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For $n=0$, the identity matrix gives the assertion. Suppose it holds at $n$. Conditioning on the state at time $n$ and using the Markov property givesThe nonnegative countable sum is justified by the law of total probability. Induction proves the $n$-step transition matrix is $P^n$.

Condition on the full sampled history through time $n$. The Markov property of $X$ implies that its future after $2n$ depends on that history only through $X_{2n}=Y_n$. Summing over the intermediate state givesThus $Y$ is a Markov chain with transition matrix $P^2$. This is a deterministic-time sampled Markov chain.

Starting at one, state two can first be reached only after an even number of steps. At every even time before that hit, the chain is again at one. By the Markov property and part (i), each successive two-step block hits two with probability $4/9$, while a failure returns to one with probability $5/9$. Therefore, for the first hitting time $H$,Equivalently $H/2$ has the geometric distribution on the positive integers with success probability $4/9$.

The function below counts the initial uniformly chosen state as the first of the $n$ visited states, and then generates $n-1$ transitions using the appropriate row of $P$:

simchain <- function(n, P) {
  if (!is.numeric(n) || length(n) != 1L || !is.finite(n) ||
      n < 1 || n != floor(n)) stop("n must be a positive integer")
  if (!is.matrix(P) || !is.numeric(P) || nrow(P) == 0L ||
      nrow(P) != ncol(P) || any(!is.finite(P)) ||
      any(P < 0) || any(abs(rowSums(P) - 1) > 1e-12)) {
    stop("P must be a finite square stochastic matrix")
  }
  k <- nrow(P)
  state <- sim1(rep(1/k, k))
  counts <- numeric(k)
  counts[state] <- 1
  if (n > 1) {
    for (step in seq_len(n - 1)) {
      state <- sim1(P[state, ])
      counts[state] <- counts[state] + 1
    }
  }
  counts / n
}

Conditional on the current state $i$, sim1(P[i, ]) selects $j$ with probability $P_{ij}$, proving the Markov property with the desired transition matrix. Initialization is uniform by the same sampler. Dividing the visit counts by $n$ gives the requested Empirical occupation proportions of a finite Markov chain, including a valid result when $n=1$.

Condition on the intermediate state at time $m$. The Markov property and time homogeneity giveThese nonnegative sums are valid on a countable state space. Starting from $P(0)=I$, $P(1)=P$, induction gives $\boxed{P(n)=P^n}$. Conditioning on the initial state now yields $\boxed{\mathrm{Pr}(X_n=i)=(\mu P^n{)}_i}$. If $\mu P=\mu$, induction gives $\mu P^n=\mu$ for every $n$, so all one-time distributions equal the stationary initial distribution.

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.

At $n=0$ the transition probability is ${\delta }_{ij}=(P^0{)}_{ij}$. If the assertion holds at $n$, condition on $X_n=k$ and use the Markov property:Induction proves $\boxed{P(X_n=j\mid X_0=i)=(P^n{)}_{ij}}$. These countable sums contain nonnegative terms and are justified by total probability.

Given $X_n=k$, the next child takes a fresh uniform choice from $1,\dots ,k$, making $X_{n+1}$ uniform on $0,\dots ,k-1$. At zero the count stays zero. This conditional law depends only on the current count, so $X_n$ is a Markov chain.

The amounts taken do not generally retain that state information. For $N\ge 3$, compare two positive-probability histories with the same current value $Y_2=1$: if $Y_1=N-2$, exactly one sticker remains and $Y_3=1$; if $Y_1=N-1$, none remains and $Y_3=0$. Thus the next amount depends on earlier history even after conditioning on the current amount, violating the Markov property.

The non-Markov assertion needs this nondegenerate range of $N$. With $N=1$ the amount process is trivial; for $N=2$ it is also Markov if time-inhomogeneous transitions are permitted. The history counterexample establishes the intended failure for every $N\ge 3$.

The displayed identity requires a correction to the process construction in the source. In a piecewise-deterministic Markov process, the instantaneous hazard function is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original Markov jump process $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted transition kernel, or even a Markov process in the state variable $X$ alone.

For an explicit counterexample, take $a(t,x)=1$, $\alpha (x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the standard normal distribution, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so nonexplosion holds. In the literal construction, the first holding time has the exponential distribution of rate $2$. Thus the probability distribution of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a normal density. The asserted identity instead assigns to this no-jump atom the masswhich differs for every $t>0$.

