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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.

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    $\begingroup$ By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) \leq f(b)$)? $\endgroup$ – Artem Mavrin Dec 17 '18 at 5:15
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Consider the set of $x$, call it $S$, where $x<a$. 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, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(a)$, then $x<a$.

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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 $\mathbb{P}(X<a) = \mathbb{P}(f(X)<f(a))$. If $f$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.

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No.
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, $P(X<a) = P(X<1) = 1 \neq 0 = P(0<0) = P(f(X) < f(a))$.
Your assumption is true for strictly increasing functions. If f is strictly increasing, $ \{X<a\} = \{f(X)<f(a)\}$.

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