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For a twice continuously differentiable scalar potential, mixed second partial derivatives commute. Taking the spatial curl of the equation from part (i) gives

\nabla \times \ddot{x} = - \nabla \times \nabla ξ = 0 .

(1)

For example, its first component is

- \partial_{2} \partial_{3} ξ + \partial_{3} \partial_{2} ξ = 0

. The curl acts on the spatial representation of the material acceleration, not directly on its expression in reference coordinates.

This is a necessary local condition. It is sufficient locally, and sufficient globally on a simply connected domain under the usual smoothness assumptions. On a domain with holes, a global scalar potential additionally requires every closed-loop line integral of the material acceleration to vanish. The homogeneous motion in part (c) admits an explicit global scalar potential, so this topological qualification causes no obstruction there.

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The pressure adjustment from part (i) can be written

T = T_{0} (t) + q (x, t) 1, q = - ρ ξ + C (t) .

(1)

Indeed, since

div (q 1) = \nabla q

, the difference of the two Cauchy momentum equations is

\nabla q = ρ \ddot{x}_{0} = - ρ \nabla ξ .

(2)

Thus

q + ρ ξ

is constant in space on each connected component. The function

C (t)

is fixed by any prescribed pressure or surface traction datum. In terms of the usual compressive pressure multiplier,

p = p_{0} - q = p_{0} + ρ ξ - C (t)

. The plus sign in

q 1

therefore does not identify

q

with the compressive multiplier

p

itself.

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Restore the particle label suppressed in the question, writing the static family as

x_{0} (X; τ)

. At each fixed

τ

, its Cauchy stress tensor satisfies

div T_{0} (τ) + ρ b = 0 .

(1)

This dynamic continuation of a static incompressible elastic deformation sets

τ = t

to form a motion. At every time its deformation gradient is the same as that of the corresponding static member. For an elastic material the elastic part of the Cauchy stress tensor depends on that deformation gradient, while incompressibility leaves an arbitrary scalar constraint reaction in hyperelastic stress. Therefore write

T_{0} = S (F) - p_{0} 1, T = S (F) - p 1, T - T_{0} = (p_{0} - p) 1 .

(2)

Here

1

is the identity matrix. Subtracting static balance of linear momentum from dynamic balance of linear momentum gives

ρ \ddot{x}_{0} = \nabla (p_{0} - p) .

(3)

Since the homogeneous incompressible material has constant mass density, define

ξ = (p - p_{0}) / ρ

. Necessarily

\ddot{x}_{0} (X, t) = - \nabla ξ (x_{0} (X, t), t) .

(4)

Conversely, suppose this acceleration potential exists. Choose

p = p_{0} + ρ ξ

. Then

div T + ρ b = div T_{0} + ρ b - ρ \nabla ξ = ρ \ddot{x}_{0} .

(5)

The added isotropic stress is symmetric, so angular momentum conservation remains satisfied. This proves both directions for the field equations and the elastic constitutive equation. The surface tractions for the motion are those of the adjusted stress; identical prescribed static tractions would impose an additional boundary restriction on

ξ

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For general

α > 0

, put

S (λ) = λ^{- 4} + 2 λ^{2}

. The reduced strain energy density and its derivative are

h (λ) = \frac{μ}{2} [S (λ)^{α} - 3^{α}], h^{'} (λ) = 2 μα S (λ)^{α - 1} (λ - λ^{- 5}) .

(1)

The same factorization as in part (i) yields

\frac{h ^{'} ( λ )}{λ ^{3} - 1} = 2 μα S (λ)^{α - 1} (λ^{- 2} + λ^{- 5}) .

(2)

This form removes the apparent singularity at

λ = 1

: its limit is

4 μα 3^{α - 1}

, finite for every fixed

α > 0

. Thus only the upper endpoint can obstruct convergence of the improper integral.

λ \to \infty

S (λ) \sim 2 λ^{2}

, giving

\frac{h ^{'} ( λ )}{λ ^{3} - 1} \sim μα 2^{α} λ^{2 α - 4} .

(3)

For

μ > 0

, the comparison test with a power integral implies convergence exactly when

2 α - 4 < - 1

. Therefore

0 < α < \frac{3}{2}

(4)

is precisely the range with finite critical pressure, the cavitation criterion for power-law strain energy. At

α = 3/2

the divergence is logarithmic; for larger

α

it is a power divergence. These materials require an unbounded tensile loading magnitude in this ideal model. The qualification

μ > 0

excludes the degenerate zero-energy material.

