# Why is $X^- = -\min\{X, 0\}$?

Why is $X^- = -\min\{X, 0\}$ ? This is how it is defined in a probability book I am self-studying.

For the function $X:\Omega \to \mathcal{R}$,

The positive part of $X$ is the function $X^+ = \max \{X, 0\}$.

The negative part of $X$ is the function $X^- = -\min \{X, 0\}$.

It seems to me that it should be: $X^- = \min\{X, 0\}$

-
I think you'll need to provide more context. – joriki Sep 21 '12 at 20:50
The reason is that you want both $X^+$ and $X^-$ to be positive functions. See below for motivation – Jean-Sébastien Sep 21 '12 at 21:07

One motivation for this is that in probability theory, one defines first the integral indicator function, then those of simple functions (taking only finitely many values), then positive functions. Signed function can be more tricky at first but since $X=X^+-X^-$ and that both these function are positive, you use this as a definition for the integral of random variables. This also explain the added condition that a random variable is integrable if $\mathbb{E}(|X|)<\infty$.