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I would first collapse the duration noises as in part (c), then use self-normalized importance sampling on the two shared parameters. Draw independent

(μ_{j}, σ_{j}) \sim ρ

, give each the importance weight

w_{j} = L (μ_{j}, σ_{j})

, and return

P_{N} = \frac{\sum _{j = 1}^{N} w _{j} π ( μ _{j} , σ _{j} ) ^{5}}{\sum _{j = 1}^{N} w _{j}} .

This 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

[0, 1]

, 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-

N

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, ess

The 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

ρ L

. 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 probability course of the University of Oxford / 2014 / 3 / c / i / Solution 2026-10-31

Represent the simple random walk by

X_{n} = ξ_{1} + \dots + ξ_{n}

, where the increments are independent and identically distributed and satisfy

P (ξ_{k} = 1) = p

P (ξ_{k} = - 1) = 1 - p

. They are integrable and have expectation

2 p - 1

, so the strong law of large numbers gives

\frac{X _{n}}{n} ⟶ μ = 2 p - 1 almost surely .

p \neq = 1/2

, this nonzero drift implies that

X_{n}

eventually leaves every fixed finite interval and tends to the appropriate infinity. Hence it visits zero only finitely many times almost surely, so zero is a transient state. By translation of the starting state, every state is transient: the chain is transient for $p \neq = 1/2$ . This also covers the deterministic endpoint cases

p = 0, 1

, without assuming Markov chain irreducibility at those endpoints.

Past exam of the probability course of the University of Oxford / 2015 / 3 / a / Solution 2026-10-31

The strong law of large numbers says that, for real independent and identically distributed random variables

Z_{1}, Z_{2}, \dots

with

E ∣ Z_{1} ∣ < \infty

P (\frac{1}{n} i = 1 \sum n Z_{i} ⟶ E Z_{1}) = 1.

The assertion is almost sure convergence of the entire sequence of sample means. Finite first absolute moment suffices; a finite variance is not required.

Past exam of the probability course of the University of Oxford / 2015 / 3 / b / i / Solution 2026-10-31

Put

q = 1 - p

and

ξ_{i} = Y_{i} - Y_{i - 1}

. The Markov property and the state-independent transition probabilities show that, conditional on the whole previous history,

ξ_{i}

+ 1

with probability

p

and

- 1

with probability

q

. Multiplying these conditional probabilities for any finite step sequence proves that the

ξ_{i}

are independent and identically distributed random variables, with

E ξ_{i} = p - q = 2 p - 1 > 0.

Because

Y_{n} = k + \sum_{i = 1}^{n} ξ_{i}

, the strong law of large numbers gives

\frac{Y _{n}}{n} = \frac{k}{n} + \frac{1}{n} i = 1 \sum n ξ_{i} a.s. μ = 2 p - 1.

In particular,

Y_{n} \to + \infty

almost surely. This stronger pathwise consequence will determine both eventual crossings and the asymptotic first-passage times.

Past exam of the probability course of the University of Oxford / 2021 / 3 / b / iv / Solution 2026-10-25

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time

k (a + b)

, return to zero requires exactly

kb

upward steps. The binomial distribution gives

P_{0} (S_{k (a + b)} = 0) = (kb k ( a + b )) (\frac{b}{a + b})^{kb} (\frac{a}{a + b})^{ka} \sim \frac{a + b}{2 πabk},

where the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence

\sum_{n \geq 0} P_{0} (S_{n} = 0) = \infty

. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let

L_{1}, L_{2}, \dots

be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives

E L_{1} = \infty

. For each positive integer

K

, the strong law of large numbers applied to

min (L_{j}, K)

gives

lim inf_{r \to \infty} \frac{L _{1} + \dots + L _{r}}{r} \geq E min (L_{1}, K) almost surely .

Take a common event of probability one for these countably many

K

. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time

n

, divided by

n

, tends to zero. Counting the initial visit changes this ratio by at most

1/ n

. Consequently,

\frac{1}{n} r = 0 \sum n - 1 1_{{S_{r} = 0}} ⟶ 0 almost surely .

The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.

