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Suppose there are two random variables $X: \Omega \rightarrow U$ and $Y: \Omega \rightarrow V$ with probability space $(\Omega, \mathcal{F}, P)$, measurable spaces $(U, \mathcal{F}_u)$ and $(V, \mathcal{F}_v)$.

  1. I was wondering how the joint distribution of $X$ and $Y$ is defined? I just realized I am not clear about its definition.
  2. If I am correct, the joint distribution is not the same as the product measure of those on $(U, \mathcal{F}_u)$ and $(V, \mathcal{F}_v)$ induced by $X$ and $Y$ from $(\Omega, \mathcal{F}, P)$ respectively. If they are the same, then $X$ and $Y$ are said to be independent.
  3. Is the joint distribution of $X$ and $Y$ defined on the $\sigma$-algebra generated from $\mathcal{F}_u \times \mathcal{F}_v$?
  4. For $S_u \in \mathcal{F}_u$ and $S_v \in \mathcal{F}_v$, is the joint distribution determined by $P([X, Y] \in S_u \times S_v) = P(\{X \in S_u\} \cap \{Y \in S_v\})$?

    How about the joint distribution probability of other sets that are more general and may not be "rectangle"-like as $S_u \times S_v$?

Thanks and regards!

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The joint distribution is a function assigning a probability to each measurable subset of $U\times V$, which is the probability that the pair $(X,Y)$ is in that subset. It must of course be a probability measure on the sigma-algebra of measurable subsets of $U\times V$.

But which subsets are measurable? That's your question #3, and the answer is yes.

I'm pretty sure the answer to #4 is yes, although I haven't thought this through in a while. Certainly it's yes when $U = V = \mathbb{R}$ with the the measurable sets being the Borel sets. For sets that are not "rectangle-like", the probabilities are determined by the probabilities for sets that are rectangle like, plus countable additivity, etc.

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  • $\begingroup$ ....and your point #2 is correct. $\endgroup$ Commented Aug 23, 2011 at 0:39
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    $\begingroup$ Re #4: The family of measurable rectangles is closed under finite intersections, and generates the product $\sigma$-algebra. By the Monotone Class Theorem, the measure is determined by its value on rectangles. That is, we have uniqueness. $\endgroup$
    – user940
    Commented Aug 23, 2011 at 1:17

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