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Write $Ph=\mathbb{E}h(X)$ and $P_{n}h=n^{-1}{\sum }_{i=1}^{n}h(X_i)$. A function bracket $[\ell ,u]$ contains the measurable functions $h$ satisfying $\ell (z)\le h(z)\le u(z)$ for every $z\in T$. Require its endpoints to be integrable and call $P(u-\ell )$ its $L^1(P)$ width. A sufficient condition is that, for every $\epsilon >0$, finitely many brackets of width at most $\epsilon$ cover the whole class $\mathcal{H}$. Under this condition the uniform strong law from finite L1 bracketing statesThe conclusion holds outside a common measurable null set. If the supremum is not initially known to be measurable, this formulation means pathwise convergence on a measurable probability-one event; a pointwise separable function class, including the application below, has a measurable supremum. Pointwise brackets also ensure the sample inequalities hold simultaneously over the class.

To prove the result, choose a finite $\epsilon$-cover $[{\ell }_r,u_r]$, $1\le r\le N$. If $h$ belongs to bracket $r$, monotonicity of the empirical measure and of expectation givesandConsequentlyApply the strong law of large numbers to these finitely many integrable endpoints. On a probability-one event, the maximum tends to zero. Repeat with $\epsilon =1/q$, $q\in \mathbb{N}$, and intersect the countably many probability-one events. The limiting supremum is bounded by $1/q$ for every $q$, hence is zero. This proves the uniform law of large numbers without a boundedness assumption on the class itself.

For the moment-generating function, use the empirical measure estimatorIf $t=0$, this estimator and $m$ are both identically one. Otherwise, $X\ge 0$ makes $e^{sX}$ increasing in $s$, and $e^{sX}\le e^{tX}$ provides an integrable envelope. The dominated convergence theorem shows that $m$ is continuous on $[0,t]$, hence uniformly continuous.

For any $\epsilon >0$, choose a partition $0=s_0<s_1<\cdots <s_N=t$ so that $m(s_r)-m(s_{r-1})\le \epsilon$ for every $r$. If $s\in [s_{r-1},s_r]$, then $e^{s_{r-1}z}\le e^{sz}\le e^{s_{r}z}$ for every $z\ge 0$. These endpoint functions form finitely many integrable function brackets with the required $L^1(P)$ widths. The just-proved uniform law of large numbers therefore givesBoth functions of $s$ are continuous; their supremum equals the supremum over a countable dense subset, so it is measurable. This proves uniform consistency of an empirical moment-generating function using only the observed sample.

Write $g_{\ast }>0$ for the upper endpoint called $a$ in the question, to distinguish it from the device amplitude. For a single realization of a finite bath,The literal claim of convergence to zero as time tends to infinity is false for every fixed finite $N$. Random-coupling spin-bath decoherence must distinguish individual realizations, ensemble averages and the large-bath limit.

First establish finite spin-bath coherence recurrence without assuming commensurate couplings. Fix a reference time $t_0>0$ and consider $Q^N+1$ points with coordinates $j{g}_k{t}_0/\pi$ modulo $1$, for $0\le j\le Q^N$. Partition the unit cube into $Q^N$ smaller cubes of side $1/Q$. Two points share a cube, so their difference provides an integer $1\le q\le Q^N$ withAt time $q{t}_0$ every cosine is arbitrarily close to $1$. As $Q$ increases, either these $q$ have an unbounded subsequence, or a bounded subsequence supplies a fixed $q$ with all distances exactly zero; its arbitrarily large multiples are then exact recurrences. In both cases there are unbounded times $t_j$ such that $z_N(t_j)\to 1$. This is the simultaneous Dirichlet approximation theorem argument; it applies equally to typical irrational random couplings. Part (i) supplies a particularly simple exact periodic counterexample.

The intended suppression is valid after ensemble averaging. Independence and the uniform density give the ensemble spin-bath coherenceThe continuous value at $t=0$ is $1$. For $t\ne 0$ its magnitude is strictly below $1$ to the power $N$, and for fixed $N$ it has an envelope of order $(2g_{\ast }t{)}^{-N}$ as $t\to \infty$. Thus the ensemble mean tends to zero with time.

The mean square distinguishes actual loss of coherence from cancellation of signs in that mean:At long times this tends to $2^{-N}$, exponentially small for $N\gg 1$, rather than exactly zero for a finite bath. At any fixed $t\ne 0$, the bracket $q(t)$ is strictly less than $1$, since ${\cos }^2(2gt)<1$ on a set of positive measure. The Markov inequality then givesThis proves small coherence for typical large baths at a fixed nonzero time. An even stronger fixed-time formulation uses the strong law of large numbers:The logarithmic singularities at isolated cosine zeros are integrable, and an exact zero has probability zero, so the law applies. Typical coherence consequently decreases exponentially with bath size.

