Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 38 / 1

Choose a localizing sequence of discrete-time stopping times ${\tau }_j\uparrow \infty$ for $X$. For fixed integer $t$, every stopped value is bounded in absolute value by the finite sumThat sum is integrable under the hypothesis. Since $X^{{\tau }_j}$ is a martingale,The dominated convergence theorem, including its conditional version, now removes the stopping. Thus $\mathbb{E}[X_t\mid {\mathcal{F}}_{t-1}]=X_{t-1}$. The process is adapted and integrable by hypothesis, so $X$ is a true discrete-time martingale. This is the integrable discrete-time local martingale is a martingale criterion. The finite sum dominating stopped values is the crucial discrete-time feature.

The stopped processes are nonnegative martingales. Their expectations equal the finite deterministic value $X_0$. The Fatou lemma gives, for each fixed integer $t$,Thus every $X_t$ is integrable. Apply the preceding discrete-time criterion to conclude $X$ is a martingale, not merely a supermartingale. This is the nonnegative discrete-time local martingale is a martingale result; its conclusion does not extend to arbitrary continuous-time nonnegative local martingales.

Predictability means $K_s$ is ${\mathcal{F}}_{s-1}$-measurable. If $|K_s|\le L$, each product $K_s(M_s-M_{s-1})$ is integrable, and the finite sum defining $Y_t$ is integrable. Pulling the bounded predictable factor out of the conditional expectation givesTherefore the bounded predictable martingale transform $Y$ is a martingale starting at zero.

Stop just before a large predictable coefficient would be used. Setwith the infimum of the empty set equal to infinity. Because $K_{t+1}$ is ${\mathcal{F}}_t$-measurable, ${\sigma }_j$ is a stopping time. The increment of the stopped martingale transform isIts coefficient is predictable and bounded by $j$: the first coefficient exceeding $j$ occurs at the step after stopping, and is never included. Part (c) makes $Y^{{\sigma }_j}$ a martingale. Since the finitely many $K_s$ on any fixed finite horizon are finite almost surely, ${\sigma }_j\uparrow \infty$ almost surely. ThusThis is predictable-coefficient localization of a martingale transform. Stopping after taking the large increment would not give the required bound.

First propagate terminal nonnegativity backwards; it is not necessary to assume nonnegative wealth at intermediate dates. Suppose $Y_t\ge 0$ and defineThese events increase to the whole space up to a null set. On $A_j$, both the old value and the coefficient are bounded, so $1_{A_j}{Y}_t$ is integrable andThe left side is nonnegative; hence $Y_{t-1}\ge 0$ on every $A_j$, and therefore almost surely. Starting from $t=T$, induction gives $Y_t\ge 0$ for every $0\le t\le T$.

The process stopped at $T$ is now a nonnegative discrete-time local martingale. Part (b) makes it a martingale with initial value zero. Consequently $\mathbb{E}{Y}_T=0$, and a nonnegative random variable with zero expectation vanishes almost surely:This is the terminal nonnegativity criterion for a finite-horizon martingale transform. A finite deterministic horizon is essential to the backward induction.