Consider the accepted answer to the following question which says
It's Fubini's theorem . The domain of the double integral is:
$(\omega,y)\in\Omega\times [0, |X(\omega)|]$, however the outer
integral is with respect to the measure of the outcomes. $\mathrm
d\mathbb P\equiv \mathbb P(\mathrm d \omega)$. So when we apply
Fubini's theorem the new inner integral is simply the cumulative
measure.Via:
$$\begin{align}\mathbb{E}|X|^p &
=\int_{\Omega}p\left(\int_0^{|X(\omega)|}y^{p-1}\,\mathrm d y \right)\,\mathbb{P}(\mathrm d\omega) \\ & = p\int_0^\infty
y^{p-1}\left(\int_{\Omega:y<|X(\omega)|} \mathbb{P}(\mathrm
d\omega)\right)\,\mathrm d y \\ & = p\int_0^\infty y^{p-1}\mathbb
P(y<|X|)\,\mathrm d y \end{align}$$
I understand the first equality, this is just rewriting $|X|$. However, I do not understand how one goes from the first to the second line. The author suggests that this is due to Fubini. But this is not immediate according to the version I have:
Theorem 4.22 (Fubini's theorem) (i) If $f$ is non-negative measurable,
then $$ \int_E f \mathrm{~d} \mu=\int_{E_1}\left(\int_{E_2}
f\left(x_1, x_2\right) \mu_2\left(\mathrm{~d} x_2\right)\right)
\mu_1\left(\mathrm{~d} x_1\right) . $$ In particular, we have $$
\int_{E_1}\left(\int_{E_2} f\left(x_1, x_2\right)
\mu_2\left(\mathrm{~d} x_2\right)\right) \mu_1\left(\mathrm{~d}
x_1\right)=\int_{E_2}\left(\int_{E_1} f\left(x_1, x_2\right)
\mu_1\left(\mathrm{~d} x_1\right)\right) \mu_2\left(\mathrm{~d}
x_2\right) . $$ This is sometimes known as Tonelli's theorem. (ii) If
$f$ is integrable, and $$ A=\left\{x_1 \in E:
\int_{E_2}\left|f\left(x_1, x_2\right)\right| \mu_2\left(\mathrm{~d}
x_2\right)<\infty\right\} . $$ then $$ \mu_1\left(E_1 \backslash
A\right)=0 . $$ If we set $$
f_1\left(x_1\right)=\left\{\begin{array}{ll} \int_{E_2} f\left(x_1,
x_2\right) \mu_2\left(\mathrm{~d} x_2\right) & x_1 \in A \\ 0 & x_1
\notin A \end{array},\right. $$ then $f_1$ is a $\mu_1$ integrable
function and $$ \mu_1\left(f_1\right)=\int_E f \mathrm{~d} \mu $$
My issue is that $E_2$ should not depend on $E_1$ as in the theorem. But in the given application the author seems to do the following
- Take their union (in 2-D)
- Reparameterise it in the form of the second line
- Claim that this is due to Fubini.
I am struggling to see how this makes sense.
Question: Could someone help me understand why
$$\begin{align}\int_{\Omega}p\left(\int_0^{|X(\omega)|}y^{p-1}\,\mathrm d y \right)\,\mathbb{P}(\mathrm d\omega) & = p\int_0^\infty
y^{p-1}\left(\int_{\Omega:y<|X(\omega)|} \mathbb{P}(\mathrm
d\omega)\right)\,\mathrm d y \end{align}$$
In particular, how does this follow from Fubini's Theorem?
Best Answer
$\newcommand{\d}{\,\mathrm{d}}$Well, "really" we integrate: $$\int_\Omega\int_0^\infty f\cdot\chi_{\{(\omega,y):y<|X(\omega)|\}}\d y\d\omega$$Where $f$ is $(\omega,y)\mapsto py^{p-1}$. Assuming you're satisfied that Fubini allows us to exchange the order we get this equal to: $$\int_0^\infty\int_\Omega f\cdot\chi_{\{(\omega,y):y<|X(\omega)|\}}\d\omega\d y$$So far, not mysterious - I hope. Our domains of integration are constant, there is no reparametrising. Now let's get rid of the indicator function: $$\int_0^\infty\int_{\{\Omega:y<|X(\omega)\}}f\d\omega\d y$$
Happy? This trick is common. Indicator functions are theoretically useful :)