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If a random variable $X_n$ converges in probability to a constant $\mu$, we know by the rules for probability limits that its inverse converges to the inverse of the constant, i.e. $X_n^{-1} \stackrel{p}{\to} \mu^{-1}$.

But does the rate of convergence transfer over? For example, if $$ X_n - \mu = O_p(a_n), $$ does it hold that $$ X_n^{-1} - \mu^{-1} = O_p(a_n)? $$

Edit: Note, I am interested in the case when $a_n = 1/\sqrt{n}$.

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  • $\begingroup$ have a look at the delta method, which provides relevant resutlts $\endgroup$ Feb 19, 2021 at 11:19

2 Answers 2

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You mention convergence in probability, but note that $X_n - \mu = O_p(a_n)$ does not imply that $\frac{X_n - \mu}{a_n}$ converges in probability to $0$. Wikipedia has the relevant definition.

The implication does not hold. As a counterexample, suppose $\mu = 1$, every $a_n = 1$, and $P(X_n = \frac{1}{n}) = 1$.

Then $P(|\frac{X_n - \mu}{a_n}| > 10) = 0$ for all $n$, so $X_n - \mu = O_p(a_n)$.

But for any $M > 0$ and any $n > M+1$, we have $P(|\frac{X_n^{-1} - \mu^{-1}}{a_n}| > M) = 1$. So it is not the case that $X_n^{-1} - \mu^{-1} = O_p(a_n)$.


Edit: proof for the case $a_n = n^{-0.5}$.

Suppose that $X_n - \mu = O_p(a_n)$, where $a_n \to 0$ and every $a_n > 0$.

And suppose that $f$ is a function for which there exists an open interval $I$ and a positive constant $K$ such that $\mu \in I$ and for all $x \in I$, $|f(x) - f(\mu)| < K|x-\mu|$. (The reciprocal function meets this condition if $\mu \ne 0$.)

We seek to show that for any $\epsilon > 0$, there exist $M, N$ such that for all $n > N$, $P(|f(X_n) - f(\mu)| > a_n M) < \epsilon$.

First pick $U, W$ large enough that for all $n > W$, $P(|X_n - \mu| > a_n U) < \frac \epsilon 2$.

Since $a_n \to 0$, we can pick $V > W$ such that for all $n > V$, for all $x \notin I$, we have $|x - \mu| > a_n U$. This means that $P(X_n \notin I) < \frac \epsilon 2$ for $n > V$.

In symbols, if $n > V$,

$P(|f(X_n) - f(\mu)| > a_n K U) \le P(|X_n - \mu| > a_n U) + P(X_n \notin I) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$.

(The intuition for this is that if $n > V$ then it's unlikely that $X_n \notin I$; but if $X_n \in I$ it's unlikely that $|f(X_n) - f(\mu)| > a_n K U$. In other words, $X_n$ is probably not far enough from $\mu$ for $f$ to behave badly (as the reciprocal function does near 0), and if $f$ doesn't behave badly then it preserves the relevant big-O behaviour.)

Let $M = KU, N = V$ and we're done.

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  • $\begingroup$ Ok thanks. What about the case when $a_n = 1/\sqrt{n}$? This is the specific case I am interested in. Maybe I should have asked explicitly about this case instead of using the notation $a_n$, which could be equal to $1$ as in your answer. If $a_n = 1/\sqrt{n}$ does the rate of convergence transfer over? $\endgroup$ Feb 15, 2021 at 14:05
  • $\begingroup$ @fblundun you state 'suppose $\mu = 1$' but I guess that the intention of the OP might have been $\mu = E(X_n)$. (I agree that the question is confusing. It speaks about convergence in probability which is not the same as big $O$ and maybe it was meant to be small $o$. Anyway, your example is not a case of convergence in probability of $X_n$ to $\mu=1$) $\endgroup$ Feb 23, 2021 at 9:40
  • $\begingroup$ @SextusEmpiricus good point. My interpretation was that the part about convergence in probability to $\mu$ was preamble rather than part of the actual problem statement. I may have misinterpreted. $\endgroup$
    – fblundun
    Feb 23, 2021 at 11:35
  • $\begingroup$ @ManUtdBloke see my edit. $\endgroup$
    – fblundun
    Feb 23, 2021 at 14:07
  • $\begingroup$ @fblundun Very nice, thanks for the clarification $\endgroup$ Feb 26, 2021 at 14:16
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Your question relates to the continuous mapping theorem which states

$$X_n \xrightarrow[]{p} a \implies g(X_n) \xrightarrow[]{p} g(a)$$

But this convergence in probability relates to small $o$ notation and not big $O$.

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  • $\begingroup$ Yes you are correct there is a small $o$ vs big $O$ ambiguity in my question. I am actually mostly interested in the transfer of the Big $O_p$ convergence rate to the inverse however. $\endgroup$ Feb 26, 2021 at 14:18

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