This question is purely out of curiosity and mainly to question my intuitions about independence of random variables.

Q: Take two non trivial random variables, with non disjoint support (see edit below) $X,Y \in \mathbb{L}_2(\Omega, \mathcal{F}, \mathbb{P})$, so that projections and covariance formulas are well defined. If $X$ is uncorrelated with any function $g$ of $Y$, i.e. $\operatorname{Corr}(X,g(Y)) = 0, \: \forall \: g$ measurable, this implies $X$ and $Y$ are independent.

Is the above statement true? I could not find any post on mathstack on this.

One way I tried to prove the above is by proving the following:

Assume that $X$ and $Y$ are dependent, then there exists a function $f$ such that the correlation between $X$ and $f(Y)$ is nonzero.

Reason why $\mathbb{L}_2$ is important:

This is also the reason why we have to take the random variables in $\mathbb{L}_2$, otherwise one could find counterexamples to the second statement by taking $X$ with undefined variance or expectation and show that the covariance can never be nonzero, as it is not well defined.


Any ideas or references? Maybe something additional must be assumed about the functions $g$? Maybe instead of this, one should assume that the correlation is zero with any random variable $Z$ which is $X$-measurable?

Thank you very much for your help and time.

Reason why non disjoint support is important:

EDIT. Here I post a counterexample that contradicts the second statement, if we do not assume that the random variables have non disjoint support, i.e.:

Assume that $X$ and $Y$ are dependent, then there exists a function $f$ such that the correlation between $X$ and $f(Y)$ is nonzero.

Take $([0,1], \mathcal{B}([0,1]), \lambda)$, where $\lambda$ is the Lebesgue measure. The key idea is that if they have disjoint support we can find a counterexample. Take: $$ X(x) = \left(x - \frac{1}{2} \right) \mathbb{1}_{[0,1/2]}(x)$$ and: $$ Y(x) = \left(x - \frac{3}{2} \right) \mathbb{1}_{[1/2,1]}(x)$$ Take any function $f$, then $f(Y(x)) = f(0)$ constant for any $x \in [0,1/2]$ thus: $$ X(x)f(Y(x)) = f(0)X(x) \mathbb{1}_{[0,1/2]}(x)$$ which, as $\int X d\lambda = 0$, implies: $$\int X f(Y) d \lambda = 0$$ for any $f$. This implies they are uncorrelated and it provides a counterexample.

  • $\begingroup$ independence is $E[f(X)g(Y)] = E[f(X)]E[g(Y)]$ for all $f,g$ so you're doing just all $g$ and identity $f$ but as a substitute for all $f$ you're doing non-disjoint support and $\mathbb L_2$ ? $\endgroup$
    – BCLC
    Commented Dec 10, 2022 at 19:42
  • $\begingroup$ I guess your condition can be restated as saying that $E[X \mid Y]$ is constant. $\endgroup$ Commented Dec 10, 2022 at 19:44

1 Answer 1


If $Y$ has only two values (for instance, if $Y$ is a Bernoulli random variable), then whenever $X,Y$ are uncorrelated, we also have $X,g(Y)$ uncorrelated for every $g$. For when $Y$ only has two values $y_1, y_2$, then we can replace $g$ with a linear function: find constants $a,b$ with $g(y_1) = a y_1 + b$, $g(y_2) = a y_2 + b$, so that $g(Y) = aY+b$ almost surely. Then $\operatorname{Cov}(X,g(Y)) = a \operatorname{Cov}(X,Y)=0$.

For an explicit example, let $Y$ be Bernoulli(p) for any p, $\xi$ Rademacher(1/2) independent of $Y$, and $X=\xi Y$.

If you assume that $g(Y)$ and $Z$ are uncorrelated for every $Z \in \sigma(X)$, or equivalently that $f(X)$ and $g(Y)$ are uncorrelated for all functions $f,g$, then $X,Y$ are independent. For if not, there exist $A_1 \in \sigma(X)$, $A_2 \in \sigma(Y)$ which are not independent, which is to say that $1_{A_1}, 1_{A_2}$ are not uncorrelated. But by definition of $\sigma(X)$, there exists a Borel set $B_1$ with $1_{A_1} = 1_{B_1}(X)$, and likewise a $B_2$ with $1_{A_2} = 1_{B_2}(Y)$. So taking $f = 1_{B_1}$, $g = 1_{B_2}$, then $f(X), g(Y)$ are not uncorrelated.

  • $\begingroup$ Great answer! Thank you! I agree with your proofs. Do you by any chance know if some of the above statements hold if we assume that $X$ and $Y$ are absolutely continuous wrt the Lebesgue measure? I believe this will still not improve the situation. $\endgroup$ Commented Dec 10, 2022 at 20:21
  • 1
    $\begingroup$ @GrandesJorasses: I don't think so. For instance, $Y \sim U(0,1)$, $Z \sim U(-1,1)$ independent of $Y$, $X=YZ$. $\endgroup$ Commented Dec 10, 2022 at 20:26
  • $\begingroup$ Thank you a lot. This really helps. I find fascinating the concept of independence of random variables because it is mathematically tricky, even though heuristically clear. $\endgroup$ Commented Dec 10, 2022 at 21:00

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