Category Archives: Probability

Probability

Mean independent and correlated variables

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Given two real random variables \(X\) and \(Y\), we say that:

  1. \(X\) and \(Y\) are independent if the events \(\{X \le x\}\) and \(\{Y \le y\}\) are independent for any \(x,y\),
  2. \(X\) is mean independent from \(Y\) if its conditional mean \(E(Y | X=x)\) equals its (unconditional) mean \(E(Y)\) for all \(x\) such that the probability that \(X = x\) is not zero,
  3. \(X\) and \(Y\) are uncorrelated if \(\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)\).

Assuming the necessary integrability hypothesis, we have the implications \(\ 1 \implies 2 \implies 3\).

The \(2^{\mbox{nd}}\) implication follows from the law of iterated expectations: \(\mathbb{E}(XY) = \mathbb{E}\big(\mathbb{E}(XY|Y)\big) = \mathbb{E}\big(\mathbb{E}(X|Y)Y\big) = \mathbb{E}(X)\mathbb{E}(Y)\)

Yet none of the reciprocals of these two implications are true.

Mean-independence without independence

Let \(\theta \sim \mbox{Unif}(0,2\pi)\), and \((X,Y)=\big(\cos(\theta),\sin(\theta)\big)\).

Then for all \(y \in [-1,1]\), conditionally to \(Y=y\), \(X\) follows a uniform distribution on \(\{-\sqrt{1-y^2},\sqrt{1-y^2}\}\), so: \[\mathbb{E}(X|Y=y)=0=\mathbb{E}(X).\] Likewise, we have \(\mathbb{E}(Y|X) = 0\).

Yet \(X\) and \(Y\) are not independent. Indeed, \(\mathbb{P}(X>0.75)>0\)  and \(\mathbb{P}(Y>0.75) > 0\), but  \(\mathbb{P}(X>0.75, Y>0.75) = 0\) because \(X^2+Y^2 = 1\) and \(0.75^2 + 0.75^2 > 1\).

Uncorrelation without mean-independence

A simple counterexample is \((X,Y)\) uniformly distributed on the vertices of a regular polygon centered on the origin, not symmetric with respect to either axis.

For example, let \((X, Y)\) have uniform distribution with values in \[\big\{(1,3), (-3,1), (-1,-3), (3,-1)\big\}.\]

Then \(\mathbb{E}(XY) = 0\) and \(\mathbb{E}(X)=\mathbb{E}(Y)=0\), so \(X\) and \(Y\) are uncorrelated.

Yet \(\mathbb{E}(X|Y=1) = -3\), \(\mathbb{E}(X|Y=3)=1\) so we don’t have \(\mathbb{E}(X|Y) = \mathbb{E}(X)\). Likewise, we don’t have \(\mathbb{E}(Y|X) = \mathbb{E}(Y)\).

Probability

Determinacy of random variables

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The question of the determinacy (or uniqueness) in the moment problem consists in finding whether the moments of a real-valued random variable determine uniquely its distribution. If we assume the random variable to be a.s. bounded, uniqueness is a consequence of Weierstrass approximation theorem.

Given the moments, the distribution need not be unique for unbounded random variables. Carleman’s condition states that for two positive random variables \(X, Y\) with the same finite moments for all orders, if \(\sum\limits_{n \ge 1} \frac{1}{\sqrt[2n]{\mathbb{E}(X^n)}} = +\infty\), then \(X\) and \(Y\) have the same distribution. In this article we describe random variables with different laws but sharing the same moments, on \(\mathbb R_+\) and \(\mathbb N\).

Continuous case on \(\mathbb{R}_+\)

In the article a non-zero function orthogonal to all polynomials, we described a function \(f\) orthogonal to all polynomials in the sense that \[
\forall k \ge 0,\ \displaystyle{\int_0^{+\infty}} x^k f(x)dx = 0 \tag{O}.\]

This function was \(f(u) = \sin\big(u^{\frac{1}{4}}\big)e^{-u^{\frac{1}{4}}}\). This inspires us to define \(U\) and \(V\) with values in \(\mathbb R^+\) by: \[\begin{cases}
f_U(u) &= \frac{1}{24}e^{-\sqrt[4]{u}}\\
f_V(u) &= \frac{1}{24}e^{-\sqrt[4]{u}} \big( 1 + \sin(\sqrt[4]{u})\big)
\end{cases}\]

Both functions are positive. Since \(f\) is orthogonal to the constant map equal to one and \(\displaystyle{\int_0^{+\infty}} f_U = \displaystyle{\int_0^{+\infty}} f_V = 1\), they are indeed densities. One can verify that \(U\) and \(V\) have moments of all orders and \(\mathbb{E}(U^k) = \mathbb{E}(V^k)\) for all \(k \in \mathbb N\) according to orthogonality relation \((\mathrm O)\) above.

Discrete case on \(\mathbb N\)

In this section we define two random variables \(X\) and \(Y\) with values in \(\mathbb N\) having the same moments. Let’s take an integer \(q \ge 2\) and set for all \(n \in \mathbb{N}\): \[
\begin{cases}
\mathbb{P}(X=q^n) &=e^{-q}q^n \cdot \frac{1}{n!} \\
\mathbb{P}(Y=q^n) &= e^{-q}q^n\left(\frac{1}{n!} + \frac{(-1)^n}{(q-1)(q^2-1)\cdot\cdot\cdot (q^n-1)}\right)
\end{cases}\]

Both quantities are positive and for any \(k \ge 0\), \(\mathbb{P}(X=q^n)\) and \(\mathbb{P}(Y=q^n) = O_{n \to \infty}\left(\frac{1}{q^{kn}}\right)\). We are going to prove that for all \(k \ge 1\), \( u_k = \sum \limits_{n=0}^{+\infty} \frac{(-1)^n q^{kn}}{(q-1)(q^2-1)\cdot\cdot\cdot (q^n-1)}\) is equal to \(0\).

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