# Determinacy of random variables

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 $$0$$.

We only need to write following equalities for all $$k \ge 0$$: \begin{align} u_{k+1} &= \sum \limits_{n=0}^{+\infty} \frac{(-1)^n q^{kn} q^n}{\prod \limits_{i=1}^n (q^i-1)} \\&= \sum \limits_{n=1}^{+\infty} \frac{(-1)^n q^{kn} (q^n – 1)}{\prod \limits_{i=1}^n (q^i-1)} + \sum \limits_{n=0}^{+\infty} \frac{(-1)^n q^{kn} \cdot 1}{\prod \limits_{i=1}^n (q^i-1)} \\& = – q^k \sum \limits_{n=0}^{+\infty} \frac{(-1)^n q^{kn}}{\prod \limits_{i=1}^n (q^i-1)} + \sum \limits_{n=0}^{+\infty} \frac{(-1)^n q^{kn} \cdot 1}{\prod \limits_{i=1}^n (q^i-1)} = (1-q^k)u_k \end{align}

As $$u_1=0$$ we get by induction, $$u_k = 0$$ for all $$k \ge 1$$. With this, we can conclude both measures sum to $$1$$ for $$k=1$$, and that $$X$$ and $$Y$$ have the same moments.

#### Byproduct: existence of a non zero integer series with all derivatives equal to zero

For a discrete random variable $$X$$ with values in $$\mathbb{N}$$ and having finite moments of all orders, the generating function $$\mathcal{G}_X(s) = \mathbb{E}\left(s^X\right)$$ is a smooth function on $$[0,1]$$ and $$\mathcal{G}_X^{(k)}(1) = \mathbb{E}(X^k)$$ for all $$k \ge 0$$.

Denote $$a_n := \mathbb{P}(X=n) – \mathbb{P}(Y=n)$$. As mentioned above, the integer series $$S(x) = \sum \limits_{n=0}^{+\infty} a_n x^n$$ and all its derivatives converge on $$[0,1]$$. Also, $$a_1 = -e^{-q}$$ so $$f$$ is non zero. Yet, for all $$k \ge 0$$, $f^{(k)}(1) = \mathbb{E}(X^k) – \mathbb{E}(Y^k) = 0.$

Note: This cannot happen if $$1$$ is in the disk of convergence. In that case, $$\sqrt[q^n]{|a_{q^n}|} \sim_{n \to \infty} 1$$ , so $$1$$ is indeed on the boundary of the disk of convergence of $$S$$.