For geometric Brownian motion, $\log S_u=\log S_0+(\mu -{\sigma }^2/2)u+\sigma W_u$. Averaging in time givesThus $\log A_t$ is normal with mean $\log S_0+(\mu -{\sigma }^2/2)t/2$ and variance ${\sigma }^{2}t/3$. The moment-generating function of a normal distribution gives

For the time integral $I_t={\int }_0^t{W}_{s}ds$, Fubini's theorem and $\mathbb{E}{W}_s=0$ give $\mathbb{E}{I}_t=0$. Brownian covariance is $\mathbb{E}[W_u{W}_v]=\min (u,v)$, soConsequently

At time zero, exercise pays $115-A_0=15$, while continuation is worthTherefore exercise at time zero is optimal, and $V_0=15$ pounds.

The discounted unexercised option-value process is 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.

Use $q=1/2$ and discount one step by $R=1.1$. At the up node, $A_1=110$ and immediate exercise pays $5$. Continuation is worth $(0+5/3)/(2R)=25/33<5$. At the down node, $A_1=100$ and immediate exercise pays $15$, while continuation is $(25/3+15)/(2R)=350/33<15$. Thus the American put option should be exercised at either time-one node, and

Retain the whole price history when calculating the arithmetic average. The Asian option payoffs at expiry areIn particular $ud$ and $du$ have the same terminal stock price but different averages and payoffs. The average is an additional state variable; the terminal stock nodes cannot be merged for this claim.

The risk-neutral probability must make the expected stock multiplier equal the bank multiplier:Both up and down paths have positive probability under this measure. A self-financing portfolio's discounted value is then a Q-martingale, since the discounted bank asset is constant and the discounted stock is a martingale. An arbitrage with zero initial wealth would have nonnegative terminal value and positive value in some possible state, giving a strictly positive Q-expectation and contradicting its zero discounted initial value. Thus the market with $e^r=1.1$ is arbitrage free.

Short one share at time zero and invest the one hundred pounds received in the bank asset. The initial net investment is zero. After one period, repurchasing the share leaves $100e^{0.2}-S_1$. Since $S_1\le 120$ and $e^{0.2}>1+0.2=1.2$, this payoff is strictly positive in both states. Hence this is an arbitrage opportunity.

The Borel-Cantelli second lemma says that independent events whose probabilities have divergent sum occur infinitely often almost surely. The events here are independent and have probabilities $\min \{1,\lambda /n\}$. Their sum diverges by the harmonic series, so

The exchange option has terminal payoffs determined by the two successive up/down moves:The two stocks span both one-step states, so this finite market is complete. The risk-neutral pricing uses the independent transition probabilities $q=1/3$ and $1-q=2/3$. Since the synthetic risk-free rate is zero,The price is approximately 7.56 in the currency units of the stock prices.