These sequences converges in distribution and in probability

Question

I have this exercise.

Let $U$ and $V$ two independents random variables with Normal Distribution(0,1).

Let, $X_{1}=U$,$X_{2}=V$,$X_{3}=U$,$X_{4}=V$,$X_{5}=U$, ...

I) The sequence $X_{1},X_{2},X_{3},..$converge in Distribution?

II) And in probabbility?

III) $\overline{X_{n}}= \frac{1}{n} \sum\limits_{i=1}^n X_{i}$ converges in probability?

Any help?

Many thanks guys!

Answer 1

All the $X_n$'s have the same distribution (namely $N(0,1)$) so they converge in distribution to $N(0,1)$.

If $X_n \to Y$ in probability then $(X_n)$ is Cauchy in probability and $P(|U-V|>\epsilon) \leq P(|X_{2n-1}-X_{2n}|>\epsilon)) \to 0$ for any $\epsilon >0$. Hence $U=V$ almost surely. But $U$ and $V$ are independent so they must be constants. This is a contradiction. Fo $X_n$ does not converge in probability.

$\frac 1 n \sum\limits_{k=1}^{n} X_k \to \frac {U+V} 2$. This can be seen by just writing out what $\frac 1 n \sum\limits_{k=1}^{n} X_k$ is. You don't need any probability theory for this.