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 $k$.

  • $\begingroup$ in the multidimensional case many polynomials will give the same distribution so you cannot determine the polynomial from moments, e.g., $(x_1x_2, x_1^2)$ and $(x_1x_2, x_2^2)$ $\endgroup$
    – user83457
    Jan 18, 2018 at 7:51
  • $\begingroup$ @michael Even in the 1-dimensional case: there you cannot distinguish $-x$ from $x$ $\endgroup$ Jan 18, 2018 at 8:00
  • $\begingroup$ Thanks guys. Just to clarify, the goal is not to uniquely determine the $\phi$ functions, but to uniquely determine the distribution. I.e., if $m_k(P_1)=m_k(P_2)$ for 1\leq k\leq K, we can conclude that $P_1=P_2$. $\endgroup$ Jan 18, 2018 at 17:30
  • $\begingroup$ Are you also interested in knowing how many statistics you need to know (not necessarily moments)? This would probably be a simpler question. $\endgroup$
    – S.Surace
    Jan 19, 2018 at 23:51
  • $\begingroup$ yes, that'd be helpful too. $\endgroup$ Jan 21, 2018 at 1:20

1 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

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

  • $\begingroup$ Thanks. yes I did this calculation as well. Any ideas for a more general case? $\endgroup$ Jan 18, 2018 at 17:32
  • 1
    $\begingroup$ @SoheilFeizi yes I suppose you could work with a formula for $E(X^k)$ as a function of $k$, and the multinomial theorem $\endgroup$ Jan 19, 2018 at 2:59

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