cumulative function

**Veve** · June 7th 2010, 01:52 PM

I have a problem, which I assume is easy to solve, but I don't know exact how to write a coherent argument...

1. For any distribution function and any $a\ge0$ , we have $\int_{-\infty}^\infty [F(x+a)-F(x)]dx=a$ .

2. For any compactly supported function $\phi:\mathbb{R}\to\mathbb{R}$ and any probability measure $\mu$ with cumulative function F, one has $\int\phi d\mu=-\int\phi '(x)F(x)dx$ .

Thanks.

**matheagle** · June 7th 2010, 10:16 PM

IF you have a density, then

$\int_{-\infty}^\infty [F(x+a)-F(x)]dx=\int_{-\infty}^\infty [P(X\le x+a)-P(X\le x)]dx$

$=\int_{-\infty}^\infty P(x<X\le x+a)dx$

$=\int_{-\infty}^\infty \int_x^{x+a}f(u)dudx$

$=\int_{-\infty}^\infty \int_{u-a}^uf(u)dxdu$

$=\int_{-\infty}^\infty af(u)du$

$=a\int_{-\infty}^\infty f(u)du=a$

So, it's correct. Next use dF(x) instead of f(x) and switch the order of integration again...

$\int_{-\infty}^\infty \int_x^{x+a}dF(u)dx$

The second one looks like parts.

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