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How do I show that if $\sqrt{n}(X_n - \theta)$ converges in distribution, then $X_n$ converges in probability to $\theta$?

Setting $Y_n = \sqrt{n}(X_n - \theta)$ , convergence in distribution (to a random variable $Y$) means: $P(Y_n \leq y)$ implieas $P(Y \leq y)$.

Convergence in probability requires that $P(Y_n \geq \epsilon) \rightarrow 0 $

My reasoning so far is the following. Given convergence in distribution I can use Porohov's theorem that $P(|Y_n|>M)< \epsilon$, for some positive $M$ and some positive $\epsilon$. Now, I need to show that this translates into $P(|Y_n| \geq \epsilon)=0$ and this will be convergence in probability. I'm quite stuck however, any hints are appreciated

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We have: $$ \frac{1}{\sqrt{n}}\to 0\implies\frac{1}{\sqrt{n}}\overset{L}\to 0\implies X_n-\theta=\frac{1}{\sqrt{n}}[\sqrt{n}(X_n-\theta)]\overset{L}{\to} 0\implies X_n-\theta\overset{P}{\to} 0. $$ Here, $\overset{L}{\to}$ indicates convergence in distribution whereas $\overset{P}{\to}$ convergence in probability. The second implication above uses Slutsky's Theorem and last implication uses the fact that convergence in distribution to a constant implies convergence in probability to the same constant.

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  • $\begingroup$ I don't understand what you are saying. You not only have convergence in probability but also convergence almost surely of the averages by SLLN. $\endgroup$ – Kavi Rama Murthy Oct 9 '18 at 23:57
  • $\begingroup$ @Xiaomi In your example $\sqrt n (\overset {-} {X_n} -\mu /2)$ does not converge in distribution. $\sqrt n (\overset {-} {X_n} -\mu )$ does. The statement in the question is correct and there is no need to know what the limit of $\sqrt n (X_n-\theta )$ is. $\endgroup$ – Kavi Rama Murthy Oct 10 '18 at 0:36
  • $\begingroup$ Thanks @kavi , I see the error now after writing out $X_i - \mu/2$ and trying to apply CLT. We end up with $- \mu$ either way. Cheers! $\endgroup$ – Xiaomi Oct 10 '18 at 0:41
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You have almost finished the proof. Let $\delta >0$. $P\{|X_n-\theta| >\delta\} \leq P\{|Y_n| >M\}$ for all $n$ such that $\delta \sqrt n >M$, hence for all sufficiently large $n$. So $P\{|X_n-\theta| >\delta\} <\epsilon$ for all sufficiently large $n$.

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  • $\begingroup$ Thanks, @kavi-rama-murthy. Can you just clarify the first inequality (transitioning from a probability statement with $\delta$ to probability statement with $M$). I don't quite get that part $\endgroup$ – Daniel Oct 10 '18 at 8:52
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    $\begingroup$ If $|X_n-\theta| > \delta$ then $\sqrt n |X_n-\theta| >\sqrt n \delta$ which is $>M$ provided $\sqrt n \delta >M$. $\endgroup$ – Kavi Rama Murthy Oct 10 '18 at 8:56
  • $\begingroup$ Thanks! One more question though. Eventually then I get $P(|X_n-\theta| < \delta ) < \epsilon$, where I guess the $\delta$ is fixed. What I need however is $P(\sqrt{n}|X_n-\theta|) \rightarrow 0$. What is the connection between the two, I don't have any $1/\sqrt{n}$ in the first expression in order to conclude it goes to zero? $\endgroup$ – Daniel Oct 10 '18 at 9:18
  • $\begingroup$ $\delta$ is fixed throughout. What we proved is that $P(|X_n-\theta| >\delta) <\epsilon)$ whenever $n$ is sufficiently large. That is exactly what we want to prove. $\endgroup$ – Kavi Rama Murthy Oct 10 '18 at 9:22
  • $\begingroup$ Ok, I'm sure it's right. It's just that this is a different convergence condition than what I have in the text book (namely $P(|X_n - \theta| > \delta) \rightarrow 0$ for $n \rightarrow \infty$) and I don't see directly the connection between the two $\endgroup$ – Daniel Oct 10 '18 at 9:50

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