# cumulative function

• Jun 7th 2010, 01:52 PM
Veve
cumulative function
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 $\displaystyle a\ge0$, we have $\displaystyle \int_{-\infty}^\infty [F(x+a)-F(x)]dx=a$.

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

Thanks.
• Jun 7th 2010, 10:16 PM
matheagle
IF you have a density, then

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

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

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

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

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

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

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

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

The second one looks like parts.