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At an interior statistical parameter $\theta$, local asymptotic normality means that there are random vectors ${\Delta }_{n,\theta }$ and a symmetric nonnegative Fisher information matrix $I_{\theta }$ such that, for each fixed $h\in {\mathbb{R}}^p$,A common stronger definition requires this expansion also for $h_n\to h$. The fixed-$h$ version suffices here. A regular identifiable model usually has positive definite Fisher information; nonsingularity is unnecessary for the contiguity conclusion.

Here are sufficient differentiability conditions for a proof by Taylor's theorem. Suppose the local probability density functions have common support, ${\ell }_{\vartheta }(y)=\log f(\vartheta ,y)$ is twice continuously differentiable near $\theta$, differentiation can pass through the normalization integral twice, the score function $s_{\theta }=\nabla {\ell }_{\theta }$ has finite second moment, and the local Hessian matrix is bounded in operator norm by an integrable envelope of a function class. These hypotheses implyThe first identity differentiates $\int f_{\theta }=1$ once; the second uses ${\nabla }^2{f}_{\theta }=f_{\theta }({\nabla }^2{\ell }_{\theta }+s_{\theta }{s}_{\theta }^T)$ and differentiates twice. The central limit theorem gives ${\Delta }_{n,\theta }=n^{-1/2}{\sum }_i{s}_{\theta }(Y_i)\stackrel{d}{\to }{N}_p(0,I_{\theta })$. Taylor's integral remainder givesThe weak law of large numbers and the continuity and integrable-envelope hypotheses make the bracket converge in probability to $-I_{\theta }/2$. Indeed, the expectation of the supremum of the Hessian matrix difference over a shrinking neighborhood tends to zero by dominated convergence, and Markov inequality controls its empirical average. This proves local asymptotic normality and explains the role of both derivatives.

Write $Q_n◃P_n$ for contiguity of probability measures: for every measurable sequence $A_n$, $P_n(A_n)\to 0$ implies $Q_n(A_n)\to 0$. Mutual contiguity requires this in both directions. A useful form of Le Cam's first lemma is the following: if the Radon-Nikodym derivative $L_n$ of the absolutely continuous part of $Q_n$ relative to $P_n$ converges in distribution under $P_n$ to $L$, with $L>0$ almost surely and $\mathbb{E}L=1$, then the two sequences are mutually contiguous. This includes the usual case $L_n=d{Q}_n/d{P}_n$. The mean-one condition ensures uniform integrability of $L_n$ and vanishing mass of any singular part; positivity ensures the reverse implication.

Take $P_n=P_{\theta }^n$ and $Q_n=P_{\theta +h/\sqrt{n}}^n$. Their likelihood ratio on the support of $P_n$ satisfies, by local asymptotic normality and the continuous mapping theorem,This limit is positive and $\mathbb{E}L=e^{-v/2}{e}^{v/2}=1$, including $v=0$. The conditions of Le Cam's first lemma therefore hold, giving mutual contiguity of the product sampling laws. The products, rather than the one-observation laws, are the relevant sequences.

For any $\epsilon >0$, let $A_n=\{\Vert {\hat{\theta }}_n-\theta \Vert >\epsilon \}$. Consistency under $P_{\theta }^n$ makes $P_{\theta }^n(A_n)\to 0$, and contiguity gives $Q_n(A_n)\to 0$. Thus the estimator remains consistent for $\theta$ under each fixed local alternative. Since $\Vert \theta +h/\sqrt{n}-\theta \Vert \to 0$, it is also consistent for the moving parameter in the sense that $\Vert {\hat{\theta }}_n-(\theta +h/\sqrt{n})\Vert \to 0$ in probability under $Q_n$.

For $\lambda \ge 0$, a Poisson distribution has mass $\mathbb{P}(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,\dots$; at $\lambda =0$ it is concentrated at zero. Its moment-generating function isFor independent $X,Y$, the moment-generating function of their sum is the product, $\exp [(\lambda +\mu )(e^s-1)]$. Uniqueness of the moment-generating function near zero therefore gives $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$. Repeating the argument gives $S_n={\sum }_{i=1}^n{X}_i\sim \mathrm{Poisson}(n)$ when every parameter is one.

The displayed sequence is $a_n=\mathbb{P}(S_n\le n)$. The classical central limit theorem states that independent identically distributed variables with finite mean $m$ and nonzero finite variance ${\sigma }^2$ have standardized sum $(S_n-nm)/(\sigma \sqrt{n})$ converging in distribution to the standard normal distribution. Here $m={\sigma }^2=1$, soConvergence of the distribution functions at the continuity point zero justifies the last step. This is the Poisson cumulative probability at its growing mean.

