Retain the binary random variables $B_j=1_{\{U_j\le X_{j-1}\}}$. Iterating the recursion givesEach binary history determines the full vector $(X_0,\dots ,X_n)$, so its sigma-algebra is finite by the sigma-algebra of a finite-valued random vector. Conversely, because $0<x<1$, the integer $2^n{X}_n-x$ uniquely determines the digits $B_1,\dots ,B_n$; also $B_j=2X_j-X_{j-1}$. ThusAll $2^n$ binary histories have positive probability. By independence of the driving uniform random variables, along a fixed finite history every intermediate state is in $(0,1)$, so each subsequent conditional branch probability, either $X_{j-1}$ or $1-X_{j-1}$, is strictly positive. It follows that ${\mathcal{F}}_n^X$ has $2^n$ nonempty fibres, each an atom of a sigma-algebra, and exactly $2^{2^n}$ elements.

Adaptedness from the preceding part gives ${\mathcal{F}}_n^X\subseteq {\mathcal{F}}_n$. For $n\ge 1$, however, ${\mathcal{F}}_n$ contains the events $\{U_1\le t\}$ for every $t\in [0,1]$. They have distinct probabilities $t$, so there are infinitely many distinct such events. Consequently the finite sigma-algebra ${\mathcal{F}}_n^X$ cannot equal ${\mathcal{F}}_n$. The inclusion is strict for every $n\ge 1$.

For every $n$, ${\mathcal{F}}_n$ is a sigma-algebra contained in $\mathcal{F}$, and including one more random variable among its generators can only enlarge it. Hence ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$; the trivial ${\mathcal{F}}_0$ is contained in every sigma-algebra. These are exactly the requirements for a filtration.

The constant $X_0=x$ is ${\mathcal{F}}_0$-measurable. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable. The pair $(X_n,U_{n+1})$ is then ${\mathcal{F}}_{n+1}$-measurable, and the event $\{U_{n+1}\le X_n\}$ is ${\mathcal{F}}_{n+1}$-measurable because comparison of two real coordinates defines a Borel set. Define its indicator function by $B_{n+1}$. The recursion becomesIt is a measurable function of that pair. Mathematical induction proves $X_n$ is ${\mathcal{F}}_n$-measurable for every $n$, so $(X_n)$ is an adapted process. The same recursion also proves $0<X_n<1$ at every finite time, a useful bound for the later integrability arguments.

Write $A=g(X)$ and $B=g(Y)$. The two conditional expectation identities supply a $\sigma (Y)$-measurable version of $B$ and a $\sigma (X)$-measurable version of $A$. Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.

Testing the first conditional expectation identity against $B$, and the second against $A$, givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations $B_m=(-m)\vee (B\wedge m)$ and $A_m=(-m)\vee (A\wedge m)$, then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore $g(X)=g(Y)$ almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.

For a counterexample to equality of the original random variables, let $X$ give probability $1/4$ to each of $-2,-1,1,2$, set $Y=-X$, and use the nonconstant function $g(t)=t^2$. Then $g(X)=g(Y)=X^2$ takes both values $1$ and $4$. It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$, while neither random variable can be zero. Thus

Put $A=\{|X|\ge Z\}$. Both $A$ and its indicator function are measurable, and $0\le Z{1}_A\le |X|$. Thus $Z{1}_A$ is an integrable random variable. Monotonicity of conditional expectation givesThe measurable-factor property of conditional expectation says that a bounded $\mathcal{G}$-measurable function $H$ satisfies $\mathbb{E}[HW\mid \mathcal{G}]=H\mathbb{E}[W\mid \mathcal{G}]$ for every integrable random variable $W$. Here the factor $Z$ need not be bounded. Apply that property first to $Z\wedge m$ and $W=1_A$, then use monotone convergence on the nonnegative products, including the conditional version, to obtainAll the left-hand conditional expectations are well defined; the final one is finite almost surely because $Z{1}_A$ is an integrable random variable. Since $Z>0$ almost surely, division now yieldsThis proves the conditional Markov inequality with a measurable threshold without assuming that $Z^{-1}|X|$ is an integrable random variable. The measurable-factor rule, with its product-integrability condition, is also stated in the standard conditional expectation theory.^[ref]

Let $I_n$ count immigration events during the first $n$ transitions, and let $A_n$ count upward steps during those transitions. Write $h(k)=\theta /(k+\theta )$ for the conditional immigration probability. If $J_m$ is the indicator that transition $m$ to $m+1$ is immigration, then $J_m-h(X_m)$ is a bounded martingale difference with respect to the history that records event types as well as population sizes. The strong law for martingales with bounded increments impliesSince $h$ is bounded, the countable-state Markov chain ergodic theorem givesApplying the same argument to upward-step indicators gives $A_n/n\to {\sum }_k{\pi }_k{u}_k$ almost surely. To evaluate this denominator, sum detailed balance over all neighboring edges:Since $u_k+d_k=1$ and ${\sum }_k{\pi }_k=1$, both sums equal $1/2$. The denominator limit is positive, so division is valid and the immigration fraction among upward steps isThis is the requested answer in terms of $\pi$. Substituting part (iii) additionally gives ${\sum }_k{\pi }_k\theta /(k+\theta )=(1-r)/2$, so the fraction simplifies to $1-r=(1-2p)/(1-p)$.

