# Applying Ito's formula to complex functions

Within my lecture notes, the following definition is given:

We say that the stochastic process $X_t$ has stochastic differential $$dX_t = b_t dt + \sigma_t dW_t$$ if and only if $$X_t = X_0 + \int_0^t b_s ds + \int_0^t \sigma_s dW_s$$

However, it then goes on to discuss Ito's formula for a complex function by defining $$f(x) := A(x) + iB(x)$$ and showing that, for a function of this form, $$df(W_t) = f'(W_t) dW_t + \frac{1}{2} f''(W_t) dt \hspace{10mm} (*)$$ So far so good; I can follow all of this.

However, it then goes on to say that $(*)$ may be expressed in integral form as $$f(W_t) - f(0) = \int_0^t f'(W_s) dW_s + \int_0^t \frac{1}{2} f''(W_s) ds$$

This is where my confusion comes from. Based on the definition of a stohastic integral, why would $(*)$ not be evaluated as $$[f(W_s)]_0^t = f(W_t) - f(W_0) = f(W_0) + \int_0^t f'(W_s) dW_s + \int_0^t \frac{1}{2} f''(W_s) ds$$ simplifying to $$f(W_t) - 2 f(0) = \int_0^t f'(W_s) dW_s + \int_0^t \frac{1}{2} f''(W_s) ds$$

• $f(x) = f(a) + \int_a^x f'(u)du$, so likewise for the stochastic integral version: $f(W_t) = f(W_0) + \int_0^t f'(W_s)dW_s + \frac{1}{2}\int_0^t f''(W_s)ds$ and since $W_0 = 0$ you get the integral formula of (*) as stated by your textbook. – byouness May 19 '18 at 17:32