First derive the survival renewal equation for a classical risk model. Before the first claim, capital rises deterministically. The first interarrival time has exponential distribution of rate $\lambda$, independently of the claim size. Conditioning on its time and size, and using the Markov property after that claim, givesA claim larger than the capital available then causes immediate ruin and contributes zero. There is almost surely a first claim because $\lambda >0$.

Put $g(v)={\int }_0^v\phi (v-x)f(x)dx$. Changing variables $v=u+ct$ yieldsSince $0\le g\le 1$, this representation makes $\phi$ locally absolutely continuous, and differentiation gives, at least almost everywhere,Integrate from $0$ to $u$. The integrands are nonnegative, so Tonelli theorem permits interchange in the double integral. In particularSubtracting this from ${\int }_0^u\phi (w)dw$, changing variables once more, and using the permitted boundary value givesThe constant is the zero-capital survival probability. Positive relative safety loading means $c>\lambda \mu$, so it is positive. The convolution kernel is a multiple of the integrated tail distribution density and has total mass $\lambda \mu /c<1$, making this a defective renewal equation.

For the individual portfolios define ${\rho }_i={\lambda }_i{\mu }_i/c_i$. Under the positive-loading convention of the question, $0<{\rho }_i<1$. At zero capital, portfolio $i$ survives with probability $1-{\rho }_i$. The independence of random variables of the entire portfolio processes makes their ultimate survival events independent. ConsequentlyThe exponential form is not needed for this zero-capital probability under positive loading; only the claim mean enters. For completeness, the company description does not separately repeat positive loading for every portfolio. If nonpositive loadings are allowed, certain ruin with nonpositive loading and finite claim variance instead givesHere the positive part sets a nonpositive factor to zero. To justify the additional case, observe capital at claim times: its independent increments are $c_{i}T-X$, with mean $c_i/{\lambda }_i-{\mu }_i$. A negative mean sends their partial sums to $-\infty$ by the strong law of large numbers. At zero mean, these increments have finite nonzero variance. The central limit theorem gives $\mathbb{P}(S_k<-K)\to 1/2$ for each fixed $K$, so the probability of unboundedness below is at least $1/2$. That event is unchanged by altering finitely many increments and hence is a tail event; the Kolmogorov zero-one law makes its probability one. Ruin therefore occurs almost surely also at zero loading.

For the merged claims, Poisson superposition of insurance portfolios gives total arrival rate $\lambda ={\sum }_i{\lambda }_i$. Each arrival is from portfolio $i$ with probability ${\lambda }_i/\lambda$, independently of other arrival labels. Thus its claim-size mixture distribution has densityThe weights sum to one, so this density integrates to one. An independent transform verification uses the aggregate for one accounting period. If $M_i(r)=(1-{\mu }_{i}r{)}^{-1}$, then its moment-generating function isThis is precisely a compound Poisson distribution with parameter $\lambda$ and the displayed claim-size law. The transform identity can safely be read at $r\le 0$; positive arguments must be below the relevant poles.

Premium incomes and initial capitals add. Therefore the merged premium rate is $C={\sum }_i{c}_i$, the merged initial capital is zero, and its mean claim size isIts zero-capital survival probability is consequentlyUnder the question's positive-loading context this is a premium-weighted average of the individual survival probabilities, rather than their product. If arbitrary individual loadings are admitted, recompute the loading of the merged portfolio; its finite-variance mixture law gives the complete formulaAn individually nonpositive loading does not force merged ruin when aggregate premiums still exceed aggregate expected claims. The survival probability under risk pooling is at least their product because a product of numbers in $[0,1]$ is at most each factor. Pooling allows one portfolio's surplus to cover another's deficit, so merged survival does not require every original portfolio to remain solvent separately.

Let $w(x)=f(x)/g(x)$ where $g(x)>0$, and set it to zero on the g-null set where $g=0$. The domination assumption makes $f=0$ there and gives $0\le w\le M$. Integrating it also gives $M\ge 1$. For iid proposals $X_i\sim g$, the importance sampling estimator isCauchy-Schwarz inequality under $f$ gives $|\theta |\le (\int {\varphi }^{2}f{)}^{1/2}<\infty$. Direct integration proves unbiasedness, ${\mathbb{E}}_g[\varphi (X)w(X)]=\theta$. MoreoverThusThe iid central limit theorem applies to these finite-variance weighted observations:If the variance is zero, this denotes the point mass at zero. This is the bounded-weight importance-sampling moment bound.