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The neo-Hookean solid corresponds to

α = 1

, with positive shear modulus

μ

. Substitution of the principal stretches gives

h (λ) = \frac{μ}{2} (λ^{- 4} + 2 λ^{2} - 3), h^{'} (λ) = 2 μ (λ - λ^{- 5}) .

(1)

The identity

λ^{6} - 1 = (λ^{3} - 1) (λ^{3} + 1)

simplifies the integrand:

\frac{h ^{'} ( λ )}{λ ^{3} - 1} = 2 μ (λ^{- 2} + λ^{- 5}) .

(2)

The two improper integrals are elementary:

P_{crit} = - 2 μ (\int_{1}^{\infty} λ^{- 2} d λ + \int_{1}^{\infty} λ^{- 5} d λ) = - 2 μ (1 + \frac{1}{4}) = - \frac{5}{2} μ .

(3)

Thus the critical tensile stress magnitude is

5 μ /2

. As a check on the finite-sphere limit, part (c) gives

P (a) = - 2 μ (λ_{b}^{- 1} + \frac{1}{4} λ_{b}^{- 4}), λ_{b} = (1 + a^{3} / B^{3})^{1/3},

(4)

which tends to

- 5 μ /2

as the cavity radius tends to zero.

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Take the inner surface to be free of traction, as appropriate to an empty cavity with no surface tension. At the outer surface the imposed pressure is compressive when positive. The outward normal vectors are

- e_{r}

r = a

and

+ e_{r}

r = b

, giving

t_{r} (a) = 0, t_{r} (b) = - P .

(1)

Integrating the ordinary differential equation from part (ii) between the two surfaces gives

- P = \int_{λ_{a}}^{λ_{b}} \frac{h ^{'} ( λ )}{1 - λ ^{3}} d λ .

(2)

The outer stretch follows from incompressibility:

λ_{b}^{3} = \frac{b ^{3}}{B ^{3}} = 1 + \frac{A ^{3}}{B ^{3}} (λ_{a}^{3} - 1) .

(3)

Hence an expression entirely in terms of the inner stretch and the reference radii is

P (λ_{a}) = \int_{λ_{a}}^{[1 + (A / B)^{3} (λ_{a}^{3} - 1)]^{1/3}} \frac{h ^{'} ( λ )}{λ ^{3} - 1} d λ .

(4)

Equivalently

P = - \int_{λ_{b}}^{λ_{a}} h^{'} (λ) / (λ^{3} - 1) d λ

. For an expanding shell,

1 < λ_{b} < λ_{a}

; when

h^{'} > 0

on this range,

P < 0

, correctly describing an outward tensile loading. If an inner pressure

P_{i}

were present, the same derivation would give

P - P_{i}

on the left. Thus the empty-cavity condition specifies the pressure reference needed for the requested answer.

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Let

W_{r}, W_{θ}, W_{ϕ}

denote derivatives of the strain energy density with respect to its three principal stretches. The principal stress formula for incompressible hyperelasticity is

t_{r} = λ_{r} W_{r} - p, t_{θ} = λ_{θ} W_{θ} - p, t_{ϕ} = λ_{ϕ} W_{ϕ} - p .

(1)

It follows by applying

T = F (\partial_{F} W)^{T} - p 1

in principal axes. Equal tangential principal stretches and material isotropy give

W_{θ} = W_{ϕ}

(λ^{- 2}, λ, λ)

. Differentiating the reduced strain energy density by the chain rule gives

h^{'} (λ) = - 2 λ^{- 3} W_{r} + W_{θ} + W_{ϕ} = 2 (W_{θ} - λ^{- 3} W_{r}) = \frac{2}{λ} (t_{θ} - t_{r}) .

(2)

The unknown pressure cancels in this difference. Substituting in part (i), and then using the supplied derivative of

λ

, gives

\frac{d t _{r}}{d r} = \frac{λ h ^{'} ( λ )}{r} = \frac{h ^{'} ( λ )}{R}, \frac{d λ}{d r} = \frac{1 - λ ^{3}}{R} .

