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Fix a probability space $(\Omega,\mathcal F,\mathbb P)$, and let $X,Y:\Omega\to\mathbb R$ be random variables with laws $\mu_X,\mu_Y$ respectively. Let $\pi$ be a probability measure on $\mathbb R^2$ with marginal distributions $\mu_X,\mu_Y$. Does there exist a random variable $Z:\Omega\to\mathbb R^2$ such that $\mu_Z=\pi$?

Intuitively this seems true, and if it is true, it seems like something that would be a classical well-known fact in probability. However, I have not read any such statement anywhere.

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Not necessarily. Let $\Omega = \{0,1\}$ have two points and have the probability measure assign equal weight to both of them. Let $X$ and $Y$ both be the inclusion function into $\mathbb R.$ Then these are two random variables with the Beroulli distribution $p=1/2.$ But it's pretty clear we can't, for instance, make a RV $Z:\Omega\to \mathbb R^2$ that has the distribution of $(Z_1,Z_2)$ where $Z_1, Z_2$ are independent Bernoulli's with $p=1/2$ since $(Z_1,Z_2)$ can take at most two values.

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    $\begingroup$ As a somewhat related question, can we construct a probability space which admits $\mathbb R^n$-valued rvs with arbitrary law? This is really all I need. $\endgroup$ Mar 16, 2018 at 23:47
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    $\begingroup$ @MonstrousMoonshine Yes. $(\mathbb R^n, \mathcal B(\mathbb R^n), \mu)$ with $\mu$ the law in question is such a probability space. (Let the $\mathbb R^n$-valued RV be the identity function.) $\endgroup$ Mar 16, 2018 at 23:51
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    $\begingroup$ I should have clarified – I meant a single probability space which can simulate all $\mathbb R^n$-valued laws. $\endgroup$ Mar 17, 2018 at 0:36
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    $\begingroup$ @MonstrousMoonshine Do you mean a single space on which for any measure on $\mathbb R^n,$ a random variable with that law exists? $\endgroup$ Mar 17, 2018 at 0:48
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    $\begingroup$ @MonstrousMoonshine Definitely for $\mathbb R.$ Lebesgue measure on $[0,1]$ should suffice where we map $x$ to $F^{-1}(x)$ (where $F^{-1}$ is the generalized inverse CDF.) For higher dimensions I want to say yes too. Measures with support on the whole space (or any other family of equivalent measures) can always be pushed between one another via the RN-derivative. But I'm not sure how to make it work in general. Sounds like a good question to ask on here (I suspect I'm missing something that is 'standard knowledge'). $\endgroup$ Mar 17, 2018 at 1:12

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