The initial time scale is also explicit. When $g_{\ast }|t|\ll 1$, $\log \cos (2g_{k}t)=-2g_k^2{t}^2+O(g_k^4{t}^4)$, so the short-time Gaussian spin-bath decoherence approximation isHere $N^{-1}\sum g_k^2\to g_{\ast }^2/3$; on the scale $t\sim (g_{\ast }\sqrt{N}{)}^{-1}$ the displayed remainder tends to zero. The coherence is therefore rapidly suppressed for a large bath and is usually tiny at later fixed times, while rare recurrences still prevent a finite-realization long-time zero limit. Ensemble decay or a specified large-$N$ limit is the correct qualification of the printed assertion, consistent with reversible global unitary time evolution.

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.

Construct a rate-$\lambda$ Poisson process, for $\lambda >0$, from independent waiting times $E_j$ with exponential distribution $\mathrm{Exp}(\lambda )$. Put $S_k=E_1+\cdots +E_k$, $S_0=0$, and $N_t=\max \{k:S_k\le t\}$. The strong law of large numbers gives $S_k/k\to 1/\lambda$, so there are only finitely many arrivals on each finite interval. Consequently the counting paths are càdlàg, and $N_0=0$ almost surely.

At any deterministic time $s$, the memorylessness of the exponential distribution says that the residual waiting time has again exponential distribution $\mathrm{Exp}(\lambda )$ and is independent of the observed history. Subsequent waiting times are fresh independent copies. Thus the Poisson process restarts independently at $s$, proving independent increments and stationary increments. Integrating the joint waiting-time densities over $0<s_1<\cdots <s_k\le t<s_{k+1}$ givesso its time values have the expected Poisson distribution.

For $h\downarrow 0$,Together with stationary increments, this gives stochastic continuity at every time, from either side where applicable. All the Lévy process requirements hold, andFor $\lambda =0$, the identically zero process gives the degenerate case.

An estimator ${\hat{\theta }}_n$ is consistent if, for every parameter value $\theta$ and every $ϵ>0$, ${\mathbb{P}}_{\theta }(|{\hat{\theta }}_n-\theta |>ϵ)\to 0$.

For the proposed root criterion, fix $ϵ>0$. Convergence in probability at the two points givesOn this event the intermediate value theorem gives a zero between the two points. Since the zero is unique, it is ${\hat{\theta }}_n$. Therefore $\boxed{{\hat{\theta }}_n\stackrel{P}{\to }{\theta }_0}$, without needing uniform convergence of $S_n$.

For independent $N({\theta }_0,1)$ observations, the log-likelihood has derivative $n({\bar{X}}_n-\theta )$ and negative second derivative $-n$. Its unique maximizer is ${\hat{\theta }}_n={\bar{X}}_n$. Apply the criterion to $S_n(\theta )=\theta -{\bar{X}}_n$: the strong law of large numbers gives the limit $S(\theta )=\theta -{\theta }_0$, with the required signs. Hence the maximum-likelihood estimator is consistent.

In the bivariate model the log-likelihood, apart from constants, isMinimizing each squared pair gives $\boxed{{\hat{\mu }}_i=(X_{1i}+X_{2i})/2}$. Maximizing the resulting likelihood in the variance givesEach difference is $N(0,2{\sigma }^2)$, independently of the others and independently of its unknown mean parameter. Thus the strong law of large numbers gives ${\hat{\sigma }}^2\to {\sigma }^2/2$ in probability, not ${\sigma }^2$. This is the Neyman-Scott incidental parameter problem: the number of nuisance means grows with the sample size, but each mean is estimated from only two observations. The usual fixed-dimensional likelihood consistency theory therefore does not apply. The corrected estimator $2{\hat{\sigma }}^2$ is consistent.

Let $C$ be an open communicating class, and fix $i\in C$. There is a path of positive probability from $i$ to a state outside $C$. Choose a shortest such path, so it does not revisit $i$ before exiting. No state reached outside $C$ can have a path back to $C$: combined with the exit path and communication within $C$, such a path would put it in the same communicating class. Thus with positive probability the chain never returns to $i$. Its return probability is less than one, making $i$ a transient state. This proves transience throughout $C$.

Conversely, suppose a finite communicating class were both closed and transient. Starting inside it, a path stays there forever. Since it has finitely many states, at least one state must be visited infinitely often. But a transient state is visited only finitely often almost surely, and the union of finitely many exceptional null events still has probability zero. This is a contradiction. Every finite transient communicating class is therefore open.

