I am looking at the 4th central moment of a weighted-sum of correlated random variables, which takes the form

$$\mu_4 = \sum_{i,j,k,l=1}^n w_i w_j w_k w_l \mu_{ijkl}$$

where $\mu_{ijkl}$ are the fourth-order co-moments of the $n$ random variables and $w_i$ are the weights. The variables I assume to be identically distributed and by correlated I mean that the dependence structure is defined by a Gaussian copula, so is a function of a correlation matrix only.

$\mu_4$ is a multivariate non-negative convex polynomial in the $w_i$ and is homogeneous of order 4. Numerically, individual cases can be written as a sum of squares, which can be obtained by solving a semi-definite program. As far as I am aware there does not exist any specific example of a non-negative convex polynomial that cannot be written as a sum-of-squares, so I believe this is always possible for $\mu_4$. I wondered if anyone might have an idea about how to prove this in general?

Thank you.


By a well-known result due to Richter and Rogosinsky (see e.g. Kemperman, Lemma 1, p. 69), there is a probability measure $\nu$ on a finite set $T\subset\mathbb R^n$ such that $$\mu_{ijkl}=\int_{T}\nu(dt)t_it_jt_kt_l=\sum_{t\in T}\nu(\{t\})t_it_jt_kt_l\tag{1}$$ for all $i,j,k,l$ in $[n]:=\{1,\dots,n\}$. So, $$\mu_4=\sum_{t\in T}\nu(\{t\})\Big(\sum_{i\in[n]}w_it_i\Big)^4.$$ So, indeed $\mu_4$ is the sum of squares of polynomials in the $w_i$'s.

Here the joint distribution of the (correlated or not) random variables in question does not matter, as long as they have finite $4$th moments. Of course, the same sum-of-squares conclusion holds for the $k$th order moment $\mu_k$ of the weighted sum of random variables for any even natural $k$.

Whereas representation (1) is of course correct, Kemperman's Lemma 1 referred to above, not even is incorrect, but does not make sense, unfortunately. Indeed, condition (ii) in that lemma mentions a "measure $\mu$ on $S$ having a finite support", where "$S$ carries no special topology and is merely a measurable space." Of course, without a topology, the notion of the support of a measure makes no sense. Also, of course there are measurable spaces with no measurable nonempty finite sets. Above, I rather carelessly rendered the finite support condition for a measure as the measure being defined on a finite set. (This is the first time I see a mistake made by Kemperman, and my apologies for following Kemperman without enough thought.)

However, representation (1) follows e.g. from the following statements by Winkler:

(i) Theorem 3.1, implying that the (convex) set of all Borel probability measures on a Polish space (such as $\mathbb R^n$) satisfying finitely many generalized moment conditions has an extreme point provided that this set of measures is nonempty;

(ii) Theorem 2.1 and Example 2.1(a), implying that each such extreme point is a finite mixture of Dirac measures.

  • $\begingroup$ Thanks for the clarification Iosif. From a practical perspective I think of this as decomposing the estimate of $\hat{\mu}_4$ to the observation level to show it is a sos. This will be a long list of polynomials; however, I don't believe it matters because as long as I know $\hat{\mu}_4$ can be written as a sos, then a more compact representation will exist in the form $x'Ax$, where $x$ is an ordered list of monomials of degree 2 in the $w_i$ and $A$ is a psd matrix that can be determined using an SDP. Thanks again. $\endgroup$
    – Brian
    Nov 18 '20 at 17:13

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