convergence in distribution does imply in probability Consider uniform measure $\mathbb P$  on $=\Omega=[0,1]$, $X_n(\omega)=\omega$ for $n$ even, and $1-\omega$ for $n$ odd. Let $X\sim U[0,1]$
Then show that 


*

*Each $X_n$ has the same distribution as $X$.

*$\mathbb P(|X_n-X|<\epsilon)=\epsilon$, and hence conclude that $X_n$ does not converges to $X$ in probability.
Could anyone help me to solve the problem?
I know: pdf of $X$ is $\mathbb P(X=x)=f(x)=\frac{1}{x}$ and cdf of $X$ is 
$F(x)=0$ for $x<0$, $F(x)=x $ for $x\in (0,1)$ and $1$ for $x\ge 1$.
 A: *

*Let $n$ be even, then we have for any borel set $A \subseteq [0,1]$ that
$$\mathbb{P}_{X_n}(A)=\mathbb{P}(X_n \in A)=\mathbb{P}(\{\omega \: | \: \omega \in A\}) = \mathbb{P}(A),$$
and therefore $\mathbb{P}_{X_n} =\mathbb{P}= U[0,1]$. Consider $n+1$ which is odd and therefore $X_{n+1}(\omega)=1-X_n(\omega)$, which means that
\begin{align*}\mathbb{P}(X_{n+1} \leq x) &= \mathbb{P}(1-X_n \leq x) \\ &=  \mathbb{P}(X_n \geq 1-x) \\&= 1 - \mathbb{P}(X_n \leq 1-x) \\  &= 1-(1-x) = x \end{align*}
for all $x \in [0,1]$, thus $X_{n+1} \sim U[0,1]$.


*There is a slight disambiguity in the second question, because we are not given any information on how $X$ is defined. In order to speak of $|X_n - X|$ we must have that $X$ is defined on $\Omega$ in some explicit way. However it does not matter, we can still disprove that $X_n$ converges in probability by proving that the sequence is not a cauchy sequence in probability. Consider an even integer $n$ and an odd integer $m$, then
\begin{align*} \mathbb{P}(|X_n - X_m| < \varepsilon) &= \mathbb{P}(\{\omega \:: |\omega-(1-\omega)| < \varepsilon\}) \\
&= \mathbb{P}(\{\omega \:: |2\omega-1| < \varepsilon\}) \\
&= \mathbb{P}(\{\omega \:: |\omega-\frac{1}{2}| < \frac{\varepsilon}{2}\}) \\
&= \mathbb{P}((\frac{1-\varepsilon}{2} , \frac{1+\varepsilon}{2})) \\
&=\frac{1+\varepsilon}{2}-\frac{1-\varepsilon}{2} = \varepsilon 
\end{align*}
for all $\varepsilon \leq 1$.
