I have this sequence: $X_n=(-1)^nX+\frac{1}{n}$ where $X$ is $U(-1,1)$ random variable. Does this converge in probability?
I have $P(\left|(-1)^nX+\frac{1}{n}-X\right|>\epsilon=0\ \forall$ even $n$, as I can choose a large $N$ corresponding to any $\epsilon$ such that $P(1>N\epsilon)=0$.
But for odd $n$, I have the condition that $P(\left|-2X+\frac{1}{n}\right|>\epsilon)$ should go to zero. As $n\to \infty$, this is the probability that $P(2X>\epsilon)$ which is nerly 0.5 for $\epsilon\to 0$.
So I conclude this sequence does not converge in probability to a uniform random variable. Is my argument correct? Could someone please confirm this?