Past exam of the martingales and financial mathematics course of the University of Oxford / 2007 / A / 3

A version of $\mathbb{E}(X\mid \mathcal{G})$ is an integrable $\mathcal{G}$-measurable random variable $Y$ such thatLet $D=Y_1-Y_2$ for two versions. It is integrable and $\mathcal{G}$-measurable, and its integral over every $\mathcal{G}$ event is zero. In particular $A=\{D>0\}$ is allowed, giving $\int D^{+}d\mathbb{P}=0$, so $D^{+}=0$ almost surely. Applying the same reasoning to $\{D<0\}$ gives $D^{-}=0$ almost surely. Therefore $Y_1=Y_2$ almost surely. Equality at every point is not required, since integrals cannot distinguish versions differing on a null set.

conditional dominated convergence theorem states that if $U_n\to U$ almost surely and $|U_n|\le H$ for an integrable $H$, thenFor an indicator function $Z=1_B$, $B\in \mathcal{G}$, the defining integrals of conditional expectation immediately give the take-out identity: testing on $A\in \mathcal{G}$ replaces $A$ by $A\cap B$. Finite linear combinations give it for simple $\mathcal{G}$-measurable $Z$. Approximating a bounded such $Z$ by uniformly bounded simple functions and using dominated convergence theorem provesfor bounded $Z$.

For the stated unbounded case, clip $Z$ to $Z_m=(-m)\vee (Z\wedge m)$. Then $X{Z}_m\to XZ$ and $|X{Z}_m|\le |XZ|$, an integrable bound by hypothesis. conditional dominated convergence theorem and the bounded identity givealmost surely. The equality with the integrable conditional expectation also proves integrability of the final product; it is not necessary to assume that separately.

A process is a supermartingale relative to the given filtration if $X_n$ is ${\mathcal{F}}_n$-measurable and integrable for each $n$, andIteration gives the equivalent inequality $\mathbb{E}(X_m\mid {\mathcal{F}}_n)\le X_n$ for every $m\ge n$. adapted process and integrability are part of the definition, not consequences of the displayed inequality alone.

The products are nonnegative and measurable. independence and the unit means give $\mathbb{E}{Y}_n={\prod }_{k\le n}\mathbb{E}{\xi }_k=1$, so they are integrable.

Let ${\mathcal{H}}_n=\sigma (Y_0,\dots ,Y_n)$ be the natural filtration. It is contained in $\sigma ({\xi }_1,\dots ,{\xi }_n)$, so ${\xi }_{n+1}$ is independent of ${\mathcal{H}}_n$ even if some products have vanished. Using the bounded or integrable nonnegative conditional-product identity,Thus $Y$ is a martingale in its natural filtration.

Doob's convergence theorem in the form needed here states that a nonnegative supermartingale converges almost surely to a finite integrable limit; more generally a submartingale with uniformly bounded first absolute moments has such an almost sure convergence. Applying it to $Y$ givesThe expected value bound follows from Fatou's lemma. The theorem does not assert $L^1$ convergence or equality of expected values without a uniform-integrability hypothesis.

There is an indexing error at the printed initial time. The dyadic sigma-algebra ${\mathcal{F}}_0$ is trivial, whereas the stated formula givesThis is not ${\mathcal{F}}_0$-measurable. Moreover $\mathbb{E}{X}_0=1/2$ and $\mathbb{E}{X}_1=1$, so the process cannot be a martingale on all of ${\mathbb{N}}_0$ as printed.

For every $n\ge 1$, its support $I_n=(1/2-2^{-n},1/2]$ is one atom of ${\mathcal{F}}_n$, so $X_n$ is adapted and integrable with expected value one. At the next dyadic level, $I_n$ splits into two equal-measure children; $X_{n+1}$ is zero on its left child and $2^{n+1}$ on its right child $I_{n+1}$. Its conditional average on the parent is $2^n$, and on every other parent it is zero. HenceThus the printed process is a martingale from time one onward. The intended full-index repair is to define $X_0\equiv 1$ and leave all $n\ge 1$ terms unchanged, since then $\mathbb{E}(X_1\mid {\mathcal{F}}_0)=1$.

If $\omega \ne 1/2$, it eventually lies outside the shrinking intervals, so $X_n(\omega )\to 0$. At the single null set $\omega =1/2$, the values tend to infinity. ThereforeThe repaired martingale is not uniformly integrable and does not converge in $L^1$. This is the time-zero correction for a half-point dyadic spike martingale.