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  1. Given a random vector $X: (\Omega, \mathbb{F}, P) \rightarrow (\prod_{i \in I} S_i, \prod_{i \in I} \mathbb{S}_i)$, is each component variable $X_i, \forall i \in I$ of the random vector $X$ always a random variable from $(\Omega, \mathbb{F}, P)$ to $(S_i, \mathbb{S}_i)$?
  2. If yes, I guess there are two ways to define the distributions for each component variable $X_i, i \in I$:

    • First the random vector $X$ induces a probability measure $P_X$ from the its domain to its codomain. Then define $P_{X_i} (A): = P_{X}(A \times \prod_{j \in I, j \neq i} S_j), \forall A \in \mathbb{S}_i$.
    • $X_i$ can induce a probability measure $P'_{X_i}$ from its domain to its codmain.

    I was wondering if $P_{X_i}$ and $P'_{X_i}$ are always the same on $\mathbb{S}_i$?

    Will the first definition of marginal probability measures make the product of all the marginal probability measures to be the same as the joint measure?

  3. Can the above definitions and their relation be generalized to arbitrary measure space $(\Omega, \mathbb{F}, \mu)$ and mearuable mapping $f:(\Omega, \mathbb{F}) \rightarrow (\prod_{i \in I} S_i, \prod_{i \in I} \mathbb{S}_i)$, to define the distribution of each component mapping $f_i $ of $f$?

    Does the first way of definition require that the measure $\mu$ must be a probability measure? How to adjust the definition when $\mu$ can be any measure? Shall one define $\mu_{f_i}$ from $\mu_f$ as $\mu_{f_i}(A_i):= \frac{\mu_f(A_i \times \prod_{j \in I, j\neq i} S_i)}{\prod_{j \in I, j\neq i} \mu_{f_j}(S_i)}$, or $\mu_{f_i}(A_i):= \mu_f(A_i \times \prod_{j \in I, j\neq i} S_i)$, $\forall A_i \in \mathbb{S}_i $ or something else?

    When $\mu$ can be any measure, if it is not possible to define in the first way, or it is possible but the two definitions are not the same, how about when $\mu$ is finite, i.e. $\mu(\Omega) < \infty$?

Thanks and regards!

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    For your two definitions ($P_{X_i}$ and $P_{X_i}'$), relate them to $P$ and you can see they are equivalent. Furthermore, it should be clear that the finiteness does not play a role here. At last, I think by `any measure' you mean any $\sigma$-finite measure, and your argument would remain valid.2011-02-21
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    @Morning: Thanks! (1) Nice to know the argument is valid for $\sigma$-finite measure. Does the first definition apply to this case without any modification? (2) By "any measure", I meant *any*. How about for any measure?2011-02-21
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    @Tim: (1) In general, the random elements (here random vectors) are defined via measurable mappings, which are with respect to $\sigma$-finite measures. The questions here concern only the mapping and I don't see how the finiteness plays a crucial role. (2) All the measures you see in the first class of measure theory are $\sigma$-finite (unless explicitly said to be outer measure). Focus on these measures should be general enough.2011-02-21
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    @Morning: Thanks! In my original post, (1) In Part 2, will the first definition of marginal probability measures make the product of all the marginal probability measures to be the same as the joint measure?2011-02-22
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    (2) In Part 3 , in order to go from joint measure to marginal measure, how to apply the first definition for probability measure case in Part 2 to non probability measure case, such as for finite or σ finite one? Shall one define $\mu_{f_i}$ from $\mu$ as $\mu_{f_i}(A_i):= \frac{\mu_f(A_i \times \prod_{j \in I, j\neq i} X_i)}{\prod_{j \in I, j\neq i} \mu_{f_j}(X_i)}$, or $\mu_{f_i}(A_i):= \mu_f(A_i \times \prod_{j \in I, j\neq i} X_i)$, $\forall A_i \in \mathbb{S}_i $ or something else?2011-02-22
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    @Tim A little confusing here. Where do you want your $\mu_i$ to be defined? Anyway, $A_i\times \prod_{j\in I,j\neq i}X_j$ does not make sense. I think you mean $A_i\times \prod_{j\in I,j\neq i}S_j$, which is better written as $\pi_i^{-1}(A_i)$, where $\pi_i:\prod_{j\in I}S_j\to S_i$ is the projection mapping. Beside, $f$ should be involved somehow. Your second guess is on the right track. I let you finish the final answer. You are almost there.2011-02-22
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    @Morning: Sorry. There was confusion in the notation I used. See my update to my original post for clearer one. Could you be more specific about "Your second guess is on the right track"?2011-02-22
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    @Morning: Perhaps, you could turn your comments into a reply?2011-02-22

1 Answers 1

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First, $\prod_{i\in I}\mathbb S_i$ should be understood as the product $\sigma$-algebra, i.e., the smallest $\sigma$-algebra on $\prod_{i\in I}S_i$ such that the projection $\pi_i:\prod_{j\in I}S_j \ni (s_j)_{j\in I}\mapsto s_i\in \mathbb S_i$ is measurable. In this case, then the answers to the first two questions are Yes.

For part (3), as explained in the comments and answers in another closely related post Measure from on product $\sigma$-algebra to on component $\sigma$-algebras, one needs to work with finite measures on the product space. In this case, without loss of generality assuming $\mu = \mathbb P$ is a probability measure and using $X$ instead of $f$, one recovers part (2).