Here is the full derivation for the intended piecewise-deterministic Markov process. Assume nonexplosion; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and setOne precise construction draws an independent unit-rate exponential random variable $E$ and takes $\tau =\inf \{u\ge s:H(u)\ge E\}$, with $\inf \varnothing =\infty$. This implements a hazard function along the flow of a vector field, rather than a fixed exponential holding time at the last reset. Its survival probability and jump-time probability density function areEquivalently, $S^{\prime }=-\alpha (z)S$ with $S(s)=1$. On $\{\tau >t\}$, the terminal state is $z(t)$. Conditional on $\tau =u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The Markov property and the tower property of conditional expectation therefore give the first-jump decomposition of a piecewise-deterministic Markov processThis proves the requested identity under the corrected construction, for every bounded Borel measurable function $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.

For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same conditional expectation argument givesThese $Q_{s,t}$ need not be the transition operators conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.

Take $\tau =t_f-t_i>0$ and split it into $M$ intervals of width $\Delta t=\tau /M$. The markov property and Gaussian convolution giveThis finite-dimensional identity is exact for every positive $M$. Define the path-integral measure by these normalized time slices:In the time-sliced diffusion path integral, the discrete exponent is ${\sum }_j[(x_{j+1}-x_j)/\Delta t{]}^2\Delta t/(4D)$. Its conventional continuum notation is thereforeThe time-slice normalization is essential, not an arbitrary multiplicative constant. Brownian paths are almost surely nondifferentiable, so the displayed action is shorthand for the Gaussian limiting measure rather than an ordinary integral of the derivative of each typical path.

Treat $p_n(t)$ as the probability of the complete coarse-grained state $n$ at time $t$, with $p_n\ge 0$ and ${\sum }_n{p}_n=1$. In the column-vector convention of the master equation, $W_{nm}$ for $n\ne m$ is the transition rate from $m$ to $n$. The short-time transition law isNonnegative transition probabilities and conservation of total probability requireFor a finite state space these conditions define a continuous-time Markov chain. The Markov property means that, conditional on the present state, no further history affects the transition probabilities. A time-independent $W$ additionally assumes time-homogeneous rates; the waiting time in a state is exponential, and multiple jumps in an infinitesimal interval contribute beyond first order. A reduced description that omits relevant slow variables can retain memory and need not obey this closed equation. Quantum coherences are likewise absent from this classical population model. For infinite state spaces one also needs conditions preventing infinitely many jumps in finite time. Neither equal forward/backward rates nor detailed balance is required merely to conserve probability.

The term $v\mathbf{n}$ is directed self-propulsion at fixed speed, while the two components of $\mathbf{u}$ generate translational Brownian motion with diffusion coefficient $D$. The angle has systematic angular velocity $\omega$ and rotational diffusion coefficient $D_r$. With independent standard Brownian motions, the precise stochastic differential equations areIndependent translational and rotational noises are the usual interpretation; the supplied autocorrelations alone would not specify arbitrary cross-correlated noises.

The complete state $(\mathbf{r},\theta \mathrm{mod}2\pi )$ is a Markov process: its future increments depend only on its present state and independent future noise. Its generator is $v\mathbf{n}\cdot \nabla +\omega {\partial }_{\theta }+D{\nabla }^2+D_r{\partial }_{\theta }^2$. Position alone generally lacks the Markov property, since an unobserved current orientation carries memory into future displacements. At long scales orientation averaging produces the effective positional diffusion equation in solution.

Condition on the intermediate parent's height and use the Markov property. The three mutually exclusive paths through Tall, Medium and Short contributeEquivalently, this is entry $(3,1)$ of the matrix power $P^2$. Hence

Let $\tau$ be the absorption time. Its expectation is finite: from every interior state, a string of at most $n$ successive left moves reaches $0$, and the probability of such a string is at least $p=3^{-n}$. Hence the Markov property gives ${\mathbb{P}}_k(\tau >jn)\le (1-p{)}^j$, a geometric tail bound, and ${\mathbb{E}}_k[\tau ]<\infty$.

For $1\le k\le n-1$, one step is always spent before absorption or the remaining walk. Apply first-step analysis and the law of total expectation to the three possible first moves. The conditional expectations of the remaining absorption time are $e_{k+1},e_k,e_{k-1}$, respectively; independence of future moves gives the same law as a fresh random walk from the new position. Thuswhich rearranges toThe term $1$ also appears when the walk stays at $k$, because staying still consumes a step.

