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For a freely jointed Gaussian chain, independent isotropic segment vectors have zero mean and coordinate variance ${\ell }^2/3$. The central limit theorem gives the end-to-end vector densityThe number of conformations in a fixed small vector-volume element is proportional to this density. Taking its logarithm gives the configurational entropyThis is the vector-constrained entropy appropriate to an anchored strand. A density for the scalar length alone also has a $4\pi R^2$ shell factor and is a different constraint. The approximation requires many segments and extensions well below $N\ell$.

If the internal energy has negligible extension dependence, the Helmholtz free energy is $F=F_0+3k_{B}T{R}^2/(2N{\ell }^2)$. Differentiation gives an entropic force with effective spring constant $3k_{B}T/(N{\ell }^2)$. In an affine isotropic network, averaging over initially oriented strands gives rubber elasticityfor an incompressible material; ${\nu }_s$ is the number of elastically active strands per volume. For example a volume-preserving stretch with principal factors $\lambda ,{\lambda }^{-1/2},{\lambda }^{-1/2}$ has strain-energy density $W=G({\lambda }^2+2/\lambda -3)/2$. Its nominal tensile stress is $G(\lambda -{\lambda }^{-2})=3G(\lambda -1)+O((\lambda -1{)}^2)$, proving the stated small-strain Young's modulus.

Chemically, increasing the effective cross-link density, for example by changing vulcanization, generally increases the modulus by shortening strands between cross-links. Dangling ends, loops and changes of network structure mean the chemical junction count need not equal ${\nu }_s$.

The reference area is $A_0=10^{-5}{\mathrm{m}}^2$. Neglect self-weight, thermal expansion and energetic elasticity, keep the same permanent network, and heat quasistatically in the rubbery regime. Then $G$ increases from $20$ to $22\mathrm{MPa}$. With a constant suspended load, the exact affine-network balance isIt gives ${\lambda }_{300}=1.01662$ and ${\lambda }_{330}=1.01509$. ThusLinear elasticity instead gives extensions $16.35$ and $14.86\mathrm{mm}$, a $1.49\mathrm{mm}$ rise, consistent at this small strain. Incompressibility makes the area $A_0/\lambda$, already included by using nominal stress. Heating increases entropic stiffness, so the loaded rubber contracts and performs work on the mass. Strong thermal expansion, changes in cross-linking or leaving the rubbery regime would require additional material data.

Let $m=\mathbb{E}X$ and $v=\mathrm{Var}(X)$, evaluated from the given polylogarithm formulas. Generate a large sample of independent and identically distributed random variables with r_sample_f, and compute its sample mean ${\overline{X}}_n$ and sample variance $S_n^2$ using mean and var. The central limit theorem gives the first standardized check:For the second check, the correct asymptotic variance of the sample variance involves the fourth central moment ${\mu }_4=\mathbb{E}[(X-m{)}^4]$:The denominator is not $\sqrt{2v^2}$ unless the probability distribution has the requisite fourth central moment, as does the normal distribution. Here the probability density function decays exponentially, so all required moments are finite.

To compute ${\mu }_4$, expand the denominator of the probability density function as a geometric series and integrate each nonnegative term. The gamma function integral gives the raw momentsThus $v=m_2-m_1^2$ and ${\mu }_4=m_4-4m_1{m}_3+6m_1^2{m}_2-3m_1^4$. The following R code implements both checks for chosen interior parameters and large n:

p <- 0.4
beta <- 1.3
n <- 100000
raw_m <- vapply(1:4, function(j) {
  -factorial(j) * polylog(j + 1, 1 - p) / (beta^j * log(p))
}, numeric(1))
m <- raw_m[1]
v <- raw_m[2] - m^2
mu4 <- raw_m[4] - 4 * m * raw_m[3] +
  6 * m^2 * raw_m[2] - 3 * m^4
x <- r_sample_f(n, p, beta)
c(mean_check = sqrt(n) * (mean(x) - m) / sqrt(v),
  variance_check = sqrt(n) * (var(x) - v) / sqrt(mu4 - v^2))

For each check separately, an absolute value below $1.96$ is consistent with its approximate $95\%$ confidence interval. Repeat with many independent samples: the two standardized statistics should each resemble a standard normal distribution, with about $5\%$ outside that range. Also vary $p$, $\beta$ and n; a systematic discrepancy that grows with $\sqrt{n}$ suggests a bias. The two checks from the same sample are generally dependent, so passing both is not itself a joint $95\%$ statement, nor can a finite test prove that a generator is correct.

