4
$\begingroup$

Let $D$ be a probability distribution on the unit interval $[0,1]$ with moments $\mu_i=\mathbb{E}_D [x^i]$. Let $\delta(x)$ be a singleton probability distribution with all weight at $x\in [0,1]$. Let $C(n)=\lfloor (n+2)/2 \rfloor$.

Then we can match the first $n$ moments of any $D$ using a convex combination of $C(n)$ singleton distributions. In other words, for any $D$ and any $n$, there exists $\alpha_i, x_i$ such that the distribution $$ \sum_{i=1}^{C(n)} \alpha_i \delta(x_i) $$ has the same first $n$ moments as $D$ (where $\alpha_i,x_i \in [0,1]$ and $\sum_i \alpha_i=1$.) For details, see this Mathoverflow post. Note that we can force $x_1\leq x_2 \leq \cdots \leq x_n$ without changing anything.

Now, consider a map $f$ from the (non-decreasing) singletons to the corresponding moments, i.e. $$f(x_1,...,x_{C(n)},\alpha_1,...,\alpha_{C(n)-1})=(\mu_1,...,\mu_n)$$ (Note that we can compute $\alpha_{C(n)}$ as $1-\sum_{i=1}^{C(n)-1}\alpha_i$.) Let $\Delta_n$ be the domain of $f$.

The function $f$ maps $\Delta \subseteq \mathbb{R}^{2C(n)-1}$ to $\mathbb{R}^n$. When $n$ is odd, $2C(n)-1=n$. However, when $n$ is even, $2C(n)-1=n+1$, so it would appear that the domain has an "extra dimension". I am trying to understand if it is possible to reparameterize the problem when $n$ is even so that the domain is $n$-dimensional in order to construct an injection.

My questions are:

  1. If $n$ is even and we consider all convex combinations of $C(n)$ singleton distributions, but if we force $x_1=0$, can we still match the first $n$ moments of any distribution on the unit interval? (I.e. does the range of $f$ match the range of $f|_{x_1=0}$ when $n$ is even?)
  2. Suppose the answer to the previous question if "yes". Let $g=f$ when $n$ is odd, and let $g$ be the restriction of $f$ to $x_1=0$ when $n$ is even. Is $g$ invertible?

I realize that these specific question may be misguided-- perhaps there is a better way to think about a space of canonical representative distributions with target moments. (Note that there exist moments that can only be represented by non-continuous distributions, so I wasn't able to use entropy maximization "out of the box".)

$\endgroup$
4
$\begingroup$

The answer to question #2 is no. Consider for example $n = 2$ with the measure $\delta(0)$. You need to choose $x_1 = 0$, but the choice of $x_2$ is arbitrary as $\alpha_2 = 0$.

The answer to question #1 is yes. For a more detailed analysis, I refer to theorem 2.2.3 in Dette/ Studden's The Theory of Canonical Moments with Applications in Statistics, Probability, and analysis. Essentially the problem lies therein, that the measure (and its support points) is uniquely determined as soon as one of corresponding canonical moments is 0 or 1. If you try to add any more support points, the corresponding weights need to be 0 and the choice of support points is ambiguous.

$\endgroup$
  • $\begingroup$ Thank you! I see how the answer to question #2 is "no". Does it also imply that the answer to question #1 is "no"? (The answer is "yes" when $n=2$, as you can see if you consider the moment curve, so the problem would have to happen for $n\geq 4$.) $\endgroup$ – Bill Bradley Aug 11 '15 at 15:01
  • $\begingroup$ So you want to know if the function is onto if you choose $x_1 = 0$? The answer is yes, as the aforementioned theorem shows. $\endgroup$ – Dominik Aug 11 '15 at 15:16
  • $\begingroup$ Ah, sorry, I misunderstood your second paragraph. Thanks (and thank you for the reference to the book!) $\endgroup$ – Bill Bradley Aug 11 '15 at 16:13
  • $\begingroup$ You probably didn't misunderstand my second paragraph, as I actually didn't answer you question in it ;) You can just prove the assertion with the same theorem. $\endgroup$ – Dominik Aug 11 '15 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.