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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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If $X$ has a density $f(x)$, can you write the integral on the left as a double integral of $xf(y)$ and interchange order of integration? –  Dilip Sarwate Nov 7 '12 at 4:16
    
Hi @DilipSarwate, I'm going to answer this question based on your hint. Would you be able to guide me to solve it? –  Eli Nov 7 '12 at 4:26
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up vote 2 down vote accepted

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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Hi @DilipSarwate, this is what I've got so far...is this the right way to proceed? –  Eli Nov 7 '12 at 4:45
    
You might want to write $$F(x+c)-F(x) = \int_{x}^{x+c} f(y)\,\mathrm dy$$ and then interchange order of integration. Drawing a sketch of the region of the plane over which you are integrating will help you in figuring out the limits and reducing the integral to $$\left.\left.\frac{1}{2}\right(E[X^2] - E[(X-c)^2]\right).$$ There are various points that you will need to be careful about such as "Is the interchange of order of integration justified?" "Is $E[X^2]$ also known to be finite?" etc. –  Dilip Sarwate Nov 7 '12 at 4:58
    
Wonderful. I really appreciate your help. –  Eli Nov 7 '12 at 5:03
    
@DilipSarwate, I edited the answer, the only remaining part is to show that why the integral is equal to the expression you mentioned. Would you be able to also guide me on that? I don't know how to thank you for all your help. –  Eli Nov 7 '12 at 5:20
    
As I said before, draw a sketch to figure out the limits change 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. –  Dilip Sarwate Nov 7 '12 at 13:49
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