The second central limit theorem follows from the central limit theorem for the sample variance: $\sqrt{n}(S_n^2-v)$ differs by a quantity tending to zero in probability from $n^{-1/2}{\sum }_i[(X_i-m{)}^2-v]$.

The central limit theorem gives $\sqrt{n}({\hat{\theta }}_n^{\mathrm{MC}}-\theta )\Rightarrow N(0,1)$, so, with $\Phi$ the cumulative distribution function of the standard normal distribution,The error has two tails. Using the supplied $2.58$ quantile, whose two-sided coverage is approximately $0.9902$, givesThe supplied $2.33$ quantile would be appropriate for a one-sided $99\%$ calculation; it would give only approximately $98.02\%$ two-sided coverage. Using the exact $0.995$ quantile instead of $2.58$ would give $66,349$ draws; we consistently use the paper's rounded value $N=66,564$ below.

For routine Monte Carlo integration, the central limit theorem calculation is preferable as a practical budget: it uses about one fifteenth as many draws as the Chebyshev's inequality bound. However, its coverage is approximate. Finite variance alone supplies no universal rate of convergence to the normal distribution; an integrand with large skewness or a heavy-tailed distribution can make the approximation poor even at a large sample count. If an unconditional finite-sample $99\%$ guarantee is required under only the stated assumptions, use the Chebyshev's inequality budget.

With a fixed budget of $n$ independent proposal trials, the accepted count $K_n$ has binomial distribution with success probability $1/M$. HenceFor $M>1$, the central limit theorem gives the one-sided approximationUse $z=2.33$, since now only the lower tail of the accepted count matters. Requiring $n-MN\ge z\sqrt{n(M-1)}$ and solving the resulting quadratic equation in $t=\sqrt{n}$ gives the one-sided normal approximation for a rejection proposal budgetEquivalently, before rounding, this is $MN+z^2(M-1)/2+(z/2)\sqrt{(M-1)[z^2(M-1)+4MN]}$. For the value in (a)(iii),When $M=1$, every proposal is accepted and $\boxed{n_0=N=66,564}$ exactly. Otherwise the positive safety margin above $MN$ allows for random fluctuations in the number accepted. The average cost in (b)(ii) describes stopping at the $N$th acceptance; this answer chooses a deterministic budget that reaches that acceptance with approximately $99\%$ probability.

The approximation requires a sufficiently accurate normal distribution approximation to the binomial distribution. An exact $99\%$ budget can instead be found from its tail probabilities. Also, a $99\%$ chance of enough accepted draws and approximate $99\%$ precision conditional on obtaining them are separate claims: their intersection is not automatically a $99\%$ event.

The definition of $a$ and (c)(i) give $a\le M$. Nonnegativity of the variance strengthens the lower bound toIn fact the variance of one weighted observation is exactlyIt is finite by the envelope. For $v_{\mathrm{IS}}>0$, the central limit theorem gives approximate two-sided coverage $2\Phi (0.01\sqrt{n/v_{\mathrm{IS}}})-1$, so the paper's rounded quantile yieldsIf $v_{\mathrm{IS}}=0$, the weighted observation equals $\theta$ almost surely, and one draw is sufficient. These are theoretical budgets: $a$ and $\theta$ may themselves be unknown and need preliminary estimation in an implementation.

Compare with the average rejection sampling cost $66,564M$. Ignoring integer rounding, importance sampling uses fewer proposal draws at the same approximate precision precisely whenFor literal integer budgets, the comparison is $\lceil 66,564v_{\mathrm{IS}}\rceil <66,564M$. The simpler strict inequality is the intended leading-order condition. The bound $a\le M$ alone does not ensure an improvement when $\theta \ne 0$; this is a proposal-budget variance comparison of importance and rejection sampling.

