Sequence of random variables and convergence in probability.

Question

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.

Answer 1

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