How can I prove the following:

Let $X_i$ and $Y_i$, $i = 1, \ldots, n$, $X$ and $Y$ be random variables defined on the probability space $(\Omega, \mathcal F, \mathbb P)$ and assume that $X_n$ converges in probability to $X$ and $Y_n$ to $Y$ (also in probability). Then:

If $Y_n \ne 0$ and $Y \ne 0$ almost surely, then $X_n/Y_n$ converges in probability to $X/Y$.

I tried the following:

Let $\epsilon > 0$. We need to show that \begin{align*} \mathbb P(|X_n/Y_n - X/Y| > \epsilon) \xrightarrow{n \to \infty} 0. \end{align*} Define $Z_n := 1/Y_n$ and $Z := 1/Y$. $Z_n$ and $Z$ are well defined, since $Y_n, Y \ne 0$ a.s. It is \begin{align*} \mathbb P(|Z_n-Z| > \epsilon) &= \mathbb P(|1/Y_n - 1/Y| > \epsilon) \\ &= \mathbb P(|Y-Y_n|/|Y_nY| > \epsilon) \\ &= \mathbb P(|Y_n-Y| > \epsilon |Y_n||Y|) \\ &\le \mathbb P(|Y_n-Y| > \epsilon (|Y|-|Y_n-Y|)|Y|) \\ &= \mathbb P(|Y_n-Y| > \epsilon (|Y|-|Y_n-Y|)|Y|, |Y_n-Y| \le |Y|/2) \\ &\quad + \mathbb P(|Y_n-Y| > \epsilon (|Y|-|Y_n-Y|)|Y|, |Y_n-Y| > |Y|/2) \\ &\le \mathbb P(|Y_n-Y| > \epsilon Y^2/2) + \mathbb P(|Y_n-Y| > |Y|/2). \end{align*} Now for any $A > 0$ \begin{align*} \mathbb P(|Y_n-Y| > \epsilon Y^2/2) &= \mathbb P(|Y_n-Y| > \epsilon Y^2/2, |Y| \ge 1/A) + \mathbb P(|Y_n-Y| > \epsilon Y^2/2, |Y| < 1/A) \\ &\le \mathbb P\left(|Y_n-Y| > \frac{\epsilon}{2A^2}\right) + \mathbb P(|Y| < 1/A) \xrightarrow{n \to \infty} \mathbb P(|Y| < 1/A) \end{align*} and similarly \begin{align*} \mathbb P(|Y_n-Y| > |Y|/2) &= \mathbb P(|Y_n-Y| > |Y|/2, |Y| \ge 1/A) + \mathbb P(|Y_n-Y| > |Y|/2, |Y| < 1/A) \\ &\le \mathbb P\left(|Y_n-Y| > \frac{1}{2A}\right) + \mathbb P(|Y| < 1/A) \xrightarrow{n \to \infty} \mathbb P(|Y| < 1/A). \end{align*} Since we can choose $A$ arbitrarily large, we obtain \begin{align*} \mathbb P(|Z_n-Z| > \epsilon) \xrightarrow{n \to \infty} 0. \end{align*} The result follows since we know that $X_nZ_n$ converges to $XZ$ in probability.

Edit: Is it complete now? Is there maybe a shorter proof? Maybe my estimations are too long or there is shorter way to do it.


The estimates look correct, but it seems not clear how to get the wanted convergence (we indeed have the probability that $|Y_n-Y|$ is greater than a positive number, but this one is random).

However, the problem can be solved in this way: fix $\delta$ a small number and $A$ such that $\mathbb P\{|Y|\leqslant A\}\lt \delta$. Then we decompose the probabilities obtained in the last estimate of the OP computations over the set $\{|Y|\leqslant A^{-1}\}$ and its complement.

For example, here is the proof that $\mathbb P\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\to 0$. We have $$\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}= (\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\cap \{Y\geqslant A\})\cup (\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\cap \{Y\lt A\})\\ \subset \{|Y_n-Y|\geqslant \varepsilon A^2\}\cup\{Y\lt A\}.$$

  • $\begingroup$ Thanks a lot, I'll try that. What does the abbreviation OP mean? $\endgroup$ – numerion Nov 30 '13 at 0:16
  • $\begingroup$ OP means "Opening Post". Don't hesitate if you need more details. $\endgroup$ – Davide Giraudo Nov 30 '13 at 9:16
  • $\begingroup$ I have troubles with the decomposition of the last estimate. Could you please provide a few more details? Thanks. $\endgroup$ – numerion Nov 30 '13 at 14:44
  • $\begingroup$ I've edited. Is it clearer. $\endgroup$ – Davide Giraudo Nov 30 '13 at 14:54
  • $\begingroup$ Why is that inclusion true? I don't see why $\{|Y_n-Y| \ge \epsilon |Y|^2\} \cap \{Y \le A\} \subseteq \{|Y_n-Y| \ge \epsilon A^2\}$. $\endgroup$ – numerion Nov 30 '13 at 15:16

Isn't this a consequence of the continuous mapping theorem?

Since Y is different to zero almost surely, the function g(x)=1/x satisfies the condition $P[Y\in C^c(g)]=0$, where $C^c(g)$ denotes the set of discontinuity points of $g$ (which in this case equals the set $\{0\}$). Hence we obtain $\frac{1}{Y_n}\to \frac{1}{Y}$ in probability.

Since $X_n\to X$ in probability and $Z_n\to Z$ in probability implies $X_nZ_n\to XZ$ in probability (this is proved in Resnick's "A Probability Path" using subsequences), the result follows taking $Z_n=\frac{1}{Y_n}$ and $Z=\frac{1}{Y}$.


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