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My previous post Unifying the treatment of discrete and continuous random variable, got successfully answered and allowed me to get further in my results. However I am facing a question that I can't seems to answer... So here it is :

Let $(\Omega_i, \sigma(\Omega_i), P_i) \; \forall i=0,1$ be two measure space.

Let $(\Omega_0 \times\Omega_1, \sigma(\Omega_0)\otimes\sigma(\Omega_1), \lambda)$ be the product measure space.

Following the answer to my previous post, if the spaces $\Omega_i$ are nice enough : \begin{align} \exists\; p_x: \sigma(\Omega_1)\longmapsto [0,1] \end{align} Such as $p_x$ is a measure $\forall \;x \in \Omega_0$, and is measurable $\forall\; B\in \sigma(\Omega_1)$, and : \begin{align} \lambda(A\times B)= \int_Ap_x(B)dP_X \end{align} Where $P_X(A) = \lambda(A\times\Omega_1)$.

I was wondering how does $P_X$ and $P_0$ compare? They are both measure on the same space. Is it required by construction of the product measure that $P_X=P_0$ ?

Thanks for any input.

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  • $\begingroup$ To declare that the product is "the product measure space" is to declare that $\lambda=P_0\otimes P_1$ then indeed $P_X=P_0$ and $p_x=P_1$ for every $x$ in $\Omega_0$, by definition. $\endgroup$ – Did Dec 22 '14 at 13:22
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You seem to mess up a couple of notions here. It certainly depends on which question are you dealing with more often, but I think it would be rather common to deal with a fixed measurable space, and consider several measures on it. That is, try to think of measurable spaces unless you really need to start using measures already.

To make abstract insinuations above a bit more clear, let's consider your particular example. If you have measurable spaces $(\Omega_i,\mathcal F_i)$, their product is again a measurable space $$ (\Omega,\mathcal F) = \left(\prod_i \Omega_i,\bigotimes_i \mathcal F_i\right). $$ Any measure $\lambda$ on $(\Omega,\mathcal F)$ factors as marginal distributions $\lambda_i$, conditional distribution etc. over the product structure. Note that for that we did not have to define any measures on $\Omega_i$ in advance.

Alternatively, we can start with measure spaces $(\Omega_i,\mathcal F_i, P_i)$. Then their product is a measure space itself, given by $(\Omega,\mathcal F,P)$ where $P = \bigotimes_i P_i$ is the product measure, which corresponds to independent random variables in case of probability measure, hence to trivial conditional distributions. In that case, $(\Omega,\mathcal F,P)$ already comes endowed with the fixed product measure, which is determined by measures $P_i$ defined over its factor spaces.

Summarizing the information above, you can answer your question is follows. The measures $P_i$ and $P_X$ (or $\lambda)$ in your notation are not related at all. When you define the product measurable space, you don't use $P_i$, and then you endow it with measure $\lambda$ which is taken out of the sky, hence no relation again. There is a subtle detail regarding whether you define a product measure space to be complete (in which case $\lambda$ must be defined over a bigger $\sigma$-algebra), but I don't think that's your matter here.

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  • $\begingroup$ I should have known you would answer this one too :). If I understood correctly, the two measures $P_X$ and $P_0$ have no relation whatsoever unless $\lambda$ has been constructed using $P_i$. Int this last case, $P_X = P_0$ and correspond to the marginal of $\lambda$ on the space $\Omega_0$. The completion part is understood I think. $\endgroup$ – user149705 Dec 22 '14 at 17:06
  • $\begingroup$ @user149705: yes, indeed $\endgroup$ – Ilya Dec 22 '14 at 20:42
  • $\begingroup$ Thank you for the quality of your answers. It is surely greatly appreciated by me and others. $\endgroup$ – user149705 Dec 22 '14 at 20:49

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