# Thread: Show integral equal to two things

1. ## Show integral equal to two things

Hi

I need to show that $\int_{\Omega}\int_{0}^{\infty} I_{\left[0,X(\omega)\right]} \, dx \, dP(\omega)$, where $I_{\left[a,b\right]}$ is the indicator function, i.e. $I_{\left[a,b\right]}(x)=1$ if $x \in \left[a,b\right]$ and zero otherwise, is equal to both

$E(X)$ and $\int_{0}^{\infty}(1-F(x)) \, dx$ , where $F(x)$ is the cdf of $X$.

Hence deducing that

$E(X)=\int_{0}^{\infty}(1-F(x)) \, dx$

I tried:

$\int_{\Omega}\int_{0}^{\infty} I_{\left[0,X(\omega)\right]} \, dx \, dP(\omega) = \int_{\Omega} \int_{0}^{X(\omega)} \, dx \, dP(\omega) = \int_{\Omega} X(\omega) \, dP(\omega) = E(X)$, since this is the books' definition of the expected value of $X$.

I´m not sure how to go about finding the other result.

Thanks

2. For the other one, $\displaystyle 1-F(x)=P(X>x)=\mathbb{E}[I_{X\in]x,\infty[}]=\int_\Omega I_{]x,\infty[} dP_X$

Then reverse the order of the integrals and you'll get the result.

3. So I should just substitute this expression into the integral. In other words,

$\int_{0}^{\infty} \int_{\Omega} I_{(x,\infty)} \, dP(\omega) \, dy$

What do you mean by reversing the order? I should move the zero to infinity integral inside the $\Omega$ integral. This would case my ingegration to go from $x$ to $\infty$ right.

4. $\int_{0}^{\infty}(1-F(x)) \, dx = \int_{0}^{\infty}\int_{\Omega} I_{X\in(x,\infty)} \, dP(\omega) \, dx = \int_{\Omega}\int_{0}^{\infty} I_{X\in(0,\infty)} \, dx \, dP(\omega) =$

$= \int_{\Omega}\int_{0}^{X(\omega)} \, dx \, dP(\omega) = E(X)$

$\int_{0}^{\infty}(1-F(x)) \, dx = \int_{0}^{\infty}\int_{\Omega} I_{X\in(x,\infty)} \, dP(\omega) \, dx = \int_{\Omega}\int_{0}^{\infty} \bold{I_{X\in(0,\infty)}} \, dx \, dP(\omega) =$
In the bold thing, it's rather $I_{\{x\in (0,X)\}}$, otherwise it's correct !