What is $d(W_t^2)$?

Answer: B – $2W\,dW + dt$

Expand $f(W) = W^2$ to second order: $\mathrm{d}f = f'\,\mathrm{d}W + \tfrac12 f''\,(\mathrm{d}W)^2 = 2W\,\mathrm{d}W + \tfrac12 \cdot 2 \cdot \mathrm{d}t$, so $\mathrm{d}(W^2) = 2W\,\mathrm{d}W + \mathrm{d}t$. The extra term survives because $(\mathrm{d}W)^2 = \mathrm{d}t$ rather than vanishing, which is the one fact separating this calculus from the ordinary kind. The factor of one half in the lemma cancels the two from the second derivative, which is where the wrong-by-two answers come from. An immediate consequence is that $W_t^2 - t$ is a martingale.

• A. That is ordinary calculus, and it drops exactly the term that makes this subject different.
• B. Correct. The second-order term is $(dW)^2 = dt$, which is why $W_t^2 - t$ is a martingale.
• C. The one half in Itô’s lemma cancels the two from the second derivative.
• D. The first-order coefficient is $2W$, not $W$.

Takeaway: Expand to second order: $(dW)^2 = dt$.