How to show that $\frac{1}{X_n}\to 1/c \, \mbox{ in probability} $?

Question

If a random variable $X_n$ converges to a non-zero constant $c$ in probability, then $ \dfrac{1}{X_n}\to \dfrac{1}{c}$ in probability. I try to prove this statement by definition.

Here we want to prove that for every $\epsilon>0$, as $n\to\infty$, $P \left( \bigg \lvert \dfrac{1}{X_n} - \dfrac{1}{c} \bigg \rvert \ge \epsilon \right) \to 0$. Note that $\bigg \lvert \dfrac{1}{X_n} - \dfrac{1}{c} \bigg \rvert = \dfrac{\lvert c - X_n \rvert}{\lvert c X_n \rvert}$.

Then $\displaystyle P \left( \bigg \lvert \frac{1}{X_n} - \frac{1}{c} \bigg \rvert \ge \epsilon \right) = P \left( \frac{\lvert c - X_n \rvert}{\lvert c X_n \rvert} \ge \epsilon \right) = P( \lvert c - X_n \rvert \ge \lvert c X_n \rvert \epsilon)$.

But it seems that we need $E \lvert X \rvert < \infty$?

Answer 1

You have to use the continuity of the $1/x$ function.

Since $c \neq 0$, the function $x \mapsto 1/x$ (either from $(-\infty, 0)$ to $(-\infty, 0)$ or from $(0, \infty)$ to $(0, \infty)$ depending on whether $c$ is negative or positive) is continuous at $c$.

So for any $\epsilon > 0$, there is $\delta > 0$ such that $|x - c| < \delta$ implies $|1/x - 1/c| < \epsilon$. Hence, by the contrapositive $|1/x - 1/c| \geq \epsilon$ implies $|x - c| \geq \delta$.

Therefore, the event of $|1/X_n - 1/c| \geq \epsilon$ must be contained in the event of $|X_n - c| \geq \delta$. Thus, $$ P\big( |1/X_n - 1/c| \geq \epsilon \big) \leq P\big( |X_n - c| \geq \delta \big) $$ And obviously $P\big( |X_n - c| \geq \delta \big) \to 0$ as $n \to \infty$ because $X_n \to_P c$.

Answer 2

I think the infinite-expectation is not an issue here.

WLOG let $c>0$. Then we choose $\varepsilon'=\min(\varepsilon c^2/2,c/2)$, and then we have

$$\mathbb{P}(|X_n-c|\le \varepsilon')\xrightarrow{n\to\infty} 1$$

Then, we have

$$|X_n-c|\le \varepsilon'\Rightarrow |1/X_n-1/c|=|X_n-c|/cX_n\le \frac{\varepsilon{c^2}/2}{c(c-c/2)}=\varepsilon$$

So $\mathbb{P}(|X_n-c|\le \varepsilon')\le \mathbb{P}(|1/X_n-1/c|\le\varepsilon)$, and therefore $$\mathbb{P}(|1/X_n-1/c|\le\varepsilon)\xrightarrow{n\to\infty} 1$$ since the probability can't exceed one.