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I'm trying to derive the PDE for the pricing function of an Asian option with payoff (at maturity $T$)$$\left ( S(T)-\frac 1 T \int_0^T S(t) \, dt \right)^{+}$$

After the establishment of a portfolio we end up using Ito's lemma. However when using Ito's Lemma on the option price $V(t, S(t), I(t))$, where $I(t)=\int_0^t S(\nu) \, d\nu$, we have:

$$dV(t, S(t), I(t))=\frac{\partial V}{\partial t} \, dt + \frac{\partial V}{\partial S} \, dS(t) + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} \, d\langle S(t)\rangle + \frac{\partial V}{\partial I} \, dI(t) + \frac 1 2 \frac{\partial^2 V}{\partial I^2} \,d\langle I(t) \rangle$$

However in most texts they never write the second derivative term of $I(t)$. I can't see how this term is zero. The second derivative itself does not strike me as immediately zero and I believe the quadratic variation of $I(t)$ is non zero as

$$dI(t)=S(t)\,dt+t\,dS(t)$$

which should imply that $d\langle I(t) \rangle =t^2 \, d\langle S(t) \rangle \neq 0$. If anyone can spot my mistake I would be very grateful.

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    How did you get $dI = Sdt + tdS$? Shouldn't it just be $dI = Sdt$? – 2017-01-06
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    If $S(t)$ is stochastic do we not need to take the derivative with respect to $S$ as well? – 2017-01-06
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    $dI = Sdt + tdS$ is wrong . Stochastic or not, we integrate a continuous process, and an infinitesimal increment of the integral is $dI = Sdt$ like any continuous function. – 2017-01-06
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    Ah ok thanks for the clarification – 2017-01-06

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