Starting at $X$, the two possible two-step returns are $X\to X\to X$ and $X\to Y\to X$. They are disjoint events, with probabilities $1/4$ and $3/16$. ThereforeEach path probability uses the specified Markov property to multiply its successive transition probabilities.

The Markov property and time homogeneity giveSince $p_{2n}>0$, conditional probability yields the Markov chain return bridge expressionThe explicit formula gives $p_n\sim \frac{3}{4}(3/4{)}^n$, soThe conditioning event has positive probability at each finite time even though those return probabilities tend to zero. Taking an undefined conditional on an infinite-time zero-probability event is not needed.

Write $h_i={\mathbb{E}}_i[J]=i/k$ and $v_i={\mathbb{E}}_i[JX]$. Since $0\le JX\le X$, the latter expected value is finite. Conditional on the first move to $i-1$ or $i+1$, total time is one plus the remaining time, while the terminal success indicator random variable is unchanged. The Markov property and linearity of expectation giveAt either absorbing boundary the remaining time is zero, so $v_0=v_k=0$.

Let $w_i=i(k^2-i^2)/(3k)$. The linear term has zero second difference and $i^3$ has second difference $6i$. Consequentlywhich is exactly the required recurrence, and $w_0=w_k=0$. Uniqueness follows as before: the difference of two solutions has zero second difference and zero endpoint values. HenceThis is the success-weighted duration underlying the duration of gambler's ruin conditioned on a chosen boundary.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.

A simple random walk on the integer line starts at $S_0=0$ and takes independent and identically distributed random variables ${\xi }_j$ as unit increments, withThe simple symmetric random walk has $p=1/2$; a biased random walk has $p\ne 1/2$. At $p=0$ or $p=1$ every step is deterministic, giving $S_n=-n$ or $S_n=n$ respectively.

Count the rightward steps using indicator functions:Thus the endpoint is a shifted and rescaled random variable with a binomial distribution. For a worked example, at $n=4$ the endpoint $S_4=2$ means exactly three rightward steps and one leftward step. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ possible orders, so its probability is $4p^3(1-p)$, equal to $1/4$ in the fair case. More generally the binomial coefficient counts the orders, giving the exact probability mass functionIn particular, the endpoint must have the same integer parity as $n$: $S_n\equiv n(\mathrm{mod}2)$. For $0<p<1$ the support of a probability distribution is exactly $\{-n,-n+2,\dots ,n\}$. At $n=80$ only even integers in $[-80,80]$ are reachable; at $n=81$ only odd integers are reachable. This restriction is easy to lose if one draws a smooth curve instead of discrete probability masses. The fair-walk counting argument is also presented in Grinstead and Snell, Introduction to Probability, section 12.1.

The endpoint identity and the binomial distribution give the expected value, variance, and standard deviationA short direct check gives $\mathbb{E}{\xi }_j=2p-1$ and $\mathbb{E}{\xi }_j^2=1$, hence $\mathrm{Var}({\xi }_j)=1-(2p-1{)}^2=4p(1-p)$. Then apply linearity of expectation and the variance of a sum of independent random variables.

The absolute width and the width relative to the number of steps answer different questions. In the fair case the standard deviation grows from $\sqrt{20}\approx 4.472$ to $\sqrt{80}\approx 8.944$, while ${\sigma }_n/n$ shrinks from $0.2236$ to $0.1118$. At $n=80,p=0.65$, the standard deviation is $\sqrt{72.8}\approx 8.532$, or $0.9539$ times the fair width, and ${\sigma }_n/n\approx 0.1067$. Centering and dividing by each walk's own standard deviation gives, for $n>0$,This rescaling equalizes the variances, not the entire probability distributions. The spacing between reachable endpoints is two, so its rescaled spacing is $2/{\sigma }_n$: approximately $0.4472$ for the fair $20$-step walk, $0.2236$ for the fair $80$-step walk, and $0.2344$ for the biased $80$-step walk.

The positions themselves are dependent. For $1\le m\le n$, $S_n=S_m+(S_n-S_m)$, where the second term is independent of $S_m$. Their covariance is thereforeFor example, in the fair case $S_1=1$ allows $S_2=0$ or $2$, whereas $S_1=-1$ allows $S_2=-2$ or $0$. Nevertheless the position process is a Markov chain: given its current position, the next fresh increment has no dependence on its past. Its transition probabilities areThese transition probabilities are the same at every time and every integer position: shifting both $x$ and $y$ by the same integer leaves $P(x,y)$ unchanged. Thus the Markov property and independent increments coexist with dependent positions.