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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?

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up vote 2 down vote accepted

Assume that $(X_n)$ converges in probability, say to $Y$. Then, $Z_n=|X_n-X_{n+1}|$ converges in probability to $|Y-Y|=0$. Since $Z_n\to2|X|$ almost surely, $Z_n\to2|X|$ in probability. The limit in probability is unique up to a null set hence $2|X|=0$ almost surely. This is absurd hence $(X_n)$ does not converge in probability.

On the other hand, the distribution of $X$ is symmetric and $\frac1n\to0$ hence $X_n$ converges in distribution to $X$.

Edit: Here is a rewriting of (what I understand from) OP's argument.

First, for every positive $\varepsilon$, $[|X_{2n}-X|\geqslant\varepsilon]=[1\geqslant n\varepsilon]=\varnothing$ for every $n$ large enough hence $\mathrm P(|X_{2n}-X|\geqslant\varepsilon)\to0$ when $n\to\infty$, that is, $X_{2n}\to X$ in probability.

Thus, if $(X_n)$ converges in probability, $X_{2n+1}\to X$ in probability. But $X_{2n+1}-X=-2X+\frac1n$ hence that would imply that $\mathrm P(|2X-\frac1n|\geqslant\varepsilon)\to0$. If $n\geqslant\frac2\varepsilon$, $[|2X-\frac1n|\geqslant\varepsilon]\supseteq[|X|\geqslant\varepsilon]$, hence one would have $\mathrm P(|X|\geqslant\varepsilon)\to0$. For $\varepsilon\lt1$, this is absurd since $\mathrm P(|X|\geqslant\varepsilon)=1-\varepsilon$.

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Thanks @Didier, that was clear. But I'd love to know is there any flaw in my reasoning... – Bravo Apr 24 '12 at 12:59
See Edit. $ $ $ $ – Did Apr 24 '12 at 16:11

If the question is, do $(-1)^n X + \frac{1}{n}$ converge in probability to $X$, which it looks like based on your answer, yeah it looks good to me.

BTW i think you mixed up the even and odd cases?

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Thanks, edited!... – Bravo Apr 24 '12 at 10:41

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