Past exam of the martingales and financial mathematics course of the University of Oxford / 2008 / A (Martingales Through Measure Theory)

A filtration is an increasing sequence of sub-sigma-algebras ${\mathcal{F}}_n\subseteq \mathcal{F}$: if $m\le n$, then ${\mathcal{F}}_m\subseteq {\mathcal{F}}_n$. It represents information available through each time.

A stopping time $T:\Omega \to {\mathbb{N}}_0\cup \{\infty \}$ satisfiesThus whether stopping has already occurred is determined by the current information.

The whole space belongs to ${\mathcal{F}}_T$, since $\Omega \cap \{T\le n\}=\{T\le n\}\in {\mathcal{F}}_n$. If $A\in {\mathcal{F}}_T$, thenIf $A_j\in {\mathcal{F}}_T$, then $({\bigcup }_j{A}_j)\cap \{T\le n\}={\bigcup }_j(A_j\cap \{T\le n\})$ is also in ${\mathcal{F}}_n$. These are precisely the sigma-algebra closure properties, so ${\mathcal{F}}_T$ is a sigma-algebra.

For every integer $m\ge 0$,Hence $\{T\le m\}\in {\mathcal{F}}_T$. Such threshold sets generate the Borel sigma-algebra on ${\mathbb{N}}_0\cup \{\infty \}$, proving $T$ is ${\mathcal{F}}_T$-measurable. This is the stopping-time sigma-algebra.

For $A\in {\mathcal{F}}_S$, writeThe first two factors together belong to ${\mathcal{F}}_k$, since $A\cap \{S=k\}$ is the difference of its intersections with $\{S\le k\}$ and $\{S\le k-1\}$. Also $\{k\le T\le n\}\in {\mathcal{F}}_n$ by the stopping time property. The filtration inclusion puts every term, and therefore the union, in ${\mathcal{F}}_n$. By definition,

If $S\le T$, then $\{S\le T\}=\Omega$. The previous result says that every $A\in {\mathcal{F}}_S$ equals $A\cap \{S\le T\}\in {\mathcal{F}}_T$. ThusInformation known at the earlier stopping time remains known at the later stopping time.

Both stopped variables are integrable, since $|X_S|,|X_T|\le {\sum }_{j=0}^N|X_j|$. Also $X_S$ is ${\mathcal{F}}_S$-measurable: intersecting $\{X_S\in B\}$ with $\{S\le n\}$ gives the finite union ${\bigcup }_{j\le \min (n,N)}\{S=j,X_j\in B\}$ in ${\mathcal{F}}_n$ for each Borel $B$.

The increment identity isFor $A\in {\mathcal{F}}_S$, the event $A\cap \{S\le k\}\cap \{T>k\}$ belongs to ${\mathcal{F}}_k$. The martingale increment has conditional mean zero given that sigma-algebra. Multiplying by its indicator and using the tower property gives zero expectation for each summand. ThereforeTogether with measurability and integrability, this is the defining property of conditional expectation, proving the bounded optional sampling theorem

A real random variable is a measurable map from $(\Omega ,\mathcal{F})$ to $\mathbb{R}$ with its Borel sigma-algebra. Equivalently, $\{X\le x\}\in \mathcal{F}$ for every real $x$. Its cumulative distribution function isIt is nondecreasing, right-continuous, and tends to zero and one at the two infinities.

Let ${\mu }_X$ be the distribution measure of $X$. The measurable change-of-variables rule for expectations givesFor the second equality, let $U$ be uniform on $(0,1)$ and $Q=F^{\leftarrow }$. The quantile function is measurable and satisfiesThis equivalence uses monotonicity and right-continuity of $F$. Consequently $\mathbb{P}(Q(U)\le x)=F(x)$: $Q(U)$ has the same distribution as $X$. In particular ${\int }_0^1|Q(u)|du=\mathbb{E}|X|<\infty$, so the signed integral is legitimate. Hence the quantile expectation formula isNo continuity or strict increase of the distribution function is required.

The variables are independent and identically distributed if they all have one common probability distribution and every finite selection satisfiesfor distinct indices and arbitrary Borel sets. The condition concerns joint distributions, not merely pairwise correlations.

The strong law of large numbers states that if $X_n$ are independent and identically distributed real random variables with $\mathbb{E}|X_1|<\infty$, thenThe first absolute moment condition is part of the standard statement.

The zero means and orthogonality give the exact second momentThus $A_N\to 0$ in $L^2$. Chebyshev's inequality gives the requested weak law of large numbers:Independence is unnecessary; the stated second-moment orthogonality is sufficient.

For an explicit almost surely convergent subsequence take $N_j=2^j$. For each positive integer $m$,The First Borel-Cantelli lemma makes only finitely many such exceedances occur. Intersecting the resulting probability-one events over $m$ gives $A_{2^j}\to 0$ almost surely.

A version of $\mathbb{E}(X\mid \mathcal{G})$ is an integrable $\mathcal{G}$-measurable random variable $Y$ satisfyingThe measurability requirement restricts $Y$ to the available information; the integral identities preserve all averages visible to that information. This is the definition of conditional expectation.

