Solution to $\int_0^t dW_s $

Question

This question is so basic that I can't find the answer anywhere. I'd think it would be just $W_t$ but since there are additional rules in stochastic calculus, I'm not 100% sure. I've only seen integrals such as $$\int_0^t W_s\,dW_s$$ or $$\int_0^t W_s\,ds$$ or $$\int_0^t s\, dW_s$$ in the resources I've found so far. Is $\int_0^t dW_s $ just equal to $W_t$?

Answer 1

Stochastic calculus doesn't exactly have additional rules; it has an alternative rule. The fundamental theorem of calculus does not apply, and instead we use Ito's rule, $$ g(W_t) = g(W_0) + \int_0^t g'(W_s)\,dW_s + \frac12\int_0^t g''(W_s)\,ds. $$ We can rewrite this using $f=g'$ and $F=g$ as $$ \int_0^t f(W_s)\,dW_s = F(W_t) - F(W_0) - \frac12\int_0^t f'(W_s)\,ds. $$ So we see that instead of the usual fundamental theorem of calculus, we have a correction term involving $f'$. In your case, you have $f(x)=1$, so $f'(x)=0$ and the correction term vanishes. Therefore, the answer is indeed $W_t$.