So an answer of this post provided an example of a sequence of random variables that converges in distribution but not in probability. There is also a post here that gives an answer on how to prove that a sequence of random variables (defined as below) does not converge in probability. I'd like to know if an alternate proof that I propose here is correct or not.
Let $U \sim \text{Unif}(0,1)$ and define
$$
X_n =
\begin{cases}
U, \ \ \ \ \ n \mod 2 = 0,\newline
1 – U, \ n \mod 2 = 1
\end{cases}
$$
If $X \sim \text{Unif}(0,1)$. Then $X_n \to X$ in distribution since $X_n$ has the same distribution function as $X$ for any $n \in \mathbb{N}$.
If $n$ is even, then
$$X_n – X = 0$$
Therefore $\mathbb{P}(|X_n – X| > \varepsilon) = 0$ since $\varepsilon > 0$. If $n$ is odd, then
$$X_n – X = 1 – 2X$$
So,
\begin{align*}
\mathbb{P}(|X_n – X| > \varepsilon) &= \mathbb{P}(1 – 2X > \varepsilon) + \mathbb{P}(1 – 2X < – \varepsilon) \newline
&= \mathbb{P}(X <(1 – \varepsilon)/2) + \mathbb{P}(X > (\varepsilon + 1)/2) \newline
&= \frac{1 – \varepsilon}{2} + 1 – \frac{1 + \varepsilon}{2} = 2
\end{align*}
Hence $X_n$ does not converge to $X$ in probability.
Is my reasoning correct?
Best Answer
This is not correct. $(X_b)$ converges in distribution to lots of r.v.'s and $X$ is not the only possible limit. To prove that this sequence does not converge in probability it is not enough to consider $X_n-X$ since $X_n$ may converge in probability to some other r.v.
$P(|X_{2n}-X_{2n-1}| >\epsilon) =P(|U-(1-U)| > \epsilon)\geq P(U>\frac {1+\epsilon} 2)$ which proves that $(X_n)$ is not Cauchy in probability. Hence, it cannot converge to any r.v. in probability.