= Solution
{c}
At time zero, exercise pays $115-A_0=15$, while continuation is worth
$$
\frac{\mathbb E_Q V_1}{R}=\frac{(5+15)/2}{1.1}=\frac{100}{11}<15.
$$
Therefore \b[exercise at time zero is optimal], and $V_0=15$ pounds.
The discounted unexercised option-value process is \b[not a Q-martingale]: already $\mathbb E_Q[V_1/B_1]=100/11<15=V_0/B_0$. It is the <Snell envelope> of the discounted exercise payoff, hence a supermartingale in general, with strict decrease of expected value when immediate exercise beats continuation. Stopping at an optimal exercise time restores the stopped martingale property; that is a different assertion from the unexercised price being a martingale.
\Image[/past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2010-asian-put-tree.png]
{title=The American arithmetic-average put: four terminal histories are retained because ud and du have different average prices. Backward induction selects immediate exercise at both time-one nodes and at time zero.}
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