Why this sequence of R.V converge in distribution, but doesnt in probability

Question

Why this sequence of R.V converge in distribution, but doesnt in probability?

Probability space $([0,1],B,m)$. ($B$ consists of all Borel sets of $[0,1]$, $m$ is the Lebesgue measure.)

Let $X_{2n}(ω)=ω$, $X_{2n−1}(ω)=1−ω$

My intuiton says that I have to show that $X_{2n}$ and $X_{2n−1}$ have the same distribution, which is an Uniform distribution. And after this, showing that they converge in distribution, which is obvious.

But How can I show this?

And also, how can I Show that $X_{n}$ does not converge in probability.

Any help guys?

Answer 1

You're right about the convergence in distribution part. To show they have a uniform distribution, calculate the CDF. For instance, for $x\in (0,1),$ $P(X_0\le x) = P(\omega < x) = m(\{\omega :\omega <x\})=m((0,x)) = x,$ and you can compute similarly for $1-\omega.$

To show that it doesn't converge in probability, observe that the sequence oscillates between $\omega$ and $1-\omega.$ The distance between the two is $1-2\omega.$ Thus for $m$ even and $n$ odd, $P(|X_n-X_m|>\epsilon) =P(|1-2\omega|>\epsilon)>0.$ Thus $X_n$ is not Cauchy in probability and therefore not convergent in probabillity.

Answer 2

Let $Y(\omega)=\omega$ and $Z(\omega)=1-\omega$. Then $X_{2n}\to Y$ in probability and $X_{2n-1}\to Z$ in probability. If $X_n$ converges in probability then we must have $Y=Z$ almost surely which is false since $P(Y=Z)=0$.