Attempting to show $P(|S_n| <1)$ for a martingale $(S_n)$

Question

Now, I am stuck on the last part of the question. I managed to find the solutions, but I don't udnerstand them completely.

What I don't understand is: How they got that indicator function, and why they are using double expectation to calculate the probability. I'd appreciate it if someone coudl help out, thanks.

Answer 1

First, we can show the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\sqrt{1+x^2}$ is such that:

1. $\forall x,f(x)\geq 1$. Also $f(x)=1\iff x=0$.
2. $x< 0 \implies 0<x+f(x)<1$. Similarly $x>0\implies -1<x-f(x)<0$

Second, to prove $$ 1_{\{|S_n|<1\}}=\frac{-X_nsign(S_{n-1})+1}{2} $$ note this: \begin{eqnarray} \frac{-X_nsign(S_{n-1})+1}{2} =1 & \iff & -X_nsign(S_{n-1})=1\\ & \iff & (X_n=-1\:\wedge\:S_{n-1}>0)\vee(X_n=1\:\wedge\:S_{n-1}<0)\\ & \implies & |S_n|<1 \end{eqnarray}

(the last line is consequence of the second property of $f(x)$ previously mentioned). On the other hand \begin{eqnarray} \frac{-X_nsign(S_{n-1})+1}{2} =0 & \iff & -X_nsign(S_{n-1})=-1\\ & \iff & (X_n=-1\:\wedge\:S_{n-1}<0)\vee(X_n=1\:\wedge\:S_{n-1}>0)\\ & \implies & |S_n|>1 \end{eqnarray}

(the last line is consequence of the first property of $f(x)$ previously mentioned).

Third, by the previous point $$ \mathbb{P}[|S_n|<1]=\mathbb{E}[1_{\{|S_n|<1\}}]=\mathbb{E}[(-X_nsign(S_{n-1})+1)/2] $$ in other words, computing the desired probability is the same as computing $ \mathbb{E}[(-X_nsign(S_{n-1})+1)/2] $. Computing this directly looks pretty tedius (you'd need to compute the distribution of $X_nsign(S_{n-1})$, etc). Luckily, using the Law of total expectation, the conditional expectation $$ Y_{n-1}:=\mathbb{E}[(-X_nsign(S_{n-1})+1)/2|\mathcal{F}_{n-1}] $$ verifies: $$ \mathbb{E}[(-X_nsign(S_{n-1})+1)/2]=\mathbb{E}[\mathbb{E}[(-X_nsign(S_{n-1})+1)/2|\mathcal{F}_{n-1}]=\mathbb{E}[Y_{n-1}] $$

Hence $$ \mathbb{P}[|S_n|<1]=\mathbb{E}[\mathbb{E}[(-X_nsign(S_{n-1})+1)/2|\mathcal{F}_{n-1}] $$ and since $S_{n-1}\in \mathcal{F}_{n-1}$ and $X_n$ is independent of $\mathcal{F}_{n-1}=\sigma(X_1,...,X_{n-1})$, by properties of conditional expectation we deduce $$ Y_{n-1}=\mathbb{E}[(-X_nsign(S_{n-1})+1)/2|\mathcal{F}_{n-1}]=\frac{\mathbb{E}[X_n]sign(S_{n-1})+1}{2}=\frac{1}{2} $$ hence $$ \mathbb{E}[(-X_nsign(S_{n-1})+1)/2]=\mathbb{E}[Y_{n-1}]=1/2 $$