# Derivative of Stochastic Integral

I am trying to take the derivative of the following stochastic integral, $$d\left(\int g(S_t) dS_t \right),$$ where $$dS(t) = \sigma S(t) dW_t$$ and $$g(.)$$ is some (smooth) deterministic function. My understanding is that we can't just apply the fundamental theorem of calculus, but instead need to account for QV. My attempt: $$d\left(\int g(S_t) dS_t \right)=g(S_t)dS_t+\frac{1}{2}g'(S_t)(dS_t)^2=g(S_t)S_t\sigma dW_t+g'(S_t)S_t^2\sigma^2dt$$ Is that right?

No. Itō’s formula helps you derive the dynamics of $$f (S_\cdot )$$ given the SDE followed by $$S$$. Here this is not the case. You simply have: $$\mathrm{d} \left[\int{g(S_t)\mathrm{d}S_t}\right] = g(S_t) dS_t$$
• Consider $g(S_t)=\frac{1}{S_t}$. Then $\int \frac{1}{S_t} dS_t =\ln{S_t} + C$ so $d \left( \int \frac{1}{S_t} dS_t \right)=d \ln{S_t}$. The formula above would give $d \left( \int \frac{1}{S_t} dS_t \right)=\frac{1}{S_t}dS_t \neq d \ln{S_t}$. What am I missing? Apr 8, 2021 at 23:01
• @chester What you are missing is that $\frac{dS}{S} \neq d \log S$. In fact by Ito's lemma $\frac{dS_t}{S_t} = d\log S_t + \frac{1}{2} \sigma^2 dt$, and if you substitute the right hand side in the integral and do things the 'long way', you'll just end up with what siou0107 gave as answer. Apr 9, 2021 at 7:13