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

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.