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I am encountering the following problem concerning existence of a limiting random variable (in distribution): assume a sequence of positive random variables $\{X_n\}_{n\geq 0}$ from which we know their moments ($\mathbb{E}[X_n^s]$), and also the limit of these moments (which we denote by $\{m_s\}_{\geq 0}$). Let me say that all these limits are finite numbers (no divergence).

We know that this sequence satisfies Carleman's criterion (namely, $\sum_{s\geq 0} m_s^{-1/(2s)}$ diverges), but the thing that we do NOT know in general is if this sequence defines a proper random variable $X$.

Is it clear that if this sequence satisfies the conditions of the Stieltjes moment problem (see here) then this sequence $\{X_n\}_{n\geq 0}$ converges to the corresponding random variable $X$ defined by the sequence $\{m_s\}_{\geq 0}$, and by Carleman's criterion $X$ is uniquely determined. In particular the sequence $X_n$ tends in distribution to $X$.

My question now is the following: it is possible that the sequence $\{m_s\}_{s\geq 0}$ does NOT define a random variable (namely, Stieltjes condition is NOT satisfied) but nevertheless there is convergence in distribution of my sequence to a certain random variable $X$?

In our particular case, we would like to prove that there is NOT convergence in distribution towards a Poisson (this is a very specific problem on random graphs), but we do not know to justify the last step (as in general, it is possible that there is not convergence of moments, but convergence in distribution could be true).

Thanks!

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    $\begingroup$ A sequence is a moment sequence of some measure if and only if the matrices $(m_{j+k})_{j,k}$ are positive definite (is this what you mean by ``Stieltjes condition?''), and this condition is trivially preserved under taking limits. $\endgroup$ – Christian Remling Nov 7 '15 at 18:48
  • $\begingroup$ yes, positive definite is the Stieljes condition (we also know that all random variables are positive). But the point that is not clear to me is the following (and this is the question, essentially): the moments of the sequence $X_n$ tends to certain sequence which we know that is NOT a sequence of moments of any r.v. (hence, there is not convergence in moments). But I want to conclude (by some argument) that there is no convergence in distribution neither... $\endgroup$ – gaussian-matter Nov 7 '15 at 19:23
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$\newcommand{\bE}{\mathbb{E}}$ As Christian Remling pointed out in his comment, the Stiltjes condition is preserved under limits. Hence there will always exist a at least one positive measure finite measure $\mu$ on $[0,\infty)$ such that

$$ m_s=\int_0^\infty x^s \mu(dx),\;\;\forall s=0,1,2,\dotsc. $$

The measure $\mu$ is a probability measure iff $m_0=1$. If the sequence $(m_s)_{s\geq 1}$ satisfies the Carleman condition, then there exists exactly one measure with these moments. If you have a sequence of nonnegative random variables $(X_n)$ such that

$$\lim_{n\to\infty}\bE\bigl[ \, X_n^s\,\bigr]=m_s,\;\; s=1,2,\dotsc, $$

and the sequence $(m_s)$ satisfies the Carleman condition, then the sequence $X_n)$ converges in law to a random variable with moments $(m_s)$.

However, if the nunmbers $m_s$ grow too fast,

$$ m_s\geq (s m_{s-1})^s,\;\;\forall s=1,2,\dotsc, $$

then there exist at least two measures $\mu_1,\mu_2$ on $[0,\infty)$ with moments $m_s$; see Sec. III.16 of Widder's book on Laplace transform. If we consider the probability measure $\mu_1\otimes \mu_2$ on $[0,\infty)^2$ and the random variables

$$X_n:[0,\infty)^2 \to[0,\infty),\;\;X_n(x_1,x_2) =\begin{cases} x_1, & n \equiv 1\bmod 2,\\ x_2, &n\equiv 0\bmod 2, \end{cases} $$

then

$$\bE\bigl[\, X_n^s\,\bigr]=m_s,\;\;\forall s,\;\;\forall n, $$

yet the sequence $(X_n)$ is not convergent in distribution.

In general, if you have a sequence of (nonnegative) random variables $(X_n)$ such that for any $s=1,2,\dotsc$ the sequence $\bE[X_n^s]$ is convergent, then the sequence $(X_n)$ contains a subsequence that is convergent in distribution.

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