Let $(X_n)$ be a sequence of IID random variables with


and let $(\mathcal{F}_n)$ be the natural filtration of


Define a sequence of random variables $(S_n)$ by

$$S_0=0 \\ S_n = S_{n-1} + X_n\sqrt{1+S^2_{n-1}}\ \ \ \ \ \mbox{ if } n\geq1 $$

(a) Show that $(S_n)$ is a martingale with respect to filtration.

(b) Show that if $n\geq1$ then $S_n$ does not take the value $0$. Deduce that if $n\geq2$ then $S_n$ does not take values $1$ or $-1$.

(c)Show that for each fixed value of $S_{n-1}$ the two possible values of $S_n$ have a product equal to $-1$

(d)Deduce that for $n\geq2$


I have done part (a) which i answered myself below and I'm stuck at part (b)

Attempted so far

(a) Clearly $S_n$ is adapted and $E|S_n|<\infty$. Since $S_n$ is bounded. For $n\geq1$

$$E[S_n| \mathcal{F}_{n-1}]=S_{n-1}+E[X_n]\sqrt{1+S^2_{n-1}}=S_{n-1}+\frac{1}{2}\sqrt{1+S^2_{n-1}}-\frac{1}{2}\sqrt{1+S^2_{n-1}}=S_{n-1}$$

Hence $S_n$ is a martingale.

(b) Stuck my attempt was to find $S_1$ and $S_2$ but that does not give me $S_n$



  1. is correct
  2. Suppose that $S_n=0$ for some $n \geq 1$. Since $$S_n = 0 \iff -S_{n-1} = X_n \sqrt{1+S_{n-1}^2},$$ this implies in particular $$S_{n-1}^2 = (X_n)^2 (1+S_{n-1}^2).$$ Show that this in turn implies $1=0$ - which is obviously a contradiction. To prove the second part of (b), note that $$S_n = 1 \iff 1-S_{n-1} = X_n \sqrt{1+S_{n-1}^2}.$$ Again this implies $$(1-S_{n-1})^2 = (X_n)^2 (1+S_{n-1}^2).$$ Show that this is equivalent to $S_{n-1}=0$ and deduce the claim from the first part of (b).
  3. Calculate $$(S_{n-1}+ \sqrt{1+S_{n-1}^2})(S_{n-1}-\sqrt{1+S_{n-1}^2}).$$
  4. For any two numbers $x,y \in \mathbb{R} \backslash \{1,-1\}$ such that $x \cdot y=-1$ it holds either $|x| >1, |y|<1$ or $|x|<1, |y|>1$. By part (b),(c), this means that exactly one of the two possible values $S_{n-1}+ \sqrt{1+S_{n-1}^2}$ and $S_{n-1}-\sqrt{1+S_{n-1}^2}$ has absolute value $>1$ (and the other one absolute value $<1$). Since both events have the same probability $\frac{1}{2}$, we get $$\mathbb{P}(|S_n|<1) = \frac{1}{2}.$$

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