What is the analog Stratonovich SDE to WdW?

Question

i have the Ito-SDE $\int \limits_0^t W(t) dW(t)$
But how can I change this SDE $\int \limits_0^t W(t) dW(t)$ into a Stratonovich-SDE? Normally I do $\underline f=f-\tfrac{1}{2}gg'$.
Is the Stratonovich SDE then
$\int \limits_0^t \tfrac{1}{2} W(t) W'(t) dt+\int \limits_0^t W(t) \circ dW(t)$.
If I calculate it in Ito-Form, I get
$ \int \limits_0^t W(s) dW(s)=\frac{1}{2} \int \limits_0^t (dW(s))^2- \frac{1}{2}\int \limits_0^t1 ds=\frac{1}{2}W(t)^2-\frac{1}{2}t.$
How can I calculate that with the chain rule in the Stratonovich-Form?
Thank you so much.

Answer 1

By definition of the Stratonovich integral, we have

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

where $\circ dW_s$ denotes the Stratonovich integral. Applying this equality to $f(x) := x$ yields

$$\int_0^t W_s \, \circ dW_s = \int_0^t W_s \, dW_s+ \frac{t}{2} \tag{1}$$

Since the Stratonovich integral satisfies the classical chain rule, we can compute the left-hand side simply by using the fundamental theorem of calculus:

$$\int_0^t W_s \, \circ dW_s = \bigg[ \frac{x^2}{2} \bigg]_0^{W_t} = \frac{W_t^2}{2}\tag{2}$$

Combining $(1)$ and $(2)$ we find

$$\frac{W_t^2}{2} = \int_0^t W_s \, dW_s + \frac{t}{2}$$