What is $\int_0^t W_s\,dW_s$?

Answer: B – $\tfrac12 W_t^2 - \tfrac12 t$

Integrate the lemma rather than the integrand. From $\mathrm{d}(W^2) = 2W\,\mathrm{d}W + \mathrm{d}t$, integrating from $0$ to $t$ gives $W_t^2 = 2\int_0^t W\,\mathrm{d}W + t$, so $\int_0^t W\,\mathrm{d}W = \tfrac12 W_t^2 - \tfrac12 t$. The $-t/2$ is the quadratic variation, and it is what makes the answer a martingale: its expectation is $\tfrac12 t - \tfrac12 t = 0$, while the naive $\tfrac12 W_t^2$ would have expectation $t/2$. Taking the expectation is the check to run on any answer of this shape.

• A. That would make the integral have non-zero expectation, and an Itô integral is a martingale starting at zero.
• B. Correct, and the check is immediate: its expectation is $\tfrac12 t - \tfrac12 t = 0$, as a martingale requires.
• C. Off by a factor of two throughout.
• D. Right magnitude, wrong sign – and the expectation test catches it.

Takeaway: The Itô integral of $W$ carries a $-t/2$ that Riemann calculus does not.