For the infinite counterexample, take the biased random walk on $\mathbb{Z}$ with independent increments $+1$ of probability $3/4$ and $-1$ of probability $1/4$. All states communicate, so $\mathbb{Z}$ is one infinite closed communicating class. The strong law of large numbers gives $X_n/n\to 1/2$ almost surely. Consequently every fixed state is visited only finitely often, and the class is transient. Finiteness is exactly what fails in the preceding argument.

Let $\mu =\mathbb{E}{X}_1$ and let the cumulant-generating function and its Legendre transform beUnder the permitted assumption that $\psi$ is finite everywhere, the right-tail form of the Cramér theorem is, for $a\ge \mu$,Use $\log 0=-\infty$. The restriction to nonnegative parameters is valid on this side of the mean: convexity gives $\psi (\theta )\ge \mu \theta$, so negative parameters cannot improve the value zero available at $\theta =0$. For $a<\mu$, the right-tail probability tends to one by the strong law of large numbers, so its logarithmic rate is zero; it is not generally $-I(a)$.

The full large deviation principle bounds probabilities of closed sets from above and open sets from below with rate $I$. For the displayed tail equality, a possible finite upper endpoint also needs attention. If $b=\mathrm{esssup}{X}_1<\infty$, then at $a=b$ the event requires every summand to equal $b$; its probability is $\mathbb{P}(X_1=b{)}^n$, and $I(b)=-\log \mathbb{P}(X_1=b)$. The identity follows by taking $\theta \to \infty$ in $\mathbb{E}{e}^{\theta (X_1-b)}$. Above $b$, both sides have infinite rate. The theorem describes exponential decay, including endpoints and impossible events.

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 $\mu$.

The printed identity needs an expectation on its right-hand side. For fixed $r\ge 1$, condition on $X_2,\dots ,X_r$ and use independence of $X_1$:The expression inside this expectation is generally random and cannot equal the unconditional probability by itself.

For each independent Monte Carlo method replicate, draw $R_j$ from its given distribution. If $R_j=r\ge 1$, draw $r-1$ independent $F$-variables, form $W_j={\sum }_{i=2}^r{X}_{j,i}$ and set $H_j=F(x-W_j)$. For $r=1$ the empty sum is $0$. If $R$ may equal zero, use $H_j=1_{\{x\ge 0\}}$ when $R_j=0$, since $S_0=0$. The law of total expectation givesSince $0\le H_j\le 1$, the estimator has statistical consistency by the strong law of large numbers. It is the conditional Monte Carlo estimator obtained by Rao-Blackwellization of the direct indicator $1_{\{S_R\le x\}}$, and the law of total variance gives $\mathrm{Var}(H_j)\le l(1-l)$.

This version assumes that the cumulative distribution function $F$ can be evaluated. An easy sampler alone does not automatically supply an easy cumulative distribution function evaluation. If only sampling is available, use the direct indicator estimator, or replace each conditional cumulative distribution function by the average of several independent indicators $1_{\{X_{j,1,k}\le x-W_j\}}$; the nested version remains an unbiased estimator. With $L$ such inner draws its variance per outer replicate is $\mathrm{Var}(H_j)+L^{-1}E[H_j(1-H_j)]$.

Use the ergodic theorem for a positive Harris recurrent Markov chain: for an invariant distribution $\pi$ and an integrable function $h$, the time average converges almost surely to $\int hd\pi$ under the usual ergodicity conditions. This is the relevant theorem for correlated Metropolis–Hastings algorithm output, rather than the strong law of large numbers. Here $h(x)=x_1$ is integrable. Indeed, puttingwe have $V(x)\ge (x_1-4{)}^2+(x_2-4{)}^2-32$, which bounds the unnormalized density by a constant times an integrable Gaussian density and also gives finite first moments.

For a proposal scale $s>0$, initialize $X_0$ anywhere in ${\mathbb{R}}^2$. At each step draw independent random variables $Z_1,Z_2$ with the standard normal distribution and an independent uniform $U$, propose $Y=X_t+s(Z_1,Z_2)$, and accept whenOtherwise set $X_{t+1}=X_t$. The normal proposal is symmetric, so its density cancels; the target normalizing constant $c$ also cancels. Include every state in the average, including repeated states after rejection. The ergodic theorem for a positive Harris recurrent Markov chain then givesA fixed discarded initial segment does not change this limit.

