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Incoming links: Conditional expectation
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 30K c i Solution Created 2026-09-24 Updated 2026-09-29
Assume (i), and let be a stopping time. The increments of the stopped process satisfyBecause , taking conditional expectation given makes the right-hand side zero. Integrability follows froma finite sum of integrable random variables. Hence is a martingale, proving (ii).
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 4 29K b Solution Created 2026-09-24 Updated 2026-10-03
The arbitrage-free value of the discounted claim is the conditional expectationConditional on the observed path , the future multiplicative increments are independent of and have the same law as a fresh stock path divided by its initial value. ThereforewhereThis is Risk-neutral valuation in the Cox--Ross--Rubinstein model for a path-dependent claim.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 201 1 a Solution 2026-09-28
For an integrable random variable and a sub-sigma-algebra , the conditional expectation is an integrable, -measurable random variable satisfyingfor every .
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 208 2 a Solution 2026-09-28
Write for all coordinates except , let be an independent random variable with the same distribution as , and let . Three equivalent forms of the Efron–Stein inequality arefor arbitrary square-integrable measurable with respect to , andThe first is the sharp choice within the second because conditional expectation is the least-squares projection. The first and third right sides are equal because two conditionally independent copies have expected squared difference twice their conditional variance.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 219 2 a Solution 2026-09-28
The vector is Jointly Gaussian. Assuming , Gaussian conditional independence givesThus the required condition is . Under it, conditioning on supplies no further information after , and the Gaussian process regression posterior isBoth the conditional expectation and conditional variance depend only on ; neither contains or .
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 216 1 c Solution 2026-09-28
For a decision in the unit ball, conditional expectation under the posterior givesThe symmetric positive semidefinite matrix has a Rayleigh quotient maximized over by any unit eigenvector corresponding to its largest eigenvalue. Bayes decision rule therefore chooses any such eigenvector for the stated utility.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 202 2 c ii Solution Created 2026-09-24 Updated 2026-10-03
Let . It is a continuous square-integrable martingale, and the assumed bracket identity givesfor every continuous square-integrable martingale . Choose and use part (i):Thus almost surely, and the conditional expectation property gives for every . Hence up to indistinguishability of stochastic processes.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 201 3 a Solution Created 2026-09-24 Updated 2026-09-25
The conditional expectation is the almost surely unique integrable, -measurable random variable such that for every .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 201 1 a Solution Created 2026-09-24 Updated 2026-09-25
The finite sigma-algebra is partitioned by the nonempty eventsThese are its atoms. Define the random variableand give it any finite value on the union of the null atoms. It is -measurable and integrable. Every is a union of atoms, soThis constructs the requested variable directly, without invoking the general existence theorem for conditional expectation.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 202 4 b Solution Created 2026-09-24 Updated 2026-09-25
With zero interest rate, the bank account is constant. The risky asset is a continuous local martingale by assumption, while the European contingent claim priceis a true martingale by the defining property of conditional expectation. Thus the original probability measure is an equivalent local martingale measure for all traded discounted prices. The fundamental theorem of asset pricing then excludes arbitrage for admissible self-financing strategies.
I would first collapse the duration noises as in part (c), then use self-normalized importance sampling on the two shared parameters. Draw independent , give each the importance weight , and returnThis is likelihood weighting: the observations affect the weights rather than forcing repeated simulation of twenty latent inequalities. Integrating out both the constraints' auxiliary randomness and the future Boolean query is Rao-Blackwellization; at fixed parameters it removes randomness without changing the relevant conditional expectations.
The weights and weighted queries lie in , and the evidence is positive. The strong law of large numbers therefore makes the numerator and denominator averages converge to their corresponding integrals, proving almost sure convergence of their ratio to part (d). The finite- ratio is generally biased; the assertion concerns consistency rather than a finite-sample unbiased estimator. Use logarithms of importance weights to avoid numerical underflow:
def weighted_answer(n, rng):
if n <= 0:
raise ValueError("n must be positive")
samples = []
for _ in range(n):
mu = rng.uniform(0.0, 60.0)
sigma = rng.uniform(1.0, 20.0)
samples.append((journey_loglik(mu, sigma),
math.exp(5.0 * normal_logcdf((mu - 30.0) / sigma))))
offset = max(logw for logw, h in samples)
if not math.isfinite(offset):
raise ArithmeticError("all weights underflowed")
scaled = [(math.exp(logw - offset), h) for logw, h in samples]
den = math.fsum(w for w, h in scaled)
answer = math.fsum(w * h for w, h in scaled) / den
ess = den * den / math.fsum(w * w for w, h in scaled)
return answer, essThe returned effective sample size of importance sampling is a concentration diagnostic, not a proof of accuracy. Repeat runs and inspect the weights. If the prior is much broader than the posterior, improve the proposal with the corresponding importance sampling density correction, or use the Metropolis–Hastings algorithm targeting . Because this particular collapsed problem is only two-dimensional, deterministic numerical integration is also an excellent independent accuracy check; it was used for the figure and numerical illustration. Straight rejection of the original traces is correct but wastes most proposals here.
Past exam of the optimisation for data science course of the University of Oxford 2023 2 c Solution 2026-10-25
There is a summation-index misprint. On the native PDF, paper page 3, equation (4) has inside a sum indexed by . The intended telescoping estimate has inside that sum; this is the estimate proved below and used in part (d). The index mismatch is present in the original PDF, rather than being introduced by the transcription.
