I'm working through example 1.4 on page 3 of http://people.maths.ox.ac.uk/xu/Martingale_convergence.pdf
I'm having trouble showing that $X_n = 2^n 1_{[0,1/2^n)}$ is a martingale. I'm not sure why my constants are wrong.
So $E[X_{n+1}| {\cal{F}}_n] = 2^{n+1}E[1_{[0,1/2^{n+1})}|{\cal{F}}_n]$
and now $1_{[0,1/2^{n+1})} = 1_{[0,1/2^n)}1_{[1/2^{n+1},1/2^n)^c}$.
By using the properties of conditional expectation, I only have to evaluate $E[1_{[1/2^{n+1},1/2^n)^c} | {\cal{F}}_n] = E[1_{[1/2^{n+1},1/2^n)^c}] = 1-\frac{1}{2^{n+1}}$.
But my constants are now messed up. Any idea where I went wrong?
I dont really see what you're trying to do, but it really helps to draw a picture of which sets are in $\mathcal{F_n}$.
Then you can see: \begin{eqnarray}\mathbb{E}[X_n|\mathcal{F_{n-1}}]\\ =\mathbb{E}[2^n \mathbb{1}_{[0,\frac{1}{2^n})}|\mathcal{F_{n-1}}]\\ =2^n\mathbb{E}[ \mathbb{1}_{[0,\frac{1}{2^n})}|\mathcal{F_{n-1}}]\\ (1) =\frac{1}{2} 2^n\mathbb{E}[ \mathbb{1}_{[0,\frac{1}{2^{n-1}})}|\mathcal{F_{n-1}}]\\ (2) =2^{n-1} \mathbb{1}_{[0,\frac{1}{2^{n-1}})}\\ =X_{n-1} \end{eqnarray}
Where by $(1)$ we used that $\frac{1}{2} \mathbb{E}[ \mathbb{1}_{[0,\frac{1}{2^{n-1}})}]= \mathbb{E}[ \mathbb{1}_{[0,\frac{1}{2^{n}})}]$, on all sets in $\mathcal{F_{n-1}}$ (draw a picture to see that there is only one non-zero). And by $(2)$ we used the measurability of $\mathbb{1}_{[0,\frac{1}{2^{n-1}})}$ in $\mathcal{F_{n-1}}$.