Why do these three conditions imply divergence in probability?

Question

This is a follow up to A martingale converging in distribution but not a.s. or in probability. Suppose we have a sequence of integer-valued RVs that satisfies

(1) $P(X_n=a~i.o.)=1$ for each $a=-1,0,1$

(2) $\sup_n|X_n(w)|<\infty$ a.s.

(3) For some $p\in(0,1/2)$, $P(X_n=1),P(X_n=-1)\to p$ and $P(X_n=0)\to 1-2p$.

How does one show this sequence of RVs doesn't converge in probability?

My Attempts: Clearly $X_n$ converges in distribution to RV $X$ with $P(X=1)=P(X=-1)=p,P(X=0)=1-2p$. Assume by contradiction that $X_n$ converges in probability, then it must be the case that $X_n\to X$ in probability. So there must be a subsequence $X_{n_k}\to X$ a.s.. But I couldn't find a way to go forward.

The condition (1) can be clarified, does it mean that for each $a\in {-1, 0,1}$ that $P[X_n=a \quad i.o.]=1$? — Michael, Feb 28 '24 at 16:40

score 2 · Accepted Answer · answered Feb 28 '24 at 16:19

I'm suspicious of this claim. Here's an example. I fold together the values $1$ and $-1$ from your situation, but that simplification is unimportant.

Example: On an appropriate probability space $(\Omega,\mathcal F,P)$ let $U$ be uniformly distributed on $(0,1)$ and let $\xi_1,\xi_2,\ldots$ be a sequence of independent $\{0,1\}$-valued random variables, independent of $U$, such that $P(\xi_n =1)=1/n$. For fixed $p\in(0,1)$ define $$ X_n =\cases{ 1,&$0<U\le p, \xi_n=0$,\cr 0,&$0<U\le p, \xi_n=1$,\cr 0,&$p<U<1, \xi_n=0$,\cr 1,&$p<U<1,\xi_n=1$.\cr } $$ Then $P(X_n=1)=p\cdot(1-{1\over n})+(1-p){1\over n}$, so $\lim_n P(X_n=1)=p$. Likewise, $\lim_n P(X_n=0)=1-p$.

You have $$ P(X_n=1 \hbox{ i.o.})=p\cdot P(\xi_n=0\hbox{ i.o.})+(1-p)\cdot P(\xi_n=1\hbox{ i.o.})=p+(1-p)=1, $$ and similarly $P(X_n=0\hbox{ i.o.})=1$.

Finally, let $Y:=1_{\{U\le p\}}$. The $P(X_n\not= Y)\le 2P(\xi_n=1)=2/n\to 0$ as $n\to\infty$. Thus $X_n$ converges in probability to $Y$.

Why do these three conditions imply divergence in probability?

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