(3)

Consequently the desired ordinary differential equation is

\frac{d t _{r}}{d λ} = \frac{h ^{'} ( λ )}{1 - λ ^{3}} .

(4)

For completeness, the supplied derivative follows from

R^{3} = r^{3} - (a^{3} - A^{3})

d R / d r = r^{2} / R^{2} = λ^{2}

, and differentiating

λ = r / R

gives

(1 - λ^{3}) / R

. At the undeformed value

λ = 1

the division must be interpreted by a limit; the spherically symmetric deformation with

a = A

has

λ \equiv 1

and should be handled directly by the radial equilibrium equation.

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The Cauchy momentum equation in static elastic equilibrium, with zero body force, is

div T = 0

. The Cauchy stress tensor is symmetric. Material isotropy and the equal tangential principal stretches imply

t_{θ} = t_{ϕ}

, so its radial equation is

\frac{d t _{r}}{d r} + \frac{2}{r} (t_{r} - t_{θ}) = 0, equivalently \frac{d t _{r}}{d r} = \frac{2}{r} (t_{θ} - t_{r}) .

(1)

All stresses and spatial derivatives in this equation are measured in the current configuration.

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Only the

k = 0

term of the polynomial contributes to the limit:

\frac{M ( τ )}{τ} = \frac{A ^{4}}{2} (μ_{1} + 3^{n} μ_{2}) + O (τ^{2}) .

(1)

Thus the requested torsional stiffness per unit twist is

α = \frac{A ^{4}}{2} (μ_{1} + 3^{n} μ_{2}) .

(2)

The small-strain shear modulus is

G = 2 \overset{w}{^}^{'} (3) = 2 (μ_{1} + 3^{n} μ_{2})

. Therefore

α = G A^{4} /4

, consistently with the usual circular-shaft torque relation after restoring the

2 π

factor. If the prescribed end rotation is the angle

φ = τL

, the physical torsional stiffness is

d N / d φ = 2 π α / L

at zero twist. This also explains why

L

does not enter the requested

α

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Using

a = A

and the end-face shear stress from part (c), the printed torque normalization gives

M (τ) = 2 τ \int_{0}^{A} [μ_{1} + μ_{2} (3 + τ^{2} r^{2})^{n}] r^{3} d r .

(1)

The binomial theorem expands the remaining integral into powers of

r

(3 + τ^{2} r^{2})^{n} = \sum_{k = 0}^{n} (k n) 3^{n - k} τ^{2 k} r^{2 k} .

(2)

Integrating each term gives an expression valid also at

τ = 0

M (τ) = \frac{μ _{1} A ^{4}}{2} τ + μ_{2} k = 0 \sum n (k n) \frac{3 ^{n - k} A ^{2 k + 4}}{k + 2} τ^{2 k + 1} .

(3)

For example,

n = 1

gives

M = (μ_{1} + 3 μ_{2}) A^{4} τ /2 + μ_{2} A^{6} τ^{3} /3

. The torque is an odd polynomial in the twist per unit reference length, as expected when reversing the twist reverses the end-face shear stress.

For a physical end-face torque

N

, integration over the azimuth normally gives

N = \int_{0}^{2 π} \int_{0}^{A} T_{θ z} r^{2} d r d θ = 2 π M .

(4)

The original PDF explicitly defines

M

without this

2 π

, so the boxed answer retains its convention. Since

T

is the Cauchy stress tensor, its shear component here is spatial

T_{θ z}

; the printed capital

Z

T_{θZ}

does not require a change to a nominal stress tensor.

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Choose

k

distinct coordinates uniformly for the starting ones; each element of

S_{k}

then has starting probability

1/ (k n)

. At each iteration choose one current one and one current zero uniformly and swap them. This implements the normalized proposal distribution from (c)(i), remains in

S_{k}

, and needs just one precomputed matrix inverse.

mymcmc <- function(x, k, sigma) {
  n <- length(x)
  stopifnot(length(k) == 1L, is.finite(k), k == floor(k),
            k > 0L, k < n, is.matrix(sigma),
            all(dim(sigma) == c(n, n)))
  sigma_inv <- invsigma(sigma)
  u <- integer(n)
  u[sample.int(n, k, replace = FALSE)] <- 1L
  draws <- matrix(0, 100L, n)
  for (iteration in seq_len(100L)) {
    ones <- which(u == 1L)
    zeros <- which(u == 0L)
    i <- ones[sample.int(length(ones), 1L)]
    j <- zeros[sample.int(length(zeros), 1L)]
    v <- u
    v[i] <- 0L
    v[j] <- 1L
    alpha <- mylogapp(x, u, v, sigma_inv)
    if (runif(1L) < alpha) u <- v
    draws[iteration, ] <- u
  }
  colMeans(draws)
}

