uniquely determining a distribution using moments

Question

Let $A$ be a parametric family of probability distributions that include all distributions in the form of $\phi(X)$ where $X\sim\mathcal{N}(0,\mathbf{I})$ is jointly Gaussian and $\phi:\mathbb{R}^d\to \mathbb{R}^d$ belongs to a parametric set of functions $G$.

To uniquely determine a distribution in $A$, how many moments do we need to know? For example, if $G$ is linear, A includes all Gaussian distributions. Thus, we need first and second moments to uniquely determine a distribution in the set $A$. I wonder if such a result can be extended for other $G$ such as polynomials with degree $k$.

Answer 1

Looks complicated in general, but here's a derivation that for $d=1$ and $k=2$, in the slightly further restricted case where $\phi(X)=aX^2+b$, three moments suffice.

Let $m_i=E(\phi(X)^k)$. Recall $E(X^4)=3$ and $E(X^6)=15$. We calculate $$m_1=a+b$$ $$m_2=3a^2+2ab+b^2$$ with solution: $$m_2=3a^2+2a(m_1-a)+(m_1-a)^2$$ $$m_2=a^2+2am_1+m_1^2-2am_1 +a^2$$ $$a^2=\frac12(m_2-m_1^2) =: u$$ So far we don't know $a$ exactly. But $$m_3=15a^3+9a^2b+3ab^2+b^3$$ so we can use our value for $a^2$ to reduce $$m_3=15a^3+9a^2(m_1-a)+3a(m_1-a)^2+(m_1-a)^3$$ to a linear equation in $a$: $$m_3=15ua+9um_1-9ua+3am_1^2-6um_1+3ua+m_1^3-3am_1^2+3um_1-ua$$ $$m_3=8ua+6um_1+m_1^3$$ and we get the somewhat appealing solution

$$\begin{eqnarray*}a&=&\frac14\frac{m_3-m_1^3}{m_2-m_1^2}-\frac34m_1\\ b&=&m_1-a\end{eqnarray*}$$