Past exam of the probability course of the University of Oxford / 2023 / 1 / a / Solution 2026-10-25

The strong law of large numbers states that, for independent and identically distributed random variables

X_{1}, X_{2}, \dots

with

E ∣ X_{1} ∣ < \infty

\frac{1}{n} j = 1 \sum n X_{j} ⟶ E X_{1} almost surely .

Thus, outside an event of probability zero, the sample mean converges to the common expected value.

Past exam of the probability course of the University of Oxford / 2023 / 1 / d / ii / Solution 2026-10-25

Taking logarithms of the product gives

W_{n} = exp (- j = 1 \sum n V_{j}) .

By part (d)(i),

V_{j} = E_{j} - lo g b

in probability distribution, where

E_{j}

has the exponential distribution of rate one. Thus

E ∣ V_{j} ∣ \leq 1 + ∣ lo g b ∣ < \infty

and

E V_{j} = 1 - lo g b

. The strong law of large numbers gives

\frac{1}{n} lo g W_{n} = - \frac{1}{n} \sum_{j = 1}^{n} V_{j} ⟶ lo g b - 1 almost surely .

b < e

, choose

δ = (1 - lo g b) /2 > 0

. Eventually

lo g W_{n} \leq - δ n

, so

W_{n} \leq e^{- δ n} \to 0

. If

b > e

, choose

δ = (lo g b - 1) /2 > 0

; eventually

W_{n} \geq e^{δ n} \to \infty

. Therefore

W_{n} \to 0 a.s. if 0 < b < e; W_{n} \to \infty a.s. if b > e .

In particular, the second range cannot have almost sure convergence to zero. This is the almost-sure growth rate of independent products. At

b = e

the logarithmic limit is zero, which by itself decides neither behavior; that boundary case is not requested.

Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / S / 1 / b / Solution 2026-10-31

For unknown normalization constants, use self-normalized importance sampling with

\tilde{w} = \tilde{p} / \tilde{q}

\tilde{μ}_{N} = \frac{\sum _{i} w ~ ( X _{i} ) f ( X _{i} )}{\sum _{i} w ~ ( X _{i} )} .

Under the same support condition,

E_{q} \tilde{w} = Z_{p} / Z_{q} > 0

and

E_{q} [\tilde{w} f] = (Z_{p} / Z_{q}) μ

. Moreover

E_{q} ∣ \tilde{w} f ∣ = (Z_{p} / Z_{q}) E_{p} ∣ f ∣ < \infty

, by Cauchy-Schwarz inequality and the supplied

E_{p} f^{2} < \infty

. The strong law of large numbers applies to numerator and denominator sample averages. Their ratio tends almost surely, hence in probability, to

μ

. If a finite sample happens to give zero denominator, it can be assigned an arbitrary value; this does not affect the eventual almost sure limit. The ratio estimator is generally biased at finite

N

Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 1 / a / i / Solution 2026-10-25

Generate independent and identically distributed random variables

X_{1}, \dots, X_{n}

with probability density function

p

, evaluate

ϕ (X_{i})

for each draw, and return the Monte Carlo estimator

θ_{n}^{MC} = \frac{1}{n} i = 1 \sum n ϕ (X_{i}) .

Linearity of expectation makes this an unbiased estimator:

E θ_{n}^{MC} = θ

. Independence and the specified variance give

Var (θ_{n}^{MC}) = 1/ n

. Its Monte Carlo standard error is therefore

n^{- 1/2}

, and the strong law of large numbers gives convergence to

θ

almost surely.

Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 2 / a / iv / Solution 2026-10-25

Use the same acceptance indicator functions and set

K_{n} = \sum_{i = 1}^{n} A_{i}

S_{n} = \sum_{i = 1}^{n} A_{i} ϕ (Y_{i})

. The rejection sampling estimator based on these

n

proposals is the mean over the accepted observations:

θ_{n}^{RS} = \frac{S _{n}}{K _{n}} (K_{n} > 0) .

For

K_{n} = 0

, any fixed convention, such as zero, is harmless asymptotically. The

A_{i}

are independent and identically distributed random variables with Bernoulli distribution and success probability

1/ M

, so the strong law of large numbers gives

K_{n} / n \to 1/ M

almost surely. In particular

K_{n} > 0

eventually almost surely. The exact identity

θ_{n}^{IS} = \frac{M K _{n}}{n} θ_{n}^{RS}

then holds whenever

K_{n} > 0

. Wherever

S_{n} \neq = 0

, it follows that

\frac{θ _{n}^{IS}}{θ _{n}^{RS}} = \frac{M K _{n}}{n} ⟶ 1 almost surely .

