Define a martingale $\{M_n\}$ with respect to filtration $\{\mathcal{F}\}_n$ such that $M_n = P(A|\mathcal{F}_n)$ for $n\geq 0$. Assume that $P(A) = 1/2$, $A\in \mathcal{F}_\infty$ and $A$ is independent of $\mathcal{F}_0$. How to show \begin{equation} \frac{3}{4}\leq E[\sup_{n\geq 0} M_n]\leq \frac{1+\log 2}{2}? \end{equation}
I have already shown that \begin{equation} P\left(\sup_{n\geq 0} M_n\geq m\right)\leq \frac{1}{2m}. \end{equation} It seems that the upper bound can be obtained by doing integral. But, how about the lower bound?
Thanks!
For the lower bound notice that on $A^{c}$ $\sup_n M_n \geq 1/2$; and on $A$, by Lévy's zero–one law, $\sup_n M_n =1$. Therefore, $$E (\sup_n M_n) \geq \frac{1}{2}\cdot \frac{1}{2} +\frac{1}{2}\cdot 1=\frac{3}{4}$$
From Doob's maximal inequality $ \mathbb{P}(\sup_n M_n \geq t )\leq \min(1,\frac{1}{2t})$ for all $t \in [0,1]$. Therefore, \begin{align} E (\sup_n M_n) &\leq \int_{0}^{1}\min(1,\frac{1}{2t})\mathop{dt} \\ &=\frac{1+\ln (2)}{2} \end{align}