Skip to content
All library documents

Why Brownian Motion’s Cubic Increments Sum to Zero in Probability

Article Quant Q&A · Author: cmd1991
Read the original

Summary

The document shows that the sum of cubed Brownian increments over a partition converges to zero in probability as the partition mesh shrinks. It assumes a partition of a fixed time interval and uses the facts that each increment is centered with zero third moment and has sixth moment equal to fifteen times the cube of its time-step length.

The proof applies Chebyshev’s inequality to the squared sum. Independence of disjoint Brownian increments removes cross terms in expectation, leaving a bound proportional to the sum of cubed time steps. This is at most fifteen times the interval length, multiplied by the squared partition mesh and divided by the squared tolerance, so the probability of exceeding that tolerance vanishes. The result is stated for cubic variation and convergence in probability; the answer does not establish almost-sure convergence or treat arbitrary higher powers in detail.

Key ideas

  • Brownian increments have zero third moment and a sixth moment determined by the cube of the time-step length.
  • Independence makes cross terms vanish when computing the second moment of the cubic-increment sum.
  • Chebyshev’s inequality bounds the probability of a nonzero-sized sum by a term that shrinks with the partition mesh.
  • The conclusion is convergence in probability for the cubic sum, not a claim of almost-sure convergence.

Tags

Statistics
Full text
# Why Variations of order higher than two vanish for Brownian motion?


# Why Variations of order higher than two vanish for Brownian motion?












Let $W_{t}$ be a Brownian Motion. Verify that variations of Brownian Motion of higher order, say, of order three, vanishes. I try to prove that $\lim_{n\rightarrow\infty}\sum^{n}_{i=1}(W_{t_{i}}-W_{t_{i-1}})^{3}=0$ but I have trouble continuing the step. Thank you!

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/17829

I will assume that the convergence is in probability and the partition $\Pi_n$ is given by \begin{align*} 0=t_0 < t_1 < \cdots < t_n = t. \end{align*} Note that $$\mathbb{E}\big((W_{t_i}-W_{t_{i-1}})^3 \big)=0,$$ and $$\mathbb{E}\big((W_{t_i}-W_{t_{i-1}})^6 \big)=15(t_i-t_{i-1})^3.$$ Let $\|\Pi_n\| = \max_{i=1}^n|t_i-t_{i-1}|$. Then, for any small $\delta>0$, by Chebyshev inequality, \begin{align*} \mathbb{P}\Big(\Big|\sum_{i=1}^n (W_{t_i}-W_{t_{i-1}})^3\Big| > \delta \Big)& \leq \frac{1}{\delta^2}\mathbb{E}\bigg(\Big( \sum_{i=1}^n (W_{t_i}-W_{t_{i-1}})^3\Big)^2\bigg)\\ &=\frac{1}{\delta^2}\mathbb{E}\bigg( \sum_{i,j=1}^n (W_{t_i}-W_{t_{i-1}})^3(W_{t_j}-W_{t_{j-1}})^3\bigg)\\ &=\frac{1}{\delta^2}\mathbb{E}\Big( \sum_{i=1}^n (W_{t_i}-W_{t_{i-1}})^6\Big)\\ &=\frac{15}{\delta^2}\sum_{i=1}^n (t_i-t_{i-1})^3\\ &\leq \frac{15\, t}{\delta^2}\|\Pi_n\|^2. \end{align*} That is, $$\lim_{n\rightarrow \infty} \mathbb{P}\Big(\Big|\sum_{i=1}^n (W_{t_i}-W_{t_{i-1}})^3\Big| > \delta \Big) = 0.$$ Therefore, for any $t>0$, $$\lim_{n\rightarrow \infty} \sum_{i=1}^n (W_{t_i}-W_{t_{i-1}})^3 = 0$$ in probability.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.