Expectation and Distribution Function? Consider X as a random variable with distribution function $F(x)$. Also assume that $|E(x)| < \infty$. the goal is to show that for any constant $c$, we have:
$$\int_{-\infty}^{\infty} x (F(x + c) - F(x)) dx = cE(X) - c^2/2$$
Does anyone have any hint on how to approach this?
Thanks
 A: Based on @DilipSarwate suggestion, we can write the integral as a double integral because:
$\int f(y)dy = F(y)$ so, we can write:
$ \int_{-\infty}^{\infty} x (F(x + c) - F(x)) dx = \int_{-\infty}^{\infty} x \{\int_{x}^{x + c} f(y)dy\} dx = \int_{-\infty}^{\infty} \{\int_{x}^{x + c} xf(y)dy\} dx  = \int_{-\infty}^{\infty} \int_{x}^{x + c} xf(y)dy\ dx = \text{based on the Fubini's Thm. since $f(y)\ge 0$ and we know that $\int |f|dp < \infty (why?) $, then we can change the order of integrals}\\ = \text{assume that we can show the integ. is eq to} = \frac{1}{2}(E(X^2)- E((X - c)^2)) = \frac{1}{2}(E(X^2) - E(X^2 + c^2 - 2Xc)) = \frac{1}{2}(-E(c^2) + 2cE(x)) = cE(x) - \frac{c^2}{2}$
The missing part here is to know how to show the integral of $\int_{-\infty}^{\infty} \int_{x}^{x + c} xf(y)dy\ dx$ is equal to $\frac{1}{2}\{E(X^2) - E(X^2 + c^2 - 2Xc)\}$ ?!
A: Here is an approach valid for every distribution: integrate the pointwise identity
$$
\int_{-\infty}^{+\infty}x\,\mathbf 1_{x\lt X\leqslant x+c}\,\mathrm dx=\int_{X-c}^{X}x\,\mathrm dx=cX-c^2/2.
$$
