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... 


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*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 $dY_t=b_t \, dt+h_t \, dW_t$.
 A: 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.