The sample mean $\overline{X}=T/n$ has mean $\theta$ and variance $\sigma (\theta {)}^2=\theta (1-\theta )/n$. The central limit theorem gives a normal approximation for interior $\theta$ when both expected binomial counts are sufficiently large. Under the null hypothesis, a level-$\alpha$ upper-tail cutoff is approximatelyTo obtain the power function, evaluate this same cutoff under the true statistical parameter $\theta$, standardizing with $\sigma (\theta )$:Here $\Phi$ is the standard normal distribution's cumulative distribution function. The formula treats the binomial threshold as continuous; exact integer cutoffs and boundary randomization are finite-sample refinements.

First establish the limiting law needed by the delta method. For fixed $t$, put $u_n=1/2+t/\sqrt{n}$ and let $N_n$ count how many of the $U_i$ are at most $u_n$. Then $N_n$ has a binomial distribution with expectation $n/2+t\sqrt{n}$ and variance $n/4-t^2$. The event $M_U\le u_n$ is exactly $N_n\ge (n+1)/2$. Its standardized threshold tends to $-2t$, so the central limit theorem for these binomial counts givesHere $\Phi$ is the cumulative distribution function of the standard normal distribution. Equivalently,Use the usual local regularity for the delta method: $f$ is continuous at the population median $\theta$ and $f(\theta )>0$. The inverse function theorem, applied at $F(\theta )=1/2$, givesThereforeThe asymptotic normality gives the requested leading variance approximationHere the variance parameter for the rescaled limiting law is $1/(4f(\theta {)}^2)$, while the source states the corresponding $1/n$ approximation for the unscaled estimator. Strict increase of $F$ alone does not force $f(\theta )>0$; that local condition is needed for this form of the asymptotic distribution of a sample median.

For $0<{\sigma }^2<\infty$, the central limit theorem givesIf $g$ is differentiable at $\mu$, the delta method applies to the first-order expansionwhere the remainder notation means that its product with $\sqrt{n}$ tends to zero in probability. HenceThe useful nondegenerate normal approximation requires $g^{\prime }(\mu )\ne 0$. When $g^{\prime }(\mu )=0$, the first-order limit is zero and a higher-order expansion may be needed.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.

Write $A=-{\sum }_i\log X_i>0$, with $\log$ the natural logarithm. The likelihood function and log-likelihood function areThe score function is ${\ell }^{\prime }(\theta )=n/\theta -A$ and the second derivative is ${\ell }^{\prime \prime }(\theta )=-n/{\theta }^2<0$. Thus the log-likelihood function is strictly concave, its stationary point is the unique global maximum, and the maximum-likelihood estimator isThe one-observation Fisher information is $I(\theta )=1/{\theta }^2$. The support of a probability distribution here is independent of $\theta >0$, so the regular asymptotic normality of a maximum likelihood estimator givesThere is also a direct check: $U_i=-\log X_i$ has an exponential distribution with rate $\theta$, since $\mathbb{P}(U_i>u)=\mathbb{P}(X_i<e^{-u})=e^{-\theta u}$ for $u>0$. Its mean and variance are $1/\theta$ and $1/{\theta }^2$. Apply the central limit theorem to $A/n$ and the delta method to $h(u)=1/u$ to obtain the same result.

Using the raw moments from part (i), the variance of one observation isThe central limit theorem for the independent statistical sample givesFor $g(x)=x/(1-x)$, its derivative at the mean is $g^{\prime }(\mu )=(1-\mu {)}^{-2}=(\theta +1{)}^2$. The delta method therefore givesEquivalently, the large-sample normal approximation isHere $v$ denotes the leading asymptotic variance, rather than an assertion about an exact finite-sample variance.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

Expanding the exponent, the log-likelihood function iswhere $C$ is independent of $\theta$. Its score function and second derivative areThe first is positive below $\bar{y}$ and negative above it, so the unique maximum-likelihood estimator isUsing $\mathbb{E}{\sum }_i{Y}_i=n\theta$, the one-observation Fisher information is $I(\theta )=-n^{-1}\mathbb{E}{\ell }^{\prime \prime }(\theta )={\theta }^{-3}$. The support does not depend on the interior parameter $\theta >0$, and the regular asymptotic normality of a maximum likelihood estimator givesThis is Inverse Gaussian mean estimation with known shape one. Equivalently, the Inverse Gaussian distribution has variance ${\theta }^3$, so the same result follows directly from the central limit theorem applied to its sample mean.

