Let $((0,1], \text{Borel}_{(0,1]}, \mu)$ be a measure space, where $\mu$ is the Lebesgue measure. I want to prove that $\{X_n\}_{n\geq 0}$ is a martingale with respect to the filtration $\mathcal{F}_n=\sigma(X_1,\ldots,X_n)$, where $X_n=2^n\textbf{1}_{(0,2^{-n}]}$. I only need help to prove the third property of martingales, that is, $\mathbb{E}(X_n\mid\mathcal{F}_{n-1})=X_{n-1}$ $\forall n\geq1$.

So far this is what I've tried:

\begin{align} \mathbb{E}(X_n\mid\mathcal{F}_{n-1})&= \mathbb{E}(2^n\textbf{1}_{(0,2^{-n}]}\mid\mathcal{F}_{n-1}) \\&=2\mathbb{E}(2^{n-1}\textbf{1}_{(0,2^{-n}]}\mid\mathcal{F}_{n-1}) \\&=2\mathbb{E}[2^{n-1}(\textbf{1}_{(0,2^{-n+1}]}-\textbf{1}_{(2^{-n},2^{-n+1}]})\mid\mathcal{F}_{n-1}] \\&=2X_{n-1}-\mathbb{E}[2^n\textbf{1}_{(2^{-n},2^{-n+1}]}\mid\mathcal{F}_{n-1}] \end{align}

If I were working with normal expectantions instead of conditional ones, I could write: $$\mathbb{E}[2^n\textbf{1}_{(2^{-n},2^{-n+1}]}\mid\mathcal{F}_{n-1}]=\mathbb{E}[2^n\textbf{1}_{(0,2^{-n}]}\mid\mathcal{F}_{n-1}]$$ because $(0, 2^{-n}]$ and $(2^{-n},2^{-n+1}]$ have the same probability under Lebesgue measure, and then my problem would be solved because I would get: $$\mathbb{E}(X_n\mid\mathcal{F}_{n-1})=2X_{n-1}-\mathbb{E}(X_n\mid\mathcal{F}_{n-1}) \Longrightarrow \mathbb{E}(X_n\mid\mathcal{F}_{n-1})=X_{n-1}$$ However I don't think I can use this technique here, because:

$$\exists A\in \mathcal{F}_{n-1} \quad \mbox{so that} \quad \int_{A}\textbf{1}_{(0,2^{-n}]}d\mu \neq \int_{A}\textbf{1}_{(2^{-n}, 2^{-n+1}]}d\mu$$ And therefore the conditional expectations are not the same.

Can someone please help me?


1 Answer 1


Since $X_n$ is $ \newcommand{\F}{\mathcal{F}}\F_n$-measurable you only need to prove that $$ \newcommand{\E}{\mathbf{E}} \forall A\in\F_n,\quad\E[X_{n+1}1_A] \overset{(\star)}= \E[X_{n}1_A].$$ Furthermore, every measurable $A\in\F_n$ is a disjoint union of intervals from the set $$\Big\{\big(0,2^{-n}\big], ~\big(2^{-n}, 2^{-n+1}\big], ~\dots~,~ \big(1/4,1/2\big],~ \big(1/2, 1\big]\Big\}$$ and so you only need to prove $(\star)$ for $A$'s chosen from this set of intervals. All equalities are clear (they are all $0=0$ except the first one which is $1=1$) so that $X_n=\E[X_{n+1}\mid\F_{n}]$


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