Integral with respect of $(dW_s)^n$

Question

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 1

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.