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I am dealing with a real-valued random function $\sigma_n(\theta)$, where $\theta\in\mathbb{R}$ is a parameter. I know that
$$ \sigma_n(\theta) \overset{p}{\rightarrow} \sigma(\theta), $$ uniformly, as $n \rightarrow \infty$. My understanding is that, due to the Continuous Mapping Theorem, we have $$ 1/\sigma_n(\theta) \overset{p}{\rightarrow} 1/\sigma(\theta) $$ as well, but not uniformly on $\mathbb{R}$, because the reciprocal function is not uniformly continuous.

My question is, given that $1/x$ is uniformly continuous on $x\in[\mu, \infty)$, with $\mu > 0$, if I impose something like $$ \text{Prob}\{\sigma_n(\theta)<\mu\}\rightarrow 0,\;\;\; \text{as} \;\; n\rightarrow \infty, $$ for any $\theta$, will this be sufficient to get uniform convergence in probability? If the answer is negative, what kind of conditions should I impose?

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  • $\begingroup$ Do you mean $\sigma_n(\theta) \overset{p}{\rightarrow} \sigma(\theta)$? $\endgroup$
    – user140541
    Commented Oct 24, 2016 at 21:26
  • $\begingroup$ Yes, off course, thanks for pointing that out. $\endgroup$
    – Jugurtha
    Commented Oct 24, 2016 at 21:33
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    $\begingroup$ Do you have that $\sigma(\theta)$ is bounded away from $0$, i.e. $\sigma(\theta)\ge \mu >0$ a.s.? $\endgroup$
    – user140541
    Commented Oct 24, 2016 at 21:35
  • $\begingroup$ That's what I would like to impose, and it's a reasonable assumption for my application, but I would like to know how to do it (in a rigorous way) as to guarantee uniform convergence in prob. $\endgroup$
    – Jugurtha
    Commented Oct 24, 2016 at 21:39
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    $\begingroup$ Look at Theorem 1.11.1 on page 67 (just ignore * notation) here books.google.ca/… $\endgroup$
    – user140541
    Commented Oct 24, 2016 at 21:52

1 Answer 1

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I suspect that, in order to get uniform convergence, you may need to require $P(\sigma_n(\theta) < \mu) \to 0$ uniformly in $\theta$. Probably there are general results to this effect, but a direct proof from scratch is also not difficult.

Lemma. Let $\sup_\theta P(\sigma_n(\theta) < \mu) \to 0$ as $n\to\infty$ and $\sigma_n(\theta) \to \sigma(\theta)$ in probability, uniformly in $\theta$. Then $1/\sigma_n(\theta) \to 1/\sigma(\theta)$ in probability, uniformly in $\theta$.

Proof. We first show that $\sigma(\theta) \geq \mu$ a.s. for all $\theta$. This is true since we have

$$\begin{align*} 0 &\leq P\bigl(\sigma(\theta) < \mu-\epsilon\bigr) \\ &= P\bigl(\sigma(\theta) < \mu-\epsilon, \sigma_n(\theta)<\mu \bigr) + P\bigl(\sigma(\theta) < \mu-\epsilon, \sigma_n(\theta) \geq \mu\bigr) \\ &\leq P\bigl(\sigma_n(\theta)<\mu \bigr) + P\bigl(\bigl|\sigma_n(\theta) - \sigma(\theta)\bigr| >\epsilon\bigr) \\ &\to 0+0 \\ &= 0, \end{align*}$$ and thus $P(\sigma(\theta) < \mu-\epsilon) = 0$ for all $\epsilon>0$, which implies $P(\sigma(\theta) < \mu) = 0$ for all $\theta$.

Now we can show the required convergence: Let $\epsilon > 0$ and choose $\delta > 0$ such that $|1/x - 1/y| \leq \epsilon$ for all $x,y\geq \mu$ with $|x-y| \leq \delta$. For fixed $\theta$ we get $$\begin{align*} P\bigl(\bigl|\frac{1}{\sigma_n(\theta)}-\frac{1}{\sigma(\theta)}\bigr| > \epsilon\bigr) &\leq P\bigl(\bigl|\sigma_n(\theta)-\sigma(\theta)\bigr| > \delta \mbox{ or } \sigma_n(\theta) < \mu \mbox{ or } \sigma(\theta) < \mu\bigr) \\ &\leq P\bigl(\bigl|\sigma_n(\theta)-\sigma(\theta)\bigr| > \delta\bigr) + P\bigl(\sigma_n(\theta) < \mu \bigr) \\ &\leq \sup_\theta P\bigl(\bigl|\sigma_n(\theta)-\sigma(\theta)\bigr| > \delta\bigr) + \sup_\theta P\bigl(\sigma_n(\theta) < \mu \bigr) \end{align*}$$ and thus $$\begin{align*} \sup_\theta P\bigl(\bigl|\frac{1}{\sigma_n(\theta)}-\frac{1}{\sigma(\theta)}\bigr| > \epsilon\bigr) &\leq \sup_\theta P\bigl(\bigl|\sigma_n(\theta)-\sigma(\theta)\bigr| > \delta\bigr) + \sup_\theta P\bigl(\sigma_n(\theta) < \mu \bigr) \\ &\to 0 + 0 \\ &= 0. \end{align*}$$ This completes the proof.

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