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Let $\Omega:=[0,1]$ with Lebesgue measure.Let's define a sequence $X_n= \frac{\omega}{n}$. Check:

a) Convergence in distribution

b) Convergence in probability

My solution: b) $P(|X_n-X| \ge \epsilon)=P(|X_n| \ge \epsilon)=P(\frac{\omega}{n} \ge \epsilon)=P(\omega \ge \epsilon n )$ And this is equal 0 for each $\epsilon >0$ and $ n \to \infty$.

a) $X \equiv 0$

We have $F_X(t)= \begin{cases} 0 &\text{for } t < 0\\1 &\text{for } t \ge 0\end{cases}$

Now we have to calculate $F_{X_n}$ and here I have problem.

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Convergence in probability implies convergence in distribution, so having done b) you do not need to do a).

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  • $\begingroup$ Ok, but I have problem with a. $\endgroup$ – pawelK Jan 17 at 19:06
  • $\begingroup$ You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t \in [0,1/n]. $\endgroup$ – Jakub Andruszkiewicz Jan 17 at 19:12

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