I'm studying probability with Shiryaev's book, and I'm trying to see if I got the meaning of the change of variable with Lebesgue Integral (so that I can study the Fubini theorem). I'm trying then to build a simple example and see how it works.
The theorem states that for two mesurable spaces $(\Omega, \mathcal{F})$ and $(E, \mathcal{E})$ and for $X=X(\omega)$ random variable $\mathcal{F}/\mathcal{E}$-mesurable with values in $E$, if i have $\mathbb{P}$ as a probability measure in $(\Omega, \mathcal{F})$, then I can created an induced measure:
$$ \mathbb{P}_X(A) := \mathbb{P}\left\{\omega: X(\omega) \in A \right\}, $$ and with that I'll have that for every Borel-function $g=g(x)$ that is $\mathcal{E}$-mesurable, holds:
$$ \int_A g(x) \mathbb{P}_X(dx) = \int_{X^{-1}(A)} g(X(\omega))\mathbb{P}(dx), A \in \mathcal{E}. $$
My example. I consider $\Omega=[0,1]$, the Lebesgue measure, $X(\omega) = \omega^3$ and I can choose $g(x)=x$ for a toy example.
Then for instance I know that:
$$ \int_{\Omega} g(X(\omega))d\mathbb{P}= \int_{0}^{1} \omega^3d\omega = 1/4, $$ because in this case the Lebesgue integral and the Riemann integral will match in this case. On the other hand, considering the induced measure:
$$ \mathbb{P}_X(\mathbb{R}) = \mathbb{P}\{ X^{-1}(\mathbb{R})\}= \mathbb{P}\left\{\omega: \omega^3 \in \mathbb{R} \right\}, $$ then how can I proceed to evaluate this?
$$ \int_{\mathbb{R}} g(x) \mathbb{P}_X(dx) = \int_{\mathbb{R}} x \mathbb{P}_X(dx)$$
I'm trying to do it by definition, to "verify" that the result of the theorem actually hold. I can't see why I can write the integral as: $$ \int_{X^{-1}(\mathbb{R})= \Omega} X(\omega) d\mathbb{P} $$
Any advice?
Best Answer
Now, we have probability space $([0,1],\mathcal{B}([0,1]),\lambda)$ and $X(\omega)=\omega^3$ and $g(y)=y$ so $$\int_{[0,1]}\omega^3\lambda(d\omega)=\int_0^1\omega^3d\omega=\frac{1}{4}$$ Let us find the probability distribution of $X$. First, $$X^{-1}((-\infty,a])=\begin{cases} \emptyset & a<0 \\ [0,a^{1/3}] & a \in [0,1] \\ [0,1] & a >1 \end{cases}$$ Therefore $$P_X((-\infty,a])=\begin{cases} 0 & a<0 \\ a^{1/3} & a \in [0,1] \\ 1 & a >1 \end{cases}$$ we have a density $$f_X(a)=\frac{1}{3}a^{-2/3} \ \ \ a \in [0,1]$$ so finally $$\int_{\mathbb{R}}x P_X(dx)=\int_0^1a \, \frac{1}{3}a^{-2/3}da=\int_0^1\frac{1}{3}a^{1/3}da=\frac{1}{4}$$