Sampling positions with sample.int and then indexing into ones or zeros also works when

k = 1

n - k = 1

. In R, a call such as sample(ones, 1) would incorrectly sample from 1:ones when ones is a single numeric value greater than one.

The code records the state after each of exactly

100

transitions, including repetitions after rejection. Its answer is the vector Monte Carlo estimator

m = \frac{1}{100} t = 1 \sum 100 U^{(t)} \approx E [U ∣ X = x] .

(1)

This estimates the posterior mean coordinate by coordinate; because each coordinate is binary, its entries estimate the posterior probabilities of inclusion. Every stored state has coordinate sum

k

, so the estimated means lie in

[0, 1]

and sum to

k

For neighbouring states, the symmetry of the proposal and the Metropolis–Hastings acceptance probability give detailed balance:

π (u ∣ x) q (v ∣ u) α (u, v) = q (v ∣ u) min {π (u ∣ x), π (v ∣ x)} = π (v ∣ x) q (u ∣ v) α (v, u) .

(2)

Thus the posterior distribution is a stationary distribution. Any two states in

S_{k}

can be connected by moving the ones that occupy the wrong positions, and the multivariate normal distribution gives strictly positive target masses. The resulting Markov chain is therefore an irreducible Markov chain, and its long-run state averages estimate the posterior mean.

The requested

100

iterations are a finite approximation, without a convergence guarantee or discarded burn-in. A randomly chosen starting subset has the prior distribution, rather than generally the posterior distribution. No aperiodicity assumption is needed for the finite-state Markov chain ergodic theorem for averages; indeed a uniform two-state target with

n = 2, k = 1

gives a period-two alternating chain. This does not invalidate either the code or the state-average estimator.

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The Cholesky decomposition gives

Σ = L L^{T}

. If

B = L^{- 1}

, reversal of factors in a matrix inverse gives

Σ^{- 1} = (L^{T})^{- 1} L^{- 1} = B^{T} B .

(1)

Hence the required R implementation is

invsigma <- function(sigma) {
  L <- mychol(sigma)
  B <- ltinv(L)
  crossprod(B)
}

Here crossprod(B) computes

B^{T} B

. Using tcrossprod(B) would compute

B B^{T}

, which generally differs from

Σ^{- 1}

. Both helper functions preserve the one-by-one matrix case.

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The source's function name suggests a log return value, and (c)(i) asks for

lo g α

, while the wording here and in (c)(iv) asks for the acceptance probability. We use the literal probability-returning convention and explicitly exponentiate the result of (c)(i):

mylogapp <- function(x, u, v, sigma_inv) {
  d <- v - u
  log_ratio <- as.numeric(crossprod(d, sigma_inv %*% (x - (u + v) / 2)))
  log_alpha <- min(0, log_ratio)
  exp(log_alpha)
}

Thus mylogapp returns

α \in [0, 1]

, and (c)(iv) compares it directly with a uniform distribution draw. Taking the minimum before exponentiating prevents overflow. The arguments

u, v

are assumed to be valid neighbouring states. A log-returning convention would instead return log_alpha and compare log(runif(1)) with it; the two conventions implement the same Metropolis–Hastings algorithm, provided the caller uses the matching comparison.

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Each state has exactly

k (n - k)

neighbours: choose one occupied coordinate and one unoccupied coordinate to swap. The normalized proposal distribution is therefore

q (v ∣ u) = \frac{1}{k ( n - k )} 1_{{v \sim u}} .

(1)

The swap relation is symmetric and the number of neighbours is constant, so

q (v ∣ u) = q (u ∣ v)

. The discrete uniform distribution used as the prior distribution is also constant on

S_{k}

, with mass

(k n)^{- 1}

. Both factors cancel in the Metropolis–Hastings acceptance probability.

