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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

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    Hi @DilipSarwate, I'm going to answer this question based on your hint. Would you be able to guide me to solve it?2012-11-07

2 Answers 2

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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. $

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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 $\frac{1}{2}\{E(X^2) - E(X^2 + c^2 - 2Xc)\}$ ?!

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    As I said before, **draw a sketch** to figure out the limits cha$n$ge when you interchange the order of integration. It is **not true** that $$\int_{-\infty}^{\infty}$x$\int_$x$^{$x$+c}f(y)\,\mathrm dy\,\mathrm dx=\int_x^{x+c}f(y)\int_{-\infty}^{\infty}x\,\mathrm dx\,\mathrm dy.$$ The **limits** change, and when you do it right and compute the inner integral, the expression will fall out directly.2012-11-07