Write $p=1/M$ and $v_f=\int {\varphi }^{2}f-{\theta }^2$. The Bernoulli central limit theorem givesIf $M=1$, all proposals are accepted and this limit is degenerate.

Conditional on $N=m$, the accepted sample is iid from $f$. Since $N/n\to p>0$, its size tends to infinity, and the ordinary target-sample CLT therefore givesAssign any fixed value to the estimator on $N=0$; the probability of that event tends to zero, so it does not change this limit. This is the random-count central limit theorem for accepted rejection samples.

For comparison at the same budget of $n$ proposals, Slutsky's theorem rescales this asThe distinction between accepted-sample size and proposal count is essential for the final efficiency comparison.

A convenient general-state-space ergodic theorem is the following. For a positive Harris recurrent Markov chain with invariant probability measure $\pi$, and a measurable function $h$ with $\pi |h|<\infty$,Here $\pi h=\int hd\pi$ is the stationary expected value. Harris recurrence means that every set of positive irreducibility measure is visited almost surely from every state; positive recurrence supplies an invariant probability rather than only an infinite invariant measure. The ergodic theorem for a positive Harris recurrent Markov chain holds from any starting state under these Harris hypotheses. aperiodicity is not necessary just for averages.

A useful sufficient form of the central limit theorem for a geometrically ergodic Markov chain adds aperiodicity, geometric ergodicity, and $\pi (|h{|}^{2+\delta })<\infty$ for some $\delta >0$. It givesThese are sufficient hypotheses, not a claim that irreducibility alone ensures a central limit theorem. The Markov chain Monte Carlo asymptotic variance uses stationary covariances ${\gamma }_h(k)={\mathrm{Cov}}_{\pi }(h(X_0),h(X_k))$, with ${\gamma }_h(0)={\mathrm{Var}}_{\pi }h$. The series is absolutely convergent under the stated sufficient assumptions. If ${\gamma }_h(0)>0$, write ${\rho }_h(k)={\gamma }_h(k)/{\gamma }_h(0)$ and $v_h={\gamma }_h(0){\tau }_{\mathrm{int}}$, where the integrated autocorrelation time is ${\tau }_{\mathrm{int}}=1+2{\sum }_{k\ge 1}{\rho }_h(k)$. When $v_h>0$, the effective sample size of a Markov chain is approximately $n/{\tau }_{\mathrm{int}}$. Negative correlations can reduce the asymptotic variance; a zero asymptotic variance gives a degenerate normal limit. For constant $h$, variance is zero and the autocorrelation normalization is undefined.

Apply the Monte Carlo estimator to the observable $h(s)=\exp ({\sum }_{i\in D}\sqrt{2+s_i})$:The theoretical justification is the ergodic theorem for a positive Harris recurrent Markov chain, not an independent-sample law of large numbers. The Markov chain has finite state space, is irreducible, and has the strictly positive posterior distribution as invariant law, so it is positive recurrent and Harris recurrent with counting measure. The observable is bounded: $e^{K^2}\le h(s)\le e^{\sqrt{3}{K}^2}$. Thus the estimator converges almost surely to the posterior expected value from any initial configuration.

Each state also has positive self-transition probability, so the chain is an aperiodic Markov chain. Finiteness then gives geometric convergence and the relevant central limit theorem; serial dependence determines the Markov chain Monte Carlo asymptotic variance if error bars are required. A fixed finite burn-in can be discarded without changing consistency, but independence of the retained states should not be assumed.

Weak convergence $X_k\Rightarrow X$ means that $\mathbb{E}f(X_k)\to \mathbb{E}f(X)$ for every bounded continuous real function $f$ on ${\mathbb{R}}^d$; it is a statement about their distributions, not their coupling.

The central limit theorem says that for independent identically distributed real variables of finite mean $\mu$ and variance ${\sigma }^2$, $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2)$. For $\sigma >0$ this is equivalent to convergence of $(\sum X_j-n\mu )/(\sigma \sqrt{n})$ to the standard normal distribution. For $\sigma =0$, every variable equals $\mu$ almost surely and the limit is the point mass at zero.