Small proposal variance gives high acceptance but tiny moves and strong serial dependence. Large variance gives more ambitious moves, but many proposals enter very low-density regions and are rejected, creating long runs at one state. An intermediate scale should be assessed by exploration and effective sample size of a Markov chain, not acceptance rate alone.

This density has two modes, at $(2+\sqrt{3},2-\sqrt{3})$ and their interchange. To see this, stationary points satisfy $x_1(1+x_2^2)=x_2(1+x_1^2)=4$, implying either $x_1=x_2$ or $x_1{x}_2=1$. The unequal solutions have $x_1+x_2=4$ and are local minima of $V$; the equal solution is a saddle. Mode switching is therefore a material part of proposal-scale selection: a chain confined to one mode can have high acceptance and a misleading finite-run estimate.

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

A renewal process has independent identically distributed, nonnegative, almost surely finite inter-arrival times $X_i$, epochs $S_0=0,S_n={\sum }_{i=1}^n{X}_i$, and count $N(t)=\max \{n:S_n\le t\}$. Allowing zero inter-arrivals is harmless provided $\mathbb{P}(X_1>0)>0$. The hypothesis $\mathbb{E}{X}_1>0$ ensures there is $\epsilon >0$ with $p=\mathbb{P}(X_1\ge \epsilon )>0$.

The strong law of large numbers for the indicators gives $S_n\ge \epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\to \infty$, so $N(t)$ is finite for finite $t$. Conversely every fixed $S_n$ is finite, so eventually $N(t)\ge n$. On their countable common probability-one event, this proves $\boxed{N(t)\to \infty \text{ as }t\to \infty \text{ almost surely}}$.

For fixed $t$, put $L=\lfloor t/\epsilon \rfloor$. The hinted comparison process has epochs $S_n^{\prime }=\epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\le S_n$, hence $N(t)\le N^{\prime }(t)$. Equivalently,If $p<1$, for $n>L$ this is bounded by $C_{t,p}{n}^L(1-p{)}^n$, by summing the first $L+1$ binomial distribution. Thus ${\sum }_n{e}^{\theta n}\mathbb{P}(N(t)=n)<\infty$ for $0<\theta <-\log (1-p)$. If $p=1$, $N(t)\le L$ deterministically. Therefore a strictly positive exponential moment exists for every fixed finite $t$, with the same allowed $\theta$ for all such $t$. Almost-sure finiteness of the inter-arrivals is implicit in a proper renewal process; allowing an infinite inter-arrival with positive probability would invalidate the divergence claim.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

Split the adjacent products into two subsequences:Within each subsequence the pairs use disjoint coordinates, so each subsequence consists of independent and identically distributed random variables. The two subsequences need not be independent of each other. By independence within a pair,Apply the strong law of large numbers from (c) separately to obtain $k^{-1}{\sum }_{j=1}^k{Y}_j\to m^2$ and $k^{-1}{\sum }_{j=1}^k{Z}_j\to m^2$ simultaneously almost surely. Sinceand both prefactors tend to $1/2$, the strong law for adjacent products gives

The maximum-likelihood estimator isUnder the score-outer-product convention stated in the question,This is the sample covariance matrix with divisor $n$. The strong law of large numbers, applied entrywise, givesThus the empirical score outer-product information consistently estimates the per-observation Fisher information.

The strong law of large numbers givesBecause $0<\lambda <1$, it follows that ${\lambda }^{S_n}\to 0$ almost surely. If this martingale were uniformly integrable, almost-sure convergence would imply convergence in $L^1$, and thereforeBut the martingale has constant expectation $\mathbb{E}[{\lambda }^{S_n}]=1$. This contradiction proves that

Fix $a<b<c<1$ and chooseUnder ${\mathbb{P}}_{\lambda }$, the increments remain independent and identically distributed, with mean ${\psi }^{\prime }(\lambda )=b$. The strong law of large numbers therefore givesPart (b) impliesLet $c\downarrow b$ and then $b\downarrow a$. Since $\lambda b-\psi (\lambda )={\psi }^{\ast }(b)$ and ${\psi }^{\ast }$ is continuous on $[0,1)$,The same argument includes $a=0$ by taking $b\downarrow 0$.

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

The strong law of large numbers gives $S_n/n=\overline{X}\to 1/\theta$ almost surely, and hence in probability. Thereforeby the continuous mapping theorem. Thus the Bayes estimator has consistency.

It differs from the maximum-likelihood estimator bySince $S_n/n\to 1/\theta$, the numerator is $O_p(n)$ and the denominator is $O_p(n^2)$, so this difference is $O_p(n^{-1})=o_p(n^{-1/2})$. The limit in part a and Slutsky theorem now give