Why Higher-Order Brownian Increment Sums Vanish
Summary
The document explains why sums of powers of Brownian motion increments of order three or higher converge to zero as partitions of a time interval become finer. It first bounds the sum of cubed increments by the largest absolute increment times the sum of squared increments. Brownian paths are continuous, so the largest increment vanishes, while the quadratic variation converges to elapsed time.
It then extends the argument by induction to higher integer powers, using the vanishing maximum increment and the preceding power's limit. This connects the result to quadratic variation and cautions that symbolic multiplication rules for differentials do not by themselves justify the integral claim. The exposition is an almost-sure partition argument for Brownian paths; it does not discuss other processes or establish broader stochastic integration results.
Key ideas
- Brownian motion has quadratic variation equal to elapsed time along refining partitions.
- The largest Brownian increment tends to zero as the partition mesh shrinks.
- The sum of cubed increments vanishes by bounding it with the largest increment and quadratic variation.
- An induction argument extends the vanishing result to higher integer powers.
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# Integral with respect of $(dW_s)^n$
# Integral with respect of $(dW_s)^n$
I know $$\int _0^t dW_s=W_t-W_0=W_t$$ Since $ dW_s dW_s=ds$ , so $$\int _0^t( dW_s)^2=\int_0^t ds=t-0=t$$ I Want to know why for $n\ge 3$ we have $$\int _0^t (dW_s)^n=0$$ My try $$(dW_s)^2 dW_s (dW_s)^{n-3}=ds dW_s (dW_s)^{n-3} =0$$ Is it true? What is relation between Ito Integarl and varation?
## Answer by user16651 (score 1, accepted)
https://quant.stackexchange.com/a/29965
The result is true, but your solution is meaningless.Let $I=\{t_0,t_1,\cdots,t_m\}$ is a sequence of partitions of $[0,t]$ and $\delta_m=\max\{t_{i+1}-t_{i}\}_{i=0}^{m}\to 0$ as $m\to \infty$. For $n=3$, we have
$$\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^3\Big|\le\underset{0\le i\le m-1}{\mathop{\max }}\Big|W_{t_{i+1}}-W_{t_{i}}\Big|\cdot\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^2\Big|\tag 1$$
Since the sample path of Wiener process is continues, thus $$\lim_{m\to \infty}\underset{0\le i\le m-1}{\mathop{\max }}\Big|W_{t_{i+1}}-W_{t_{i}}\Big|=0\tag 2$$
Moreover
$$\lim_{m\to \infty}\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^2\Big|=\lim_{m\to \infty}\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^2=t<\infty\tag 3$$
$(1)$ and $(2)$ and $(3)$
$$\lim_{m\to \infty}\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^3\Big|=\Big|\lim_{m\to \infty}\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^3\Big|\le 0$$ therefore $$\lim_{m\to \infty}\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^3=0$$ In other words $$\int_{0}^{t}(dW_s)^3=0$$ Let the theorem is true for $n=k$, $k>3$, i.e $$\int_{0}^{t}(dW_s)^k=\lim_{m\to \infty}\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^k=0\tag 4$$
For $n=k+1$, we have
$$\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^{k+1}\Big|\le\underset{0\le i\le m-1}{\mathop{\max }}\Big|W_{t_{i+1}}-W_{t_{i}}\Big|\cdot\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^{k}\Big|\tag 5$$
$(2)$ and $(4)$ and $(5)$
$$\lim_{m\to \infty}\Big|\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^{k+1}\Big|=\Big|\lim_{m\to \infty}\sum_{i=0}^{m-1}(W_{t_{i+1}}-W_{t_{i}})^{k+1}\Big|\le 0$$ In other words $$\int_{0}^{t}(dW_s)^{k+1}=0$$
> The theorem was proved by mathematical induction.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.