Skip to main content
11 events
when toggle format what by license comment
Sep 25 at 20:11 vote accept user2103480
Sep 22 at 23:07 comment added Thomas Kojar @user2103480 I see. Yes, for your particular bound you indeed need a slightly different $A_t$. So we need to replace the triangle inequality by the convexity inequality $$|a+b|^r<2^{r-1}(|a|^r+|b|^r), r\geq 1.$$
Sep 22 at 22:56 comment added user2103480 @ThomasKojar I didn't mean to doubt the validity for general $p$. However, as it stands, Theorem 4.10 in Revuz-Yor would yield an upper bound of $\Bbb E[\tau^{p/2}]$ instead of $\Bbb E[\tau^{p(1/2-\alpha)}]$ for $A_T$ as defined in your answer
Sep 22 at 21:03 vote accept user2103480
Sep 22 at 21:03
Sep 22 at 21:02 comment added user2103480 Dear Thomas, thank you for the helpful reference and elucidating comments. I think you have a small typo here though! I think it should be $$A_T = \left(\sup_{0 \leq s < t \leq T} \frac{|B_t-B_s|}{|t-s|^\alpha}\right)^{\frac{1}{2(1/2-\alpha)}}?$$ For $F(x) = x^{2p(1/2-\alpha)}$, the desired inequality then follows (for each $p$)
Sep 21 at 18:07 history edited Thomas Kojar CC BY-SA 4.0
added 233 characters in body
Sep 21 at 17:58 history edited Daniele Tampieri CC BY-SA 4.0
Minor formatting
Sep 21 at 17:50 comment added Thomas Kojar @JohnDawkins I should have added an inequality, not equality. I fixed it, thank you.
Sep 21 at 17:50 history edited Thomas Kojar CC BY-SA 4.0
added 3 characters in body
Sep 21 at 17:09 comment added John Dawkins Could you supply a little more detail in the first step of your claim that $A_{T+S} -A_S =A_T\circ\theta_S$? Clearly $$ A_{T+S} = \max\left( A_S, \sup_{0\le s< S, S\le t\le T+S}{|B_t-B_s|\over |t-s|^\alpha}, A_{T}\circ\theta_S\right), $$ but after that ...?
Sep 20 at 0:21 history answered Thomas Kojar CC BY-SA 4.0