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The Quasi-Poisson regression retains the same conditional mean but permitsThis is a mean–variance function specification through quasi-likelihood; it does not assign a full probability distribution to each count. For independent observations, the quasi-score equation is proportional to ${\sum }_i{x}_i(Y_i-{\mu }_i)=0$, so the mean statistical parameter estimates equal those from Poisson regression. The output estimates $\hat{\varphi }=3.515351$ through the Pearson dispersion estimator, and inflates the standard errors by approximately $\sqrt{\hat{\varphi }}=1.875$.

The large residual deviance relative to 728 residual statistical degrees of freedom also signals substantial overdispersion. Daily weather, traffic and other omitted conditions may produce greater count variation than a homogeneous Poisson distribution allows. The Quasi-Poisson regression accounts for that extra marginal variance. It still requires a correct conditional mean and an appropriate independence assumption; a common dispersion parameter alone does not repair serial correlation.

A scoring rule in the reward convention gives $S(P,x)$ for announcing the probability distribution $P$ and then observing $x$. Under genuine belief $Q$, the expected score isA proper scoring rule satisfies $s(Q,Q)\ge s(P,Q)$ for every admissible $P,Q$. A strictly proper scoring rule additionally requires equality only when the distributions coincide:The expected scores must be well defined. A loss convention reverses the inequalities, but the question uses rewards.

Condition on the count and use the independence of random variables of the claim sizes. For $n\ge 1$ their joint exponential transform factors, while for $n=0$ the empty sum contributes $1$. Consequently the law of total expectation givesThis random sum of independent claims transform uses both independence assumptions: the count must be independent of the entire claim-size sequence, and the sizes must be mutually independent with the same probability distribution. The probability generating function is interpreted through its defining nonnegative series. The identity holds as a finite moment-generating function wherever that series is finite; outside that domain the expectation and series can agree at $+\infty$. In particular a positive argument may take $M_X(r)$ beyond $1$, so finiteness does not follow merely from the usual unit-disk domain of a probability generating function.

The moment-generating function of a mixture distribution is the mixture of its component transforms, soFor the aggregate, positivity of every claim implies $S=0$ exactly when $N=0$. ChooseIf $q>0$, take $\tilde{N}$ to have the zero-truncated claim-count distribution, namely the conditional law of $N$ given $N>0$. ThenChoose this count independently of a fresh independent claim-size sequence and put $Z={\sum }_{i=1}^{\tilde{N}}{X}_i$. Its probability distribution is that of $S$ conditional on being positive. Thus the hurdle decomposition of a positive random sum givesand an independent Bernoulli random variable $B$ of success probability $q$ realizes $S\stackrel{d}{=}BZ$. This establishes the distributional representation, including that $Z$ is itself a positive random sum of independent claims. If $p=1$, the aggregate is identically zero; set $q=0$ and choose any positive $Z$, for example one claim. Conditioning the count on positivity is then unnecessary and would be undefined.

For the specified geometric distribution on the nonnegative integers,An exponential distribution of expected value $\mu$ has transform $M_X(r)=(1-\mu r{)}^{-1}$. Substitution yieldsHence $Z$ is exponential with rate $1/(2\mu )$ and expected value $2\mu$. This is the geometric sum of exponential variables with a rescaling of the claim mean. Identification can also use the uniqueness theorem for Laplace transforms of nonnegative random variables by taking $r\le 0$.

The resulting distribution function isIts jump of size $1/2$ at zero is important: the aggregate law is not a purely continuous exponential distribution.

In an ontological model of a quantum system, a preparation of $|\psi ⟩$ gives a probability distribution ${\mu }_{\psi }$ over a physical state $\lambda$. A fixed quantum measurement has response probabilities ${\xi }_j(\lambda )\ge 0$ with ${\sum }_j{\xi }_j=1$, reproducing the Born rule after averaging over ${\mu }_{\psi }$. The PBR theorem says that, assuming preparation independence and the quantum predictions, distributions for distinct pure states cannot overlap with positive probability. Thus the quantum state is determined by the physical state in the psi-ontic model sense.

Preparation independence says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded psi-epistemic model hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.

