Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 29K (Stochastic Financial Models)

The stock tree has $S_0=5$, first-period values $6,4$, and terminal successors $7,5$ from $6$ and $5,3$ from $4$. When $r=1/6$, the risk-free growth factor is $R=7/6$. At the upper time-one node, investing $6$ in the risk-free asset produces $7$ at time two, while one share bought for $6$ produces only $7$ or $5$.

Use the following predictable self-financing portfolio. Hold nothing initially. If $S_1=6$, short one share and invest the proceeds $6$ in the risk-free asset; if $S_1=4$, continue to hold nothing. Its terminal payoff ison the upper branch, and zero on the lower branch. It costs zero, is never negative, and is positive with positive probability, so it is an arbitrage.

Now $R=1+r=9/8$. The local risk-neutral up probabilities solve $RS=q{S}_u+(1-q)S_d$. They areThe terminal payoff of the European put option with strike $5$ is $0,0,2$ at stock prices $7,5,3$. Backward option pricing gives time-one valuesand henceThe first-period stock holding in the replicating portfolio in a binomial market isThus the hedge initially shorts $2/9$ of a share.

The European call option with strike $5$ pays $2,0,0$ at terminal stock prices $7,5,3$. Its lower-node value is zero, while its upper-node value isA second step of backward option pricing givesBy the fundamental theorem of asset pricing, this discounted expectation under the unique risk-neutral probabilities is the unique no-arbitrage price.

For the American put option, compare exercise with continuation at time one. At $S_1=6$, both values are zero. At $S_1=4$, immediate exercise pays $1$, whereas the continuation value is the European value $4/9$, so exercise is optimal. At time zero, immediate exercise pays zero and continuation isThe optimal policy is therefore to continue at time zero, exercise at time one if $S_1=4$, and otherwise continue to maturity. This nodewise maximum is the Snell envelope value.