Past exam of the probability, measure and martingales course of the University of Oxford / 2023 / 2 / b

For this part and part (c), take $X$ to be $[0,1]$-valued everywhere. The uniform distribution gives $\mathbb{P}(X=1)=0$, but the paper's expression for $Z_n$ is undefined there because $X_n+2^{-n}>1$. Define $Z_n=0$ on $\{X=1\}$. Keep $X_n=\lfloor 2^{n}X\rfloor /2^n$ exactly as printed, including $X_n=1$ on that null set; this preserves the exact sigma-algebra equality in part (ii). Every slope formula below is asserted on $\{X<1\}$, hence almost surely.

The floor function gives the deterministic boundfor every outcome, including $X=1$. Thus all three convergence assertions hold: there is pointwise, and hence almost sure convergence, as well as convergence in probability and convergence in L2. In detail, for $\epsilon >0$ the probability $\mathbb{P}(|X-X_n|>\epsilon )$ is zero as soon as $2^{-n}\le \epsilon$; alsoThe uniform distribution even gives the exact second moment of the error by integrating over the dyadic partition:

Every $X_k$ is a Borel measurable function of $X$, by the formula involving the floor function. ThusConversely, part (i) gives the everywhere identity $X={\lim }_{k\to \infty }{X}_k$. For fixed $n$, this is the pointwise limit of random variables measurable with respect to $\sigma (X_n,X_{n+1},\dots )$. Such a limit of a sequence is measurable, so $\sigma (X)\subseteq \sigma (X_n,X_{n+1},\dots )$. ConsequentlyThese are increasing-precision observations even when the first finitely many are discarded; they do not behave like the independent random variables whose tail sigma-algebra appears in Question 1(b).

A family $({\xi }_i{)}_{i\in I}$ of integrable random variables is uniformly integrable whenThe Lipschitz condition gives, on $\{X<1\}$,Our chosen zero value on $\{X=1\}$ satisfies the same bound. Hence the random variables $(Z_n)$ are integrable, and for $K\ge L$ every tail expectation in the definition vanishes. Therefore the family is uniformly integrable.

For $1\le j\le n$, repeated use of the floor function identity givesThus the filtration satisfies ${\mathcal{F}}_n=\sigma (X_n)$. Each positive-probability atom of a sigma-algebra here has the formwith probability $2^{-n}$; $\{X=1\}$ is a possible additional null set atom. Averaging over each dyadic interval gives a versionFor a globally defined version of the right side, assign any value, for example zero, on $\{X=1\}$.

Indeed, the displayed random variable is ${\mathcal{F}}_n$-measurable and bounded by $\sup |h|$. On each $A_{n,k}$ its integral isby the uniform distribution. Every member of ${\mathcal{F}}_n$ is a union of the displayed sets and the possible zero-probability atom. Summing these identities therefore verifies the defining conditional expectation property on all of ${\mathcal{F}}_n$.