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We are given that Yn converges to Y in distribution and that $P(Y \le t)$ is continuous. We want to show $\frac{Yn}{\sqrt{n}}$ converges to 0 in probability.

I've attempted to solve this question, but I'm not sure if I'm correct.

I said, we are given $\lim_{n\to\infty}$$P(Yn \le t)=P(Y \le t)$.

From there I say that $\lim_{n\to\infty}$P($\frac{Yn}{\sqrt{n}} \le t)=P(\frac{Y}{\sqrt{n}} \le t)$

Then I get $P(|\frac{Yn}{\sqrt{n}}-\frac{Y}{\sqrt{n}}|>\epsilon) = P(|Yn-Y|>\sqrt{n}\epsilon) $

From there, n approaches infinity, so I say $P(|Yn-Y|>\infty)= 0$, as the probability that anything is greater than infinity is 0. Am I correct?

Looks like I made a mistake, my new approach I say

$P(|\frac{Yn}{\sqrt{n}}-0|\ge\epsilon)$ = $P(|Yn|\ge\epsilon\sqrt{n})$ = $P(|Y|\ge \sqrt{n}\epsilon)$, by given information.

Then I say $P(|Y|\ge \sqrt{n}\epsilon) \le \frac{E(Y)}{\sqrt{n}\epsilon}$ by markov. The right side goes to 0 as n apppoaches infinity, so by squeeze theorem, we get what we want.

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    You should be showing $Y_n/\sqrt{n}$ tends to zero in probability, not $Y$. – 2017-02-24
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    how do you know $Y_n/\sqrt{n}$ converge to $Y/\sqrt{n}$ in distribution? – 2017-02-24
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    You're right, it should converge to 0 not Y. I changed my answer a bit... – 2017-02-24
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    @Adam You cannot use that $Y$ (or $Y_n$) has finite expectations. It is not presumed. And $P(|Y_n|>\varepsilon\sqrt{n})$ does not equal to $P(|Y|>\varepsilon\sqrt{n})$ in general. – 2017-02-24

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$P(|Y_n| \geq \varepsilon\sqrt{n})=P(Y_n \leq -\varepsilon\sqrt{n})+P(Y_n \geq \varepsilon\sqrt{n})$.

For all $\varepsilon>0$ and for all $M>0$ there exists $n_{\varepsilon,M}$ such that for any $n\geq n_{\varepsilon,M}$ hold: $$ -\varepsilon\sqrt{n}<-M, \quad \varepsilon\sqrt{n} > M. $$ By monotonicity of probability, for $n\geq n_{\varepsilon,M}$ $$ P(Y_n \leq -\varepsilon\sqrt{n})\leq P(Y_n \leq -M), \quad P(Y_n \geq \varepsilon\sqrt{n})\leq P(Y_n > M) $$ R.h.s.'s in this inequalities tend to $P(Y\leq-M)=F_Y(-M)$ and $P(Y>M)=1-F_Y(M)$ respectively. Then for arbitrary $M$ and $\varepsilon$ $$\limsup_n P(|Y_n| \geq \varepsilon\sqrt{n}) \leq F_Y(-M)+ 1-F_Y(M). $$ Since $M$ is arbitrary positive number, we can take the limit of r.h.s. as $M\to\infty$. Any CDF tends to $0$ at $-\infty$ and to $1$ at $+\infty$, therefore the r.h.s. above tends to $0$. Then $\lim P(|Y_n| \geq \varepsilon\sqrt{n})=0$.