cumulative function

May 2008
50
1
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.
 

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
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.
 
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