(Zero-strike Asian call). Consider a zero-strike Asian call whose payoff at time $T$ is
$$
V(T)=\frac{1}{T} \int_0^T S(u) d u .
$$
(i) Suppose at time $t$ we have $S(t)=x \geq 0$ and $\int_0^t S(u) d u=y \geq 0$. Use the fact that $e^{-r u} S(u)$ is a martingale under $\widetilde{\mathbb{P}}$ to compute
$$
e^{-r(T-t)} \tilde{\mathbb{E}}\left[\frac{1}{T} \int_0^T S(u) d u \mid \mathcal{F}(t)\right] .
$$
Call your answer $v(t, x, y)$.
(ii) Verify that the function $v(t, x, y)$ you obtained in (i) satisfies the BlackScholes-Merton equation (7.5.8) and the boundary conditions (7.5.9) and (7.5.11) of Theorem 7.5.1. (We do not try to verify $(7.5 .10)$ because the computation of $v(t, x, y)$ outlined here works only for $y \geq 0$.)
(iii) Determine explicitly the process $\Delta(t)=v_x(t, S(t), Y(t))$, and observe that it is not random.
(iv) Use the Itô-Doeblin formula to show that if you begin with initial capital $X(0)=v(0, S(0), 0)$ and at each time you hold $\Delta(t)$ shares of the underlying asset, investing or borrowing at the interest rate $r$ in order to do this, then at time $T$ the value of your portfolio will be
$$
X(T)=\frac{1}{T} \int_0^T S(u) d u
$$