Let

Q = Σ^{- 1}

and

d = v - u

. The displayed multivariate normal distribution presupposes that the covariance matrix

Σ

is a positive-definite matrix, so

Q

is a symmetric matrix. Expanding the difference of quadratic forms gives

lo g \frac{π ( v ∣ x )}{π ( u ∣ x )} = - \frac{1}{2} {(x - v)^{T} Q (x - v) - (x - u)^{T} Q (x - u)} = d^{T} Q (x - u) - \frac{1}{2} d^{T} Q d = d^{T} Q (x - \frac{u + v}{2}) .

(2)

Consequently

lo g α (u, v) = min {0, (v - u)^{T} Σ^{- 1} (x - \frac{u + v}{2})} .

(3)

The truncation at zero is essential: a log acceptance ratio can be positive, but the logarithm of an acceptance probability cannot.

For the particular proposed swap, let

a = i_{1}

be the coordinate changing from zero to one and

b = i_{2}

the coordinate changing from one to zero. Then

d = e_{a} - e_{b}

, and the untruncated scalar simplifies further to

(Q (x - u))_{a} - (Q (x - u))_{b} - \frac{1}{2} (Q_{aa} + Q_{bb} - 2 Q_{ab}) .

(4)

Only two rows and four entries of

Q

are needed for this evaluation. This is Metropolis sampling on fixed-cardinality subsets.

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The proposal's normalization of a probability distribution is

Z_{q} = \int_{R} e^{- ∣ x ∣} d x = 2, q (x) = \frac{1}{2} e^{- ∣ x ∣} .

(1)

This is the Laplace distribution with location zero and scale one. For example, generate its magnitude from the exponential distribution with rate one and give it an independent fair sign. The target is the standard normal distribution, with

Z_{p} = 2 π

To obtain the tight envelope, let

t = ∣ x ∣

. By completing the square,

\frac{p ( x )}{q ( x )} = e^{- t^{2} /2 + t} = e^{1/2} e^{- (t - 1)^{2} /2} \leq e^{1/2} .

(2)

Equality occurs at

x = \pm 1

, so

M_{*} = e

is the smallest valid envelope constant. It maximizes the acceptance probability for this fixed proposal. The supplied lower bound must be clipped at zero to satisfy the algorithm's nonnegativity requirement:

h (x) = max {0, 1 - x^{2} /2} .

(3)

This is valid because

e^{- s} \geq 1 - s

for

s \geq 0

, while

p

is everywhere positive. Its support is

[- 2, 2]

Here is the fully specified squeeze rejection sampling procedure. Draw

Y

with probability density function

q (y) = e^{- ∣ y ∣} /2

, independently draw

U \sim Unif (0, 1)

, and compute

H = max (0, 1 - Y^{2} /2)

and

B = e e^{- ∣ Y ∣}

. If

U \leq H / B

, return

Y

immediately. Otherwise evaluate

P = e^{- Y^{2} /2}

, independently draw

V \sim Unif (0, 1)

, and return

Y

V \leq (P - H) / (B - H)

. On rejection restart with fresh draws. In this example

B > H

everywhere, so the residual denominator is positive.

Avoiding a target evaluation means accepting at the first test. Its unconditional probability for one proposal is

P (no target evaluation) = \int_{R} q (y) \frac{h ( y )}{M q ( y )} d y = \frac{1}{2 e} \int_{- 2}^{2} (1 - y^{2} /2) d y = \frac{2 2}{3 e} \approx 0.571843.

(4)

For comparison, the overall acceptance probability is

2 π / (2 e) = π / (2 e) \approx 0.760173

. The first-stage saving

0.571843

is the fraction of proposals requiring no target evaluation, including both eventual accepted and rejected proposals in the denominator; it is not conditional on acceptance.

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For a measurable set

C

, (i) gives the joint acceptance probability

P (Y \in C, accept) = \int_{C} \frac{q ( y )}{Z _{q}} \frac{p ( y )}{M q ( y )} d y = \frac{1}{M Z _{q}} \int_{C} p (y) d y .

(1)

Divide by

Z_{p} / (M Z_{q})

from (ii). The resulting conditional probability is

P (Y \in C ∣ accept) = \frac{1}{Z _{p}} \int_{C} p (y) d y = \int_{C} p (y) d y .

