Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 40 / 3

A complete market permits claim replication for every finite maturity and every bounded claim measurable at that maturity: there are an initial capital and a predictable self-financing portfolio whose terminal wealth equals the claim almost surely. In a finite-state market this is equivalent to replicating every terminal payoff, since all such payoffs are bounded. Trading is allowed dynamically in the existing $n$ assets; adding a new security is not part of the definition.

We prove the stronger finite branching bound in a complete market: modulo null sets, ${\mathcal{F}}_t$ is generated by at most $n^t$ atoms. Here an atom is a positive-probability event which cannot be split into two positive-probability measurable pieces.

At time zero there is one atom because ${\mathcal{F}}_0$ is trivial. Suppose ${\mathcal{F}}_{t-1}$ has at most $n^{t-1}$ atoms, and fix one such atom $A$. Any predictable process holdings over $(t-1,t]$ are constant on $A$, so the restriction to $A$ of every time-$t$ replicable payoff lies ina vector space of dimension of a vector space at most $n$. If $A$ contained $n+1$ disjoint positive-probability ${\mathcal{F}}_t$ events, their indicators would have linear independence on $A$. Market completeness would replicate each indicator, contradicting that dimension bound.

For clarity, the absence of $n+1$ disjoint positive pieces implies that $A$ is a union of at most $n$ atoms: start with $A$ and repeatedly split any non-atom into two positive pieces. Each split increases the count by one. The process must stop before the count exceeds $n$, and at termination every piece is an atom. Thus each time-$(t-1)$ atom has at most $n$ successors. Induction gives at most $n^t$ atoms at time $t$.

Each disjoint positive-probability ${\mathcal{F}}_t$ event contains at least one distinct atom. ThereforeThe argument includes $t=0$ and uses the ability to replicate the indicator claims at every intermediate maturity.

Use the conditional second-moment matrix $V_t=\mathbb{E}[P_t{P}_t^{\top }\mid {\mathcal{F}}_{t-1}]$ from the PDF. By part (b), the filtration is finite-state at every finite time. On a time-$(t-1)$ atom let $p_1,\dots ,p_m$ be the successor price vectors and $q_1,\dots ,q_m>0$ their conditional probabilities. Then $m\le n$ andPositive definiteness gives rank $n$, so $m\ge n$. Hence $m=n$, and the $n$ successor vectors have linear independence.

We use the finite-horizon fundamental theorem of asset pricing in deflator form: an arbitrage-free finite market admits a strictly positive adapted process $D$, with $D_0=1$, such that $DP$ is a martingale. Equivalently, on every step,Fix a finite horizon containing the step in question. On the parent atom, write $d_i=D_t/D_{t-1}>0$ on successor $i$ and $p_0=P_{t-1}$. ThenThe proposed values $z_i=p_i^{\top }{V}^{-1}{p}_0$ satisfy precisely the same equation:Because the $p_i$ have linear independence and the $q_i$ are positive, this linear system has a unique solution. Thus $z_i=d_i>0$, provingThis is the positive regression deflator in a complete finite market. It also proves the suggested conclusion: with $Y_0=1$ and $Y_t={\prod }_{u=1}^t{Z}_u$,Hence $Y$ is a strictly positive martingale deflator. Finite-state structure makes these expectations integrable on each finite horizon. Positive definiteness alone would not ensure positivity of the regression factor; market completeness and the positive deflator are essential.