My idea is using the following inequality, $p(|g(Xn,Yn)-g(0,Y)|>\alpha) \le p(|Xn-Yn|>\delta)+p(|Yn-Y|>\delta)+p(|Y|>N)$. However, I have a problem to find that $\delta$.
$g(Xn,Yn)$ converges in probability to $g(o,Y)$
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probability-theory
1 Answers
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This is a version of the so-called Slutsky's Theorem.
Since $X_n$ converges in distribution to a degenerate limit, it does so also in probability. Then apply the continuous mapping theorem. Proofs of the latter result can be found, for example, here.
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0Thanks to your answer. I can understand the proof on the link you shared, but in the case I posted: $g$ is not a continuous functions on $R^2$ and under the assumption of $g$, I can't find suitable $\delta$ to get the similar inequality. – 2017-01-23
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0Take a closer look at the assumptions of the theorem in the link. The continuity is only required in a certain subset of the domain of the function. Can you identify this subset in your case? – 2017-01-23
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0There is a counterexample that I got from somewhere. Let $g(x,y)=\frac{2xy}{x^2+y^2}$ if $(x,y) \neq (0,0)$ and $g(0,0)=0$. Let $X_n$ be a sequence which converges to $0$ in probability such that $P(X_n=0)=0$, and also let $Y_n=X_n$.Then $g(X_n,Y_n)$ doesn't converges to $g(0,0)=0$ in probability. – 2017-01-31