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This is a homework problem. If $X_n$ converges in probability to 1, show $X_n^{-1}$ converges in probability to 1.

My attempt:

$$\begin{align*} P(|X_n^{-1}-1| > \epsilon) &= P(|X_n^{-1}-X_n + X_n-1| > \epsilon)\\ &\leq P(|X_n^{-1} - X_n| > \epsilon/2) + P(|X_n - 1| > \epsilon/2)\\ &= \end{align*} $$

I know I can bound the 2nd term, but I am not sure how to bound the first term. Perhaps another approach is necessary. Any suggestions would be appreciated.

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  • $\begingroup$ I would try rewriting it as $P(|X_{n} - 1| > \epsilon |X_{n}|)$ instead and split this probability into two cases, e.g. when $|X_{n}| > 1 + \epsilon$ and its complement. $\endgroup$
    – johnny
    Nov 4, 2013 at 21:19
  • $\begingroup$ I'm not seeing which two cases to split your probability statement into, could you elaborate a little. $\endgroup$
    – bdeonovic
    Nov 4, 2013 at 22:23
  • $\begingroup$ I would observe that, given $|X_n -1| < \epsilon < 1$, there is an $\eta$ (what function of $\epsilon$ is it?) such that $|1/X_n-1| < \eta$ and that as $\epsilon \to 0$, $\eta \to$ (what), and continue on from there. $\endgroup$
    – jbowman
    Nov 5, 2013 at 1:35

2 Answers 2

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Fix $\varepsilon\gt 0$. We have for each positive $A$, $$\{|X_n^{-1}-1|\gt \varepsilon\}=\{|X_n-1|\gt |X_n|\varepsilon\}\subset\{|X_n-1|\gt A\varepsilon\}\cup\{|X_n|\leqslant A\},$$ this because we wrote $S=(S\cap \{|X_n|\gt A\})\cup (S\cap\{|X_n|\leqslant A\})$.

Take $A:= 1/2$; then $\{|X_n|\leqslant 1/2\}\subset\{|X_n-1|\geqslant 1/2\}$ (because $(-1/2,1/2)\subset (-\infty,1/2)\cup(3/2,\infty)$). We thus obtained

$$\{|X_n^{-1}-1|\gt \varepsilon\}\subset\{|X_n-1|\gt \varepsilon/2\}\cup\{|X_n-1|\gt 1/2\}.$$ The probability of the last two events goes to $0$ as $n\to\infty$.

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Consider the case $0<\epsilon<1$. Defining $\delta = \epsilon / (1+\epsilon)$, from the figure we have $$ |x-1|<\delta \Rightarrow \left| \frac{1}{x}-1\right| < \epsilon \, . $$ Hence, $$ P\left(\left| \frac{1}{X_n}-1\right| < \epsilon\right) \geq P\left(\left| X_n-1\right| < \delta\right) \to 1 \, , $$ when $n\to\infty$. The case $\epsilon\geq 1$ is easy, because $$ P\left(\left| \frac{1}{X_n}-1\right| < \epsilon\right) \geq P\left(\left| \frac{1}{X_n}-1\right| < \frac{1}{2}\right) \, , $$ and we can use the previous case.

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    $\begingroup$ I like the clarity of your answer. $\endgroup$
    – bdeonovic
    Nov 5, 2013 at 13:18

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