probability, $$P(S)$$. Any expression that lead to $$S$$ produces the exact same probability, $$b$$, irrespective of its decleration. If $$f(x)$$ is a monotonic (strictly) increasing function, $$x directly implies $$f(x) and vice versa, i.e. if $$f(x), then $$x.
If $$f$$ is strictly increasing then you have:
\begin{equation} \begin{aligned} \{ X < a \} &= \{ \omega \in \Omega | X(\omega) < a \} \\[6pt] &= \{ \omega \in \Omega | f(X(\omega)) < f(a) \} \\[6pt] &= \{ \omega \in \Omega | f(X)(\omega) < f(a) \} \\[6pt] &= \{ f(X) < f(a) \}, \\[6pt] \end{aligned} \end{equation}
which means that $$\mathbb{P}(X. If $$f$$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.
Assuming that you call monotonically increasing a function that is non-decreasing (for all $$x$$ and $$y$$ such that $$x\leq y$$, one has $$f ( x ) \leq f ( y )$$), consider $$X$$ following a uniform distribution on $$[0, 1]$$, $$f=0$$ and $$a=1$$.
Then, $$P(X.
Your assumption is true for strictly increasing functions. If f is strictly increasing, $$\{X.