# Expected value of the negative part of a random variable

I want to prove that, if $X$ is a real valued random variable with finite expected value, then:

$$\mathbb{E}[X]=\displaystyle \int_{0}^{\infty} \mathbf{P}(X \geq t)dt - \int_{-\infty}^{0} \mathbf{P} (X \leq t)dt.$$

We have that $$\mathbb{E}[X] = \mathbb {E} (X^{+}) - \mathbb{E} (X^{-})$$ and I know how to prove that if $Y$ is a non-negative r.v., then its expected value can be expressed as $$\mathbb{E}[Y]=\displaystyle \int_{0}^{\infty} \mathbf{P}(Y \geq t)dt.$$

I am having trouble expressing the second integral as the expectation of the negative part, $X^{-}$.

Can anyone help me with that?

• Isn't this statement false for a Cauchy distribution: $P(x) = 1/(1+x^2)$? – eyeballfrog May 28 '17 at 14:51
• Apply what you know to $X^+$ and to $X^-$, then use $X=X^+-X^-$ hence $E(X)=E(X^+)-E(X^-)$ with $$E(X^+)=\int_0^\infty P(X^+>x)dx=\int_0^\infty P(X>x)dx$$ and $$E(X^-)=\int_0^\infty P(X^->x)dx=\int_0^\infty P(X<-x)dx=\int_{-\infty}^0P(X<x)dx$$ – Did May 28 '17 at 15:10
• That was very clear, thank you. – dimvolt May 28 '17 at 15:40
• Even though this post is slightly different, I’d like to link it to the current choice of mother post. Also see the meta post for (abstract) duplicates. – Lee David Chung Lin Nov 13 '18 at 13:35

Rewrite the second integral with the change of variables $Y=-X$.