For the two given qubit preparations, consider the following PBR exclusion measurement for zero and plus, written in the ordered basis $00,01,10,11$:Direct inner products show that these four vectors are an orthonormal basis. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated projective measurement satisfiesThis is an example of antidistinguishable quantum states: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.

To prove the contradiction without requiring deterministic quantum measurement responses, choose a common dominating measure for ${\mu }_0,{\mu }_{+}$ and write their densities $m_0,m_{+}$. If they overlap, $\nu (\lambda )=\min (m_0,m_{+})$ has mass $\epsilon >0$. By preparation independence, every one of the four product densities dominates $\nu ({\lambda }_1)\nu ({\lambda }_2)$. Its total mass is ${\epsilon }^2$. The zero Born rule probability for outcome $xy$ implies that its nonnegative response ${\xi }_{xy}({\lambda }_1,{\lambda }_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore ${\mu }_0$ and ${\mu }_{+}$ are mutually singular.

For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|{\Psi }_i⟩=|{\psi }_i{⟩}^{\otimes n}$. The tensor-power reduction of PBR overlap givesEach block therefore has an effective two-dimensional Hilbert space. Explicitly,are orthonormal and $|{\Psi }_2⟩=(|e_0⟩+|e_1⟩)/\sqrt{2}$. Embed the four-vector exclusion basis above into the tensor product of these two block spaces. To obtain a complete quantum measurement on all $2n$ qubits, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.

If the single-copy preparation distributions overlap with common mass $\epsilon >0$, their $2n$-fold products all dominate the common measure ${\nu }^{\otimes 2n}$ of mass ${\epsilon }^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. ThusThe argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

Use the intrinsic definition of a Lévy process: $X_0=0$ almost surely, its increments over disjoint time intervals are independent, their probability distributions depend only on interval length, and $X$ has stochastic continuity. In symbols,with times restricted to the half-line. Such a stochastic process has a càdlàg modification, and is usually represented by that version. Requiring càdlàg paths in the definition is a common equivalent convention at the level of modifications; it is important to distinguish this from a claim about the paths of an arbitrary supplied version.

For the characteristic function, writeHere $\psi (0)=0$ and $\psi$ is the positive-time-sign characteristic exponent of a Lévy process. The independent increments and stationary increments give ${\varphi }_{s+t}(\theta )={\varphi }_s(\theta ){\varphi }_t(\theta )$; stochastic continuity gives continuity in time and ${\varphi }_0=1$. This continuous multiplicative semigroup has the stated exponential form. Its exponent has the Lévy–Khintchine formulawhere $b\in \mathbb{R}$, ${\sigma }^2\ge 0$, and the Lévy measure $\nu$ satisfies $\int (1\wedge y^2)\nu (dy)<\infty$. The truncation convention fixes the drift coefficient $b$; the displayed sign convention agrees with $\Psi =-\psi$.

A unit-root autoregressive process retains shocks permanently, whereas a causal time series with $|\varphi |<1$ reverts towards its mean. Testing the unit root determines whether stationary autoregressive analysis is appropriate or differencing is needed. For the model without an intercept or trend, use the Dickey–Fuller test against the lower-sided alternative $\varphi <1$ near the null. PutThis is the ordinary regression statistic for a zero coefficient when $\Delta X_t$ is regressed on $X_{t-1}$, but its null probability distribution is not the usual Student law. Under the standard unit-root initialization $X_1=o_P(\sqrt{n})$ and innovations independent of the starting value,where $W$ is standard Brownian motion. If $d_{\alpha }$ is the lower $\alpha$-quantile of $D$, the asymptotic level-$\alpha$ critical region isThe deterministic terms and null initialization must match the critical-value table. For exact finite-sample size of a statistical test, calibrate the statistic from its Gaussian random walk null with the specified initial condition and noise scale; for a zero starting value its distribution is scale-free. The printed two-sided recurrence alone specifies neither an initial law nor a universal finite-sample critical value. Ordinary normal quantiles do not give the intended size of a statistical test.

