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Conditional expectation makes $Y_n$ measurable with respect to ${\mathcal{F}}_n$, and conditional Jensen gives $E|Y_n|\le E|Y|<\infty$. The tower property of conditional expectation for nested sigma-algebras givesThus this is a martingale. Its finite limit exists almost surely for every integrable $Y$ under the stated increasing filtration: the martingale convergence theorem applies to the uniform bound on $E|Y_n|$. In fact the family is uniformly integrable. One way to see this is to approximate $Y$ in $L^1$ by a bounded variable, whose conditional expectations remain bounded, while the conditional expectations of the error have uniformly small $L^1$ norm. Consequently convergence is also in $L^1$, with limit $E[Y\mid \sigma ({\bigcup }_n{\mathcal{F}}_n)]$. Mere martingale status without a suitable bound would not imply convergence.

Work with a probability measure-preserving system, so $\mu (X)=1$. PutThe maximal ergodic theorem, also called the maximal ergodic lemma, says that for real $f\in L^1(\mu )$,No ergodic transformation or invertibility assumption is needed for this maximal ergodic lemma.

Here is a proof of the maximal ergodic lemma. Set $M_0=0$ and $M_N=\max (0,S_{1}f,\dots ,S_{N}f)$. These are integrable: $0\le M_N\le {\sum }_{j=0}^{N-1}|f|\circ T^j$. Splitting off the first summand givesOn $E_N=\{M_N>0\}$, this implies $f=M_N-M_{N-1}\circ T\ge M_N-M_N\circ T$. Outside $E_N$ the right side is $-M_N\circ T\le 0$. ConsequentlyIntegrating and using the measure-preserving transformation property gives ${\int }_{E_N}fd\mu \ge 0$. Since $E_N$ increases to $E$, the dominated convergence theorem, with dominating function $|f|$, proves the infinite-supremum assertion of the maximal ergodic lemma.

For example, applying the maximal ergodic lemma to $|f|-\epsilon$ yields the useful maximal boundIndeed on the latter set $G$, the maximal ergodic lemma gives $\epsilon \mu (G)\le {\int }_G|f|d\mu$.

The pointwise ergodic theorem states that for a probability measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The conditional expectation is invariant and has integral $\int fd\mu$. If $T$ is an ergodic transformation, this conditional expectation is the constant $\int fd\mu$. The assertion applies to real or complex integrable functions; it is the Birkhoff ergodic theorem.

For a nonnegative measurable $f$, if $\int fd\mu <\infty$, the pointwise ergodic theorem already proves the required conclusion. For the remaining case, let $f_K=\min (f,K)$, $K=1,2,\dots$. Each $f_K$ is integrable on the probability measure-preserving system. The pointwise ergodic theorem and ergodic transformation property give, on a single set of full measure obtained by a countable intersection,Since $f\ge f_K$, every such $x$ satisfies ${\mathrm{liminf}}_n{A}_{n}f(x)\ge \int f_{K}d\mu$ for all $K$. The monotone convergence theorem gives $\int f_{K}d\mu \uparrow \int fd\mu =+\infty$. Thus the nonnegative ergodic averages with infinite integral tend to $+\infty$, rather than merely having an unbounded subsequence. Combining the two cases,Nonnegativity is what makes truncation a lower bound. The probability normalization is also essential to the displayed value: for a finite measure of total mass $c>0$, the corresponding constant is $c^{-1}\int fd\mu$.

Work on a filtered probability space $(\Omega ,\mathcal{F},({\mathcal{F}}_n{)}_{n\ge 0},\mathbb{P})$. A submartingale is a real-valued stochastic process satisfying the following three conditions: $Y_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|Y_n|<\infty$, andThus it is adapted, has integrable random variables at every time, and its conditional expected next value is at least its current value. Repeated conditioning gives $\mathbb{E}[Y_m\mid {\mathcal{F}}_n]\ge Y_n$ for every $m\ge n$. The inequality concerns conditional expectation; individual sample paths need not increase. If no filtration is specified, the definition can be applied to the natural filtration generated by the process.

The overshoot hypothesis as printed cannot hold for every $R>0$. For $R\le 1$, $T_R=0$ and $M_{T_R}=1$; choosing $R<1/\tau$ contradicts $M_{T_R}\le \tau R$. The intended and sufficient assumption is the overshoot bound for $R\ge 1$, on the event $\{T_R<\infty \}$. We use those thresholds below; when $R\le 1$, the requested stopped-time bound is trivial because $T\wedge T_R=0$.