(2)

Thus every accepted value has target probability density function

p

. Repeating rejected trials with fresh independent random variables does not change this law: the contribution from acceptance on trial

j

is multiplied by

(1 - Z_{p} / (M Z_{q}))^{j - 1}

, and summing that geometric series gives the same conditional law. Independent repetitions produce independent target draws.

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Average the conditional probability in (i) over

Y \sim q = q / Z_{q}

P (accept) = \int_{Ω} \frac{q ( y )}{Z _{q}} \frac{p ( y )}{M q ( y )} d y = \frac{1}{M Z _{q}} \int_{Ω} p (y) d y = \frac{Z _{p}}{M Z _{q}} .

(1)

This is exactly the acceptance mass for an unnormalized rejection envelope. In normalized form the envelope is

p \leq (M Z_{q} / Z_{p}) q

, giving the same acceptance probability. Here

M

bounds the ratio of the unnormalized functions, so its value need not equal the envelope constant for the normalized probability density functions.

The squeeze changes the number of expensive target evaluations, while preserving the per-proposal acceptance probability. The expected number of proposals per accepted draw is

M Z_{q} / Z_{p}

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Fix a proposal

y

with

q (y) > 0

, and abbreviate

B = M q (y)

H = h (y)

and

P = p (y)

. The envelope says

0 \leq H \leq P \leq B

. Assuming fresh independent uniform distribution draws at both stages, the law of total probability gives

P (accept ∣ Y = y) = \frac{H}{B} + (1 - \frac{H}{B}) \frac{P - H}{B - H} = \frac{H}{B} + \frac{P - H}{B} = \frac{p ( y )}{M q ( y )} .

(1)

The second term includes both reaching the residual test and passing it. If

H = B

, the envelope forces

P = B

and the first test accepts with probability one; the second-stage

0/0

ratio is never evaluated. This is squeeze rejection sampling.

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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) .

(1)

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}

(2)

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 .

(3)

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.

 Read the full article

Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 2 / a / iii / Solution by

Codex  0 2026-10-25

 View more

For the proposed joint probability density

ρ

, define the acceptance indicator function

A_{i} := 1_{{V_{i} \leq w (Y_{i}) / M}} .

(1)

On the proposal's support the importance weight is

π (Y_{i}, V_{i}) / ρ (Y_{i}, V_{i}) = M A_{i}

. Therefore the requested augmented-space importance sampling estimator is

θ_{n}^{IS} = \frac{M}{n} i = 1 \sum n A_{i} ϕ (Y_{i}) .

(2)

Its expected value is

θ

, provided

E_{p} ∣ ϕ (X) ∣ < \infty

. This is not the ordinary estimator

n^{- 1} \sum_{i} w (Y_{i}) ϕ (Y_{i})

from question 1. Taking the conditional expectation of this augmented estimator given all

Y_{i}

produces that ordinary estimator, because

E [A_{i} ∣ Y_{i}] = w (Y_{i}) / M

. This is the Rao-Blackwell identity for a fixed-denominator rejection estimator.

 Read the full article

 Pinned article: Introduction to the OurBigBook Project

Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.

Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!

Video 1.

Intro to OurBigBook

. Source.

We have two killer features:

topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculus
Articles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.
Figure 1.
Screenshot of the "Derivative" topic page
. View it live at: ourbigbook.com/go/topic/derivative
Video 2.
OurBigBook Web topics demo
. Source.
local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
Figure 2.
You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website
.
Figure 3.
Visual Studio Code extension installation
.
Figure 4.
Visual Studio Code extension tree navigation
.
Figure 5.
Web editor
. You can also edit articles on the Web editor without installing anything locally.
Video 3.
Edit locally and publish demo
. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension.
Video 4.
OurBigBook Visual Studio Code extension editing and navigation demo
. Source.
Internal cross file references done right:
Infinitely deep tables of contents:
Figure 6.
Dynamic article tree with infinitely deep table of contents
.
Live URL: ourbigbook.com/cirosantilli/chordate
Descendant pages can also show up as toplevel e.g.: ourbigbook.com/cirosantilli/chordate-subclade

All our software is open source and hosted at: github.com/ourbigbook/ourbigbook

Further documentation can be found at: docs.ourbigbook.com

Feel free to reach our to us for any help or suggestions: docs.ourbigbook.com/#contact

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 Pinned article: Introduction to the OurBigBook Project