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Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ be the natural filtration. An integer-valued random variable $M$ is a stopping time with respect to $(X_i)$ whenThus whether one has stopped by time $n$ can be decided from the first $n$ observations, without seeing future variables. Equivalently, $\{M\ge i\}\in {\mathcal{F}}_{i-1}$.

Write the stopped sum asThe event $\{M\ge i\}$ depends only on $X_1,\dots ,X_{i-1}$, so its indicator and $X_i$ are independent random variables. Since $\mathbb{E}|X_1|<\infty$ and $\mathbb{E}M<\infty$, Tonelli theorem givesWe may consequently exchange expectation and summation. Using the tail-sum formula for $M$ proves Wald's equation:

For the renewal process, put $S_n=X_1+\cdots +X_n$ and $N_t=\max \{n:S_n\le t\}$. The random variable $N_t+1$ is a stopping time. Since $S_{N_t+1}>t$, Wald's equation givesand hence

For $c>0$, set $Y_i=X_i\wedge c$ and ${\mu }_c=\mathbb{E}{Y}_i$. Wald's equation still applies to the stopping time $N_t+1$ and the independent variables $Y_i$. Because $S_{N_t}\le t$,ThereforeBy the monotone convergence theorem, ${\mu }_c\uparrow \mu$ as $c\to \infty$. Combining the bounds proves the elementary renewal theorem

For a fixed word over $q$ equally likely symbols, the waiting time for a word in independent uniform symbols has mean $\sum q^k$, where $k$ runs through the lengths of the borders of a word, including the full word. The word lava' has no proper nonempty border, so $$ \boxed{\mathbb E\tau_{\mathtt{lava}}=100^4=100\,000\,000}. $$ The word aa' has borders of lengths $1$ and $2$, soFor a direct check, let $E_0$ be the expected remaining time with no trailing a' and $E_1$ with one trailing a'. Thenwhich gives $E_0=10100$.

By the Tonelli theorem and the planar Brownian transition density,For $s\ge 1$,while for $0\le s\le 1$ it is at most $\Vert f{\Vert }_{\infty }$. Hencefor $t\ge 1$, and consequently

For $g\ge 0$, the Tonelli theorem and conditioning on $N_t$ giveSince $\mathbb{E}[N_t]=\lambda t$,with both sides allowed to be $+\infty$.

For compact $K\subset \mathbb{H}$, letBy the Tonelli theorem and part (ii),Every point in the SLE range has conformal radius tending to zero and therefore belongs to every $E_{ϵ}$. Hence the range inside $K$ has zero expected Lebesgue measure, and so has zero measure almost surely. Exhausting $\mathbb{H}$ by countably many compact sets proves

By Tonelli theorem,If $p<p_c$, part (a) bounds the summand by $e^{-c\Vert x{\Vert }_{\infty }}$. There are only polynomially many vertices at each radius, so the series converges and $\chi (p)<\infty$.

Conversely, suppose $\chi (p)<\infty$. Then $p<1$. Choose $q>p$ so close to $p$ thatUse the standard sprinkling coupling for Bernoulli percolation: first expose the $p$-open clusters, then independently open each remaining edge with probability $\alpha =(q-p)/(1-p)$. Explore the $q$-cluster of the origin cluster by following sprinkled edges. Each discovered $p$-cluster has at most $2d$ times its number of vertices as many incident edges, so the exploration is dominated by a Galton-Watson process of mean at most $2d\alpha \chi (p)<1$. This process dies out almost surely, and hence there is no infinite $q$-open cluster. Thus $q\le p_c$, and $p<q$ implies $p<p_c$. Therefore

Let ${\pi }_0$ be the probability density function of the unknown and $\rho$ that of the noise, both relative to Lebesgue measure. By independence of random variables, the conditional distribution of the measurement given $u$ has density $\rho (m-Au)$. Thus its likelihood function and the joint density areThe Bayesian model evidence is the marginal data densityBayes theorem gives the posterior density solving this Bayesian inverse problem:This formula applies when $0<Z(m)<\infty$. Since $\int Z(m)dm=1$ by Tonelli theorem, these conditions hold almost everywhere under the marginal data law. Values of a conditional distribution at exceptional data are not determined by the joint law. The posterior mean or a maximum a posteriori estimate can then give a point estimate, while the full posterior density describes the remaining uncertainty.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

Let $K(t,\cdot )$ be the Markov kernel of the Metropolis–Hastings algorithm. Away from the possibility of remaining at $t$, its transition density isMultiplication by the target density givesThis is detailed balance for every pair of distinct states.

Writefor the rejection probability, soFor every measurable $B\subseteq \Theta$, detailed balance and the Tonelli theorem giveHence $\mu$ is an invariant distribution of the chain.

For $u\ge 0$, let ${\mathcal{C}}_u$ be the collection of interval components of the superlevel set $\{r:g(r)\ge u\}$. Two times $s,t$ lie in the same member of ${\mathcal{C}}_u$ exactly when $u\le m_g(s,t)$. ThereforeFor arbitrary real ${\lambda }_1,\dots ,{\lambda }_n$, the Tonelli theorem now givesThus $(m_g(s_i,s_j){)}_{i,j}$ is a positive semidefinite matrix.

