# Independence, Fubini's Theorem and Product Spaces $(\Omega_1\times\Omega_2, \mu_1\otimes \mu_2)$

Following is an excerpt from Lecture 7 of these notes, about the independence of random variables. I have some very basic questions about what's going on. Please scroll down to read them.

Independence can always be viewed in a canonical way. Let $$(\Omega,\mu)$$ be a product space $$(\Omega_1\times\Omega_2, \mu_1\otimes \mu_2)$$ where $$\mu_1$$ and $$\mu_2$$. Suppose $$X$$ and $$Y$$ are random variables on $$\Omega$$ for which the value $$X(\omega_1,\omega_2)$$ depends only upon $$\omega_1$$, while $$Y(\omega_1,\omega_2)$$ depends only upon $$\omega_2$$. Then any integral (that converges appropriately) can be written as the product of integrals by Fubini's theorem. $$\mathbb E[f(X)g(Y)] = \int f(X(s)) g(Y(t))\ \mathrm d\mu_1\otimes\mu_2 (s,t)$$ $$= \int f(X(s))\ \mathrm d \mu_1(s) \int g(Y(t))\ \mathrm d\mu_2(t)= \mathbb E[f(X)] \mathbb E[g(Y)]$$

1. What do we mean by "canonical" here? I didn't get much on searching on Google.
2. What is $$\mu_1\otimes\mu_2$$? I have never seen this notation before. It'd be helpful if someone could explain the product space (or point me to some references) to me.
3. Where is Fubini's theorem being used here? From what I know, Fubini's theorem is a prescription for swapping integrals.
4. Inside the integrals, I see that the author chose to write $$\mathrm d\mu_1(s)$$ instead of just $$\mathrm d\mu_1$$ (similarly in two other places). Is there a specific reason for this? I haven't seen such notation being used before. When integrating some function $$f$$ with respect to a measure $$\mu$$, we typically just write $$\int_X f\ \mathrm d \mu$$.
• Maybe start with the wiki enty regarding product measures: en.m.wikipedia.org/wiki/Product_measure May 15 at 17:56
• Canonical in this case means that the probability space constructed as a product space admits explicit random variables that have prescribed distributions and that are independent. May 15 at 17:56
• @Riquelme Thanks for the link. Q2 is answered now! May 15 at 17:58
• @OliverDiaz I don't really understand it still. Do you have any examples or non-examples? What is a non-canonical way? May 15 at 17:59
• @epsilon-empereor: Let me give you another example of a space that you can consider canonical. A sequence of i.i.d Bernoulli random variables $\{X_i\}$ can be realize by looking at the probability space $((0,1),\mathscr{B}(0,1),\lambda)$ where $\mathscr{B}(0,1)$ is the Borel $\sigma$-algebra in the interval $(0,1)$, and $\lambda$ is the Lebesgue Measure restricted to $(0,1)$. For any $t\in(0,1)$ consider its binary expansion $t=\sum_n\frac{a_n(t)}2^{n}$. The functions $t\mapsto a_n(t)$ happen to form an i.i.d sequence of random variables in $(0,1)$ that are Bernoulli (to be continued) May 15 at 18:25

## 1 Answer

1. In mathematics, "canonical" is a word, that describes a construction or a representation of an object, which is in some sense obvious, very handy with respect to calculation and - in many cases - independent of a choice. Maybe have a look into this post.

Here, it refers to the construction of a product of measurable spaces. It obvious in the sense, that the product of the sets $$\Omega_1$$ and $$\Omega_2$$ is just the cartesian product $$\Omega_1\times\Omega_2$$. The niceness of the construction is the property, that the projections $$\pi_i:\Omega_1\times\Omega_2\to \Omega_i,\quad \pi_i(\omega_1,\omega_2):=\omega_i \quad\text{ for }(\omega_1,\omega_2)\in\Omega_1\times\Omega_2$$ are measurable as functions on the measurable spaces $$(\Omega_1\times\Omega_2,\mathcal{A}_1\otimes\mathcal{A}_2)\to (\Omega_i,\mathcal{A}_i).$$ Here, $$\mathcal{A}_i$$ denotes the $$\sigma$$-algebra on $$\Omega_i$$ and $$\otimes$$ refers to the product $$\sigma$$-algebra on $$\Omega_1\times\Omega_2$$.

2. As Riquelme posted, $$\mu_1\otimes\mu_2$$ is the product measure of the two measures. It is defined on the product $$\sigma$$-algebra $$\mathcal{A}_1\otimes\mathcal{A}_2$$.

3. Fubini's theorem is used in the second equation, where the product measure is split into the two measures. Since the functions $$(s,t)\mapsto f(X(s))$$ and $$(s,t)\mapsto g(Y(t))$$ only depend on one of the variables, one can integrate each function on its own. Note, that the projections $$\pi_1(s,t):=s$$ and $$\pi_2(s,t):=t$$ are themselves only dependent on one of the two variables, and that we could write $$(s,t)\mapsto f(X(\pi(s,t)))$$ for the first function (and the second respectively). In this sense, the projection functions are - by construction - the independent functions on a product space.

4. This simply denotes the dependence of the integrant on the measure. In the simple case of $$\int_\Omega f(\omega)\,d\mu(\omega)=\int_\Omega f\, d\mu,$$ one omits the dependence of $$f$$ on $$\omega$$ because of lazyness. If you want to omit the $$(s,t)$$ in the example above, you could again use the projections $$\int_{\Omega_1\times\Omega_2}f(X(\pi_1))g(Y(\pi_2)) \:d(\mu_1\otimes\mu_2).$$ Whether you like this notion or not is up to personal preference, I guess.