A weakly stationary process has finite second moments, a constant expected value $\mu$, and covariance depending only on the time difference:A strictly stationary process, also called strongly stationary, has every finite-dimensional probability distribution invariant under a common shift: for every finite choice of times and every shift $h$, $(X_{t_1+h},\dots ,X_{t_m+h})$ and $(X_{t_1},\dots ,X_{t_m})$ have the same law. strict stationarity with finite second moments implies weak stationarity. weak stationarity alone does not determine the full joint law, and strict stationarity alone does not guarantee finite moments.

Integrating the Weibull distribution density gives $F(x)=1-\exp (-x^{\alpha }/\beta )$ for $x>0$. Inverse transform sampling therefore givesReplacing $1-U_1$ by $U_1$ produces the same probability distribution. Both uniforms belong to $(0,1)$ with probability one. Here the conventional Weibull scale is ${\beta }^{1/\alpha }$, rather than $\beta$.

The number present is an M-M-s queue: births occur at rate $\lambda$, deaths at rate $\mu \min (n,s)$. Both rates are bounded, so the chain is nonexplosive. For $\lambda ,\mu >0$ and an integer $s\ge 1$, the detailed-balance weights with ${\pi }_0=1$ areTheir sum is finite precisely when $\lambda <s\mu$, giving positive recurrence and a stationary probability distribution after normalization.

For completeness, the birth–death recurrence criterion is ${\sum }_{n\ge 0}({\lambda }_n{\pi }_n{)}^{-1}=\infty$. It follows by solving the harmonic hitting-probability recurrence: successive differences are proportional to $({\lambda }_n{\pi }_n{)}^{-1}$, and divergence makes the probability of escaping to infinity before reaching zero vanish. The geometric tail of the weights makes this sum diverge when $\lambda \le s\mu$ and converge when $\lambda >s\mu$. Thus $\lambda =s\mu$ gives null recurrence, since the chain is recurrent but has no finite invariant normalization, and $\lambda >s\mu$ gives transience. The critical case is not positive recurrent merely because arrival and maximum service rates balance.

Order the prescribed lengths as $l_1\le \cdots \le l_N$. Construct a prefix code on the binary tree. At depth $l_i$, previously assigned codewords exclude ${\sum }_{j<i}{2}^{l_i-l_j}$ of the $2^{l_i}$ vertices. The Kraft inequality impliesso an available vertex can be assigned to the next symbol. None of the assigned words is a prefix of another. Reading each word up to the next assigned vertex gives unique decoding, so this constructs the required uniquely decodable code.

For completeness the same length inequality is necessary even for a uniquely decodable code which is not a prefix code. Write $K={\sum }_a{2}^{-l(a)}$ and $L={\max }_{a}l(a)$. For $r$ successive symbols, distinct symbol strings have distinct concatenated codewords. At total length $j$ there are at most $2^j$ such strings. ThereforeTaking $r$th roots and letting $r\to \infty$ gives $K\le 1$. This is the McMillan inequality.

Set $q(a)=2^{-l(a)}/K$. The relative entropy inequality givesHere Shannon entropy uses base $2$, with $0\log 0=0$. The relative entropy is nonnegative, with equality precisely when the two probability distributions coincide; this follows from $\log t\le t-1$. Consequently equality holds exactly whenIn particular all alphabet symbols must have positive probability for equality with finite prescribed lengths.

A weak solution of a stochastic differential equation consists of a filtered probability space, an adapted continuous process $X$, and a Brownian motion $B$ relative to that filtration, with specified initial distribution, such that for every finite $t$ the coefficient integrals are finite andIn one dimension the integrability conditions are ${\int }_0^t\sigma (X_s{)}^{2}ds<\infty$ and ${\int }_0^t|b(X_s)|ds<\infty$. The noise integral is the Itô integral. A solution on a prescribed stochastic basis with prescribed $B$ is distinguished from this weak construction.

A strong solution of a stochastic differential equation uses the prescribed $B$ and initial variable, and is adapted to their augmented natural filtration; thus no additional randomness is needed. For deterministic $X_0=x$, each $X_t$ is determined by the driving Brownian motion through time $t$. Merely being adapted to a larger filtration is insufficient for this definition.

Uniqueness in law means that any two weak stochastic solutions with the same initial distribution have the same law of $X$ on continuous path space. Pathwise uniqueness means that two solutions on one stochastic basis, with the same $B$ and almost surely equal initial values, are indistinguishable stochastic processes. The latter compares paths under the same noise, rather than just comparing their probability distributions.