For the proof when $\sigma >0$, put $Y=(X-\mu )/\sigma$. Its characteristic function has expansion ${\phi }_Y(t)=1-t^2/2+o(t^2)$. To justify this using only a second moment, subtract the first two Taylor terms of $e^{itY}$ and use dominated convergence theorem, with the second-order remainder bounded by a constant times $t^2{Y}^2$. Independence gives the characteristic function of the normalized sum as $[{\phi }_Y(t/\sqrt{n}){]}^n\to e^{-t^2/2}$. The Lévy continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies weak convergence. Since $e^{-t^2/2}$ is the characteristic function of $N(0,1)$, the result follows.

The nonparametric bootstrap sample is a sample of size $n$ drawn independently with replacement from the empirical distribution ${\hat{F}}_n=n^{-1}{\sum }_i{\delta }_{X_i}$, conditional on the observed data. To simulate it, take independent uniform $U_j$ and set $X_j^b=X_{1+\lfloor n{U}_j\rfloor }$, assigning $U_j=1$ to the last index if it occurs in a finite-precision generator.

Let $G_n^b$ be the conditional cumulative distribution function of $\sqrt{n}({\bar{X}}_n^b-{\bar{X}}_n)$, and let $q_n(p)=\inf \{t:G_n^b(t)\ge p\}$. A basic bootstrap confidence interval isIn practice the conditional quantiles are estimated from many simulated bootstrap means; the asymptotic argument below describes exact bootstrap quantiles or a simulation count tending to infinity.

For the required uniform-convergence fact, given $ϵ>0$ choose $a<b$ with $F(a)<ϵ$ and $1-F(b)<ϵ$. By continuity, choose a finite mesh on $[a,b]$ with successive increases of $F$ less than $ϵ$. Pointwise convergence at the mesh points makes $|F_n-F|<ϵ$ there for large $n$. Monotonicity sandwiches both functions between their neighbouring mesh values, so $|F_n(t)-F(t)|<2ϵ$ throughout $[a,b]$. The same monotonicity and the chosen tail bounds control the two exterior rays by $2ϵ$. Hence ${\sup }_t|F_n(t)-F(t)|\to 0$.

The bootstrap consistency theorem for the sample mean states that, for iid observations with $0<{\sigma }^2<\infty$, the conditional law of $\sqrt{n}({\bar{X}}_n^b-{\bar{X}}_n)$ converges weakly, in probability, to $N(0,{\sigma }^2)$. By continuity of that limiting distribution and the preceding monotonicity argument, the convergence of conditional distribution functions is uniform in probability. Since the limit is strictly increasing, $q_n(p)\to \sigma {\Phi }^{-1}(p)$ in probability for $0<p<1$. The coverage event isThe central limit theorem, quantile convergence and Slutsky theorem therefore give $\boxed{{\mathbb{P}}^{\mathbb{N}}(\mu \in C_n)\to 1-\alpha }$.

Positive variance is needed for this exact coverage conclusion. If the finite variance allowed in the wording is zero, all observations equal $\mu$ almost surely and the interval collapses to $\{\mu \}$, with coverage one. The bootstrap distribution is still consistent with the degenerate limit, but a nonrandomized usual interval cannot then have limiting coverage $1-\alpha$ for $0<\alpha <1$.

For fixed $\theta >0$, the central limit theorem givesThe weak law of large numbers gives $\overline{X}\to \theta$, so Slutsky theorem permits replacing $\theta$ in the standard error by $\overline{X}$. With $z=z_{0.025}\approx 1.96$, an approximate 95% confidence interval isThe lower endpoint may be truncated at zero because the parameter is nonnegative. This is a large-sample normal approximation for fixed positive $\theta$; an all-zero or very sparse sample does not justify the plug-in normal approximation. At the degenerate parameter value zero, every observation is zero almost surely.

First, the absolute values in the printed limit are an error: the correct long-run variance of a stationary process is the signed sum of autocovariances. Directly,soEach coefficient tends to $1$ and has absolute value at most $1$. Absolute summability and the dominated convergence theorem therefore giveThe sum of absolute values is an upper bound, not the general limit. For a concrete counterexample to the printed assertion, take $X_t={\eta }_t-{\eta }_{t-1}$ with unit-variance iid noise. Then $\gamma (0)=2$, $\gamma (\pm 1)=-1$, and other covariances vanish. The absolute sum is $4$, but ${\overline{X}}_T=({\eta }_T-{\eta }_0)/T$, so $T\mathrm{Var}({\overline{X}}_T)=2/T\to 0$.

