Calculation with Ito processes, what is $ds \, dt$, $dW_t \, ds$ and $dW_s \, dW_t$?

I am working on an exercise and I am not sure how to deal with these 3 cases...

• For example, is $ds \, dt=0$? I know $(dt)^2=0$, but I am not sure when it is 2 different variables.

• And what about $dW_s \, dW_t$? I know $(dW_t)^2=dt$ and that $W_t W_s=\min(t,s)$, but what happens when $dW_s \, dW_t$, is it zero or $\min(dt,ds)$?

• And what is $dW_t \, ds$?

Thanks a lot!

Extra info:

$X_t$ is an Ito process given by $X_t = X_0 + \int_0^t \mu_s ds+\int_{0}^{t} \sigma_s \, dW_s$, where $X_0\in\mathbb{R}$, and $Y_t = Y_0 + \int_0^t b_s \, ds + \int_0^t h_s \, dW_s$, where $Y_0\in\mathbb{R}$. Therefore, $dX_t=\mu_t \, dt+\sigma_t \, dW_t$ and $dY_t=b_t \, dt+h_t \, dW_t$.

• You haven't given enough context. Do you have two different independent Brownian motions, one with time index $s$ and the other with time index $t$? If so, then none of these quantities are zero.
– Ian
Aug 11 '15 at 0:33
• I've edited the post. I believe that they are not zero then? My professor has not said if they are independent Brownian motions though. Aug 11 '15 at 0:42
• As it is the question does not make much sense, are you asking what is the quadratic variation of small increments $\Delta t$? and $\Delta W_t$ ? The extra info are irrelevant to the question. Aug 11 '15 at 3:04

The rules $$(dt)^2 = 0 \qquad dW_t \, dt = 0 \qquad (dW_t)^2 = dt \tag{1}$$ are heuristic rules to simplify calculations when applying Itô's formula. Mind that this is the only application; do not use them anywhere else.

In Itô's formula expressions of the form

$$\int_0^T f(X_t,Y_t) dX_t \, dY_t$$

pop up. Using $(1)$, we get

$$dX_t \, dY_t = (\mu_t dt + \sigma_t dW_t) (b_t \, dt + h_t \, dW_t) = \sigma_t h_t \, dt,$$

i.e.

$$\int_0^T f(X_t,Y_t) dX_t \, dY_t = \int_0^T f(X_t,Y_t) h_t \sigma_t \, dt.$$

As already mentioned above, this is the only application. When applying Itô's formula, you will never encounter

$$\int_0^T f(X_s,Y_t) \, dX_s \, dY_t$$

or similar things; therefore, it is enough to have $(1)$ and there is no need to ask for further rules.

Beware of the following mistake: Suppose that $$X_t = \int_0^t \mu_s \, ds \qquad \text{and} \qquad Y_t := \int_0^t b_s \, ds,$$ i.e. $dY_t = b_t \, dt$ and $dX_t = \mu_t \, dt$. Now it might be tempting to do the following calculation:

$$dX_t dY_t = \mu_t b_t (dt)^2 =0$$

and conclude that $X_t \cdot Y_t = 0$. This reasoning is not correct. To show that $X_t Y_t = 0$ we have to prove $d(X_t Y_t)=0$ and, in general, $d(X_t Y_t) \neq dX_t dY_t$. Just think of some examples where the processes can be calculated explicitly, e.g. set $\mu_t = b_t = t$. Then $$X_t = Y_t = \frac{t^2}{2}$$ and so $X_t \cdot Y_t = 0$ does, obviously, not hold true.

Remark: This question might be also of interest.