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Without using the continuous mapping theorem, I want to show that, given $\{X_n\}$ is a sequence of random variables converging in probability to $1$, $\{1/X_n\}$ converges in probability to $1$.

The place where I am stuck is: how do I know that probabilistically, $1/X_n$ converges? Because after I know that if $1/X_n\overset{p}\to a$ where $a$ is a real number, then I can use the fact that if $X_n$ and $Y_n$ converge in probability to $x$ and $y$ respectively, then $X_nY_n\overset{p}\to xy$. Then taking $Y_n=1/X_n$ the result will be immediate.

But how do I know that the new sequence $1/X_n$ converges at all? I would love to use results in real analysis but I am sure they cannot be applied here.

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  • $\begingroup$ suppose $P(X_n=1)=p_n$ and $P(X_n=0)=1-p_n$ for $p_n=1-\frac{1}{n}$. Then $X_n$ converges in probability to $1$, but $\frac{1}{X_n}$ is not properly defined, hence cannot converge in probability to $1$. $\endgroup$ – drhab Jan 24 '15 at 13:51
  • $\begingroup$ The question is from Rohatgi. I am not sure a book as famous as his will make this grievous mistake. Anyhow, suppose an additional information that $1/X_n$ is bounded is given. This is an assumption I am making up. Can we try this then? $\endgroup$ – Landon Carter Jan 24 '15 at 13:54
  • $\begingroup$ You received nice answers. When it comes to my comment: see the remark in the answer of @saz. $\endgroup$ – drhab Jan 24 '15 at 14:14
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We have to show that $$\mathbb{P} \left( \left| \frac{1}{X_n} - 1 \right|> \varepsilon \right) \to 0 \qquad \text{as} \, \, n \to \infty$$ for any $\varepsilon>0$. To this end, we note that

$$\left| \frac{1}{X_n} - 1 \right|> \varepsilon \iff |X_n-1|> \varepsilon \cdot |X_n|.$$

This implies

$$\begin{align*} \left\{ \left| \frac{1}{X_n} - 1 \right|> \varepsilon \right\} & \subseteq \left\{|X_n-1|> \varepsilon \cdot |X_n|; |X_n| \geq \frac{1}{2} \right\} \cup \left\{|X_n-1|> \varepsilon \cdot |X_n|; |X_n|< \frac{1}{2} \right\} \\ &\subseteq \left\{|X_n-1|> \frac{\varepsilon}{2} \right\} \cup \left\{|X_n| < \frac{1}{2} \right\}. \end{align*}$$

Hence,

$$\mathbb{P} \left( \left| \frac{1}{X_n} - 1 \right|> \varepsilon \right) \leq \mathbb{P} \left( |X_n-1|> \frac{\varepsilon}{2} \right) + \mathbb{P}\left(|X_n|< \frac{1}{2} \right).$$

The first term on the right-hand side converges to $0$ as $n \to \infty$ since $X_n \to 1$ in probability. For the second one, we note that

$$\mathbb{P} \left( |X_n| < \frac{1}{2} \right) \leq \mathbb{P} \left( |X_n-1|> \frac{1}{2} \right) \to 0$$

as $n \to \infty$. This finishes the proof.

Remark: Since $X_n(\omega)$ might equal $0$, we have to use the convention $\frac{1}{0} = \infty$. This means that the random variable $\frac{1}{X_n}$ takes values in $\mathbb{R} \cup \{\infty\}$.

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You are given $\displaystyle \lim_{n \to \infty} P(\{|X_n - 1| \ge \epsilon\}) = 0$ for every $\epsilon > 0$. If $0 < \epsilon < 1$ then $$\left| \frac{1}{X_n} - 1 \right| \ge \epsilon \iff -\epsilon \le \frac{1}{X_n} - 1 \le \epsilon \iff \frac{1}{1 + \epsilon} \le X_n \le \frac{1}{1 - \epsilon}$$ and $$ \frac{1}{1 + \epsilon} \le X_n \le \frac{1}{1 - \epsilon} \iff -\frac{\epsilon}{1 + \epsilon} \le X_n - 1 \le \frac{\epsilon}{1 - \epsilon}.$$ But $\displaystyle \frac{\epsilon}{1 + \epsilon} < \frac{\epsilon}{1 - \epsilon}$, so that $$-\frac{\epsilon}{1 + \epsilon} \le X_n - 1 \le \frac{\epsilon}{1 - \epsilon} \implies \left| {X_n} - 1\right| \le \frac{\epsilon}{1 - \epsilon}.$$ Thus $$P\left( \left\{ \left| \frac{1}{X_n} - 1 \right| \ge \epsilon \right\} \right) \le P \left( \left\{ |X_n - 1| \ge \frac{\epsilon}{1 - \epsilon} \right\} \right) \to 0.$$

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