# Think independent random variables defined on a duplicated product space

Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a probability space on which defines two random variables

$$X \colon \Omega \to \mathbb{R}\\ Y \colon \Omega \to \mathbb{R}$$

Now suppose that $$X$$ and $$Y$$ are independent meaning that $$\mathbb{P}(\omega \in \Omega \colon X \in A, Y \in B) = \mathbb{P}(\omega \in \Omega \colon X \in A)\, \mathbb{P}(\omega \in \Omega \colon Y \in B)$$ for every $$A,B \in \mathcal{B}$$.

Now let us duplicate the space $$(\Omega, \mathcal{F}, \mathbb{P})$$ yielding $$(\Omega', \mathcal{F}', \mathbb{P}')$$ such that $$\Omega = \Omega'$$, $$\mathcal{F}=\mathcal{F}'$$, and $$\mathbb{P} = \mathbb{P}'$$. We consider the product space $$(\Omega \times \Omega', \mathcal{F}\otimes \mathcal{F}', \mathbb{P}\,\mathbb{P}')$$.

We then extend those random variables to that product space by

$$\overline{X}(\omega,\omega')\colon \Omega \times \Omega' \to \mathbb{R} = X(\omega), \\ \overline{Y}(\omega,\omega')\colon \Omega \times \Omega' \to \mathbb{R} = Y(\omega')$$

We can verify that $$\overline{X}$$ and $$\overline{X}$$ are random variables on that product space, and they are independent.

Am I on a correct path to think about the independence of $$X$$ and $$Y$$ in this product measure sense? In short, suppose we have two indepdendent random variables on a probaiblity space, can I think them as the sliced random variables on their product measure space?

• If you start with two independent random variables on some probability space $(\Omega, \mathscr{F}, \mathbf{P})$ then they are already independent and there is nothing left to do. I believe you want the opposite, how to construct two independent copies of a given distribution. Check my answer below. Jun 14, 2021 at 20:40
• @WillM. Thank you for your answer, but my question seems a bit poorly written. I know what independence mean in terms of distribution. What I want to know is: Given two independent random variables on some probability space, is it "legal" to form a product probability space not state space?
– anon
Jun 15, 2021 at 5:43
• My answer and comment are very relevant then. You are completely lost: you are confusing a random variable with its distribution. A random variable is by definition a measurable function. As such, if you are given two independent random variables, that means that you are given two functions already, nothing to define. If, in contrast, you want to create two independent random variables with a given distribution, then my answer is the construction. Jun 15, 2021 at 6:04
• Are you confused by phrases such as "take an independent copy of $X$"? Because those phrases mean my construction in the answer. Also, notice that when such a phrase occurs, is probably something relating to distributional properties of the random variable and not the actual random variables as functions. Jun 15, 2021 at 6:06
• Also, random variables do not have state space, they are real-valued always! So, it seems you are confusing many different probability concepts. Jun 15, 2021 at 6:11

If you want to create two independent copies of a given radom variable with distribution $$F,$$ do this: start with a distribution, say $$F$$ on $$\mathbf{R}.$$ This distribution function induces a measure $$\mu_F$$ on the Borel set of $$\mathbf{R}.$$ If you want a pair of random variables with distribution $$F$$ each of them and that are independendent, then you will consider the probability space $$(\mathbf{R}^2, \mathscr{B}_{\mathbf{R}^2}, \mu_F \otimes \mu_F)$$ and to define the random variables you will then set $$X(\omega) = x$$ and $$Y(\omega) = y$$ where $$\omega = (x, y).$$ By definition, $$X$$ and $$Y$$ are independent and they both have distribution $$F.$$ Mutatis mutandis you can do the same with two probability spaces $$(\mathrm{X}, \mathscr{X}, \mu)$$ and $$(\mathrm{Y}, \mathscr{Y}, \nu)$$ and consider the product space $$\mathrm{Z} = \mathrm{X} \times \mathrm{Y},$$ $$\mathscr{Z} = \mathscr{X} \otimes \mathscr{Y}$$ and $$\rho = \mu \otimes \nu$$ and the random object is $$T(x, y) = (x, y)$$ which is a $$\mathrm{Z}$$-valued random object with independent coordinates and whose marginal laws are $$\mu$$ and $$\nu,$$ respectively.
If you start with two random variables $$X$$ and $$Y$$ defined on respective probability spaces $$(\Omega_X, \mathscr{F}_X, \mathbf{P}_X)$$ and $$(\Omega_Y, \mathscr{F}_Y, \mathbf{P}_Y)$$: then you can construct the product space $$(\Omega, \mathscr{F}, \mathbf{P}) = (\Omega_X \times \Omega_Y, \mathscr{F}_X \otimes \mathscr{F}_Y, \mathbf{P}_X \otimes \mathbf{P}_Y)$$ and on it then define a random vector $$Z(\omega) = Z(\omega_x, \omega_y) = (X(\omega_x), Y(\omega_y)) \in \mathbf{R}^2.$$ Consider the projections $$\mathbf{R}^2 \to \mathbf{R}$$ given by $$\pi_1(x,y) = x$$ and $$\pi_2(x,y) = y.$$ In this construction, it follows that $$\pi_1(Z)$$ and $$\pi_2(Z)$$ are independent random variables defined on $$\Omega$$ with respective probabilities $$\mathbf{P}_X$$ and $$\mathbf{P}_Y.$$ These are your $$\bar{X}$$ and $$\bar{Y}.$$
Do every pair of independent random variables come from a product space? No. Consider uniform distribution on $$[0, 1].$$ Given a random number of this distribution, it can be expanded uniquely as $$\sum\limits_{k = 1}^\infty b_k 2^{-k}$$ where $$b_k$$ is either $$0$$ or $$1$$ and the expansion is unique by considering finite representations (that is, there is no infinite sequence of contiguous 1s starting some index until infinity). The random bits $$(b_k)$$ are independent Bernoulli with parameter $$\dfrac{1}{2}$$ and they do not come from a product space. However, in general is good idea to think of independent random vectors as defined on product spaces.
• You second part is very close to my question ^. Given a pair of independent rvs, can we always form a product probability space (not the space that rvs map to)? Your examples shows that this is not true. But, how about those extended rvs $\overline{X}$ and $\overline{Y}$ as shown in the questions?
• You have not understand yet. If you start with $X$ and $Y$ two indepedent random variables, there is nothing to do, they are already defined (by whatever means). Jun 15, 2021 at 15:07