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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?

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    $\begingroup$ 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. $\endgroup$
    – Lutz Lehmann
    Feb 1 at 10:38
  • $\begingroup$ 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. $\endgroup$
    – user6247850
    Feb 1 at 16:25

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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 $m$.

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  • $\begingroup$ 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. $\endgroup$
    – Lutz Lehmann
    Feb 4 at 14:57
  • $\begingroup$ I misread the coefficients. The sum does not converge pointwise, only in distribution. I will correct my answer. $\endgroup$
    – Yuval Peres
    Feb 4 at 17:39

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