\leq b y^{-p}$, hence the first of these contributions is at most $$3^r ({3t \over 4})^r b ({t \over 4})^{-p} .$$ Using (\[eq:parts\]) again, we bound the second of the three contributions by $$3^r ({t \over 4})^r \P (\DD_k \geq {t \over 4} \| \F_k) 
   + 3^r \int_{t/4}^\infty r y^{r-1} \P (\DD_k \geq y) \, dy$$ which is at most $$3^r b ({t \over 4})^{r-p} + 3^r {r \over p-r} ({t \over 4})^{r-p} .$$ Lemma \[lem:Lp\] implies $\E (|M_t - M_{k+1}|^r \| \F_{k+1}) \leq 
c_p b t^{r/2}$, while $\one_{T=k} \in \F_{k+1}$ and has conditional expectation at most $b (t/4)^{-p}$ given $\F_k$. Therefore the third contribution is bounded by $$3^r b ({t \over 4})^{-p} c_p b t^{r/2} .$$ Summing the three contributions gives $$\label{eq:ub3}
\E ((M_t^+)^r \one_{T = k , M_t > t} \| \F_k) \leq c_3 t^{r-p}$$ where $c_3 = 3^r 4^p b (c_p b + {p \over p-r} + 3^r)$.

Now we bound the second term, by decomposing according to the value of $T$. $$\begin{aligned}
\E \left( (M_t^+)^r \one_H \right) &
= & \sum_{k=0}^{\lfloor t/2 \rfloor} 
\E \left( (M_t^+)^r \one_H \one_{T = k} \right)
   \nonumber \\
& = & \sum_{k=0}^{\lfloor t/2 \rfloor} \E \left [ \E ((M_t^+)^r \one_H \one_{T 