Almost sure convergence of sum

Question

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.

Answer 1

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.