exercise 1.21 of chapter 1 of Revuz and Yor's

Question

This is the exercise 1.21 of chapter 1 of Revuz and Yor's:

Let $X=B^+$ or $|B|$ where $B$ is the standard linear BM, $p$ be a real number $>1$ and $q$ its conjugate number ($q^{-1}+p^{-1}=1$).

1. Prove that the r.v. $J_p=\sup_{t\geq0}(X_t-t^{p/2})$ is a.s. strictly positive and finite and has the same law as $\sup_{t\geq0}(X_t/(1+t^{p/2}))^q$.

2. Using time-inversion, show that $$\sup_{t\geq0}(X_t/(1+t^{p/2}))\overset{(d)}{=}\sup_{u\leq1}\left(\frac{1}{1+u^{p/2}}\right)\left(\frac{X_u}{u^{1-p/2}}\right)$$ and conclude that $E[J_p]<\infty$.

3. Prove that there exists a constant $C_p(X)$ such that for any positive r.v. $L$ $$E[X_L]\leq C_p(X)||L^{1/2}||_p.$$

4. Let $L_\mu$ be a random time such that $$X_{L_\mu}-\mu L_{\mu}^{p/2}=\sup_{t\geq0}(X_t-\mu t^{p/2}).$$ Prove that $L_\mu$ is a.s. unique and that the constant $C_p(X)=p^{1/p}(qE[J_p])^{1/q}$ is the best possible.

5. Prove that $$E[X_{L_1}|J_p]=qJ_p \text{, }E\left[L_1^{p/2}|J_p\right]=\frac{q}{p}J_p.$$

By the hints and the reference cited in the notes of the book, I almost solved this question. My only question is : In the proof of 4, I think I need $EL_\mu^{p/2}<+\infty$ (or $EX_{L_\mu}<+\infty$) firstly, but can't figure it out. Any help, thanks!

Answer 1

Indeed, one needs to verify that $E[L^{p/2}_\mu]<+\infty.$ Consider the inequality $$ E[L^{p/2}_\mu]\leq 1+\sum^\infty_{n=0}2^{(n+1)p/2}P(2^n<L_\mu\leq 2^{n+1}). $$ To bound the probability $P(2^n<L_\mu\leq 2^{n+1})$ few crude estimates are enough. The event $\{2^n<L_\mu\leq 2^{n+1}\}$ implies that the supremum of the function $t\mapsto X_t-\mu t^{p/2}$ is attained on $(2^n,2^{n+1}].$ Since this supremum is positive, it follows that for some $t\in (2^n,2^{n+1}]$ $$ X_t>\mu t^{p/2}. $$ Further it follows that $X_t>\mu 2^{np/2}$ and, finally, $$ \sup_{0\leq t\leq 2^{n+1}}|B_t|>\mu 2^{np/2}. $$ Consequently, $$ P(2^n<L_\mu\leq 2^{n+1})\leq P\bigg(\sup_{0\leq t\leq 2^{n+1}}|B_t|>\mu 2^{np/2}\bigg)\leq 4P\bigg(B_1>\frac{\mu}{\sqrt{2}}2^{(p-1)\frac{n}{2}}\bigg). $$ Using tail estimates for the normal distribution it is easy to verify that the following series converges: $$ \sum^\infty_{n=0}2^{\frac{pn}{2}}P\bigg(B_1>\frac{\mu}{\sqrt{2}}2^{(p-1)\frac{n}{2}}\bigg)<\infty. $$ These estimates show that $L_\mu$ actually has all moments.