Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 29K (Stochastic Financial Models)

In the one-period building block of the Cox--Ross--Rubinstein model, the money-market account grows by a deterministic factor $R>0$, while the stock grows independently at each time byThus $S_t^0=R^t{S}_0^0$ and $S_t^1=S_0^1{\prod }_{j=1}^t{Y}_j$. Equivalently, the discounted stock has multipliers $u=U/R$ and $d=D/R$.

The model has no arbitrage precisely when the risk-free return lies strictly between the stock returns:Indeed, failure of either strict inequality gives a one-period arbitrage, while under the strict inequalities the probabilitylies in $(0,1)$. Taking the moves independent under $Q$ with up probability $q$ makes the discounted stock a martingale, since $qu+(1-q)d=1$. This is the unique Risk-neutral probability in the Cox--Ross--Rubinstein model, hence the unique equivalent martingale measure.

The arbitrage-free value of the discounted claim is the conditional expectationConditional on the observed path $(S_0^1,\dots ,S_t^1)=(x_0,\dots ,x_t)$, the future multiplicative increments are independent of ${\mathcal{F}}_t$ and have the same law as a fresh stock path divided by its initial value. ThereforewhereThis is Risk-neutral valuation in the Cox--Ross--Rubinstein model for a path-dependent claim.

Let $N=T-t$ and use the discounted multipliers $u,d$. Under $Q$, the number $K$ of up moves in the remaining $N$ periods has the binomial distribution $\mathrm{Binomial}(N,q)$. Sincethe conditional expectation in part (b) becomesFor an undiscounted stock, replace $u,d$ by $U,D$ and discount the resulting payoff by the appropriate power of $R$.

Given the present state $(S_t^1,M_t)=(x,m)$, take an independent fresh copy of the stock path and denote its initial value, terminal value after $T-t$ steps, and running maximum by $(y,z,w)=(S_0^1,S_{T-t}^1,M_{T-t})$. Scaling the fresh path to start at $x$ makes its terminal value $xz/y$ and its future running maximum $xw/y$. Including the maximum already attained givesHence defineIndependent increments then give the Running-maximum state reduction in the Cox--Ross--Rubinstein modelwith