Under $H_0$, each $Z_i=-\log X_i$ has expected value $1/2$ and variance $1/4$. The central limit theorem therefore approximatesFor the supplied data this standardized value isUsing the provided approximation $\Phi (1.6)\approx 0.95$, the lower-tail 5% rule rejects when $S\le 12.5-1.6(2.5)=8.5$. The observed $6.25$ is in this region, with approximate p-value $\Phi (-2.5)\approx 0.005$. Reject $H_0:\theta =2$ at the 5% level in favor of the specified alternative $\theta =3$. The unusually large geometric mean is consistent with the direction of that alternative. This is evidence in this comparison, not a proof that the parameter is exactly three.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

If a density has polynomially decaying tails on both half-lines, its moment-generating function is infinite at every nonzero real parameter: exponential growth defeats polynomial decay on the positive tail for a positive parameter, and on the negative tail for a negative parameter. Thus no open real tilting domain exists for a saddlepoint density approximation, even when the distribution has finite mean and variance and a central limit theorem applies.

For a zero-mean stationary velocity process of variance ${\sigma }_v^2$ and normalized autocorrelation $(1+|s|{)}^{-\alpha }$, integrating velocity twice gives the displayed expression, exactly rather than just asymptotically. For $\alpha >1$ the long-time diffusion coefficient is ${\sigma }_v^2/(\alpha -1)$. At $\alpha =1$, the displacement variance is asymptotic to $2{\sigma }_v^{2}t\log t$. For $0<\alpha <1$, it is asymptotic to $2{\sigma }_v^2{t}^{2-\alpha }/[(1-\alpha )(2-\alpha )]$. Linear variance growth does not by itself imply a central limit theorem or a normal distribution for displacement.

Probability theory describes uncertainty through a probability space: possible outcomes form a sample space, observable questions are events, and a probability measure assigns them numbers between zero and one. The Kolmogorov axioms make those assignments consistent. They let us combine information about several events, update it through conditional probability, and study random variables without listing every possible outcome separately.

For example, roll two dice whose results are independent random variables, each with a discrete uniform distribution on $\{1,\dots ,6\}$. There are 36 equally likely ordered outcomes. Only $(1,6),(2,5),\dots ,(6,1)$ have sum seven, so that event has probability $6/36=1/6$. The sum itself is not uniformly distributed: sum two has only one outcome. Choosing a probability distribution for the elementary outcomes is part of the model; equal likelihood does not follow merely from counting the named possibilities.

The subject becomes especially useful when complete outcome lists are enormous or infinite. Independence turns a joint calculation into simpler pieces, expected value summarizes a random variable, and the law of large numbers explains how averages can become predictable even when individual observations are uncertain. The central limit theorem describes the scale and shape of many remaining fluctuations. These are distinct claims: a small variance of an average does not make every observation close to its expected value.

There is a satisfying unity here: dice, hidden causes, and Brownian motion all use the same rules for events and conditional probability. A model still needs justification. Its exact deductions describe what its assumptions imply; comparison with observations belongs to statistical inference.

For an estimating equation with population root $\theta =T(F)$, let $A={\mathbb{E}}_F{\partial }_{\theta }\psi (X,\theta )$ be nonsingular and $B={\mathbb{E}}_F[\psi {\psi }^T]$. Differentiating the contaminated population equation gives influence function $-A^{-1}\psi (x,\theta )$. A consistent sample root with differentiable local expansion satisfies $\sqrt{n}(\hat{\theta }-\theta )=-A^{-1}{n}^{-1/2}{\sum }_i\psi (X_i,\theta )+o_p(1)$, so the central limit theorem gives the displayed covariance. In the scalar case this is $\mathbb{E}{\psi }^2/A^2$.

In a second-order sense a weakly stationary process has short memory when its autocovariance is absolutely summable. Then its spectral density of a stationary process is continuous and its sample-mean variance has the usual finite long-run variance of a stationary process. A central limit theorem still needs additional conditions.