I do not have a good title for this. Anyone has a better title idea, feel free to edit. Below is my question:

Suppose you have a sequence $b_n\overset{d}{\to}N(0,1)$, and also you have another sequence $a_n\overset{a.s.}{\to}0$ if $|b_n|<c$ where $0<c<\infty$. Does this imply that $a_n\overset{p}{\to}0$?

I know this is somewhat a weird setup. I encounter this while I was trying to show $a_n\overset{a.s.}{\to}0$ and $b_n$ is some term in the Taylor expansion of $a_n$ and my proof need $|b_n|$ to be bounded by some constant in order to show $a_n\overset{a.s.}{\to}0$. I guess maybe I can show a weaker result, that is, $a_n\overset{p}{\to}0$. I have a sketch of proof but I do not know if it is correct.

For any $\epsilon>0$, and for any $0<c<\infty$,

$$pr\bigg(|a_n|>\epsilon\bigg)<pr\bigg(|b_n|<c\bigg)pr\bigg(|a_n|>\epsilon\bigg| |b_n|<c\bigg)+pr\bigg(|b_n|>c\bigg)$$

For any fixed $c$, the first term goes to zero since $a_n\overset{a.s.}{\to}0$ given $|b_n|<c$, and applying the fact that $b_n\overset{d}{\to}N(0,1)$ we have

$$\lim_{n\to\infty} pr\bigg(|a_n|>\epsilon\bigg)\leq\lim_{n\to\infty}pr\bigg(|b_n|>c\bigg)=\Phi(|x|>c),$$

where $\Phi(\cdot)$ is the cdf for standard normal. Then, letting $c\to\infty$, we have

$$\lim_{n\to\infty} pr\bigg(|a_n|>\epsilon\bigg)=0.$$

Is my reasoning correct? I am not confident with the "letting $c\to\infty$" part of proof.

Thanks in advance for your time.

  • $\begingroup$ It is impossible for a sequence of random variables $\{B_n\}_{n=1}^{\infty}$ to converge in distribution to a Gaussian distribution if they are always bounded by a constant $c$. So you would need to modify or clarify the assumption $|B_n|\leq c$. $\endgroup$ – Michael Feb 12 '18 at 17:56
  • $\begingroup$ True, but I feel I could at least show that my $a_n$ converge with high probability? For a large $c>0$, $pr(|b_n|<c)$ is high, and conditional on the event $B:|b_n|<c$, we have $a_n\to 0$. I was wondering if we could show $a_n\overset{p}{\to}0$. Since you could let that $c$ to be arbitrarily large. $\endgroup$ – Bayesric Feb 12 '18 at 18:01
  • $\begingroup$ It seems then that your $B_n$ variables are irrelevant? If $\{A_n\}_{n=1}^{\infty}$ is a sequence of random variables that converge almost surely to a random variable $X$, then they also converge "in probability" to $X$ (and also in distribution, to the distribution of $X$, with some caveats/reminders about continuity points for the CDF of $X$). In this case, it seems your random variable $X$ is the always-constant variable $X=0$. $\endgroup$ – Michael Feb 12 '18 at 18:03
  • $\begingroup$ $b_n$ appears in the higher order term in the Taylor expansion of $a_n$. In order to prove $a_n\to 0$, we need that higher order term goes away hence I need an upper bound on $b_n$. But I do not have that since $b_n$ converge in distribution to a normal. $\endgroup$ – Bayesric Feb 12 '18 at 18:06
  • $\begingroup$ I think you should clarify the assumptions and connections between $A_n$ and $B_n$. $\endgroup$ – Michael Feb 12 '18 at 18:07

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