I would greatly appreciate any advice on attempting this question. References to any textbook would also be helpful.
Consider the probability space $(\Omega,\mathcal{F},P)$ with $\Omega = [0,1]$, $\mathcal{F}$ the $\sigma$-algebra of Borel sets, and $P$ the Lebesgue measure. Define the sequence $X_n$ by
$X_{2n} = (0 \text{ if } \omega < \frac{1}{2} ; 1 \text{ if } \omega \ge\frac{1}{2})$ and
$X_{2n+1} = (1 \text{ if } \omega < \frac{1}{2}; 0 \text{ if } \omega \ge \frac{1}{2})$.
Show that $X_n$ converges in distribution, but does not converge in probability.
I apologize for the bad typesetting, I am not fully proficient in LaTex.
Best Answer
$X_{2n}$ is the indicator of $[\frac{1}{2},1]$ and $X_{2n+1}$ is the indicator of $[0,\frac{1}{2})$, then
$$P(X_{2n}=1)=P \left(\{\omega:\omega\in[\frac{1}{2},1]\}\right)=\frac{1}{2}=P(X_{2n}=0)$$ $$P(X_{2n+1}=1)=P \left(\{\omega:\omega\in[0,\frac{1}{2})\}\right)=\frac{1}{2}=P(X_{2n+1}=0)$$
Now, let $F_{2n}(x)$ be the distribution function, then $F_{2n}(x)=P(X_{2n}\leq x)=1$ if $x=1$, and is $\frac{1}{2}$ if $x\in[0,1)$. For $F_{2n+1}(x)$ you have the same distribution function. Therefore $F_n$ does not depend on $n$, then is constant, which implies convergence in distribution.
Now, to proof it does not converge in probability, assume the contrary, i.e. there exists $X$ such that $\lim P(|X_n-X|>\epsilon)\to 0$. Because the limit is unique and convergence in distribution implies convergence in probability, you have that the distribution function of $X$ is the one you obtained before, therefore $X$ only takes two values, $0$ and $1$, and both with the same probability ($\frac{1}{2}$).
Note that convergence in probability implies that $X_{2n}$ and $X_{2n+1}$ both converge in probability to $X$, so $X_{2n}-X_{2n+1}$ converges to $0$ in probability, however, $X_{2n}-X_{2n+1}$ does not converge in distribution to $0$ which is a contradiction because convergence in probability implies convergence in distribution.