A qualification is missing from the printed claim: the quotient must be defined eventually. A convenient sufficient assumption is

E_{p} ∣ ϕ ∣ < \infty

and

θ \neq = 0

. Indeed, another application of the strong law of large numbers gives

S_{n} / n \to θ / M

, ensuring that

S_{n}

is eventually nonzero. This proves the stated quotient limit under that assumption.

As printed, the assertion is false as a universally defined quotient. For example,

ϕ \equiv 0

makes both estimates zero for every

n

. Even excluding that example is insufficient: take

p = q

to be the uniform distribution on

(0, 1)

M = 1

, and

ϕ (y) = 1

for

y < 1/2

and

- 1

otherwise. Then

S_{n}

is a simple symmetric random walk, which returns to zero infinitely often almost surely, leaving the quotient undefined infinitely often.

With integrability but no assumption

θ \neq = 0

, the exact multiplicative identity remains valid, and both estimates converge to

θ

almost surely. In particular

θ_{n}^{IS} - θ_{n}^{RS} \to 0

almost surely. One may alternatively define the quotient's removable

0/0

cases as

M K_{n} / n

; that explicitly extended quantity always converges to one. These are valid repairs of the source's unrestricted statement.

Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2008 / B / 6 / ii / Solution 2026-10-31

planar Brownian motion is recurrence of planar Brownian motion when, almost surely, it returns to every nonempty open neighborhood infinitely often, at arbitrarily large times. This does not mean that it hits each specified point; a prescribed point is polar for planar Brownian motion.

recurrence of planar Brownian motion and continuity imply that all the circle-hitting times

S_{k}, T_{k}

in the question are finite. Starting on radius

ϵ

, it eventually visits the exterior of radius

2 ϵ

and must cross that circle; it later visits the interior of radius

ϵ

and must cross back. Each duration

Δ_{k} = T_{k} - S_{k}

is strictly positive by path continuity, because the two radii are separated.

S_{k}

, the Strong Markov property restarts Brownian motion at its current point on the inner circle. rotational invariance of Brownian motion makes the law of the subsequent first exit time from the radius-

2 ϵ

disk independent of that starting angle. Its conditional distribution given the past is therefore the same fixed law at every cycle, proving that the

Δ_{k}

are independent and identically distributed.

To prove divergence of the conformal clock for any nonconstant entire

f

, choose a point

z_{0}

with

f^{'} (z_{0}) \neq = 0

. By continuity, there exist

ρ, c > 0

such that

∣ f^{'} (z) ∣^{2} \geq c

throughout the closed disk

∣ z - z_{0} ∣ \leq 2 ρ

. Use the same recurrence of planar Brownian motion circle cycles centered at

z_{0}

, with radii

ρ, 2 ρ

. On each segment from the inner circle to the outer circle, the path remains in that disk, giving

\int_{0}^{\infty} ∣ f^{'} (Z_{s}) ∣^{2} d s \geq c \sum_{k \geq 0} (T_{k} - S_{k}) .

The durations are positive iid variables. If their mean has not been computed, apply the strong law of large numbers to the bounded iid variables

min (Δ_{k}, 1)

: their mean is strictly positive, so their sum, and hence

\sum Δ_{k}

, diverges almost surely. Therefore

\int_{0}^{\infty} ∣ f^{'} (Z_{s}) ∣^{2} d s = \infty almost surely .

Centering the disks at a noncritical point avoids an unjustified positive lower bound on

∣ f^{'} ∣

around the original origin.

Write

f = u + i v

. Harmonicity and Itô's formula make

u (Z_{t}), v (Z_{t})

continuous local martingales. The Cauchy-Riemann equations give equal brackets

\int_{0}^{t} ∣ f^{'} (Z_{s}) ∣^{2} d s

and zero mutual bracket. Consequently Lévy's characterization of Brownian motion after time change of a continuous process makes

W

a planar Brownian motion started at

f (Z_{0})

, with no finite clock lifetime because the clock just proved is unbounded. The clock is strictly increasing: otherwise

f^{'}

would vanish along a nonconstant path interval, contradicting its isolated zeros.