Let $Y,Z$ be two versions. For every $A\in \mathcal{G}$, their defining identities imply ${\int }_A(Y-Z)d\mathbb{P}=0$. The set $A_m=\{Y\ge Z+1/m\}$ belongs to $\mathcal{G}$, and thereforeEvery $A_m$ is null. Their union is $\{Y>Z\}$, so it too is null. Reversing $Y,Z$ proves the same for $\{Z>Y\}$. Hence versions of conditional expectation are equal almost surely.

A martingale is an adapted process $X_n$ with $\mathbb{E}|X_n|<\infty$ for every $n$ andAdaptedness means that $X_n$ is ${\mathcal{F}}_n$-measurable. Iterating the tower property gives the analogous identity for any two deterministic times $m\le n$.

The supermartingale property and linearity of conditional expectation give, for $n\ge 1$,Each $G_n$ is integrable and ${\mathcal{F}}_{n-1}$-measurable, while $G_0=0$. Thus $Z_n$ is integrable andMoreover, all terms $G_j$, $j\le n+1$, are ${\mathcal{F}}_n$-measurable, so $Z_{n+1}\in m{\mathcal{F}}_n$. This is predictability of the decreasing component in the Doob decomposition theorem.

The process $Y_n=X_n-Z_n$ is adapted and integrable. Since $Z_{n+1}=Z_n+G_{n+1}$ is already ${\mathcal{F}}_n$-measurable,Therefore $Y$ is a martingale, and $X=Y+Z$ is the Doob decomposition in discrete time. Its normalization is $Z_0=0$, so the initial value is included in $Y_0=X_0$.

In any proposed decomposition, $Z_n^{\prime }=X_n-Y_n^{\prime }$ is integrable. Its increment is ${\mathcal{F}}_{n-1}$-measurable by predictability. Taking conditional expectation of the difference equation givesThe martingale increment has conditional mean zero, and a measurable integrable variable equals its own conditional expectation. Starting from $Z_0^{\prime }=0$, summation gives $Z_n^{\prime }={\sum }_{j=0}^n{G}_j=Z_n$. Hence $Y_n^{\prime }=X_n-Z_n^{\prime }=Y_n$. Thus the normalized predictable decomposition is unique, with a common probability-one event available for all integer times by countability.

A Borel measurable function has $f^{-1}(B)\in \mathcal{B}$ for every Borel set $B$. Thus the upper-ray condition is necessary.

For sufficiency, note that the rays generate the stipulated sigma-algebra: $(a,b]=(a,\infty )\setminus (b,\infty )$, and $(y,\infty )={\bigcup }_{m>y}(y,m]$, with integer $m$. The familyis a sigma-algebra because inverse images preserve complements and countable unions. If it contains all upper rays, it contains their generated Borel sigma-algebra. Therefore upper-ray preimages suffice to prove Borel measurability.

Every open interval is Borel, since $(a,b)={\bigcup }_{m\ge 1}(a,b-1/m]$, omitting empty pieces. Open subsets of the line are countable unions of open intervals and hence Borel. Conversely, each half-open interval is obtained from open sets by sigma-algebra operations, so this is the usual Borel sigma-algebra.

A continuous function has open, and therefore Borel, inverse images of open sets. Since open sets generate the target Borel sigma-algebra, it is a Borel measurable function. A discontinuous example is $1_{[0,\infty )}$, whose only level sets are built from the Borel half-lines; it jumps at zero.

For a lower semicontinuous function, each upper-ray inverse image is open, hence Borel, so part (a)(i) applies directly. For an upper semicontinuous $h$, the function $-h$ is lower semicontinuous becauseis open. Negation is continuous and Borel measurable, so $h=-(-h)$ is Borel measurable too. Thus both semicontinuity classes are Borel measurable.

Boundedness of $f$ makes $\phi$ finite and nonnegative. Suppose $\phi (x_0)<a$. By its infimum definition, some $n$ satisfies ${\phi }_n(x_0)<a$. If $|x-x_0|<1/(2n)$, thenThe supremum over the smaller interval cannot exceed that over the larger one. HenceThus $\{\phi <a\}$ is open for every $a$, proving $\phi$ is upper semicontinuous. Measurability of the original $f$ was never needed: these are purely pointwise oscillation bounds.

If $f$ is continuous at $x_0$, choose a small interval on which $|f(s)-f(x_0)|<ϵ/2$. The triangle inequality makes its oscillation less than $ϵ$, so $\phi (x_0)=0$.

Conversely, if $\phi (x_0)=0$, then for every $ϵ>0$ some $n$ has ${\phi }_n(x_0)<ϵ$. As $x_0$ itself belongs to the interval used in that supremum, $|f(x)-f(x_0)|<ϵ$ for $|x-x_0|<1/n$. Thus $f$ is continuous at $x_0$.

The continuity set of a function is thereforeEach set in the intersection is open by part (i), so this is a G-delta set and in particular a Borel set, even if $f$ itself is not measurable.