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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.

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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 $$\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}= (\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\cap \{Y\geqslant A\})\cup (\{|Y_n-Y|\geqslant \varepsilon |Y|^2\}\cap \{Y\lt A\})\\ \subset \{|Y_n-Y|\geqslant \varepsilon A^2\}\cup\{Y\lt A\}.$$

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  • $\begingroup$ 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
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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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