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Sorry, I just had to correct this typo in the title
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Mateusz Kwaśnicki
Mateusz Kwaśnicki
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Moments of the Hölder norm of brownianBrownian process

typo + top-level tag
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Mateusz Kwaśnicki
Mateusz Kwaśnicki
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It is well known that for a brownian process $B(t),t\geq 0$, it holds $$ \sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)}{|t-s|^\alpha}<\infty $$$$ \sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}<\infty $$ almost surely, for any $T>0$ and $\alpha<1/2$.

My question is: how about $$ \mathbb{E}\left[\left(\sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}\right)^p\right] $$ for $p\geq 1$ ?

It is well known that for a brownian process $B(t),t\geq 0$, it holds $$ \sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)}{|t-s|^\alpha}<\infty $$ almost surely, for any $T>0$ and $\alpha<1/2$.

My question is: how about $$ \mathbb{E}\left[\left(\sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}\right)^p\right] $$ for $p\geq 1$ ?

It is well known that for a brownian process $B(t),t\geq 0$, it holds $$ \sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}<\infty $$ almost surely, for any $T>0$ and $\alpha<1/2$.

My question is: how about $$ \mathbb{E}\left[\left(\sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}\right)^p\right] $$ for $p\geq 1$ ?

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LUIGI M
LUIGI M
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Moments of the Hölder norm of brownian process

It is well known that for a brownian process $B(t),t\geq 0$, it holds $$ \sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)}{|t-s|^\alpha}<\infty $$ almost surely, for any $T>0$ and $\alpha<1/2$.

My question is: how about $$ \mathbb{E}\left[\left(\sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}\right)^p\right] $$ for $p\geq 1$ ?