I am now working on R. Durrett's Probability Theory and Examples.
In his book, I am asked to construct a martingale $(X_n)$ satisfying the following three conditions.
(1) $P(X_n=a$ i.o.$)=1, a=-1,0,1$
(2) $\sup_n|X_n|<\infty$
(3) For some preassigned $p \in (0,1/2)$, $P(X_n=1),P(X_n=-1)\rightarrow p, P(X_n=0) \rightarrow 1-2p$
So that this martingale converges in distribution but not a.s. or in probability.
I know there already exists an answered question regarding (1)&(2), but so far I haven't found any identical question. So I guess this question is not a duplicate.
I have been spending 2 days and could not come up with an example. Especially, condition (3) is hard to manage since I cannot make use of the similar trick used in the answer of Martingale oscillating between three values
Any hint would be appreciated! Thanks and regards.
Below are the details for the nice example outlined in D. Thomine's comment:
Let $(Y_n)$ be independent r.v. with values in $\{0, 1\}$ and with $\mathbb{P}(Y_n = 0) = 1-1/n$. Also, independently from the $(Y_n)$, let $(Z_n)$ be i.i.d. r.v. with values in $\{-1, 0, 1\}$ according to the limit distribution. Starting from $X_0 = 0$, we iteratively define $X_{n}$ by the following transition rule:
\begin{array}{|c|c|c|} \hline & Y_n = 0 & Y_n = 1 \\ \hline |X_{n-1}| \leq 1 & X_{n-1} & X_{n-1} \pm 3 \\ \hline |X_{n-1}| > 1 & Z_n & nX_{n-1} - (n-1)Z_n \\ \hline \end{array}
(Here $\pm 3$ means with equal chance of adding or subtracting 3, independently from everything else.) It's easy to verify this is a martingale. Also, observe that $|X_n| > 1$ if and only if $Y_n = 1$.
To verify condition (2), notice that $|X_n| > 4$ only if $Y_{n-1} = Y_n = 1$. It follows from the convergence of $\sum \frac{1}{n^2}$ and the Borel-Cantelli Lemma that this a.s. will not happen infinitely often.
To verify condition (1), notice that for $a = -1, 0, 1$, $X_n = a$ if $Y_{n-1} = 1, Z_n = a$, provided $Y_n \neq 1$. If follows from the divergence of $\sum \frac{1}{n}$, the 2nd Borel-Cantelli Lemma, and the previous paragraph that this a.s. will happen infinitely often.
Finally, to verify condition (3), we just compute the distributions of $X_n$. Notice that $X_n$ are integer-valued by construction, and $P(|X_n| > 1) = P(Y_n = 1) = 1-1/n \rightarrow 0$. It remains to show $P(X_n = a) \rightarrow p_a$ for $a = -1, 0, 1$, where $p_a$ are pre-assigned. Using the transition rule, we have \begin{align} P(X_n = a) &= P(Y_n = 0) [P(X_{n-1} = a) + P(|X_{n-1}| > 1) P(Z_n = a)] \\ &= (1-1/n) (P(X_{n-1} = a) + p_a/(n-1)). \end{align} Define $p_n = \frac{P(X_n = a)}{1-1/n}$, this becomes \begin{align} p_n &= [1-1/(n-1)] p_{n-1} + p_a/(n-1) \\ p_n - p_a &= [1-1/(n-1)] (p_{n-1} - p_a) \\ &= \prod_{k=1}^n [1-1/(k-1)] (p_0 - p_a) \\ &\rightarrow 0. \end{align} Thus, $P(X_n = a) = (1-1/n) p_n \rightarrow p_a$, and we are done showing condition (3).