Let $X_1, X_2,\ldots$ be a sequence of iid normal random variables with zero mean and unit variance. I read the following as a trivial example: (1) $X_n \to X_1$ in law, (2) $X_n \not\to X_1$ in probability.

So for the first one, I suppose since $F_{X_n}(x) = F_{X_1}(x)$, for every $n$, there is nothing to show as $n\to \infty$ . Is that right?

I do not know how to show the second (does not converge in probability). I know the definitions and some ideas but not sure if they are true. If somebody gives an idea of how to prove second, that would be wonderful. Thanks!

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    $\begingroup$ Are the random variables supposed to be independent? Because, in general, $X_n \not \to X_1$ in probability does not hold. (Simply consider $X_n := X_1$ for some random variable $X_1 \sim N(0,1)$.) $\endgroup$ – saz Jun 7 '14 at 5:52
  • $\begingroup$ Exactly, thanks, they should be iid. $\endgroup$ – user48547 Jun 7 '14 at 8:42

You are right for convergence in distribution.

In order to prove that convergence in probability does not hold, notice that $X_n-X_1$ is Gaussian and its variance is $2$. Therefore $\mathbb P\{|X_n-X_1|>2\delta\}=\mathbb P\{|N|>\delta\}>0$ where $N$ is normally distributed with mean zero and unit variance. Therefore, there is not convergence in probability to $X_1$ (or any other random variable).

  • $\begingroup$ Hi Davide. I was just trying to check that $\mathbb P\{|X_n-X_1|>2\delta\}=\mathbb P\{|N|>\delta\}$. It turns out, for the right part of the equality, I find an $\epsilon > 0$ which concludes the same proof as above but $\epsilon \neq \delta$ in my case. In other words, I find $P\{|N|>\epsilon\}$ but $\epsilon \neq \delta$. There is another multiplicative factor. Am I wrong? Just checking. thanks! $\endgroup$ – user48547 Jun 9 '14 at 17:36
  • $\begingroup$ I don' think it's problematic. How is this $\varepsilon$ related to $\delta$? $\endgroup$ – Davide Giraudo Jun 9 '14 at 17:44
  • $\begingroup$ $\epsilon = \frac{\sqrt{2}}{2}\delta$ in my calculation. $\endgroup$ – user48547 Jun 9 '14 at 17:50
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    $\begingroup$ You are probably right (I admit I didn't check carefully, I gave the idea). $\endgroup$ – Davide Giraudo Jun 9 '14 at 17:55
  • $\begingroup$ OK since I respect you a lot, I want to check :) Thanks a lot. I calculated again and I think that it is probably correct. $\endgroup$ – user48547 Jun 9 '14 at 17:56

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