For $0<\kappa \le 4$, the trace meets the real boundary only at its starting point, so the intersection has zero Lebesgue measure. Suppose $4<\kappa <8$. A boundary point swallowed by an interval need not itself lie on the trace. The stated fact that $\mathbb{P}({\tau }_x={\tau }_y)>0$ for $0<x<y$, combined with the Strong Markov property and Scaling invariance of SLE at successively nested swallowed intervals, implies that a fixed deterministic $x\ne 0$ is swallowed in an interval rather than hit by the trace with probability one. Equivalently,Applying the Tonelli theorem to the random indicator of the boundary trace givesThe nonnegative random measure is therefore zero almost surely.

By Tonelli theorem for the nonnegative indicators,If this is finite, then $N$ is finite almost surely: alternatively, Markov inequality givesSince $\{N=\infty \}\subseteq \{N\ge m\}$ for every $m$,This is the first Borel-Cantelli lemma.

The masses ${\mu }_n(\mathbb{R})={\psi }_n(0)$ converge and are therefore bounded. Part b and Tonelli theorem giveThe integrands are uniformly bounded. By pointwise convergence and the continuity of $\psi$ at zero, the right side can be made uniformly small for all sufficiently large $n$ by taking $\lambda$ large; finitely many remaining measures are individually tight. Thus $({\mu }_n)$ is tight. Applying Prokhorov's theorem after normalizing the masses, or adjoining missing mass at one fixed point, gives a weakly convergent subsequence.

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

Set $g_{ϵ}(x)=x/\sqrt{{ϵ}^2+x^2}$ and $g(x)=\mathrm{sgn}(x)$, with $g(0)=0$. For every $s>0$, the normal distribution of $B_s$ has no atom at zero, so $g_{ϵ}(B_s)\to g(B_s)$ almost surely. Since $|g_{ϵ}-g|\le 2$, the dominated convergence theorem and Tonelli theorem giveThe Itô isometry followed by the Doob L2 maximal inequality now yieldsThus the stochastic integrals converge ucp to ${\int }_0^t\mathrm{sgn}(B_s)d{B}_s$.

For fixed $y\ne x$, part (i), translated by $y$, and then $\epsilon \downarrow 0$ show that planar Brownian motion has probability zero of ever hitting $y$ before leaving any fixed large disk. Letting the disk radius tend to infinity shows ${\mathbb{P}}_x(y\in B([0,1]))=0$. By Tonelli theorem,The nonnegative random area is therefore zero almost surely.

Let $(X_n)$ be Cauchy in the normChoose a subsequence $(X_{n_k})$ for whichTonelli theorem impliesalmost surely. The subsequence therefore converges uniformly on $[0,\infty )$, outside one null event, to a continuous process $X$. For each $t$, $X_t$ is the almost-sure limit of ${\mathcal{F}}_t$-measurable variables; completeness of the filtration makes the chosen version adapted. The same summable bound and monotone convergence theorem show that $\mathbb{E}{\sup }_t|X_t|<\infty$ and that $X_{n_k}\to X$ in norm.

Since the original sequence is Cauchy, the usual triangle argument upgrades convergence of the subsequence to $X_n\to X$ in norm. Thus the space of indistinguishability classes is a Banach space.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.

The spectral characterization of the Reproducing-kernel Hilbert space of a stationary Gaussian process says that its elements are precisely the functions whose Fourier transforms satisfywith $c$ determined only by the Fourier transform convention. Here $\hat{K}(u)=(1+u^2{)}^{-1}$, soThis is an equivalent norm for the Sobolev space $H^1(\mathbb{R})$, and hence the RKHS equals $H^1(\mathbb{R})$ as a set.

Since $\hat{K}$ is integrable, $K$ is continuous and the process has a jointly measurable separable version. For every finite Borel measure $\nu$ on $\mathbb{R}$, Tonelli theorem givesThus $X\in L^2(\mathbb{R},\nu )$ almost surely and is a Borel random variable there. Every continuous linear functional of $X$ is a centered normal random variable: approximate its $L^2(\nu )$ integral by finite linear combinations of process values and pass to the $L^2$ limit. Therefore the induced law is a Gaussian Borel measure on $L^2(\mathbb{R},\nu )$.

Approximate $1_A$ by relative almost periodic functions with summable squared global errors. Tonelli theorem makes their conditional errors tend to zero almost everywhere. Choose a positive base set on which the conditional mass of $A$ is bounded below, then use the Egorov theorem to make those errors uniform on a smaller positive set $C$. Localization of relative almost periodicity to base sets and conditional-norm closure prove $1_B$ is relatively almost periodic and $\mu (B)>0$.

For $s\ge 0$ and $\epsilon >0$, integrate over $0\le K\le s$ to obtain the displayed identity. Tonelli theorem then expresses the moment of a nonnegative random variable as the same integral of expected call payoffs, even when the moment is infinite. This is a static representation across strikes rather than a dynamic replicating strategy in a restricted finite-asset market.