Let $X_n$ and $X$ be random variables for which $EX_n^p \rightarrow EX^p <\infty$ for $p\in \{1, \ldots, P\}$. $\{X_n\}$ is bounded in probability (this can be shown noting that $EX_n^2 = O(1)$ and applying Markov's inequality), hence by Prokhorov's theorem, each subsequence has a further subsequence, say $\{Y_n\}$, that converges weakly to a limit $Y$. We can show that the moment convergence is also true for $Y_n$: $EY_n^p \rightarrow EY^p <\infty$ for $p\in \{1, \ldots, P\}$.

The claim is that "the moments of $Y$, for $p\in \{1, \ldots, P\}$, are identical to the moments of $X$". How can I show the veracity of this claim?

  • $\begingroup$ What if we have $\{X_n\}$ iid and two other variables $X$ and $Z$ that have the same first $P$ moments as $X_1$, but $E[X^{P+1}]\neq E[Z^{P+1}]$? So $Y$ cannot have the same $P+1$ moment as both $X$ and $Z$. $\endgroup$ – Michael Jul 5 '18 at 16:51
  • $\begingroup$ @Michael, I edited the question to address the issue you raised. $\endgroup$ – math_enthuthiast Jul 5 '18 at 17:01
  • $\begingroup$ I don't think understand the question. $(Y_n)_n$ is a subsequence of $(X_n)_n$? That is, there exists an increasing $\varphi$ such that $Y_n = X_{\varphi(n)}$ for every $n$? If so,for every $p$ we get that $\mathbb{E} Y_n^p$ is a subsequence of $\mathbb{E} X_n^p$, don't we? $\endgroup$ – Clement C. Jul 5 '18 at 17:15
  • $\begingroup$ @Javad : Your edit makes more sense. In addition to Clement's further question, it is not clear if the subsequence indices $n[k]$ are chosen deterministically, or after looking at the random sequence outcome, to form the subsequence $\{X_{n[k]}\}_{k=1}^{\infty}$. If deterministically, we have $\lim_{k\rightarrow\infty}E[X_{n[k]}^p]= E[X^p]$ for all $p \in \{1, ..., P\}$, else we do not necessarily have that (since, for example, we can have $\{X_n\}$ iid Bernoulli and choose $X_n$ to be part of the subsequence only when $X_n=0$). $\endgroup$ – Michael Jul 5 '18 at 17:24
  • $\begingroup$ @ClementC. Indeed! Your simple observation that I overlooked is actually the answer: $EY_n^p$ is itself a subsequence of $EX_n^p$ and hence converges to the same limit, therefore $EY_n^p \rightarrow EX^p = EY^p$. $\endgroup$ – math_enthuthiast Jul 5 '18 at 17:28

Your Answer

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

Browse other questions tagged or ask your own question.