Consider the measurable space $(\mathbb{R}, \mathcal{B})$ and a probability space $(\Omega, \mathcal{F}, P)$. Define a finite sequence of random variables $X_1,\ldots,X_n: \Omega \to \mathbb{R}$. Each $X_i$ induces a probability distribution measure on $\mathcal{B}$ given by $\mu_i(B_i) = P(X_i^{-1}(B_i))$ for $B_i \in \mathcal{B}$. This part is clear to me.

Now place these random variables into a vector, $X = (X_1,\ldots,X_n)$ and consider the target space $(\mathbb{R}^n, \mathcal{B} \otimes \cdots \otimes \mathcal{B})$.

Here's one part I'm not sure about. We now have $X: \Omega \to \mathbb{R}^n$, and I guess since each $X_i$ is $\mathcal{F}$-measurable we get that $X$ is somehow $\mathcal{F}$-measurable and so we may define a "joint" probability distribution measure on $\mathcal{B} \otimes \cdots \otimes \mathcal{B}$ given by $\mu(B) = P(X^{-1}(B))$ for $B = B_1 \times \ldots \times B_n \in \mathcal{B} \otimes \cdots \otimes \mathcal{B}$.

However, as I understand, a measure on a product space is called a product measure and is defined such that $\mu(B) = \mu_1(B_1)\cdots \mu_n(B_n)$. But then $P(X^{-1}(B)) = P(X_1^{-1}(B_1))\cdots P(X_n^{-1}(B_n))$ and so the random variables are independent automatically.

I've clearly misunderstood and/or overlooked something, so I would appreciate my errors being pointed out!

  • $\begingroup$ A measure defined on a product of measurable spaces does not have to be a product measure, as Michael Hardy has pointed out. However, in this case, you can check that $P^X$ will indeed be the product of the individual measures $P^{X_i}$ if and only if the $X_i$ are independent. $\endgroup$ – Shalop Jun 29 '15 at 23:12

A measure on a product space need not be a product measure. Suppose $$ X = \left. \begin{cases} 0 & \text{with probability }1/4, \\ 1 & \text{with probability }3/4, \end{cases} \right\}\text{ and }Y = \left. \begin{cases} 0 & \text{with probability }1/3, \\ 1 & \text{with probability }2/3. \end{cases} \right\} \tag 1 $$ The following two marginal distributions are consistent with infinitely many joint distributions only one of which is the product measure, which is the measure for which the two random variables are independent: $$ \begin{array}{|c|c|c|} \hline & X=0 & X=1 \\ \hline Y=0 & 3/12 & 1/12 \\ \hline Y=1 & 0/12 & 8/12 \\ \hline \end{array} \qquad \begin{array}{|c|c|c|} \hline & X=0 & X=1 \\ \hline Y=0 & 0/12 & 4/12 \\ \hline Y=1 & 3/12 & 5/12 \\ \hline \end{array} $$

  • Both of the joint distributions displayed in these two tables result in the two marginal distributions given in $(1)$.
  • In both of these two joint distributions, $X$ and $Y$ fail to be independent.
  • Every joint distribution consistent with the two marginal distributions is a weighted average, or convex combination, of these two.
  • In only one of those joint distributions, $X$ and $Y$ are independent. That one is the product measure. (That one is $1/3$ of the first one plus $2/3$ of the second.)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.