# convergence in distribution and stochastic boundness for difference of precesses

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.
• 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? – LrM May 4 at 8:08
• 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. – E-A May 4 at 23:30
• 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. – LrM May 5 at 8:20