For the intended random walk model, take $S_0$ to be deterministic, or independent of all the independent and identically distributed random variables $Y_j=S_j-S_{j-1}$. If $\nu$ is their common probability distribution, then $Y_{t+1}$ is independent of ${\mathcal{F}}_t=\sigma (S_0,Y_1,\dots ,Y_t)$. This supplies the Markov property that the backward dynamic programming argument needs. Define the transition operatorand the deterministic optimal stopping value function recursively byAt any state where the conditional law is evaluated, independence givesBackward induction, starting with $U_T=f(S_T)$, proves the intended representationIntegrability along the actual random walk makes the recursion finite at the states used by that process, up to null sets. If finite real-valued value functions on all of $\mathbb{R}$ are intended, a convenient sufficient convention is statewise integrability: $\mathbb{E}|f(s+Y_1+\cdots +Y_k)|<\infty$ for every $s$ and $0\le k\le T$. Indeed each stopping value is bounded in absolute value by the expectation of the sum of the absolute rewards. This convention makes the displayed recursion finite and measurable everywhere; the source only assumes integrability from the given starting process.

The printed independence of the increments alone does not suffice if $S_0$ can reveal future increments. Here is a bounded finite-state counterexample, also with a convex reward. Let $T=2$, let ${\epsilon }_1,{\epsilon }_2$ be independent fair signs, and setThe two increments are exactly ${\epsilon }_1,{\epsilon }_2$, hence independent and identically distributed. But ${\mathcal{F}}_1$ already knows both signs, so $U_1=\max \{S_1^{+},S_2^{+}\}$. On the two positive-probability histories with $S_1=0$, its values are respectively $0$ and $1$. Thus no deterministic $V(1,0)$ can work. Adding independence of $S_0$ from the increments repairs this missing Markov property hypothesis.

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.

Use the positive exponent in the PDF. The TeX introduces a minus sign in the numerator while retaining the positive-sign normalizing constant, which would not define the stated probability distribution. The Curie–Weiss model assigns equal weights to $s$ and $-s$, since $({\sum }_j{s}_j{)}^2$ is unchanged by simultaneous reversal. This spin inversion symmetry pairs every configuration with one having the opposite value of $S_i$, so

Specify an initial value or initial probability distribution. A strong solution of a stochastic differential equation uses a prescribed Brownian motion on a prescribed filtered probability space. It is continuous, adapted to the completed filtration generated by that Brownian motion and the initial value, and satisfiesalmost surely for every $t$, with ${\int }_0^t(|b(X_s)|+\sigma (X_s{)}^2)ds<\infty$ almost surely. The given initial variable is independent of future Brownian motion increments. Some courses allow a larger prescribed filtration in their definition of a strong solution of a stochastic differential equation; the Brownian-generated convention states explicitly what absence of extra randomness means.

A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion relative to its filtration, and a continuous adapted process $X$ satisfying the same integrability and integral equation. Here the space and noise are part of the unknown; the filtration may contain randomness beyond the initial variable and the driving path.

Pathwise uniqueness means that on any common filtered probability space, any two solutions with the same driving Brownian motion and the same initial value are indistinguishable. Uniqueness in law means that all weak solutions with the same initial probability distribution have the same law as path-valued random variables, even on different spaces. Pathwise uniqueness compares shared-noise paths; uniqueness in law compares path distributions. Neither uniqueness notion by itself asserts existence.

For finite $G$, choose a root $r$ and an ordering of the other graph vertices. Begin with a tree consisting only of $r$. At each stage take the first graph vertex not already in the tree, run an independent simple random walk from it until its first hit of the existing tree, erase the loops chronologically, and add the resulting graph path. In chronological loop erasure, revisiting a graph vertex already in the current list deletes all list entries after that graph vertex; otherwise append the new graph vertex. The retained final graph path is simple, with new internal graph vertices and exactly one endpoint in the old tree. Adding it therefore preserves connectedness and creates no graph cycle. Graph vertices already included are skipped. On a finite connected graph each walk hits the existing tree almost surely; after finitely many attachments all graph vertices are present.