Every step changes the population by one, so $X_n$ has the parity of $n$ when $X_0=0$. Hence $\mathbb{P}(X_{2m+1}=0)=0$. A return in two steps has positive probability from every state, and all return times are even, so the period of a state in a Markov chain is exactly two.

Each transition exchanges the even and odd classes. Since $\pi$ is an invariant distribution, their masses are equal: $\pi (\text{even})=\pi (\text{odd})=1/2$. The chain sampled at times $2m$, restricted to the even states, is irreducible and positive recurrent, and it is aperiodic because its diagonal entries are positive. Its stationary distribution is $2{\pi }_k$ on even $k$. The countable-state Markov chain convergence theorem applied to this sampled chain yieldsSince ${\pi }_0>0$, the full sequence has no limit. Its time average nevertheless tends to ${\pi }_0$ by the countable-state Markov chain ergodic theorem.

Assume $p<1/2$, and put $a=\theta /p$ and $r=p/q\in (0,1)$. The detailed balance equations giveMultiplying from $0$ to $k-1$, the last factor telescopes, while the product involving $a$ is evaluated by the Gamma function:Consequently the requested invariant distribution can be expressed using the paper's power series asThe assumed convergence of $g_{p,\theta }(r)$ ensures that this normalization is finite and positive.

For an explicit normalizing constant, the binomial series givesThe termwise differentiation of a power series is valid for $|r|<1$, soThe last equality uses $r=p/q$ and $a=\theta /p$: $ar+\theta (1-r)=\theta /q+\theta (q-p)/q=2\theta$. ThereforeThese positive terms sum to one and satisfy the detailed balance equations, so part (a)(ii) verifies stationarity. Irreducibility makes the invariant distribution unique.

We use the stationary weights of a reflected birth-death chain, also verifying why they decide the question. If a stationary distribution $\pi$ exists, its equation at zero is ${\pi }_0={\pi }_1{d}_1={\pi }_0{u}_0$. At $k\ge 1$, stationarity givesThus the edge flow is constant, and its value at the boundary is zero. Every stationary distribution must therefore satisfy detailed balance on each edge. Since irreducibility forces ${\pi }_0>0$, it must be proportional toConversely, if ${\sum }_k{w}_k<\infty$, normalizing these positive weights gives a probability distribution satisfying detailed balance, hence a stationary distribution. Existence of a stationary distribution for an irreducible Markov chain on a countable state space is equivalent to positive recurrence.

The ratio isFor $p<1/2$, the limit is less than one, so the ratio test gives summability. For $p\ge 1/2$, observe that $u_k=p+\theta q/(k+\theta )>p\ge 1/2$, whereas $d_{k+1}<q\le 1/2$. Thus $w_{k+1}/w_k>1$ for every $k$, so the weights cannot be summable. This also resolves $p=1/2$ without making an inconclusive appeal to the limiting ratio. Therefore

Under the specified evolution rule, conditional on the entire history, the next event's probability distribution depends only on the current population size. This is the time-homogeneous Markov property. Birth and immigration both increase the size by one, while death decreases it by one. Writing $q=1-p$, the transition probabilities on ${\mathbb{Z}}_{\ge 0}$ areand all other entries are zero. In particular $P(0,1)=1$ and $d_0=0$. The row sum is $[kp+\theta +kq]/(k+\theta )=1$. Every neighboring pair communicates, so this is an irreducible Markov chain, specifically a birth-death-immigration jump chain.

For the countable-state Markov chain ergodic theorem, let the Markov chain be irreducible and positive recurrent, with its unique stationary distribution $\pi$. From any initial state, for every real $f$ satisfying ${\sum }_j{\pi }_j|f(j)|<\infty$,In particular, $n^{-1}{\sum }_{m=0}^{n-1}{1}_{\{X_m=j\}}\to {\pi }_j$ almost surely. This time-average conclusion requires no aperiodicity.

The countable-state Markov chain convergence theorem adds the hypothesis of an aperiodic Markov chain and concludesIt also follows that ${\mathbb{P}}_{\nu }(X_n=j)\to {\pi }_j$ for every initial probability distribution $\nu$. For an irreducible recurrent chain, the occupation-frequency theorem can also be stated as a limit $1/{\mathbb{E}}_j{T}_j^{+}$, interpreted as zero for a null recurrent state; in the positive recurrent case this equals ${\pi }_j$.

The assumed identities are the detailed balance equations. Summing over $i$ and using the row sums of the transition matrix gives, for every $j$,All summands are nonnegative, so these countable sums are justified. Thusand $\pi$ is an invariant distribution.

A probability distribution on the countable state space $\mathcal{S}$ is precisely a row vector satisfyingA transition matrix is a stochastic matrix, meaningEvery row therefore gives the conditional probability distribution of the next state given the present state. These conditions do not require $\pi$ to be an invariant distribution of $P$.