For the notation in the source,

∣\nabla f ∣^{2}

is interpreted as the conformal scale

∣ f^{'} ∣^{2} = ∣\nabla u ∣^{2} = ∣\nabla v ∣^{2}

. If it instead means the sum of the squared magnitudes of both complex coordinate derivatives, it is

2∣ f^{'} ∣^{2}

, and the resulting

W

runs at half the usual Brownian speed. This convention does not change divergence of the integral. The mechanism is recurrent disk occupation makes an entire conformal clock infinite.

Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2010 / A / 2 / b / Solution 2026-10-31

For positive times the transformed process is centered and jointly Gaussian, and

E (X_{s} X_{t}) = s t min (1/ s, 1/ t) = min (s, t) .

Its paths are continuous away from zero. To verify continuity at zero, we need

B_{u} / u \to 0

almost surely as

u \to \infty

. The strong law of large numbers applied to the independent increments

B_{n + 1} - B_{n}

gives

B_{n} / n \to 0

. On the unit interval after

n

, the Doob L2 maximal inequality gives

P (sup_{0 \leq v \leq 1} ∣ B_{n + v} - B_{n} ∣ > ε n) \leq \frac{4}{ε ^{2} n ^{2}} .

The Borel-Cantelli first lemma, first for each positive rational

ε

, shows that these between-integer fluctuations divided by

n

tend to zero. Hence

B_{u} / u \to 0

for all real

u \to \infty

, and

t B_{1/ t} \to 0

t ↓ 0

The Gaussian-process characterization of Brownian motion now proves $X$ is standard Brownian motion, with respect to its natural filtration. This is time inversion of Brownian motion; the transformed process need not be adapted to the original forward filtration.

Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 6 / b / iv / Solution 2026-10-25

A general-purpose high-dimensional choice is Monte Carlo integration. Draw independent vectors

W_{r}

uniformly from the unit cube and evaluate

H (W_{r}) = \sum_{i = 1}^{100} e^{∥ W_{r} - a_{i} ∥^{2}}

. The cube has volume

1

, so its integral is the expected value of

H (W_{1})

. The iterative estimator is

y_{N} = \frac{1}{N} \sum_{r = 1}^{N} H (W_{r}), y_{N + 1} = y_{N} + \frac{H ( W _{N + 1} ) - y _{N}}{N + 1} .

The integrand is continuous on a compact set, hence bounded with finite variance. The strong law of large numbers gives almost sure convergence, and the square root of the mean squared error is exactly

σ / N

, where

σ^{2} = Var (H (W_{1}))

. Thus the standard stochastic accuracy rate is $O (N^{- 1/2})$ ; the dimension affects the variance and evaluation cost, not this exponent.

There is also an important structure-specific improvement here:

e^{∥ x - a_{i} ∥^{2}} = \prod_{j = 1}^{100} e^{(x_{j} - a_{ij})^{2}}

is a separable integrand. The Fubini theorem reduces the requested value exactly to

y = \sum_{i = 1}^{100} \prod_{j = 1}^{100} \int_{0}^{1} e^{(t - a_{ij})^{2}} d t .

One can therefore replace high-dimensional sampling by

10000

one-dimensional composite Simpson's rule calculations. For each factor use even

n

h = 1/ n

, and

Q_{n} (g) = \frac{h}{3} [g (0) + g (1) + 4 \sum_{r = 1, r odd}^{n - 1} g (r h) + 2 \sum_{r = 2, r even}^{n - 2} g (r h)],

then iterate

n \leftarrow 2 n

and form the sum of the products. These smooth factors give $O (n^{- 4})$ deterministic discretization error for fixed centres. Thus Monte Carlo integration is a valid generic answer, while exploiting separability is the more efficient specialized alternative; a large dimension alone does not force a tensor-product grid.

