# 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 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 at 7:13