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Suppose that $X_n, Yn$ ($Y_n\neq 0$ a.s) converge to $X,Y$,respectively, in probability.

I need to show

1) $X_nY_n \rightarrow XY$ in probability.

2) $X_n/Y_n \rightarrow X/Y$ in probability.

My try

1) $P\{|X_nY_n-XY|> \epsilon \}=P\{|X_nY_n-X_nY+X_nY-XY|> \epsilon \}$ $\geq P\{|X_n||Y_n-Y|>\epsilon/2 \} +P\{|Y||X_n-X|>\epsilon/2 \}$

How to take $|X_n|$ and $|Y|$ away..??

2) Is $1/Y_n $ converges to $1/Y$ in probability? if Yes, how to show it ?

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  • $\begingroup$ The statement about $X/Y$ is not correct without additional assumptions on $X$ and $Y$. Take $X_n = 1$ a.s., $Y_n = \frac{1}{n}$ a.s. and thus $X = 1$ a.s., $Y = 0$ a.s. $\endgroup$ – Hans Engler Apr 22 '15 at 18:58
  • $\begingroup$ For the second part see this question: math.stackexchange.com/q/1117666 $\endgroup$ – saz Apr 22 '15 at 19:07
  • $\begingroup$ For $X/Y$, I forgot to write the condition $P(Y_n=0)=P(Y=0)=0$. $\endgroup$ – Leonardo Apr 22 '15 at 19:17
  • $\begingroup$ Sorry saz , I don't know how to mimic the proof in the link . $\endgroup$ – Leonardo Apr 22 '15 at 19:19
  • $\begingroup$ @Leonardo Why mimic it? The linked proof shows that $1/Y_n \to 1/Y$ in probability. Combining this with the first part of your question yields $X_n/Y_n = X_n \frac{1}{Y_n} \to X \frac{1}{Y} = X/Y.$ $\endgroup$ – saz Apr 22 '15 at 19:23

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