First take integers $1\le k<n$. For independent copies $X_1,\dots ,X_n$ of $X$, defineThe two blocks involve disjoint families of independent random variables, so $A$ and $B$ are independent. The assumed strictly stable distribution gives $A\stackrel{d}{=}X$, $B\stackrel{d}{=}X$, andTherefore, on any fixed probability space carrying $X$ and an independent copy $X^{\prime }$, their joint probability distribution givesNow fix an arbitrary $0<c<1$ and choose rationals $r_m=k_m/n_m\in (0,1)$ tending to $c^{\alpha }$. Continuity of the real powers gives, on that fixed probability space,The left sides all have the probability distribution of $X$. Since almost sure convergence implies convergence in distribution, their limit has that same probability distribution. HenceThis proves strict stability implies self-decomposability. The limit argument uses no moment-generating function or finite moments, which is essential for general stable distributions.

For independent and identically distributed random variables $X_j\sim \mathcal{N}(0,{\sigma }^2)$, the linear combination of independent normal random variables $n^{-1/2}{\sum }_{j=1}^n{X}_j$ has expected value zero and variance $n^{-1}n{\sigma }^2={\sigma }^2$. ThusThis is exactly the requested $2$-stability. The paper's definition is a strictly stable distribution, since it allows no centering term.

For $X\sim \mathrm{Exp}(\lambda )$, integration gives $M_X(t)=\lambda /(\lambda -t)$ for $t<\lambda$. For $0<c<1$, the candidate remainder moment-generating function isThis is the moment-generating function of the mixture distributionConstruct it by taking an exponential random variable $E$ of rate $\lambda$ and a Bernoulli random variable $J$ with $\mathbb{P}(J=1)=1-c$, with $E,J,X$ mutually independent, and setting $R_c=JE$. Then $R_c$ is independent of $X$, andon a neighborhood of zero. By the uniqueness theorem for moment-generating functions, $cX+R_c\stackrel{d}{=}X$, so the exponential distribution is a self-decomposable distribution.

Fix $|t|<t_0$. Since $|ct|<t_0$, $0<M_X(ct)<\infty$. Using independence of random variables and the Tonelli theorem for the positive exponential integrands gives the extended-real identityThe left side is finite and the first factor is finite and strictly positive. Consequently $M_R(t)$ is finite andThe same reasoning applies to $R^{\prime }$, so their moment-generating functions are finite and equal throughout $(-t_0,t_0)$. The uniqueness theorem for moment-generating functions yieldsThis proves uniqueness of a self-decomposition remainder while also checking the required finiteness; no independence between $R$ and $R^{\prime }$ is needed.

Fix $0<c<1$ and choose $R_c$, independently of $X$, with the normal distributionThe linear combination of independent normal random variables $cX+R_c$ is normal. Its expected value is $c\mu +(1-c)\mu =\mu$ and its variance is $c^2{\sigma }^2+(1-c^2){\sigma }^2={\sigma }^2$. Hence $cX+R_c\stackrel{d}{=}X$ for every $c$, proving that every specified normal distribution is a self-decomposable distribution.

Under independence of random variables, the law of a random variable $Z$ is the mixed probability distribution $(1-p){\delta }_0+p\mathrm{Unif}(-1,1)$. For negative $z$, only the uniform component can contribute to $\mathbb{P}(Z\le z)$; at zero the probability mass $1-p$ is included. Thus its cumulative distribution function isThe sketch consists of a horizontal segment at height zero before $-1$, a line of slope $p/2$ ending at the open point $(0,p/2)$, a jump of size $1-p$ to the closed point $(0,1-p/2)$, another line of slope $p/2$ ending at $(1,1)$, and a horizontal segment thereafter. The cumulative distribution function is right-continuous at the jump.

Under the literal assumptions, write $H(z)=\mathbb{P}(U\le z\mid B=1)$. The general answer is $F_Z(z)=(1-p)1_{\{z\ge 0\}}+pH(z)$, so the continuous part of the sketch is undetermined. The jump at zero is still exactly $1-p$, because $\mathbb{P}(U=0)=0$. In the dependent counterexample from (ii), $F_Z(z)=0$ for $z<0$, $F_Z(z)=(1+z)/2$ for $0\le z<1$, and $F_Z(z)=1$ for $z\ge 1$.

If $B$ and $U$ are independent random variables, conditioning on the Bernoulli random variable $B$ giveswith the continuous value $M_Z(0)=1$. Indeed $Z=0$ when $B=0$, while the conditional probability distribution of $Z$ given $B=1$ is the original uniform distribution.

Under the literal assumptions alone, the law of total expectation gives onlyThis is not determined by $M_U$ and $p$. For an explicit counterexample, put $p=1/2$ and $B=1_{\{U>0\}}$. Then $B$ has the required Bernoulli distribution, butwhich differs from the independent-case formula.

The uniform distribution has probability density function $1/2$ on $(-1,1)$. Therefore, for $t\ne 0$,The value at zero follows either directly from the definition of the moment-generating function or by continuity. This answer needs no assumption about $B$.