in two ways the quantity $\E \left | M_{\tau \wedge S_x} \right |^p$. First, since $\{ M_{\tau \wedge S_x \wedge n} : n \geq 1 \}$ is a martingale, $|x|^p$ is convex, and $\tau \wedge S_x \geq 1$ is a stopping time bounded above by $x$, we have $$\label{eq:upside}
\E \left | M_{\tau \wedge S_x} \right |^p \leq \E \left | M_x \right |^p .$$ Using Lemma \[lem:Lp\] gives $\E |M_x - M_1|^p \leq c_p b x^{p/2}$, and since $\E |M_1|^p \leq b$, this yields $$\label{eq:upside2}
\E |M_x|^p = ||M_x||_p^p \leq (||M_1||_p + ||M_x - M_1||_p)^p 
   \leq \left ( b^{1/p} + (c_p b x^{p/2})^{1/p} \right )^p 
   \leq c'(p,b) x^{p/2}$$ with $c'(p,b) := c_p b (1 + c_p^{-1})^p$. On the other hand, on the event $\{\tau > S_x\}$ we have $M_{\tau \wedge S_x}
= M_{S_x} \ge x$, so that $$x^p \P (\tau > S_x) \leq \E |M_{\tau \wedge S_x}|^p ,$$ and combining this with (\[eq:upside\]) and (\[eq:upside2\]) gives $$\P (\tau > S_x) \leq x^{-p} c'(p,b) x^{p/2}$$ which proves the result for $x \ge 1$. Finally, for $x < 1$ we use $\P( \tau > S_x ) \le 1$, so the lemma follows with $C'(p,b) := \max(1, \, c'(p,b))$. $\Cox$

[Proof of 