Take the expectation of the supplied conditional expectation bound (3), using the tower property:Rearranging and forming a telescoping sum givesBecause is a global minimizer, for every realisation. Hence the corrected equation (4) isAll expectations at a finite iteration are finite here: a deterministic starting point and finitely many choices of the sampled index produce only finitely many possible iterates. The bound controls accumulated squared gradients, with an additive contribution from the stochastic gradient noise floor.
Past exam of the optimisation for data science course of the University of Oxford 2023 3 d Solution 2026-10-25
The supporting inequality in part 2(a) bounds below by an affine function plus a positive squared Euclidean norm, making it a coercive function. Its continuity gives a global minimizer , which is unique by strict convexity. Let , and . Here is the block size, so its role differs from the number of component functions in Question 2. Use randomized block coordinate descent with uniformly sampled blocks of size , independently at each update. Sampling uniformly from a fixed partition into equal-sized blocks gives the same calculation, since each coordinate has inclusion probability .
Let be a common upper bound on the block Lipschitz constants. Part (c) permits . With step size , part (a) givesSumming squared partial derivatives according to their inclusion probabilities,For a strongly convex function, part 2(a) also yields the strong convexity gradient inequalityIndeed, substitute into the supporting inequality and complete the square in : the expression is at least for every .
Combining the last three estimates gives a contraction in conditional expectation:By the tower property and repeated application,An optimization epoch counts coordinate partial derivative evaluations: it comprises block updates or one full gradient update. With sampling with replacement, an optimization epoch is this work measure; it need not visit every coordinate. Consequently the block method needsto bring the expected objective function gap below , for , up to integer rounding.
For the steepest descent method, use a global Lipschitz gradient constant . Part (c) with the full block permits . The descent lemma with step size , together with the strong convexity gradient inequality, givesAn exact line search along the negative gradient does at least as well in objective function decrease per update. Ignoring its additional work, the resulting epoch complexity isThe common coordinate-based guarantees favour blocks by a factor in epochs. In fact, with the conservative constants and , both methods have the same guaranteed contraction per update, while a block update uses only coordinate partial derivatives.
This compares the bounds justified by the given constants. For a particular objective function, sharper constants give the comparison against ; an actual speedup of is not universal. Computing a block coordinate gradient also need not cost exactly of a full gradient evaluation. The epoch complexity comparison assumes that coordinate partial derivative counts are a useful measure of work, and does not itself establish a wall-clock comparison.
For , the conditional density is obtained by dividing the joint density by the positive marginal density:Consequentlywhich integrates to one. Its conditional expectation isThis uses the usual continuous conditional-density version, defined by the density ratio at a point where is positive; conditioning does not require the event itself to have positive probability.
Write and . The two conditional expectation identities supply a -measurable version of and a -measurable version of . Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.
Testing the first conditional expectation identity against , and the second against , givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations and , then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.
For a counterexample to equality of the original random variables, let give probability to each of , set , and use the nonconstant function . Then takes both values and . It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives , while neither random variable can be zero. Thus
Put . Both and its indicator function are measurable, and . Thus is an integrable random variable. Monotonicity of conditional expectation givesThe measurable-factor property of conditional expectation says that a bounded -measurable function satisfies for every integrable random variable . Here the factor need not be bounded. Apply that property first to and , then use monotone convergence on the nonnegative products, including the conditional version, to obtainAll the left-hand conditional expectations are well defined; the final one is finite almost surely because is an integrable random variable. Since almost surely, division now yieldsThis proves the conditional Markov inequality with a measurable threshold without assuming that is an integrable random variable. The measurable-factor rule, with its product-integrability condition, is also stated in the standard conditional expectation theory.[ref]
The adapted process is bounded by , so every is an integrable random variable, indeed a square-integrable random variable. Independence of from and its continuous uniform distribution giveThis calculation with a random threshold is justified by conditioning on the -measurable value of : for a fixed , the Lebesgue integral equals . More formally, independence integrates the bounded Borel measurable function over its second coordinate. Thereforewhich verifies the martingale property for this filtration.
Since , the conditional expectation of the squared martingale difference isThe law of total expectation yields . In particular, for every .
The L2-bounded martingale convergence theorem states that a martingale with converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because . Thus there is almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound gives convergence in L2 by the dominated convergence theorem.[ref]
Probability: Theory and Examples (fifth edition) by Rick Durrett. §4.4, Theorem 4.4.6, printed p. 237: the Lp martingale convergence theorem
The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from givesTaking expected values and iterating provesBy bounded convergence theorem, . The random variable inside this expected value is nonnegative, so it is zero almost surely; on this forces . Again by bounded convergence theorem, . HenceThis is the Bernoulli polarization martingale with update size . It also has an exact mean squared error: the terminal representation and the equality case for conditional second moments give
The first entrance time is a stopping time, because . Before this time ; at entrance, the bounded increments give . Thus the stopped martingaleis a nonnegative martingale. Here the stopped process is a martingale since its increment is , whose conditional expectation given is zero.
The martingale convergence theorem states that a nonnegative martingale converges almost surely to a finite integrable random variable; indeed its expectations are constant, giving the required bound in the L1 norm. Apply this theorem to . On the stopped process equals the original one at every time. HenceThe overshoot bound is why the assumption of bounded increments matters.