Is there a function of a random variable, uncorrelated with the random variable itself? Besides the constant function.

Consider a non-negative discrete random variable $X$, with known probability mass function $p_X(x;\theta)$.

I am looking for a function $f$ that satisfies the following equation:

$$ \mathbb{E}_\theta[f(X) X] = \mathbb{E}_\theta[f(X)]\mathbb{E}_\theta[X] \tag{1}$$

It is immediate that a constant function (with probability $1$) satisfies the previous condition. That is, $f(X) = c \quad \forall X \in \mathcal{X}$, with $\mathbb{P}[X \in \mathcal{X}] = 1$, satisfies $(1)$.

What restrictions does $(1)$ impose on $f$?

Can I safely assume that $f(X)$ cannot depend on $X$?

Note 1: $(1)$ must be valid for any possible value of $\theta$.

Note 2: $p_X(x;\theta )$ is a member of the one-parameter exponential family.

Note 3: $X$ is a complete minimal sufficient statistics of $\theta$.

  • $\begingroup$ Consider the population $\mathcal{P}$ of uniform distributes random variables with parameter $\theta>0$, i.e. $\mathcal{P}=\{\frac{1}{2\theta}\mathbb{1}_{(-\theta,\theta)}\,dx\}$. Define $f(x)=x^2$. Then $E_\theta[f(X)X]=0=E_\theta[f(X)]E_\theta[X]$ for all $\theta>0$ ($X\sim \mu_\theta$, where $\mu_\theta\in\mathcal{P}$. $\endgroup$
    – Mittens
    Commented Dec 7, 2022 at 18:39
  • $\begingroup$ @OliverDíaz Thank you for your answer. And you are right! I didn't make it explicit, but in my problem the support of $X$ does not depend on $\theta$. I still think that, if we add that restriction, the only $f$ satisfying the equality is the constant function. Maybe one day I'll return to this problem and try to complete the proof. However that is not likely to happen in 2023...! Thanks again $\endgroup$
    – G Frazao
    Commented Dec 8, 2022 at 19:13
  • $\begingroup$ Your first problem only indicates parametric family of pdf's $p_{\theta}$. In a parametric model, the support of $\mu_\theta$ may depend on $\theta$. From uniform distributions, you can construct other examples. $\endgroup$
    – Mittens
    Commented Dec 8, 2022 at 19:33


You must log in to answer this question.