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$F$ is CDF (probability distribution function)

$\int _{\mathbb{R}}[F(x+c) - F(x)] dx $

$= \int _{\mathbb{R}}F(x+c) dx -\int _{\mathbb{R}} F(x) dx$ ( (by linearity of the integral)

$=\int _{\mathbb{R}}F(x) dx -\int _{\mathbb{R}}F(x) dx $ ( By change of variables theorem )

$=0$

For change of variables we used the known theorem (see below) with $\Omega = \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ ( Lebesgue measure)

Question: Why is the above Proof incorrect? How can we show it is not correct analytically?

My thought:

I am not sure how to show it, maybe because the density does not exist? ( Even assumption of continuity would not imply the existence of the density) ( e.g. Cantor function is a continuous CDF which does not have a density with respect to Lebesgue measure)

Below is the known Theorem that we used 

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Why is the above proof wrong ?

In other words why is this step $=\int _{\mathbb{R}}F(x) dx -\int _{\mathbb{R}}F(x) > dx $ ( By change of variables theorem ) Incorrect ?
We use the change of variable theorem (see above) with $\Omega = \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ ( Lebesgue measure)

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  • $\begingroup$ Why is this step $=\int _{\mathbb{R}}F(x) dx -\int _{\mathbb{R}}F(x) dx $ **( By change of variables theorem )** Incorrect ? For change of variables we used the known theorem (see above) with $\Omega = \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ ( Lebesgue measure) $\endgroup$
    – xiao
    Dec 10 '20 at 7:02
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Since $F$ is a CDF, $\displaystyle\lim_{x\rightarrow\infty}F(x)=1$. Therefore, the integral $\int_{\mathbb{R}}F(x)\,\mathrm{d}x$ diverges. When you get the expression $$\int _{\mathbb{R}}F(x+c)\, \mathrm{d}x -\int _{\mathbb{R}} F(x)\, \mathrm{d}x$$ this becomes $\infty-\infty$, which is an indeterminate form, so it cannot be simplified to $0$.

With regard to how one would correctly evaluate this integral, there is a simple way to do it using Fubini's Theorem.

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  • $\begingroup$ How? should I apply Fubini on each integral? So the reason the given proof is incorrect is that we can't use the Change of variable theorem since the integral diverges? The actual solution is $\int _{\mathbb{R}}[F(x+c) - F(x)] dx = c $ . $\endgroup$
    – xiao
    Dec 10 '20 at 8:18
  • $\begingroup$ No you can use the change of variable... What you cannot use is linearity. Conditions for using it are not fulfilled $\endgroup$
    – Gono
    Dec 10 '20 at 8:21
  • $\begingroup$ @xiao The problem is not with the change of variables; the problem is with splitting up the integral—you want to leave it as a single integral. First, you want to rewrite the integrand as $F(x+c)-F(x)=\int_{F(x)}^{F(x+c)}\mathrm{d}y$, and then you can use Fubuni's Theorem to change the order of integration. $\endgroup$
    – Sandejo
    Dec 10 '20 at 8:26
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For any CDF $F$ we have $\int_{\mathbb R} F(x)dx=\infty$. This is because $F(x) \to 1$ as $ x \to \infty$. If we choose $a$ such that $F(x)>\frac 1 2$ for all $x >a$ then we get $\int_{\mathbb R} F(x)dx \geq \int_a^{\infty} \frac 1 2 dx=\infty$. So splitting the integral into two terms gives you $\infty-\infty$.

For an example where the integral is not $0$ consider $exp(1)$ distribution and take $c=1$ You can easily compute the integral in this case and the answer is not $0$.

Actually the value of the integral is $c$. Let $\mu$ the probability measure corresponding to $F$. Then (by Fubini/Tonelli's Theorem) $\int_{\mathbb R} [F(x+c)-F(x)]dx=\int_{\mathbb R} \int_{(x,x+c]} d\mu(y)dx=\int \int_{y-c}^{c}dx d\mu (y)=\int c d\mu(y)=c$.

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  • $\begingroup$ So we can't split the integral into two terms because that gives us $\infty-\infty$.? which means the densities do not exist? $\endgroup$
    – xiao
    Dec 10 '20 at 8:20
  • $\begingroup$ Whether densities exist or not you cannot split it into two parts. $\int_{\mathbb R} (1-1)dx=0$ but $\int_{\mathbb R} 1 dx -\int_{\mathbb R} 1 dx $ is not defined. This has nothing to do with probability theory. @xiao $\endgroup$ Dec 10 '20 at 8:25

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