$\forall p>0$ we get $\displaystyle \lim_{n\to \infty} n^{a_{p}}\sum _{i=0}^{n}|B_{\frac{i+1}{n}}-B_{\frac{i}{n}}|^{p}=c_{p}$

Question

This is a Homework question, so please do not answer it.

Find real constants $a_{p},c_{p}$ s.t. $\forall p>0$ we get $\displaystyle \lim_{n\to \infty} n^{a_{p}}\sum _{i=0}^{n}|B_{\frac{i+1}{n}}-B_{\frac{i}{n}}|^{p}=c_{p}$ a.s, where $B_{t}$ is a Brownian motion.

Did I make a mistake somewhere? :

$\displaystyle \lim_{n\to \infty} n^{a_{p}}\sum _{i=0}^{n}|B_{\frac{i+1}{n}}-B_{\frac{i}{n}}|^{p}\stackrel{shift~ invariance}{=}\displaystyle \lim_{n\to \infty} n^{a_{p}}\sum _{i=0}^{n}|X_{\frac{1}{n}}|^{p}=\displaystyle \lim_{n\to \infty} n^{a_{p}+1}|X_{\frac{1}{n}}|^{p}\stackrel{time~ inversion}{=}\displaystyle \lim_{n\to \infty} n^{a_{p}+1}|n^{-1}Y_{n}|^{p}$.

To use Law of large numbers we let $a_{p}=-1$

$\displaystyle \lim_{n\to \infty} |n^{-1}Y_{n}|^{p}=e^{p \displaystyle \lim_{x\to 0} log(x)}=0$

The reason I think I made a mistake is that here $a_{p},c_{p}$ don't depend on p.

Answer 1

The problem is that you work with equalities in distribution, and that $\sum_{i=1}^n|B_{(i+1)/n}-B_{i/n}|^p$ is not even equal in distribution to $n|X_{1/n}|^p$. It is rather equal in distrubution to $\sum_{i=1}^nY_{i,n}$, where $(Y_{n,i})_{i=1}^n$ are i.i.d. and $Y_{n,i}$ is normally distributed, with mean $0$ and variance $1/n$.

We can use the following strategy: first identify the limit in distribution. This will give the good constants $a_p$ and $c_p$. Then we show that the convergence is almost sure.