The dual is $\boxed{D:\min b^{T}y\text{ subject to }{A}^{T}y\ge c,y\ge 0}$. For feasible points $c^{T}x\le y^{T}Ax\le y^{T}b$, so its feasible objectives bound the primal objective from above. This is weak duality with the given maximization convention.

For $j\ge 1$ the only entrance is from $j-1$, so stationarity requires ${\pi }_j={\pi }_{j-1}\lambda /j$. Induction gives ${\pi }_j={\pi }_0{\lambda }^j/j!$, and normalization using the exponential series gives ${\pi }_0=e^{-\lambda }$. HenceThe remaining zero-state equation also holds: ${\sum }_i{\pi }_i\lambda /(i+1)=e^{-\lambda }(e^{\lambda }-1)=1-{\pi }_0$, so ${\sum }_i{\pi }_i(1-\lambda /(i+1))={\pi }_0$. Thus this normalized Poisson law really is stationary, not just a solution of the positive-state equations. Irreducibility makes the stationary probability law unique.

Every upward transition has positive probability. From any positive state the transition directly to zero has positive probability, since $1-\lambda /(i+1)>0$. Thus any state can reach zero and then reach any other state by successive upward steps, proving irreducibility.

If $\lambda <1$, the holding probability at zero is positive, so its period and hence the common period of the irreducible chain is one. At $\lambda =1$ that holding probability vanishes, but zero has positive-probability return paths $0\to 1\to 0$ of length two and $0\to 1\to 2\to 0$ of length three. Their lengths have greatest common divisor one. Thus $\boxed{\text{the chain is irreducible and aperiodic for all }0<\lambda \le 1}$.

After the first step to $+1$, the right-hand walk moves away with probability $3/4$ and toward zero with probability $1/4$. Part (ii) therefore gives return probability $1/3$ from that side. After a first step to $-1$, reverse the sign of the coordinate: movement away from zero has probability $2/3$ and toward it probability $1/3$, giving return probability $1/2$. Weight by the probabilities of the first step from zero:There is no holding step at zero, so every positive-time return is included by this decomposition.

For a finite upper barrier $L>i$, let $h_i^{(L)}$ be the probability of hitting zero before $L$. It satisfies $h_0^{(L)}=1$, $h_L^{(L)}=0$, and $h_i^{(L)}=p{h}_{i+1}^{(L)}+q{h}_{i-1}^{(L)}$. For $0<q<p$, solving the second-order recurrence with roots $1$ and $r=q/p$ givesAs $L\to \infty$, the events increase to eventual hitting of zero: any finite path reaching zero has a finite maximal height. Since $r<1$, this yields $\boxed{h_i=(q/p{)}^i}$, with $h_0=1$. In the endpoint case $q=0$, a positive initial state only moves upward and the hitting probability is zero, agreeing with the positive-state formula. The finite-barrier argument selects the correct solution rather than the also-harmonic constant one.

For $i\notin A$, condition on the first step. The Markov property makes the chance of eventual entry after reaching $j$ equal to $h_j$, including $h_j=1$ when $j\in A$. Total probability givesThe countable sum is justified by its nonnegative terms. This first-step identity alone need not uniquely determine a hitting probability on an infinite state space; one must choose the minimal nonnegative solution or supply the appropriate boundary behavior.

For fixed $\theta$, Taylor expansion at zero gives $e^h-1-h=h^2/2+O(h^3)$. With $h=\theta /\sqrt{\lambda t}$, the logarithm of the moment generating function tends to ${\theta }^2/2$. Thus $M_{U_t}(\theta )\to e^{{\theta }^2/2}$ on a neighborhood of zero. The continuity theorem for moment generating functions identifies the limit asThis is the centered Poisson normal limit with its exact variance scaling.

Since a Poisson count has mean and variance $\lambda t$, substitute the centering and scaling into its moment generating function:This is finite for every real $\theta$ when $t>0$.

The total is the independent random sum $Z_t={\sum }_{j=1}^{N_t}{Y}_j$, with the empty sum zero. Conditional on $N_t=n$, its moment generating function is $M_Y(\theta {)}^n$. The Poisson probability generating function therefore givesAssuming a moment generating function in a neighborhood of zero, write $\mu =\mathbb{E}[Y]$ and $s^2=\mathrm{Var}(Y)$. Conditional expectation and variance yieldThe second contribution to the variance is due to randomness in the number of arriving coaches, not just randomness in the occupancy of each coach.

Under the intended independent-marking interpretation, the coach colors are independent also of the arrival process. Let $K$ be the index of the first white coach. Then $P(K=k)=p(1-p{)}^{k-1}$, $k\ge 1$, independently of the exponential interarrival times. For $p>0$, $T={\sum }_{j=1}^K{X}_j$, so conditioning on $K$ givesThus $\boxed{T\sim \mathrm{Exponential}(\lambda p)}$, the waiting time in the thinned Poisson process. If $p=0$, no white coach arrives and $T=+\infty$. Independence of colors from arrival times is needed for this distribution; mutual independence of colors alone would not rule out marks chosen using the interarrival durations.

If $X$ is connected, its continuous surjective image under $q$ is connected by (a)(ii). Conversely suppose $X/R$ is connected and $A\subseteq X$ is clopen. By (i) both $A$ and its complement are saturated. The quotient definition then makes $q(A)$ open, and its complement $q(X\setminus A)$ open too. Thus $q(A)$ is clopen in the connected quotient, so is empty or the whole quotient. Pulling back gives $A=\varnothing$ or $X$. By the clopen criterion $X$ is connected. Hence $\boxed{X\text{ connected}⟺X/R\text{ connected}}$.

Let $A$ be clopen and $C$ an equivalence class. The intersections $C\cap A$ and $C\setminus A$ are complementary relatively open subsets of $C$. Connectedness of $C$ means one must be empty. Thus every equivalence class is entirely in $A$ or entirely outside it, proving $\boxed{q^{-1}(q(A))=A}$. This is precisely saturation.

The set $X/R$ consists of the equivalence classes, with canonical surjection $q:x\mapsto [x]$. The quotient topology declares $V\subseteq X/R$ open exactly when $q^{-1}(V)$ is open in $X$. This makes $q$ continuous and is the finest topology on the quotient set doing so.

Suppose $f(X)=U\cup V$ were a separation into two disjoint nonempty relatively open sets. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$, and they are disjoint, nonempty and cover $X$. This contradicts connectedness of $X$. Hence $\boxed{f(X)\text{ is connected}}$.

A subset is connected if it has no decomposition into two disjoint nonempty sets both relatively open in it. Equivalently its only relatively clopen subsets are itself and the empty set. Connectedness here refers to the subset's induced topology.

An isolated point outside $B$ cannot belong to its closure. Thus the only possible extra closure point is $a$. The balls at $a$ meet $B$ exactly when arbitrarily small original-metric balls at $a$ do so. ConsequentlyIn the first case $a$ is already included. In the second case a ball smaller than the positive distance separates it from $B$; in the last case every ball meets $B$. If $B$ is empty its closure is empty, with distance convention $d(a,\varnothing )=+\infty$.