Assume that we have two sequence of random elements,$X_{n}$ and $Y_{n}$, taking values from some Hilbert space $H$ and defined on the same probability space. Then, assume that for some $a \in H$ we have $X_{n}\overset{a.s.}{\to}0$, $Y_{n}\overset{a.s.}{\to}0$ and $$ n^{s}X_{n}\overset{d}{\to} \tilde{X} $$ and $$ n^{s}Y_{n} \overset{d}{\to} \tilde{Y} $$

The question: is $n^{s}(X_{n} - Y_{n})\overset{a.s.}{\to}0$ or is $n^{s}(X_{n} - Y_{n})$ at least stochastically bounded?


Let me answer the easier part of your problem about a.s. convergence. Let's just remove $Y$ off of this problem by taking $Y_n = 0$. I assume you took $s > 0$, otherwise, the answer is "yes". Let $\Omega = [0,1]$ with the Borel sigma algebra, and let $X_n = 1/n^{-s} $ if $\omega \in [H_n, H_{n+1}]$, where $H_n = \sum_{i=1}^n \frac{1}{n} - \lfloor \sum_{i=1}^n \frac{1}{n} \rfloor$. (Really, any sum that tends to infinity whose differences are going to zero works; traditional counter-examples are by harmonic numbers.) You can check that $X_n \rightarrow 0$ almost surely, and when multipled by $n^s$, will still converge in distribution to $0$, but will not do so almost surely.

As for the stochastic dominance, if $n^s (X_n - Y_n)$ converges in distribution, the tail is kind of well-controlled by the distribution of $\tilde{X} - \tilde{Y}$. From this, you can argue that $F(x) := \inf_n P(n^s (X_n - Y_n) \leq x)$ is a valid CDF.

  • $\begingroup$ what do you mean by "sum converges in distribution"? $\endgroup$ – LrM May 3 at 23:13
  • $\begingroup$ Sorry; no sum. Just a difference. $\endgroup$ – E-A May 4 at 1:36
  • $\begingroup$ I think the second part is true, because each $n^{s}X_{n}$ and $-n^{s}Y_{n}$ are stochastically bounded. Therefore, the sum of $n^{s}X_{n}$ and $-n^{s}Y_{n}$ is also stochastically bounded. Do you agree? $\endgroup$ – LrM May 4 at 8:08
  • $\begingroup$ No, and you made me realize a gap in my argument; namely, without knowing how $X_n$ and $Y_n$ are coupled, we cannot really say much about the distribution of their sums. $\endgroup$ – E-A May 4 at 23:30
  • $\begingroup$ I do not need the distribution of $n^{s}X_{n} - n^{s}Y_{n}$! I just need to know if $n^{s}X_{n} - n^{s}Y_{n} = O_{p}$ or not. It seems that the statement is correct. $\endgroup$ – LrM May 5 at 8:20

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