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Let $X_1,X_2$ be distributions on $[0,1]$ and let $X=(X_1,X_2)$ be the joint distribution of $X_1,X_2$. Let $\mathcal{X}$ be the set of all such joint distribution $X$.

Question 1: Does $\mathcal{X}$ equals to the set of all measurable functions $f$ on $[0,1]^2$ such that $\int_{[0,1]^2} f=1$ and $f(x,y)\geq 0$ for every $(x,y)\in [0,1]$?

Question 2 (weak* compactness): Let $C_b([0,1]^2)$ be the set of all bounded continous functions on $[0,1]^2$. Does the following holds?:

For any sequences of functions $f_1,f_2,\dots \in\mathcal{X}$, there exist its sub-sequence $f_{i_1},f_{i_2},\dots$ and $f_0\in\mathcal{X}$, such that for any $g\in C_b([0,1]^2)$, we have: $$\lim_{n\to\infty}\int f_{i_n}\cdot g\to\int f_{0}\cdot g.$$

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Question 1:

No, there need not exist such $f$. Consider the simple case of $X_1 = X_2$ random variables where $X_1$ has say, pdf $g$.

Then being equivalent to $f$ on $[0, 1]^2$ such that

$$\int_{[0, 1]^2} f = \int_0^1 \int_0^1 f(x, y) \, dx \,dy = 1$$

is saying that $(X_1, X_2)$ admits a joint pdf $f$ with respect to Lebesgue measure. This $f$ would satisfy, for Borel subsets $E \subset [0, 1]^2$,

$$\mathbb P[(X_1, X_2) \in E \,] = \int_E f(z) d\mathcal L^2 (z),$$

with $\mathcal L^2$ denoting the $2$-dimensional Lebesgue measure.

But the correct integral is

$$\mathbb P[(X_1, X_2) \in E \,] = \int_{E} g(D^{-1}(z)) \, d (D_* \mathcal L^1)(z),$$

where $D: [0, 1] \to [0, 1]^2$ is the diagonal map $z \to (z, z)$ and $D_* \mu$ is the pushforward measure of a measure $\mu$ by $D$.

The integration measures are mutually singular, and so cannot be converted to each other by multiplication by any Radon-Nikodym derivative, hence your statement doesn’t hold.

Intuitively, $(X_1, X_2)$ is supposed on the “lower dimensional” diagonal $\Delta := \{(x, x) \, | \, x \in [0, 1]\}$.

Question 2:

On the other hand, the space of joint probability distributions is weak-star compact, and you can find a proof in any standard introductory probability text, eg Billingsley or Durrett.

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  • $\begingroup$ Thanks for your answer! So, does the set of all measurable functions on $[0,1]^2$ a subset of $\mathcal{X}$? And does $\mathcal{X}$ equals to the space of all probability distributions on $[0,1]^2$? $\endgroup$
    – tom jerry
    Commented Oct 11 at 12:43
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    $\begingroup$ Yes to both! The set of all measurable functions as you wrote has a natural interpretation as the set of distributions admitting a joint density. @tomjerry $\endgroup$
    – Nate River
    Commented Oct 11 at 14:01
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    $\begingroup$ Thank you, River! And thus, is it wrong to view $\mathcal{X}$ as a subset of $L_1([0,1]^2)$ since there exist the counterexample as you give in your answer? $\endgroup$
    – tom jerry
    Commented Oct 11 at 14:22
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    $\begingroup$ That is also right I believe. @tomjerry $\endgroup$
    – Nate River
    Commented Oct 11 at 14:30
  • $\begingroup$ May I ask a question: For a distribution which is supposed on a 0-measured subset of $[0,1]^2$, how could we calculate its integer on a measureable subset of $[0,1]^2$? i.e., let $f:[0,1]^2\to [0,1]$ be a measureable function with $\int f=0$. Could we consider the distribution as something like $f(x,y)/(\int f)$ and calculate its integer on $[0,0.5]\cdot [0,0.5]$? Thank you a very much for your time! $\endgroup$
    – tom jerry
    Commented Nov 15 at 9:42

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