Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given random variables ${(X_\alpha)_{\alpha \in A}}$ defined on a probability space, how shall one construct random variables ${(Y_\beta)_{\beta \in B}}$ which induce prespecified probability measures ${(\mu_\beta)_{\beta \in B}}$, are both independent within ${(Y_\beta)_{\beta \in B}}$ themselves and independent with ${(X_\alpha)_{\alpha \in A}}$? Obviously, ${(Y_\beta)_{\beta \in B}}$ must also be defined on the same probability space as ${(X_\alpha)_{\alpha \in A}}$.

My question comes from Terrence Tao's blog

The product measure construction allows us to extend Lemma 4 (Creating a random variable with a specified distribution):

Exercise 18 (Creation of new, independent random variables) Let ${(X_\alpha)_{\alpha \in A}}$ be a family of random variables (not necessarily independent or finite), and let ${(\mu_\beta)_{\beta \in B}}$ be a collection (not necessarily finite) of probability measures ${\mu_\beta}$ on measurable spaces ${R_\beta}$. Then, after extending the sample space if necessary, one can find a family ${(Y_\beta)_{\beta \in B}}$ of independent random variables, such that each ${Y_\beta}$ has distribution ${\mu_\beta}$, and the two families ${(X_\alpha)_{\alpha \in A}}$ and ${(Y_\beta)_{\beta \in B}}$ are independent of each other.

My question is the same as how such ${(Y_\beta)_{\beta \in B}}$ are constructed (via the product measure construction or some other methods) in the exercise?

We can consider some simpler questions first.

  • Given a random variable $X$, how shall one construct another random variable $Y$ which has a prespecified probability measure $\mu$ and is independent from $X$?

  • Given a random variable $X$, how shall one construct two random variable $Y_i, i=1,2$, such that $Y_1$ and $Y_2$ have prespecified probability measure $\mu_i, i=1,2$ respectively, $Y_1$ and $Y_2$ are independent from $X$, and $Y_1$ and $Y_2$ are independent between each other?

Thanks and regards!

share|cite|improve this question
"i.i.d" often refers to a single random variable. Re Monte Carlo: computers are deterministic finite state machines that can only generate pseudorandom numbers that approximate i.i.d (one would need to sample external input like thermal noise to get randomness). – alancalvitti Feb 3 '13 at 14:51
@alancalvitti: Thanks! (1) I think i.i.d. applies to more than one random variables, and not to a single one. Think of the definition of independence between several r.v.s in terms of the sigma algebras they generate. (2) For constructing i.i.d. Bernoulli random variables defined from [0,1] to {0,1}, see "Proposition 2.8 (Daniell)" on page 4 of…. Does it belong to Monte Carlo simulation? – Tim Feb 3 '13 at 17:25
@alancalvitti: I now realize that I messed up constructing independent random variables and simulating independent random variables. They are different tasks. So I edit my post. Your comment addressed simulating independent random variables. I move that part of my question to stat.SE here. – Tim Feb 3 '13 at 19:31
You can do this using the product measure construction. – Michael Greinecker Feb 3 '13 at 19:38
@alan: I would be very curious to see any legitimate example where i.i.d. was used to refer to a single random variable. The very definition of the first two words implies a reference to a collection of them. – cardinal Feb 3 '13 at 21:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.