I am stuck on the 1st equation of the solution where the Wiener process $W_{t_i}^2$ is expanded so that the Itô integral (in terms of infinite sums) looks like the RHS of the first equation of the solution. I can follow the rest of what they did.

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If the problem were to solve $\int_0^t W_s\text dW_s$ instead then I think the analogous piece would be $$W_{t_i} = \frac{1}{2}\left(W_{t_{i+1}} + W_{t_i}\right) - \frac{1}{2}\left(W_{t_{i+1}} - W_{t_i}\right).$$

What's the trick to getting this solution started?


An alternative way is using the Stratonovich integral. By definition, we have

$$\int_0^t X_s \, \circ dW_s = \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{X_{t_i} +X_{t_{i-1}}}{2}\left( W_{t_i} -W_{t_{i-1}}\right) \; \; (1)$$

One can then show that for a deterministic smooth functions $f$ and $g$ we have:

$$ \int_0^t g'(W_s)\, \circ dW_s = g(W_t)- g(W_0)\; \; (2) $$ and $$\int_0^t f(W_s)\, \circ dW_s =\int_0^t f(W_s) \, dW_s + \frac{1}{2} \int_0^t f'(W_s) \, ds \; \; (3). $$

Using (1), we get:

$$\int_0^t W_s \, \circ dW_s = \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{W_{t_i} +W_{t_{i-1}}}{2}\left( W_{t_i} -W_{t_{i-1}}\right) $$ $$= \frac{1}{2} \lim_{n\rightarrow \infty} \sum_{i=1}^n \left( W_{t_i}^2 -W^2_{t_{i-1}}\right) = \frac{1}{2} W_t^2 $$

Using (2) with $g(x) = 1/2x^2, g'(x) = x$, we get the same result $$\int_0^t W_s \, \circ dW_s = \frac{1}{2} W_t^2. $$

From (3) with $f(x)=x, f'(x) = 1$, we can now get the Ito integral:

$$\int_0^t f(W_s) \, dW_s = \int_0^t f(W_s)\, \circ dW_s - \frac{1}{2} \int_0^t f'(W_s) \, ds $$ $$ = \frac{1}{2} W_t^2 - \frac{1}{2} t $$

We can repeat the recipe above for calculating:

$$ \int_0^t W_s^2 \, dW_s $$

The convenience of Stratonovich integral definition strikes again:

$$\frac{1}{3}W_t^3 \stackrel{(2)}{=}\int_0^t W_s^2 \, \circ dW_s \stackrel{(1)}{=} \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{W_{t_i}^2 +W_{t_{i-1}}^2}{2}\left( W_{t_i} -W_{t_{i-1}}\right) $$ $$\stackrel{algebra}{=} \lim_{n\rightarrow \infty} \sum_{i=1}^n W_{t_{i-1}}^2 \left( W_{t_i} - W_{t_{i-1}}\right) + \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{1}{2} \left( W_{t_i}^2 - W_{t_{i-1}}^2\right) \left( W_{t_i} - W_{t_{i-1}}\right) $$

$$ = \int_0^t W_s^2dW_s + \frac{1}{2} [W^2, W]_t = \int_0^t W_s^2dW_s + \int_0^t W_s ds $$

Note that the algebra needed is much nicer than @Kevin's :):

$$ \frac{1}{2}(a+b)(x-y) = b(x-y)+ \frac{1}{2} (a-b)(x-y) $$

and it is the basis of the fundamental result behind (3):

$$\int_0^t X_s \, \circ dW_s = \int_0^t X_s dW_s + \frac{1}{2} [X,W]_t $$

  • $\begingroup$ @BobJansen Done. Thank you. $\endgroup$ – ir7 Mar 15 at 12:32

Integrating $W_t$

Consider the partition $t_i=it/n$ with $t_0=0$ and $t_n=t$. Then, by definition, \begin{align*} \int_0^t W_s\text{d}W_s &= \lim_{n\to\infty} \sum_{i=0}^{n-1} W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right). \end{align*}

You can do the limit by using the identity $$ W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right)=\frac{1}{2}\left(W_{t_{i+1}}^2-W_{t_i}^2-\left(W_{t_{i+1}}-W_{t_i}\right)^2\right).$$

Integrating $W_t^2$

Using the same partition as before, \begin{align*} \int_0^t W_s^2\text{d}W_s &= \lim_{n\to\infty} \sum_{i=0}^{n-1} W_{t_i}^2\left(W_{t_{i+1}}-W_{t_i}\right). \end{align*} For this case, you can use the identity $$ W_{t_i}^2\left(W_{t_{i+1}}-W_{t_i}\right)=\frac{1}{3}(W_{t_{i+1}}^3-W_{t_i}^3)-W_{t_i}(W_{t_{i+1}}-W_{t_i})^2-\frac{1}{3}(W_{t_{i+1}}-W_{t_i})^3.$$ The rest is as usual.

Itô's Lemma

You get the same result much quicker if you set $f(t,x)=\frac{1}{3}x^3$. Then, \begin{align} \text df(W_t) = W_t\text{d}t + W_t^2\text dW_t, \end{align} which implies $$\frac{1}{3}W_t^3=\int_0^tW_s\text{d}s+\int_0^tW_s^2\text{d}W_s.$$

  • $\begingroup$ It sounds like the trick is memorizing the identity which is what the solution I pasted shows. It's fine if that's all there is to it. I guess I hoped there were a simple repeated pattern that led to those 3 terms and could be used for other similar problems. $\endgroup$ – Soran Mar 15 at 3:19
  • 2
    $\begingroup$ @Soran Yes, I fear you're right? I don't know whether there's a general decomposition of $W_{t_i}$ that works for all integrals, $\int_0^t f(s,W_s)\text{d}W_s$. The trick is to think what you want and need. For example, it's clear that $W_{t_{i+1}}^3-W_{t_i}^3$ are needed to get the usual $W_t^3$ from the integral (telescoping sums). Then, you kind of need to look what term is left over and how you can write them in terms of Brownian increments. It's a bit tedious. Yes. That's why we normally use Ito's Lemma (or other tricks like Statanovich, as shown by @ir7). Hope that helps! $\endgroup$ – Kevin Mar 15 at 11:42

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