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Assume the integral exists. Show that for any distribution function $M$ with density function $m$ exists and $y\geq 0$, $\int_{-\infty}^{\infty} [M(x+y) - M(x)] dx = y$,

My attempt: I see that $F(x+a) - F(x) = P(x< X < x+a)$ for $X$ is a random variable. But this means the indefinite integral is just the sum of all the probabilities, so why is it $a$, but not $1$? I think I made a serious mistake somewhere in this argument.

Could someone give me some help on this problem? Would really appreciate any of your help.

ghjk
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  • I think you need to assume the expectation of $X$ exists and is finite. Then, you might try integration by parts. – Michael Sep 15 '16 at 03:10
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    A more clever way (with no assumptions on expectation of $X$ being finite) is to find an expression for $\int_{x=-\infty}^{\infty} 1_{{X \in (x, x+y]}}dx$ and then use a Fubini-Tonelli argument on $E[\int_{x=-\infty}^{\infty} 1_{{X \in (x, x+y]}}dx]$. – Michael Sep 15 '16 at 03:19
  • @Michael: thank you so much for your great help! I got it. Could you please give this problem a try as well? http://math.stackexchange.com/questions/1927392/pairwise-independent-events-but-not-mutually-independent – ghjk Sep 15 '16 at 04:13

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If $M$ is a cumulative distribution function, $M: \mathbb{R} \rightarrow [0,1]$.

The integral $\int_{-\infty}^{\infty} [M(x+y) - M(x)] dx$ represents the area between $M(x+y)$ and $M(x)$. And that integral can be evaluated using the Lebesgue integral from $0$ to $1$ (since the codomain of $M$ is $[0,1]$) of $x+y -x= y$ (which is the shift you mention). Therefore,

$\int_{-\infty}^{\infty} [M(x+y) - M(x)] dx = \int_{0}^{1} [x+y-x] dl = \int_{0}^{1} y ~ dl = y.$

jDAQ
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