Application of Fubini's Theorem

Let $(\Omega, \mathcal{F}, \lambda)$ be a probability space, and $X, Y : \Omega \rightarrow \mathbb{R}$ be independent random variables.

Define the distributions of $X$ and $Y$ by

Hence, both $\mu$ and $\nu$ are probability measures on $(\mathbb{R}, \mathcal{B})$. Denote $\mu \times \nu$ the product measure of $\mu$ and $\nu$ on $(\mathbb{R}^2, \mathcal{B}^2)$.

By independence of $X$ and $Y$, $\forall A, B \in \mathcal{B}$,

Since $\mathcal{S} \doteq \{ A \times B: A, B \in \mathcal{B} \}$ generates $\mathcal{B}^2$, i.e., $\sigma(\mathcal{S}) = \mathcal{B}^2$, applying $\pi$-$\lambda$ theorem, we get

It follows that $\mu \times \nu$ is the distribution of $(X, Y)$.

We list some interesting corollaries based on Fubini's theorem.

Corollary One. For each $M \in \mathcal{B}^2$, it holds that

$$\Pr[ (X, Y) \in M ] = \int_{\mathbb{R}^2} \Pr[ (x, Y) \in M ] \, \mu( dx ).$$

Similarly, for each $A \in \mathcal{B}, M \in \mathcal{B}^2$,

$$\Pr[ X \in A, (X, Y) \in M ] = \int_{A} \Pr[ (x, Y) \in M ] \, \mu( dx ).$$

Proof. Via Fubini's theorem,

where $\Pr[ (x, Y) \in M ]$ is a shorthand for

Substituting the function $\mathbb{I}[ (x, y) \in M ]$ with

proves the second claim.

$\square$

Another corollary is a direct application of the Fubini's theorem.

Corollary Two. Assume that $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is a function measurable $\mathcal{B}^2 / \mathcal{B}$. Define

$$g(x) \doteq \mathbb{E}[ f(x, Y) ] = \int_{\mathbb{R}} f(x, y) \,\, \nu (dy), \quad \forall x \in \mathbb{R}.$$

Then we have:

$$\mathbb{E}[ g(X) ] = \mathbb{E} [ f(X, Y) ],$$

i.e.,

$$\int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(x, y) \,\, \nu (dy) \right) \, \, \mu (dx) = \int_{\mathbb{R}^2} f(x, y) \, \, (\mu \times \nu)(d (x,y))$$

Remarks:

1. If $X$ and $Y$ are not independent, then the above claim may not hold, since the product measure $\mu \times \nu$ may not be the joint distribution of $(X, Y)$.

2. In the above, we operate on the product space $(\mathbb{R} \times \mathbb{R}, \mathcal{B} \times \mathcal{B}, \mu \times \nu) = (\mathbb{R}^2, \mathcal{B}^2, \mu \times \nu)$, instead of $(\Omega \times \Omega, \mathcal{F} \times \mathcal{F}, \lambda \times \lambda) = (\Omega^2, \mathcal{F}^2, \lambda^2)$.

3. There are various interpretations of $\mathbb{E} [ f(X, Y) ]$:

• $f(X, Y)$ can be viewed as a function from $\Omega$ to $\mathbb{R}$, given by $\omega \rightarrow f( X(\omega), Y(\omega) )$. It is easy to see that it is measurable $\mathcal{F} / \mathcal{B}$, and therefore is a random variable. Then

  $$\mathbb{E} [ f(X, Y) ] = \int_\Omega f( X(\omega), Y(\omega) ) \, \, \lambda (d \omega).$$

• Alternative, we can view $f(x, y)$ as a measurable function from the probability space $(\mathbb{R}^2, \mathcal{B}^2, \mu \times \nu)$, to $\mathbb{R}$. It is a random variable, and

  $$\mathbb{E} [ f(X, Y) ] = \int_{\mathbb{R}^2} f(x, y) \, \, (\mu \times \nu)(d (x,y)).$$

• It can be proven that the two integrals are equal:

  • If $f(x, y) = \mathbb{I}[ (x, y) \in A \times B]$, for some $A, B \in \mathcal{B}$, then by independence of $X$ and $Y$, both integrals equal $\mu(A) \cdot \nu(B)$.

  • Since $\mathcal{S} \doteq \{ A \times B: A, B \in \mathcal{B} \}$ generates $\mathcal{B}^2$, i.e., $\sigma(\mathcal{S}) = \mathcal{B}^2$, applying $\pi$-$\lambda$ theorem, if $f(x, y) = \mathbb{I}[ (x, y) \in M]$ for some $M \in \mathcal{B}^2$, the two integral are equal.

  • By linearity of integration, if $f(x, y)$ is $\mathcal{B}^2$ simple function, the two integral are equal.

  • By monotone convergence theorem, if $f(x, y)$ is a non-negative measurable function, the two integral are equal.

  • By decomposing $f$ into non-negative and negative parts, if $f(x, y)$ is an integrable function, the two integral are equal.

Reference

[1] P. Billingsley, Probability and measure, 3rd ed. New York: Wiley, 1995.