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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

The strong law of large numbers givesA continuous one-to-one real function is strictly monotone and has a continuous inverse on its image. Hencewhenever ${\hat{\mu }}_n$ lies in that image; an arbitrary extension outside the image handles the remaining finite-sample values. The continuous mapping theorem gives ${\hat{\theta }}_n\to {\theta }_0$ almost surely, so this is a consistent estimator.

Here the total holding rate isand therefore the jump chain has state-independent transitionsIt is an upward biased random walk on $\mathbb{Z}$. By the strong law of large numbers, $Y_n/n⟶1/3$ almost surely, so $Y_n\to +\infty$. The jump chain, and hence $X$, is transient.

Counting measure is invariant for this jump chain: for every $j$, its incoming mass is $1/3+2/3=1$. The invariant-measure transfer between a jump chain and a CTMC therefore gives an invariant measure proportional to $q_i^{-1}=3^{-|i|-1}$. Sincethe normalized invariant distribution of a continuous-time Markov chain isThis does not contradict transience because an invariant distribution of an explosive chain need not imply positive recurrence.

It remains to verify explosion. Starting from zero, the probability that the biased walk ever reaches $-k$ is $2^{-k}$. Once it reaches any state, its probability of returning after departure isso its expected total number of visits to that state, conditional on reaching it, is $1/(1-2/3)=3$. If $V_i$ is the number of visits to $i$, thenConsequently the expected sum of all holding times isThe nonnegative random variable $T_{\infty }$ is therefore finite almost surely. By explosion by summable holding times, $X$ makes infinitely many jumps by time $T_{\infty }$, so it is explosive.

The strong law of large numbers states that if $(X_i{)}_{i\ge 1}$ are independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, thenalmost surely. Under the stronger fourth-moment hypothesis, it has a short proof. Put $\mu =\mathbb{E}{X}_1$, $Y_i=X_i-\mu$, and $S_n={\sum }_{i=1}^n{Y}_i$. Independence and centering make every term in the expansion of $S_n^4$ vanish unless each index occurs at least twice, soFor every $\epsilon >0$, the Markov inequality givesThe probabilities are summable, so the first Borel-Cantelli lemma says that $|S_n|>\epsilon n$ occurs only finitely often almost surely. Apply this simultaneously to $\epsilon =1,1/2,1/3,\dots$ to obtain $S_n/n\to 0$ almost surely, which is the claimed law.

By the strong law of large numbers,almost surely. Brownian scaling and a maximal inequality show that changing Brownian time by $o(n)$ changes its value by $o_{\mathbb{P}}(\sqrt{n})$; explicitly, first restrict to $|T_n-n{\sigma }^2|\le \delta n$, bound the Brownian maximum over a time interval of length $2\delta n$, and then let $\delta \downarrow 0$. Consequentlyin probability.

Butfor every $n$. Since $B_{T_n}$ has the law of $S_n$, Slutsky theorem proves the Central limit theorem from the Skorokhod embedding:

The strong law of large numbers states that for independent identically distributed integrable random variables,To prove it, set $Y_n=X_n{1}_{\{|X_n|\le n\}}$. The tail-sum formula gives ${\sum }_n\mathbb{P}(X_n\ne Y_n)<\infty$, so the Borel-Cantelli lemmas make the two sequences eventually equal. AlsoThe Kolmogorov convergence theorem implies that ${\sum }_n(Y_n-\mathbb{E}{Y}_n)/n$ converges almost surely, and Kronecker lemma yieldsFinally $\mathbb{E}{Y}_n\to \mu$, so the Cesàro mean of these expectations tends to $\mu$.

Using Bayes' theorem and the fact that the prior is proper,Thus $\hat{I}$ is an unbiased estimator of $Z^{-1}$, and the Harmonic mean estimator of Bayesian model evidence is $\hat{Z}=1/\hat{I}$. The reciprocal is not itself generally unbiased, though it is consistent when the strong law of large numbers applies.

Convergence in distribution makes $(X_n)$ tight. Hence for every $\eta >0$ there is $M$ such that $\mathbb{P}(|X_n|>M)<\eta$ for all sufficiently large $n$. Since $Y_n-c\stackrel{P}{\to }0$,Letting $\eta \downarrow 0$ gives $X_n(Y_n-c)\stackrel{P}{\to }0$. Also, the continuous-mapping theorem gives $c{X}_n\stackrel{d}{\to }cX$. Sincethe Slutsky theorem yields

Now write the requested statistic asThe central limit theorem givesSince $\mathbb{E}[Z_i^2]=1<\infty$, the strong law of large numbers givesalmost surely, and therefore its reciprocal converges in probability to one. Applying the product result provesThis is a self-normalized central limit theorem with a second-moment denominator.

Assume that $f(x)=0$ wherever $h(x)=0$ on the part of the space relevant to $g$. The importance sampling algorithm draws independent and identically distributed random variables $Y_1,\dots ,Y_m$ from $h$, forms the weightsand returnsA change of measure givesHence the estimator is unbiased whenever the expectation exists, and, if ${\mathbb{E}}_h|g(Y)w(Y)|<\infty$, the strong law of large numbers gives ${\hat{\mu }}_m\to {\mathbb{E}}_f[g(X)]$ almost surely.

Conditionally on $Y_1,\dots ,Y_m$, the cumulative distribution function of $X^{\ast }$ is the weighted empirical distributionFor fixed $t$, the strong law of large numbers gives, almost surely,whileThe ratio therefore converges almost surely to $F(t)$. This is precisely the claimed convergence in distribution of the self-normalized importance sampling resample.