I am studying variation of Brownian motion, and in this context I need to prove that $$S_n=\sum_{k=1}^{[2^nt]}\frac{1}{2^n}Y^2_{k,n}-t$$ converges almost surely to $0$. $Y_{k,n}$ is standard normally distributed for all $k$ and $n$, so my idea was show that $$\sum^{\infty}_{n=1}P(|S_n|>\varepsilon)<\infty$$ by using $$P(|S_n|> \varepsilon)\leq \varepsilon^{-1}E|S_n|\geq \varepsilon^{-1}\left|\frac{[2^nt]}{2^n}-t\right|$$ and somehow show that $$\sum^{\infty}_{n=1}\left|\frac{[2^nt]}{2^n}-t\right|< \infty.$$ Is that the way to go? Thanks a bunch.

  • $\begingroup$ How do you compute $\mathbb E|S_n|$? $\endgroup$ – Davide Giraudo Jan 11 '16 at 23:01
  • $\begingroup$ $$ES_n=\sum_{k=1}^{[2^nt]}\frac{1}{2^n}EY^2_{k,n}-t=\sum_{k=1}^{[2^nt]}\frac{1}{2^n}-t=\frac{[2^nt]}{2^n}-t$$ seeing as each $Y_{k,n}$ is $\mathcal{N}(0,1)$ distributed and $EY^2_{k,n}=1$. However it does seem like I forgot about Jensens inequality $\endgroup$ – user128836 Jan 12 '16 at 23:22
  • $\begingroup$ Question is corrected now $\endgroup$ – user128836 Jan 12 '16 at 23:33
  • $\begingroup$ What you wrote in last term of the sixth line is $|\mathbb ES_n|$, not $\mathbb E|S_n|$. However, you can fix it by evaluating $\mathbb E[S_n^2]$ using independence of increments of Brownian motion. $\endgroup$ – Davide Giraudo Jan 13 '16 at 15:07
  • $\begingroup$ Yes I realise that was wrong. I get $ES_n^2=Var(S_n)+(ES_n)^2$ where $$Var(S_n)=Var\left(\sum_{k=1}^{[2^nt]}\frac{1}{2^n}Y^2_{k,n}-t\right)=\frac{1}{4^n} \sum_{k=1}^{[2^nt]}VarY^2_{k,n}=\frac{2[2^nt]}{4^n}$$ $$(ES_n)^2=\left(\frac{[2^nt]}{2^n}-t\right)^2=\frac{[2^nt]^2}{4^n}+t^2-2t\frac{[2^nt]}{2^n}$$ How do you use that? Thanks $\endgroup$ – user128836 Jan 15 '16 at 15:40

Note that $$S_n=\sum_{k=1}^{\left[2^nt\right]}2^{-n}\left(Y_{k,n}^2-1\right)+2^{-n}\left[2^nt\right].$$ Since $\left[2^nt\right]\leqslant 2^nt\lt \left[2^nt\right]+1$, we have $2^{-n}\left[2^nt\right]\leqslant t\lt 2^{-n}\left[2^nt\right]+2^{-n}$, which gives $$2^{-n}\lt 2^{-n}\left[2^nt\right]-t\leqslant 0.$$ It thus suffices to show that $T_n:=\sum_{k=1}^{\left[2^nt\right]}2^{-n}\left(Y_{k,n}^2-1\right)\to 0$ almost surely. Since each random variable in the sum is centered, and the terms are independent, we have $$\mathbb E\left[T_n^2\right]=\sum_{k=1}^{\left[2^nt\right]}2^{-2n}\mathbb E\left[\left(N^2-1\right)^2\right]\leqslant \mathbb E\left[\left(N^2-1\right)^2\right]2^{-2n}2^nt,$$ where $N$ has a standard normal distribution. We derive $\sum_{n\geqslant 1}\mathbb E\left[T_n^2\right]<\infty$, hence by the Borel-Cantelli lemma, that $T_n\to 0$ almost surely.

  • 1
    $\begingroup$ Thats a neat trick, thanks a lot! $\endgroup$ – user128836 Jan 16 '16 at 20:36

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