3

I just came across the following remark: If $(B_t)_{t\geq0}$ is a one dimensional Brownian motion and if we have a subdivison $0=t_0^n<...<t_{k_n}^n=t$ such that $\sup_{1\leq i\leq k_n}(t_i^n-t_{i-1}^n)$ converges to $0$ is $n$ converges to $\infty$ then $\lim_{n\to\infty}\sum_{i=1}^{k_n}(B_{t_i^n}-B_{t_{i-1}^n})^2=t$ in $L^2$. If $\sum_{n\geq 1}\sum_{i=1}^{k_n} (t_i^n-t_{i-1}^n)^2<\infty$, then we also have almost sure convergence

Now I'm trying to find a subdivison so that $\lim_{n\to\infty}\sum_{i=1}^{k_n}(B_{t_i^n}-B_{t_{i-1}^n})^2$ does not converge almost surely and I'm kinda stucked.

My idea was the following: First note that the subdivision needs to fulfill the following: $\sum_{n\geq 1}\sum_{i=1}^{k_n} (t_i^n-t_{i-1}^n)^2=\infty$. Next I thought about using Borel-Cantelli, which says: If $\sum_{n=1}^{\infty} \mathbb{P}(E_n)=\infty$ and if the events $E_n$ are independent, we have that $\mathbb{P}(\limsup E_n=\infty)=1$.

Now if we can create/formulate events $E_n$ s.t. $\mathbb{P}(E_n)=\sum_{i=1}^{k_n}(t_i^n-t_{i-1}^n)^2$ and such that $\limsup E_n=lim_{n\to\infty}\sum_{i=1}^{k_n} (B_{t_i^n}-B_{t_{i-1}^n})^2$, then Borel-Cantelli would tell us that $\sum_{i=1}^{k_n} (B_{t_i^n}-B_{t_{i-1}^n})^2$ would not converge almost surely.

My problem is that I can't see how in this case $E_n$ needs to be defined. (And I'm not sure whether this is a good approach...)Can anyone help me? Thanks in advance!

Fribar
  • 33
  • There is an error in your statement - the relevant condition for a.s. convergence is $$\sum_{n\geqslant1}\sup_{1\leqslant i\leqslant k_n}\left(t_i^n-t_{i-1}^n\right)<\infty. $$ – Math1000 Jul 09 '16 at 09:38
  • @Math1000 either is a sufficient condition; the one OP posted is proven in Morters and Peres' book – user217285 Jun 30 '18 at 05:57

1 Answers1

0

Consider the sequence of partitions $$\pi(n) = \bigcup_{i=0}^n \left\{\frac in t \right\},\ n\geqslant1 $$ of $[0,t]$, that is, $t_i^n = \frac int$ for $0\leqslant i\leqslant n$. Then $$\sup_{1\leqslant i\leqslant n}\left(t_i^n-t_{i-1}^n\right) = \frac1n\stackrel{n\to\infty}\longrightarrow0, $$ but $$\sum_{n=1}^\infty\frac1n=\infty. $$ From here we may use the second Borel-Cantelli lemma to show that $\sum_{i=1}^n \left(B_{\frac itn}-B_{\frac{i-1}tn}\right)^2$ does not converge to $t$ almost surely. However, this is somewhat of a moot point, considering that $L^2$ convergence implies convergence in probability, which in turn implies a.s. convergence along a subsequence.

Math1000
  • 38,041
  • I guess there is a typo in your answer. Thanks for the hint, but I'm stil stuck: My attempt was: $X_n:=\sum_{i=1}^n \left( B_{\frac{i}{n}t}-B_{\frav{i-1}{n}t}$ then I tried the following: $\mathbb{P}[E_n]=\mathbb{P}((X_n-t)^2 >0)\geq \frac{\mathbb{E}[(X_n-t)^2]^2}{\mathbb{E}[(X_n-t)^4]$ by the second moment methode. And this expression should be bigger than $\frac{1/n}$. But I kinda don't get any useful estimation... – Fribar Jul 10 '16 at 14:22
  • 1
    Ok, I think I found an estimation: We know that $X_n$ is chi-squared distributed and can show that $\mathbb{E}[X_n]=t$. This means that $\mathbb{P} ((X_n-t)^2>0)\geq \frac{\mathbb{E}[(X_n-t)^2]^2}{\mathbb{E}[(X_n-t)^4]}=\frac{\mathbb{E}[(X_n- \mathbb{E} [X_n])^2]^2}{\mathbb{E}[(X_n-\mathbb{E}[X_n])^4}=\frac{1}{Kurt[X_n]}=\frac{n}{12+3n}\geq \frac{1}{12+3n}$. The problem is now, that I also have to show that the events are independent... How do I do so? Or is there a better way of estimation? – Fribar Jul 10 '16 at 16:31