Fix $R\ge 1$ and set $S=T\wedge T_R$. By part (a), the stopped martingale satisfies $0\le M_{S\wedge n}\le \tau R$: before stopping it is below $R$, at $T$ it is zero, and at $T_R$ the assumed overshoot bound applies. Thus the square of this stopped process is integrable. For $\Delta M_n=M_{n+1}-M_n$,The event $\{S>n\}$ is ${\mathcal{F}}_n$-measurable and contained in $\{T>n\}$. Conditional expectation of the linear term is zero, while the variance assumption makes the remaining conditional expectation nonnegative. The stopped increments here are bounded by the stopping and overshoot bounds, so all conditional expectations are legitimate. ConsequentlySince $X_0=1$, we have ${\sigma }^2\mathbb{E}(S\wedge n)\le \mathbb{E}{M}_{S\wedge n}^2-1$. Bounded-time optional stopping theorem also gives $\mathbb{E}{M}_{S\wedge n}=1$, soApply the monotone convergence theorem to $S\wedge n$. This proves the stronger estimate $(\tau R-1)/{\sigma }^2$ and hence the requested absorption-time bound from conditional variance and overshoot:In particular this stopping time is finite almost surely. For $0<R\le 1$, $S=0$ and $X_n=1$ is a constant submartingale, while the stopping-time expectation is zero.

For the simple predictable process from part (a), define its stochastic integral byThe time-zero value contributes nothing. Write $D_i=M_{t\wedge t_{i+1}}-M_{t\wedge t_i}$. If $i<j$ and $t>t_j$, then $h_i{D}_i{h}_j$ is ${\mathcal{F}}_{t_j}$-measurable and $\mathbb{E}[D_j\mid {\mathcal{F}}_{t_j}]=0$. If $t\le t_j$, then $D_j=0$. Boundedness ensures integrability, so all off-diagonal terms vanish by conditional expectation. ConsequentlyThe quadratic variation theorem says $M^2-[M]$ is a local martingale. A bounded continuous martingale is square-integrable, and localization gives the conditional second-moment identityMultiplication by $h_i^2$ and summation therefore give the Itô isometryFor Brownian motion, $[M{]}_s=s$. More generally the Itô isometry holds for every predictable process with finite right-hand side, against a continuous square-integrable martingale. Indeed, on a fixed horizon the quadratic-variation measure ${\mu }_M(E)=\mathbb{E}{\int }_0^t{1}_{E}d[M]$ is finite. Part (c) approximates the integrand by simple predictable processes; the displayed equality makes the elementary integrals Cauchy in $L^2$ and defines their unique limit. The finite sum is exactly the elementary case of the Itô isometry.

For the Gaussian linear mixed model, temporarily regard $\beta ,D,R$ as known, with $R={\sigma }^{2}I$ here and $V=ZD{Z}^T+R$. The joint multivariate normal distribution isBy the conditional multivariate normal distribution formula,A nonsingular Gaussian density has its mode at its mean. Thus the conditional mode of Gaussian random effects equals the conditional expectation. Replacing the unknown parameters by fitted values givesFor positive-definite $D$ and $R$, an equivalent derivation minimizes the negative conditional log density,Differentiation yieldsThe inverse-covariance penalty shrinks furnace intercepts and slopes toward the population effects. For furnace $j$, set $Z_j=(1,t_j)$ and $D_j=\mathrm{diag}({\tau }_0^2,{\tau }_1^2)$; the same formula provides its two-component conditional mode from that furnace's observations, conditional on the fitted common parameters. If a variance component is zero, the covariance form remains meaningful and its corresponding random effect is zero, whereas the formula with $D^{-1}$ needs a limiting interpretation. With known covariance parameters these are best linear unbiased predictors; plug-in variance estimates give their empirical version.

Here the innovation process consists of linear innovations, the linear one-step prediction errors: ${\epsilon }_t=X_t-{\mathrm{proj}}_{{\mathcal{H}}_{t-1}}{X}_t$, where ${\mathcal{H}}_{t-1}$ is the closed linear span of the past $X$'s in $L^2$. For Gaussian processes this is also the conditional expectation prediction error. For general non-Gaussian strong white noise these two notions can differ.

The given ${\eta }_t$ is not the linear innovation process. The moving-average factor has its zero at $1/3$, inside the unit disk, and is noninvertible as a causal moving-average filter. The identityis the moving-average root reflection that places this zero outside the unit disk. Thus the causal invertible representation has innovation variance $9$, rather than the given variance $1$.

For an explicit verification, defineThe filter has constant squared modulus $9$, so $\epsilon$ is weak white noise with variance $9$. The new moving-average factor has root $3$ and is invertible, while its autoregressive factor is causal. Hence the past spans of $X$ and $\epsilon$ agree, and $X_t-{\epsilon }_t$ belongs to that past span. Orthogonality of ${\epsilon }_t$ to past $\epsilon$ therefore identifies it as the linear innovation process. The variance difference proves that it cannot be ${\eta }_t$. If the original noise is Gaussian, the new linear innovations are independent Gaussian variables; without Gaussianity they need only be uncorrelated random variables.

