Martingale Rademacher exercise Let $(a_n)_{n\in\mathbb{N}}$ be a bounded, real valued sequence and let $Y_n$ be i.i.d. random variables with $$\mathbb{P}(Y_n=1)=\mathbb{P}(Y_n=-1)=1/2$$ Define $X_n=\sum_{k=0}^nY_ka_k$ and $\mathcal{F}_n=\sigma\{X_0,\dots,X_n\}$.


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*Show that $X_n$ converges in $L^2$ if and only if $\sum_{k=0}^\infty a_k^2<\infty.$

*Compute $\displaystyle\left<X \right>_n:= \sum_{k=0}^{n-1}\mathbb{E}\left(X_{k+1}^2-X_k^2\mid \mathcal{F}_k\right).$

*Let $\sum_{k=0}^\infty a_k^2 = \infty$ and define $T_c=\inf{\{n\geq0:|X_n|\geq c \}}$ for $c>0.$ Show that $\mathbb{P}(T_c=\infty)=0$. Hint: Consider the process $X_n^2-\left<X\right>_n$.

*Let again $\sum_{k=0}^\infty a_k=\infty$. Show that $\mathbb{P}\left(\lim\sup_nX_n=\infty = -\lim\inf_nX_n\right)=1$, i.e. $X_n$ oscillates between $\infty$ and $-\infty$ almost surely.
I was able to do the first and the second one. For the first I used a property of Cauchy sequences and for the second I was able to show that the value of the sequence must be constant. However I have troubles solving third and fourth problems. On 3. I tried to apply the doob-decomposition, but failed. I guess I should apply the stopping theorem and maybe borel cantelli but I have no idea how to get there.
 A: 3.For brevity of notation set $T:=T_c$. According to my calculations we have
$$\langle X \rangle_n = \sum_{k=0}^{n-1} a_k^2.$$
Since $X_n^2 - \langle X \rangle_n$ is a martingale, the optional stopping theorem gives
$$\mathbb{E}(X_{T \wedge n}^2) = \mathbb{E}(\langle X \rangle_{T \wedge n}) = \mathbb{E} \left( \sum_{k=0}^{(T \wedge n)-1} a_k^2 \right).$$
Using that $|X_{T \wedge n}| \leq M:= c+\sup_{k \in \mathbb{N}} |a_k| < \infty$, we get
$$\mathbb{P}(T=\infty) \sum_{k=0}^{n-1} a_k^2 \leq \mathbb{E} \left( \sum_{k=0}^{(T \wedge n)-1} a_k^2 \right) \leq M^2.$$
Use this to conlude that $\mathbb{P}(T=\infty)=0$.

4.Apply your favourite 0-1 law to show that
$$p := \mathbb{P}(\limsup_{n \to \infty} X_n = \infty) \in \{0,1\}.$$
Since the sequence $(-X_n)_n$ equals in distribution $(X_n)_n$, we have
$$p = \mathbb{P}(\liminf_{n \to \infty} X_n = - \infty).$$
On the other hand, part 3 yields readily
$$\mathbb{P} \left( \sup_{n \in \mathbb{N}} |X_n| = \infty \right)=1.$$
Combining the 3 considerations gives $p=1$. 
