# If $X_n - \mu = O_p(a_n)$ does that imply that $X_n^{-1} - \mu^{-1} = O_p(a_n)$?

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 $$a_n = 1/\sqrt{n}$$.

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

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.

• 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? Feb 15, 2021 at 14:05
• @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 $\mu=1$) Feb 23, 2021 at 9:40
• @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. Feb 23, 2021 at 11:35
• @ManUtdBloke see my edit. Feb 23, 2021 at 14:07
• @fblundun Very nice, thanks for the clarification Feb 26, 2021 at 14:16

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 $$O$$.

• 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. Feb 26, 2021 at 14:18