A martingale with the specified terminal value must equal its conditional expectation. If ${\hat{S}}_t=0$, default has already occurred and the terminal European call option payoff is zero. If ${\hat{S}}_t>0$, put $h=T-t$. Conditional on survival to $T$,where $Y$ has the standard normal distribution; the conditional probability of this survival is $e^{-\lambda h}$. The conditional expectation of the payoff in the enlarged filtration used in part (c) is consequentlyThis expression is already measurable with respect to the natural filtration of $\hat{S}$, so the tower property of conditional expectation gives the same value there. The survival factor cancels the compensating growth factor:The separate zero branch avoids division by zero. At $t=T$ it gives the required payoff, using $F(0,m)=(1-m{)}^{+}$; integrability follows from $({\hat{S}}_T-K{)}^{+}\le {\hat{S}}_T$.

Backward induction makes every $U_t$ adapted and integrable: the conditional expectation of an integrable random variable is integrable, and $|\max (a,b)|\le |a|+|b|$. The Snell envelope recursion directly gives $U_t\ge \mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]$, so $U$ is a supermartingale.

If $Z$ is a submartingale, iterating its defining inequality gives $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]\ge Z_t$. Starting at $U_T=Z_T$ and applying the tower property of conditional expectation backwards showsThis last conditional expectation is a true martingale. The conclusions areIn the second case waiting until $T$ attains the optimal stopping value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a martingale.

In a loss network with fixed routing, calls of type $r$ arrive as independent Poisson processes with rates ${\nu }_r$. Every accepted call occupies $A_{jr}$ units of resource $j$ throughout its holding time. The link-route incidence matrix $A$ has nonnegative integer entries, $C$ is the resource-capacity vector, $n_r$ counts calls currently present, and $e_r$ is the $r$th coordinate unit vector. The feasible occupancy space isThe inequalities are coordinatewise. A call that would leave this space is rejected immediately, without queuing or changing $n$. With independent unit-rate exponential distributions for holding times, the memoryless property makes $n$ a continuous-time Markov chain, whose nonzero off-diagonal transition rates areThere are $n_r$ independent clocks of rate one for the second transition. Rejected arrivals are not jumps and do not appear in the off-diagonal transition rates.

Let $a(n)={\prod }_r{\nu }_r^{n_r}/n_r!$. For every feasible adjacent pair,Thus detailed balance for a continuous-time Markov chain gives the product-form stationary distribution of a loss network:In this convention $G(C)$ is the reciprocal normalizing sum. If every route consumes some resource, finite capacities make $S(C)$ finite. Even zero-resource routes cause no normalization difficulty: the sum is at most $\exp ({\sum }_r{\nu }_r)$. With positive arrival rates the reachable feasible class is an irreducible Markov chain, since calls can depart to the empty state and any feasible occupancy can be built from it. The stationary distribution is therefore unique on that class.

The vector $m=C-An$ records the free units of each resource. Its probability distribution ${\pi }^{\prime }(m)$ sums $\pi (n)$ over all occupancies with the same free resources. The map from $n$ to $m$ generally loses information: different call mixtures may consume exactly the same resources.

For ${\pi }^{\prime }(m)>0$, the conditional expectation can be computed by removing one call. The factorial identity $n_r\pi (n)={\nu }_r\pi (n-e_r)$ givesIndeed, $n\mapsto n-e_r$ is a bijection from the occupancies in this sum with $n_r>0$ to occupancies using $C-m-A{e}_r$ units. Conversely, adding one call to any such occupancy is feasible because its final resource use is $C-m$. Consequentlyand the conditional expectation is zero otherwise. Conditioning the identity ${\sum }_r{A}_{jr}{n}_r=C_j-m_j$ and multiplying by ${\pi }^{\prime }(m)$ now yieldsAlternatively, multiply the preceding unnormalized factorial identity by $A_{jr}$ and sum over $r$ directly; this proves the same formula even when ${\pi }^{\prime }(m)=0$, without dividing by a zero probability. This gives both the conditional expectation argument and the direct summation argument.

The vertices visited by a simple symmetric random walk form an integer interval: nearest-neighbour steps cannot skip a vertex. At the stopping time $T_k$, write that interval as $[\ell ,r]$, with $r-\ell +1=k$. The walk is at one of its endpoints, because the $k$th new vertex must extend the earlier interval. This also covers $k=1$, when both endpoints coincide.

The Strong Markov property restarts the walk from this endpoint. Visiting a new vertex is exactly exiting $[\ell ,r]$, or reaching $\ell -1$ or $r+1$. Translate these absorbing boundaries to $0,k+1$. The starting point becomes either $1$ or $k$, so the expected duration of symmetric gambler's ruin is $k$ in both cases. Taking conditional expectations and then expected values gives the expected time to expand a random-walk range:

The conditional density givesusing either integration by parts or the expected value of an exponential distribution. HenceThis is a finite value for almost every realization of $X$, but not an integrable random variable: $\mathbb{E}[1/X]={\int }_0^{\infty }{e}^{-x}/xdx=\infty$. Since $Y\ge 0$, its conditional expectation remains well-defined in the nonnegative, extended-expectation sense; the tower property of conditional expectation then gives $\mathbb{E}Y=\infty$.