# cumulative function

#### Veve

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

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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.

Last edited:
• mr fantastic and Veve