For correlated data the central limit theorem can still hold under suitable strong mixing of a stationary process and moment conditions, but the limiting variance is the long-run variance of a stationary process, rather than the one-observation variance. When it is positive,Positive serial dependence usually increases the standard error, while negative dependence can reduce it. The approximate effective sample size of a stationary sample is $T{\gamma }_X(0)/{\sum }_h{\gamma }_X(h)$ when that denominator is positive. A long-memory time series can require a different normalization or a different limit law. Absolute covariance summability alone does not prove a CLT: if $X_t=V{\eta }_t$ with a common independent random scale $V$ taking values $1$ and $2$ with equal probabilities and iid ${\eta }_t$ with the standard normal distribution, the off-diagonal covariances are zero, yet $\sqrt{T}{\overline{X}}_T$ has the nonnormal scale-mixture law $VN(0,1)$. The example is not an ergodic stationary process. If the long-run variance of a stationary process is zero, the usual nondegenerate square-root-$T$ CLT is unavailable.

For the causal AR(1), $|\varphi |<1$ andThe conditional distribution follows because the current innovation is independent of the past.

For a fixed known $\varphi$, condition on the observed $X_1$ and use the $n=T-1$ transitions $t=2,\dots ,T$. Their conditional maximum likelihood criterion is, up to constants,Differentiating in $\mu$ yieldsSince $X_t-\varphi X_{t-1}=(1-\varphi )\mu +{\epsilon }_t$,Thus it is an unbiased estimator, even conditionally on $X_1$, andIt has statistical consistency with mean-square convergence, and the iid-noise strong law of large numbers also gives almost-sure statistical consistency. Its conditional distribution is exactly normal with the displayed mean and variance. If $X_0$ is also observed and all $T$ transitions are used, replace $T-1$ by $T$.

The fixed-$\varphi$ qualification is necessary for the exact finite-sample claims. If $\varphi$ is jointly estimated, conditional likelihood function is linear regression with an intercept: writing ${\bar{X}}_{-}=n^{-1}{\sum }_{t=2}^T{X}_{t-1}$ and ${\bar{X}}_{+}=n^{-1}{\sum }_{t=2}^T{X}_t$, the unconstrained estimators areThis ratio is not generally an unbiased estimator and does not have the preceding finite-sample variance. Under the usual stationary regression conditions it has statistical consistency; its asymptotic variance is ${\sigma }_{\epsilon }^2/(1-\varphi {)}^2$. Indeed, it differs from ${\bar{X}}_{+}$ by $\hat{\varphi }(X_T-X_1)/(n(1-\hat{\varphi }))$, an asymptotically negligible endpoint term. Profiling an unknown innovation variance does not change the fixed-$\varphi$ estimate of $\mu$.

Let $Y_i$ be iid with density $f$, set $H_i=h(Y_i)$, and put $I=E_{f}h(Y)$ and ${\tau }^2={\mathrm{Var}}_f(h(Y))$. The finite-second-moment assumption gives, for ${\tau }^2>0$,The sample variance $s_N^2=(N-1{)}^{-1}{\sum }_i(H_i-{\hat{I}}_N{)}^2$ consistently estimates ${\tau }^2$. The central limit theorem and Slutsky theorem therefore give the asymptotic confidence intervalIts confidence probability is approximately $1-\alpha$, conventionally written $100(1-\alpha )\%$; the printed $(1-\alpha )\%$ omits the factor $100$. For ${\tau }^2=0$, $h(Y)=I$ almost surely and the estimator has zero error. This interval concerns iid Monte Carlo estimators; correlated simulation output requires a long-run variance of a stationary process estimate instead of the iid sample variance.

The empirical characteristic function replaces the expected value in a characteristic function by the sample average:Its centered fluctuation at the central limit theorem scale is the empirical characteristic process:In particular, ${\phi }_n(0)=1$ and ${\mathcal{C}}_n(0)=0$.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.

The observations have finite mean and variance. The strong law of large numbers gives ${\overline{X}}_n\to \mathbb{E}{X}_1$ almost surely, so ${\tilde{\theta }}_n\to \theta$ almost surely and therefore is a consistent estimator. The central limit theorem gives

Represent $Y_n$ as a sum of $n$ independent $\mathrm{Poisson}(1)$ variables. The central limit theorem givesSince zero is a continuity point of the standard normal distribution,

Ignoring the multinomial coefficient, the likelihood isThe score equation isand strict concavity of the log-likelihood gives