Let there be two random variables $X$ and $Y$ with a certain joint copula. Is it always true that there is another random variable $Z$ independent from $Y$ such as the vectors (X,Y) and (X,Z) have the same law?


1 Answer 1


It depends on what you mean by "exist". Taken completely literally, no. For instance, if the sample space is $\Omega = \{0,1,2,3\}$ with each point having probability $1/4$, and $X$ and $Y$ are the random variables

$$X(0) = X(1) = 0, X(2) = X(3) = 1, \\ Y(0) = Y(2) = 0, Y(1) = Y(3) = 1,$$ then any random variable $Z$ such that $(X,Y)$ and $(X,Z)$ have the same law must satisfy $Z=Y$.

However, what I suspect you mean to ask is "given some joint law $\mu$, can we always find random variables $X$, $Y$ and $Z$ defined on some shared probability space such that $(X,Y)$ and $(X,Z)$ both have joint law $\mu$, and $Y$ and $Z$ are independent?"

If you do not know measure theory, the TL;DR version is yes, this is always possible. If you interested in the measure-theoretic proof, then proceed.

To prove this, let $\lambda$ be the marginal distribution of $X$ and $\nu_x$ be the distribution of $Y$ conditional on $X=x$. (What I am really doing here is disintegrating the measure $\mu$; see the Wikipedia article on disintegrations for details.) So specifically, we have

$$ \mu(A\times B) = \int_A \nu_x(B)\lambda(dx) $$

for all Borel sets $A,B$. Now, all we have to do is consider the Borel probability measure $\mathbb P$ on $\mathbb R^3$ which satisfies

$$ \mathbb P(A\times B\times C) = \int_A \nu_x(B)\nu_x(C) \lambda(dx).$$

The joint law of both the first and second components, and the first and third components, is $\mu$, and the second and third components are independent. Explicitly, let $\Omega = \mathbb R^3$ with its Borel $\sigma$-field and probability measure $\mathbb P$ defined above, and define

$$ X(x,y,z) = x, \qquad Y(x,y,z) = y, \qquad Z(x,y,z) = z. $$

Then by construction, $(X,Y)$ and $(X,Z)$ both have law $\mu$, and $Y$ and $Z$ are independent.


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