Product Spaces and Fubini

Product Spaces

The product space of $\Omega_1, \Omega_2$ is

$$\Omega_1 \times \Omega_2 = \{(\omega_1, \omega_2) : \omega_1 \in \Omega_1,\ \omega_2 \in \Omega_2\}$$

For $A \subseteq \Omega_1 \times \Omega_2$, the section of $A$ at $\omega_1$ is

$$A_{\omega_1} = \{\omega_2 \in \Omega_2 : (\omega_1, \omega_2) \in A\}$$

For $X : \Omega_1 \times \Omega_2 \to S$, the section of $X$ at $\omega_1$ is the function $X_{\omega_1} : \Omega_2 \to S$, $X_{\omega_1}(\omega_2) = X(\omega_1, \omega_2)$.

Sections commute with set operations. For each $\omega_1$,

$$(\mathbf{1}_A)_{\omega_1} = \mathbf{1}_{A_{\omega_1}}, \qquad (A^c)_{\omega_1} = (A_{\omega_1})^c, \qquad \Big(\bigcup_n A_n\Big)_{\omega_1} = \bigcup_n (A_n)_{\omega_1}$$

and if $X_n \to X$ pointwise then $(X_n)_{\omega_1} \to X_{\omega_1}$ pointwise.

Product $\sigma$-algebra

Let $(\Omega_1, \mathcal{B}_1), (\Omega_2, \mathcal{B}_2)$ be measurable spaces. A measurable rectangle is a set $A_1 \times A_2$ with $A_1 \in \mathcal{B}_1$ and $A_2 \in \mathcal{B}_2$. The class of all measurable rectangles

$$\mathrm{RECT} = \{A_1 \times A_2 : A_1 \in \mathcal{B}_1,\ A_2 \in \mathcal{B}_2\}$$

is a semialgebra. The product $\sigma$-algebra is the $\sigma$-algebra on $\Omega_1 \times \Omega_2$ generated by the measurable rectangles,

$$\mathcal{B}_1 \otimes \mathcal{B}_2 = \sigma(\mathrm{RECT})$$

If $A \in \mathcal{B}_1 \otimes \mathcal{B}_2$, then $A_{\omega_1} \in \mathcal{B}_2$ for $\omega_1 \in \Omega_1$ and $A_{\omega_2} \in \mathcal{B}_1$ for $\omega_2 \in \Omega_2$. If $X : (\Omega_1 \times \Omega_2, \mathcal{B}_1 \otimes \mathcal{B}_2) \to (S, \mathcal{S})$ is measurable, then $X_{\omega_1}$ is $\mathcal{B}_2/\mathcal{S}$-measurable for $\omega_1 \in \Omega_1$.

Measures on Product Spaces

A transition function is a function $K : \Omega_1 \times \mathcal{B}_2 \to [0,1]$ such that

1. for every $\omega_1 \in \Omega_1$, $K(\omega_1, \cdot)$ is a probability measure on $(\Omega_2, \mathcal{B}_2)$,
2. for every $A_2 \in \mathcal{B}_2$, $K(\cdot, A_2)$ is $\mathcal{B}_1$-measurable.

Let $P_1$ be a probability measure on $(\Omega_1, \mathcal{B}_1)$ and $K$ a transition function. There is a unique probability measure $P$ on $(\Omega_1 \times \Omega_2, \mathcal{B}_1 \otimes \mathcal{B}_2)$ such that for every measurable rectangle $A_1 \times A_2$

$$P(A_1 \times A_2) = \int_{A_1} K(\omega_1, A_2) \, P_1(d\omega_1)$$

Product Measure For probability measures $P_1, P_2$ on $(\Omega_1, \mathcal{B}_1), (\Omega_2, \mathcal{B}_2)$, the product measure $P_1 \otimes P_2$ is the unique probability measure on $(\Omega_1 \times \Omega_2, \mathcal{B}_1 \otimes \mathcal{B}_2)$ such that

$$P_1 \otimes P_2(A_1 \times A_2) := P_1(A_1)P_2(A_2) \quad \forall A_1 \in \mathcal{B}_1,\ A_2 \in \mathcal{B}_2$$

It exists by the theorem above with $K(\omega_1, A_2) := P_2(A_2)$.

Under $P_1 \otimes P_2$, the $\sigma$-algebras $\mathcal{B}_1^{\#} = \{A_1 \times \Omega_2 : A_1 \in \mathcal{B}_1\}$ and $\mathcal{B}_2^{\#} = \{\Omega_1 \times A_2 : A_2 \in \mathcal{B}_2\}$ on $\Omega_1 \times \Omega_2$ are independent.

Smoothing Theorem

Let $P_1$ be a probability measure on $(\Omega_1, \mathcal{B}_1)$, $K$ a transition function, and $P$ the probability measure on $\mathcal{B}_1 \otimes \mathcal{B}_2$ they determine. Let $X$ be $\mathcal{B}_1 \otimes \mathcal{B}_2$-measurable with $X \ge 0$ or $X \in L^1(P)$. Define

$$Y(\omega_1) := \int_{\Omega_2} X(\omega_1, \omega_2) \, K(\omega_1, d\omega_2)$$

Then $Y$ is $\mathcal{B}_1$-measurable and

$$\int_{\Omega_1 \times \Omega_2} X \, dP = \int_{\Omega_1} Y \, dP_1$$

If $X \ge 0$, $Y \ge 0$ is defined for all $\omega_1$. If $X \in L^1(P)$, $Y$ is finite for $P_1$-a.e. $\omega_1$ (set $Y := 0$ on the exceptional null set) and $Y \in L^1(P_1)$.

Fubini-Tonelli

Let $P_1, P_2$ be probability measures on $(\Omega_1, \mathcal{B}_1), (\Omega_2, \mathcal{B}_2)$, $P = P_1 \otimes P_2$, and $X$ $\mathcal{B}_1 \otimes \mathcal{B}_2$-measurable with $X \ge 0$ or $X \in L^1(P)$. Then

$$\int X \, dP = \int_{\Omega_2} \Big[\int_{\Omega_1} X \, dP_1\Big] dP_2 = \int_{\Omega_1} \Big[\int_{\Omega_2} X \, dP_2\Big] dP_1$$

This also holds for products of $\sigma$-finite measures, e.g. $P \otimes \lambda$ with $\lambda$ Lebesgue measure.

Consequences

If $X_1, X_2$ are independent and either both $\ge 0$ or both in $L^1(P)$, then

$$E[X_1 X_2] = E[X_1] E[X_2]$$

If $X_1, X_2$ are independent with dfs $F_1, F_2$, then for all $a \in \mathbb{R}$

$$P(X_1 + X_2 \le a) = \int_{\mathbb{R}} F_1(a - x) \, F_2(dx)$$

This is the convolution $F_1 \ast F_2$.

Tail Integral For an RV $X$,

$$E X^+ = \int_0^\infty P(X > x) \, dx$$

so if $X \ge 0$ then $E X = \int_0^\infty P(X > x) \, dx$.