constructing a zero-mean martingale $\to -\infty$ a.s. I'm reading a book over martingale theory and there is a exercise where I'm interested in a solution.
Let $X_{1},X_{2},...$ be independent random variables with
\begin{align}
\mathbb{P}(X_{n} = -1) &= 1 - 2^{-n}\\
\mathbb{P}(X_{n} = 2^{n} - 1) &= \frac{1}{2^{n}}
\end{align}
Now the task is to construct with thiss sequence a martingale $Y_{n}$ with $\mathbb{E}(Y_{n}) = 0$ and $\lim\limits_{n \to \infty} Y_{n} = -\infty$ alsmost surely.
 A: Sum martingale $$Y_n = \sum_{i=1}^{n} X_i$$
Check:


*

*$$E[Y_n]=E[X_n]=0$$

*$$\lim Y_n = -\infty \ a.s. \ \because \lim X_n = -1 \ a.s. \ (\text{Check with Borel-Cantelli Lemmas!}) (*) $$

*$Y_n$ is integrable, adapted to $\mathscr F_n = \mathscr F_n^X := \sigma(X_1, X_2, ..., X_n)$ and $\forall m < n,$ $$E[Y_n | \mathscr F_m]=E[\sum_{i=1}^{n} X_i | \mathscr F_m]=\sum_{i=1}^{m} X_i + E[\sum_{i=m+1}^{n} X_i | \mathscr F_m]=\sum_{i=1}^{m} X_i + E[\sum_{i=m+1}^{n} X_i]=\sum_{i=1}^{m} X_i = Y_m$$

(*)
Pf: By BCL1 & BCL2, we have
$$P(\liminf (X_n = -1))=1$$
$$P(\limsup (X_n = -1))=1$$
The latter is of course redundant. The former by itself gives us only, for now. $P(\lim A_n) = \lim P(A_n) = 1$ where $A_n = (X_n = -1)$. Now, by contrapositive of important inequalities (ii) and (iv) (**), we have, resp,
$$\limsup X_n \le -1$$
$$\liminf X_n \ge -1$$
QED

(**)
Important inequalities
Williams - Probability with Martingales



Deduced similarly:
(iii) If $\liminf x_n > z$, then
$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)
(iv) If $\liminf x_n < z $, then
$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)