The Wilson algorithm theorem states that this tree is a uniform spanning tree, independently of the chosen root and order. The required hypotheses here are an undirected connected finite graph and transition probabilities proportional to conductance; unit conductances give simple random walk and the uniform law. This is the sampling result being described, not a claim that arbitrary walk transition probabilities would produce a uniform tree.

For an infinite connected locally finite recurrent graph, choose a fixed finite root $r$ and enumerate all graph vertices. A connected locally finite graph is countable, since its finite-radius balls are finite. Run the same algorithm successively. Recurrence and irreducibility imply that a walk from any graph vertex hits $r$ almost surely, hence hits the current tree. Every attached graph path is therefore finite. The countable intersection of these probability-one events has probability one, and the increasing union is an acyclic connected subgraph containing all graph vertices: each graph vertex receives a finite route to $r$.

This gives the uniform spanning tree of a recurrent infinite graph. It is not a uniform counting measure on all infinite spanning trees. Equivalently, its joint probability distributions on finite collections of edges are the common limits of uniform spanning trees along any graph exhaustion, using the approximations defining the free uniform spanning forest and the wired uniform spanning forest. To check that equivalence, fix a finite edge set $F$ and run the chosen enumeration through a finite prefix containing all its endpoints. Once those graph vertices are in the tree, later attachments cannot add an edge with both endpoints already included, so membership of $F$ is settled permanently. Only finitely many random walks have been used; each has a finite trajectory almost surely. All their visited graph vertices and all their graph neighbours lie in a sufficiently large exhaustion set. Couple finite-volume and infinite-volume walks with the same steps there. Whether the exterior is deleted or wired to one boundary graph vertex, these initial trajectories and their loop erasures then agree. This proves convergence of the probabilities of cylinder sets determined by those edges.

The limiting law is unchanged by the root and ordering because the laws on finite graphs have this invariance. The same coupling shows that changing the exhaustion also leaves the law unchanged. ThusThe result is a single spanning tree because the graph is recurrent. On a transient graph, infinite-volume spanning-tree limits may instead be uniform spanning forests; recurrence cannot be omitted from this conclusion.

First take $x\ne y$. Sample a uniform spanning tree using Wilson's algorithm rooted at $y$, with $x$ processed first. The first walk runs from $x$ until its first hit of $y$, so its loop erasure is exactly the unique $x$-to-$y$ graph path of the resulting tree. Later attachments cannot alter that graph path. This is the uniform-spanning-tree path representation of loop-erased random walk.

Instead root Wilson's algorithm at $x$ and process $y$ first. The resulting tree still has the same uniform law, while its first attachment has the law of loop-erased random walk from $y$ to $x$. A tree's graph path between two fixed graph vertices has the same unoriented edge set in both directions. Pushing the common tree law through this graph path map therefore givesHere $\mathcal{E}$ means the set of unoriented edges. In fact, reversing the complete graph vertex sequence from the first loop-erased walk has the law of the second, giving reversibility of loop-erased random walk. If $x=y$ and first hitting allows time zero, both graph paths are empty and the equality is immediate.

This is a distributional assertion, not a pathwise identity between chronological loop erasure and reversal. For example, the valid walk trajectory $x,a,b,a,c,b,y$ in a graph containing the triangle on $a,b,c$ and the edges $xa,by$ erases forward to $x,a,c,b,y$. Erasing the reversed trajectory and then reversing the result gives $x,a,b,y$, a different graph path. Wilson's root-independent tree law is what proves the stated equality of probability distributions.

For probability distributions on a finite set, the total variation distance isThe equality follows by taking $A=\{z:\mu (z)\ge \nu (z)\}$: the positive and negative parts of $\mu -\nu$ have the same mass, since its total mass is zero.

Use the stationary distribution identity $\pi P^t=\pi$. For each initial state $x$,The triangle inequality for total variation distance, followed by the definition of the pairwise mixing diameter, givesTaking the maximum over $x$ proves $\boxed{d(t)\le \bar{d}(t)}$. This particular inequality only needs a stationary distribution; it does not require an irreducible Markov chain or an aperiodic Markov chain.