Another formula for Expectation

Question

I came across the following question:

If $ X $ is a continuous random variable $ (-\infty<X<\infty) $ having distribution function $ F(x) $, show that $$ E(X)=\int_{0}^{\infty}[1-F(x)-F(-x)]\,\mathrm{d}x, $$ provided $ x[1-F(x)-F(-x)]\to0 $ as $ x\to\infty $.

I went through a similar problem posted here, but cannot seem to generalise the result as required. Any hint will be appreciated.

Write $X = X^+ - X^-$ where $X^+ = \max(X, 0)$ and $X^- = \max(-X, 0)$ and apply the linked answer. — Giorgos Giapitzakis, Nov 29 '23 at 09:54
$P(X^+ > x) = P(X > x)$ and $P(X^- > x) = P(X < -x)$ for $x > 0$. — Giorgos Giapitzakis, Nov 30 '23 at 04:31
@GiorgosGiapitzakis Still can't get it. I am getting $ F^{+}(t)=F(t)-F(0) $ and $ F^{-}(t)=F(0)-F(-t) $. And how do I use the added condition? — Subhajit Paul, Nov 30 '23 at 04:32
I don't think the other condition is really needed but I guess if you assume that $X$ has a density then you can write $1-F(x)-F(-x) = x'[1 - F(x) - F(-x)]$ and do integration by parts on the RHS. — Giorgos Giapitzakis, Nov 30 '23 at 04:34
@GiorgosGiapitzakis Thank you. Why don't you post this as an answer so that I can accept it. — Subhajit Paul, Nov 30 '23 at 04:39

score 1 · Accepted Answer · answered Nov 30 '23 at 04:46

The key is to write $X = X^+ - X^-$ where $X^+ = \max(X, 0)$ and $X^- = \max(-X, 0)$. Now notice that for $x > 0$ we have $P(X^+ > x) = P(X > x)$ and $P(X^- > x) = P(X < -x)$. Now apply the formula of the linked answer separately for $X^+$ and $X^-$ to get the required expression. Notice that the condition $x[1-F(x)-F(-x)] \to 0$ as $x\to \infty$ was not needed.

Alternatively, if you assume $X$ has a density, you can write $$\int_0^\infty (1-F(x)-F(-x))\,dx = \int_0^\infty x'(1-F(x)-F(-x))\,dx$$ and integrate by parts to get $E(X)$ but then you would need to use the additional condition.

Another formula for Expectation

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