Solution to (general) Truncated Moment Problem

We are going to solve the truncated moment problem in this post. The theorem we are going to establish is more general than the original problem itself. The following theorem is a bit abstract, you can skip to Corollary 2 to see what the truncated moment problem is and why it has a generalization in the form of Theorem 1.

Theorem 1 Suppose ${X}$ is a random transformation from a probability space ${(A,\mathcal{A},\mathop{\mathbb P})}$ to a measurable space ${(B,\mathcal{B})}$ where each singleton set of $B$ is in $\mathcal{B}$ . Let each ${f_i}$ be a real valued (Borel measurable) function with its domain to be ${B}$ , ${i=1,\dots,m}$ . Given

$\displaystyle (\mathbb{E}f_i(X))_{i=1,\dots,m}$

and they are all finite, there exists a random variable ${Y\in B}$ such that ${Y}$ takes no more than ${m+1}$ values in ${B}$ , and

$\displaystyle (\mathbb{E}f_i(Y))_{i=1,\dots,m} = (\mathbb{E}f_i(X))_{i=1,\dots,m}.$

(If you are not familiar with terms Borel measurable, measurable space and sigma-algebras $\mathcal{A}, \mathcal{B}$ , then just ignore these. I put these term here just to make sure the that the theorem is rigorous enough.)

Let me parse the theorem for you. Essentially, the theorem is trying to say that given ${m}$ many expectations, no matter what kind of source the randomness comes from, i.e., what ${X}$ is, we can always find a finite valued random variable (which is ${Y}$ in the theorem) that achieves the same expectation.

To have a concrete sense of what is going on, consider the following Corollary of Theorem 1. It is the original truncated moment problem.

Corollary 2 (Truncated Moment Problem) For any real valued random variable ${X\in {\mathbb R}}$ with its first ${m}$ moments all finite, i.e., for all ${1\leq i\leq m}$

$\displaystyle \mathop{\mathbb E}|X|^i < \infty,$

there exists a real valued discrete random variable ${Y}$ which takes no more than ${m+1}$ values in ${{\mathbb R}}$ and its first ${m}$ moments are the same as ${X}$ , i.e.,

$\displaystyle (\mathbb{E}Y,\mathbb{E}(Y^2),\dots, \mathbb{E}(Y^m) )=(\mathbb{E}X,\mathbb{E}(X^2),\dots, \mathbb{E}(X^m)).$

This original truncated moment problem is asking that given the (uncentered) moments, can we always find a finite discrete random variable that matches all the moments. It should be clear that is a simple consequence of Theorem 1 by letting ${B={\mathbb R}}$ and ${f_i(x) = x^{i},, i=1,\dots,m}$ .

There is also a multivariate version of truncated moment problem which can also be regarded as a special case of Theorem 1.

Corollary 3 (Truncated Moment Problem, Multivariate Version) For any real random vector ${X=(X_1,\dots,X_n)\in \mathbb{R}^n}$ and its all ${k}$ th order moments are finite, i.e.,

$\displaystyle \mathop{\mathbb E}(\Pi_{i=1}^n|X_{i}|^{\alpha_i}) <\infty$

for any ${{1\leq \sum \alpha_i\leq k}}$ . Each ${\alpha_i}$ here is a nonnegative integer. The total number of moments in this case is ${n+k \choose k}$ . Then there is a real random vector ${Y \in \mathbb{R}^n}$ such that it takes no more than ${{n+k \choose k}+1}$ values, and

$\displaystyle (\mathop{\mathbb E}(\Pi_{i=1}^nX_{i}^{\alpha_i}))_{1\leq \sum \alpha_i\leq k} = (\mathop{\mathbb E}(\Pi_{i=1}^nY_{i}^{\alpha_i})) _{1\leq \sum \alpha_i\leq k}.$

Though the form of Theorem 1 is quite general and looks scary, it is actually a simple consequence of the following lemma and the use of convex hull.

Lemma 4 For any convex set ${C \in \mathbb{R}^k}$ , and any random variable ${Z}$ which has finite mean and takes value only in ${C}$ , i.e,

$\displaystyle \mathop{\mathbb E}(Z) \in \mathbb{R}^k, \mathop{\mathbb P}(Z\in C) =1,$

we have

$\displaystyle \mathop{\mathbb E} (Z) \in C.$

The above proposition is trivially true if ${C}$ is closed or $Z$ takes only finitely many value. But it is true that ${C}$ is only assumed to be convex. We will show it in this post.

We are now ready to show Theorem 1.

Proof of Theorem 1: Consider the set

$\displaystyle L = \{ (f_i(x))_{i=1,\dots,m}\mid x\in B \},$

The convex hull of this set ${L}$ is

$\displaystyle \text{conv}(L) = \{ \sum_{j=1}^l \alpha _j a_j\mid \alpha_j \geq 0 ,\sum_{j=1}^l \alpha_j =1, a_j\in L, l \in {\mathbb N}\}.$

Now take the random variable ${Z=(f_i(X))_{i=1,\dots,m}}$ which takes value only in ${L\subset \text{conv}(L)}$ , by Lemma 4 of convex set, we know that

$\displaystyle \mathop{\mathbb E} Z \in \text{conv}(L).$

Note that every element in ${\text{conv}(L)}$ has a FINITE representation in terms of ${a_j}$ s!

This means we can find ${l\in {\mathbb N}}$ , ${\alpha_j\geq 0, \sum_{j = 1}^l \alpha_j =1}$ and ${a_j \in L, j=1,\dots,l}$ such that

$\displaystyle \sum_{j=1}^l \alpha_ja_j = \mathop{\mathbb E} Z = (\mathop{\mathbb E} f_i(X))_{i=1,\dots,m}.$

Since each ${a_j = (f(x_j))_{i=1,\dots,m}}$ for some ${x_j \in B}$ , we can simply take the distribution of ${Y}$ to be

$\displaystyle \mathop{\mathbb P}(Y = x_j) = \alpha_j, \forall i =1,\dots,l.$

Finally, apply the theorem of Caratheodory to conclude that ${l\leq m+1}$ . $\Box$

4 thoughts on “Solution to (general) Truncated Moment Problem”

Latrice

March 31, 2018 at 3:44 am

Good info over again. Thumbs up;)

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Lola

November 15, 2018 at 10:29 am

certainly like your web site however you have to check the spelling on several
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- ding1ijun
  
  November 15, 2018 at 3:42 pm
  
  Hi Lola, thank you for reading the blog and sorry for the spelling errors. Some of them are prepared in a hurry and I did not get time to correct the spellings. Can you point out some spelling errors the next time you read?
  
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Maple

December 13, 2018 at 1:23 am

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