# Brownian motion and geometric series convergence

I'm confused with the property of Brownian motion that it is of infinite variation. Consider the sequence $$\sum_{i=0}^{t-1} a^{t-i} (B_{i+1} - B_i)$$ where $$a \in [0,1)$$ and $$B_t$$ is Brownian motion.

Would this sequence converge for $$t \rightarrow \infty$$ like the geometric series? Or does it not converge due to infinite variation of Brownian motion?

• So you are asking about the exponential moving average (EMA) of a sequence of independent identically distributed (i.i.d.) normal variables. No, there will be no convergence. Feb 1 at 10:38
• This doesn't really have anything to do with the infinite variation of Brownian motion. That usually refers to the fact that if we look at partitions $0 = t_1 < t_2 < ... < t_N = T$ of the interval $[0,T]$, then $\sum_{i=1}^N |B_{t_{i+1}}-B_{t_i}| \rightarrow \infty$ as the partition gets finer. That is, the infinite variation property is about adding up the increments over a fixed period of time, rather than over an infinite time frame. Feb 1 at 16:25

## 1 Answer

The variables $$S_m= \sum_{i=0}^{m-1} a^{m-i} (B_{i+1} - B_i)$$ converge in distribution to the law of $$\sum_{j=1}^\infty a^j Z_j \,,$$ where $$Z_j$$ are i.i.d. standard normal variables.

The variables $$S_m$$ do not converge a.s. because $$E[S_m S_{m+k}] \to 0$$ as $$k \to \infty$$ for each $$m$$.

• Are you really sure? The variance of a single sum is $\frac{a^2}{1-a^2}(1-a^{2t})$, so the sum over the variances of the first $N$ sums behaves like $\frac{a^2}{1-a^2}(N-\frac{a^2}{1-a^2}(1-a^{2N}))$ which certainly diverges. Feb 4 at 14:57
• I misread the coefficients. The sum does not converge pointwise, only in distribution. I will correct my answer. Feb 4 at 17:39