Quantile proof of the Glivenko-Cantelli theorem 2026-10-14

For a cumulative distribution function

G

, let

Q (u) = in f {x : G (x) \geq u}

0 < u < 1

. Right continuity gives

Q (u) \leq x

exactly when

u \leq G (x)

, including at atoms. Thus applying

Q

to independent uniform variables

U_{i}

constructs an independent sample with distribution function

G

. Its empirical distribution function is

G_{n} (x) = F_{n} (G (x))

, where

F_{n}

is the uniform-sample empirical distribution function. Consequently

sup_{x} ∣ G_{n} (x) - G (x) ∣ \leq sup_{t \in [0, 1]} ∣ F_{n} (t) - t ∣

. For the uniform sample, monotonicity and a grid of mesh

1/ m

bound the last supremum by the largest error at grid points plus

1/ m

. The strong law of large numbers on the countable union of all such grids proves the almost-sure limit. This construction proves the theorem for distributions with atoms as well as continuous distributions.

Rare-jump counterexample to a strong law with convergent first-moment averages 2026-10-31

Take independent random variables equal to

1 + n

and

1 - n

with probability

1/ (2 n lo g n)

each, and otherwise equal to one. Their first absolute deviations are

1/ lo g n \to 0

, so

X_{n} \to 1

in L1 convergence. The same triangle-inequality argument gives convergence in

L^{1}

of their arithmetic averages to one. For

p > 1

, however,

E ∣ X_{n} - 1 ∣^{p} = n^{p - 1} / lo g n \to \infty

The exceptional jumps occur infinitely often almost surely by the Borel-Cantelli second lemma, since

\sum 1/ (n lo g n) = \infty

. If the arithmetic averages

S_{n} / n

converged to any finite value along a sample path, then

X_{n + 1} / n = S_{n} / n - [(n - 1) / n] [S_{n - 1} / (n - 1)]

would tend to zero. At each exceptional jump its magnitude is at least one. Thus the averages have no finite almost sure limit. Independence and convergence of first-moment averages do not alone give a strong law of large numbers.

Repeat-until-success coding over an erasure channel 2026-10-25

With channel feedback, resend each message bit through a binary erasure channel until a non-erased copy arrives, then move to the next bit. The channel decoder deletes erasure symbols, giving the message without error whenever transmission completes. For

q < 1

, the waiting times have an independent and identically distributed geometric distribution with expectation

1/ (1 - q)

, so every finite message completes almost surely. The strong law of large numbers gives forward code rate

n / \sum_{j = 1}^{n} M_{j} \to 1 - q

, matching channel capacity. At

q = 1

, transmission never completes. Independent erasure indicators, rather than independent message bits, are the essential hypothesis.

Short-time Gaussian spin-bath decoherence 2026-10-07

For initially equatorial spins and uniform couplings on

[0, g_{*}]

, expand

lo g cos (2 g_{k} t) = - 2 g_{k}^{2} t^{2} + O (g_{k}^{4} t^{4})

. The strong law of large numbers then gives

\sum g_{k}^{2} ≃ N g_{*}^{2} /3

. On the scale

t = O ((g_{*} N)^{- 1})

, the total fourth-order remainder tends to zero, giving the Gaussian coherence envelope. This approximation concerns early times and does not remove finite spin-bath coherence recurrence.

Stationary workload supremum 2026-10-14

For independent and identically distributed arrivals with finite mean below service

c

, the strong law of large numbers makes the net cumulative input tend to minus infinity, so the supremum is finite almost surely. Applying the same formula to the past input gives a stationary solution of the Lindley recursion. This representation turns queue workload overflow into an event involving some lookback horizon.

Strong law for adjacent products Created 2026-10-05 Updated 2026-10-06

For independent and identically distributed random variables

X_{j}

with

E ∣ X_{1} ∣ < \infty

and

m = E X_{1}

, the sample mean of

X_{j} X_{j + 1}

tends to

m^{2}

almost surely. Split into odd and even indexed products; each subsequence consists of independent and identically distributed random variables that are integrable random variables. Apply the strong law of large numbers separately.

Strong law for Poisson arrival times 2026-10-14

For a homogeneous Poisson process of rate

λ > 0

N (t) / t \to λ

almost surely: unit-time increments are independent Poisson variables and the strong law of large numbers applies, then monotonicity fills the intervals between integers. At its

n

th point

Y_{n}

, simplicity gives

N (Y_{n}) = n

. Consequently

Y_{n} / n \to 1/ λ

. Rate and mean interarrival time are reciprocals.