Assume $\forall y\in\mathbb{R}$, $P(Y_n\leq y)\rightarrow\Phi(y)$ and if $n\rightarrow\infty$, $x_n\rightarrow x$, prove $P(Y_n\leq x_n)\rightarrow\Phi(x)$ if $x_n\rightarrow\infty$. It seems very intuitive, but I have no idea how to prove it.
CLT for sequences
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sequences-and-series
central-limit-theorem
probability-limit-theorems
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1What is $x$? Did you mean $x_n \to x$ as $n \to \infty$? – 2017-02-20
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0ah thanks! I didnt see it, its corrected – 2017-02-20
1 Answers
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In your case, the normal distribution function $\Phi(x)$ is a continuous function, hence by Polya's theorem, $\mathsf{P}(Y_n\le y)\to \Phi(y)$ uniformly in $y$, that is $$ \lim_{n\to \infty}\sup_{y\in\mathbb{R}}|\mathsf{P}(Y_n\le y)- \Phi(y)|=0. $$ Therefore, \begin{align} |\mathsf{P}(Y_n\le x_n)- \Phi(x)|&\le |\mathsf{P}(Y_n\le x_n)-\Phi(x_n)|+|\Phi(x_n)-\Phi(x)|\\ &\le \sup_{y\in\mathbb{R}}|\mathsf{P}(Y_n\le y)- \Phi(y)|+|\Phi(x_n)-\Phi(x)|.\tag{1} \end{align} From (1) it is easy to get what you want.