Let $\delta$ be the Skorohod integral on $L^2(\Omega\times [0, T])$, i.e. for $\xi\in L^2(\Omega\times [0, T])$ $$\delta (\xi) = \int_0^T \xi(t, \omega) \delta W_t.$$
Let $S_t = \exp\{\theta t + \sigma W_t\}$ be a stochastic process with given constants $\theta$ and $\sigma$ and a Brownian motion $W$. I wanted to know how to compute $$\delta \left(\frac{S.}{\int_0^T S_t dt} \right).$$
Remark: This question is an intermediate step in computing greeks of an exotic option, see page 334 of the book " The Malliavin Calculus and Related Topics" by Nualart 2006, google books.
By setting $$G_t = \int_t^T S_r dr,$$ we can write our question as $$I = \delta (G_0^{-1} S),$$ and my goal is to find a representation of $I$. Next we use a formula $$\delta (Fu) = F \delta (u) - \langle DF, u \rangle,$$ where $F$ is a random variable and $u$ is a random process with some nice properties, $DF$ is a malliavin derivative which gives a random process, and $\langle u, v\rangle = \int_0^T u(t) v(t) dt.$ Now we have $$I = G_0^{-1} \delta(S) - \langle D G_0^{-1}, S \rangle.$$ Since $S$ is an adapted process, $\delta(S)$ is indeed Ito integral and we do not need to improve further on it. We want to compute $D_t G_0^{-1}$, the value at time $t$ of the process $D G_0^{-1}$: $$D_t G_0^{-1} = - G_0^{-2} D_t \int_0^T S_r dr = - G_0^{-2} \int_0^T D_t S_r dr.$$ In the above, we used chain rule and switch the order of integral and derivative. By using $D_t f(W_r) = f'(W_r) I_{[0, r]}(t)$, we have $D_t S_r = \sigma S_r I_{[0, r]}(t)$, and this leads to $$D_t G_0^{-1} = - \sigma G_0^{-2} G_t.$$ Due to the fact $\int_0^T \int_t^T f(t)f(s) ds dt = \frac{1}{2} (\int_0^T f(t) dt)^2$, we can obtain after simplifications $$\langle DG_0^{-1}, S\rangle = - \frac \sigma 2.$$ Therefore, the answer is $$I = \frac{\int_0^T S_t dW_t}{\int_0^T S_t dt} + \frac \sigma 2,$$ which is enough for the purpose of the greek computation.