Let $X, X_1, X_2, \ldots : \Omega \to (0,\infty)$ be random variables such that $X_n \xrightarrow{\mathbb{P}} X$. I'd like to show from the definition that $\frac{1}{X_n} \xrightarrow{\mathbb{P}} \frac{1}{X}$.

Since the proof provided by my text is very confusing, I asked for an alternative proof over a month ago. Now, after becoming a bit more familiar with the topic I revisited the textbook-proof hoping to understand it yet to no avail.

So my question is: how does one understand the following proof:

We need to show that $$\forall a>0 \ \ \forall \varepsilon>0 \ \ \exists n_0 \ \ \forall n \geq n_0 : P\left(\left|\frac{1}{X_n}-\frac{1}{X}\right|>a \right)<\varepsilon$$

Let $a, \varepsilon>0.$ Choose such $s >0$ that $P(X \geq 2s)=1-\varepsilon/3.$ Due to $X_n \xrightarrow{\mathbb{P}} X$ there is $\hat{n}_0$ such that $P(|X_n-X| > b) < \varepsilon/3$. On the rest of the set holds $|X_n-X| \leq b$. On this set also hold both $X\geq s$ and $X_n\geq s$. And now we're done because $\frac{1}{x}: \Omega \to [s,\infty)$ is uniformly continuous.

All the $\varepsilon$-fine-tuning is useless if the big picture isn't provided, and the proof seems to obscure it.

Okay, we use the uniform continuity, therefore we'd have something like $$\exists b>0: \left\{ |X_n-X| \leq b \right\}\subset \left\{ \left|\frac{1}{X_n}-\frac{1}{X}\right|\leq a \right\}$$


\begin{align*} P\left\{ \left|\frac{1}{X_n}-\frac{1}{X}\right|\leq a \right\} & = P\left( \left\{ \left|\frac{1}{X_n}-\frac{1}{X}\right|\leq a \right\} \cap \{ |X_n-X| \leq b \} \right) \\ & + \ P\left( \left\{ \left|\frac{1}{X_n}-\frac{1}{X}\right|\leq a \right\} \cap \{ |X_n-X| > b \} \right) \\ & \ldots \\ & \geq 1-\varepsilon \end{align*}

What happens in between?


A way to get an intuition of the result is the following: write $$\frac 1{X_n}-\frac 1X=\frac{X-X_n}{XX_n}.$$ We know how to control $X_n-X$, but $X$ and $X_n$ can take small values hence dividing by them may produce a large quantity. However, we can, for $\delta$ positive, write $$\left\{\left|\frac 1{X_n}-\frac 1X\right|\gt a\right\}\subset \color{red}{\left( \left\{\left|\frac 1{X_n}-\frac 1X\right|\gt a\right\}\cap\{|X_n|\gt \delta\}\cap\{|X|\gt \delta\}\right)}\cup\{|X_n|\leqslant \delta\}\cup \{|X|\leqslant \delta\}.$$ (we will take $\delta$ small) The probability of the red event can be bounded by $\mathbb P\{|X_n-X|\gt \delta^2a\}$, which converges for each $a$. Since $\{|X_n|\leqslant\delta\}\subset\{|X|\leqslant 2\delta\}\cup\{|X_n-X|\gt \delta\}$.

  • $\begingroup$ So you didn't use the uniform continuity? And thanks a lot! $\endgroup$ – Leo Mar 20 '14 at 0:13
  • $\begingroup$ Not explicitely (or say, I used uniform continuity on $[\delta,infty)$ for each $\delta$). $\endgroup$ – Davide Giraudo Mar 20 '14 at 22:00

Since function $$f(t) = \frac{1}{t}\ , \quad 0<t<\infty$$ is continuous you know that you can make $f(x)$ be as close to $f(y)$ as you like, if you just make sure that $x$ is sufficiently close to $y\ .$

Thus if you want $f(X_{n})$ to be $\varepsilon$-close to $f(X)$ you can make this happen if $|X_{n}-X| < \delta(\varepsilon,X)\ ,$ where the number $\delta(\varepsilon,X)$ tells you when $X_{n}$ is sufficiently close to $X\ .$ $$\mathbb{P}(|X_{n}-X| < \delta(\varepsilon,X)) \leq \mathbb{P}(|f(X_{n})-f(X)|<\varepsilon) \ .$$ Working with the complementary events (for convenience) you know that $$0\leq \mathbb{P}(|f(X_{n})-f(X)|>\varepsilon) \leq \mathbb{P}(|X_{n}-X| > \delta(\varepsilon,X))\ .$$ Since the sequence $\{X_{n}\}$ converges to $X$ in probability, this last probability can be made as close to $0$ as you like if you just make sure that $n$ is large enough. Thus you have shown that the sequence $\{f(X_{n})\}$ converges in probability to $f(X)\ .$


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