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For sufficiently large $n$, $p_n=\lambda /n$ lies in $(0,1]$. The waiting time has the geometric distribution on $1,2,\dots$, so summing its geometric series giveswhenever $(1-\lambda /n)e^{t/n}<1$. Fix $t<\lambda$. The expansions $e^{t/n}=1+t/n+O(n^{-2})$ show that the numerator is $\lambda /n+O(n^{-2})$ and the denominator is $(\lambda -t)/n+O(n^{-2})$. ThereforeThe convergence theorem for moment-generating functions says that pointwise convergence of finite moment-generating functions on a fixed open interval about zero to the moment-generating function of a probability distribution implies convergence in distribution to that distribution. For example take $|t|<\lambda /2$; all large $n$ satisfy the convergence condition throughout that interval. The limit is the moment-generating function from part (a), proving

Use the characteristic function to establish convergence in distribution by the Lévy continuity theorem, especially when exponential moments do not exist; a Cauchy distribution, for example, has a characteristic function but no finite moment-generating function on a neighborhood of zero.

Use the moment-generating function for an exponential-tail estimate: for $t>0$ with $M_X(t)<\infty$, Markov inequality givesOptimizing over $t$ produces a Chernoff bound. The modulus-one characteristic function does not directly give this positive exponential bound.

Convergence in distribution to $Y$ means convergence of cumulative distribution functions at every continuity point of the limiting distribution function:Convergence at a jump of $F_Y$ is not required. Only the laws are involved, so this definition does not require a particular coupling of the variables.

The implication convergence in probability implies convergence in distribution follows by comparing nearby events. Fix a continuity point $x$ of the cumulative distribution function $F_X$ and let $\epsilon >0$. If $X\le x-\epsilon$ and $|X_n-X|\le \epsilon$, then $X_n\le x$. Conversely, $X_n\le x$ and $|X_n-X|\le \epsilon$ imply $X\le x+\epsilon$. ThusThe error probability tends to zero by convergence in probability. Taking lower and upper limits and then letting $\epsilon \downarrow 0$, continuity at $x$ squeezes both bounds to $F_X(x)$. This is exactly convergence in distribution.

Convergence in distribution means that the cumulative distribution functions satisfyat every $x$ where the cumulative distribution function of $X$ is continuous. Values at its discontinuities need not converge. This definition uses only the probability distributions, so the random variables need not be defined on the same probability space.

For $x>0$, the event $M_n/n\le x$ says that each of the independent random variables is at most $\lfloor nx\rfloor$. Hence, for all sufficiently large $n$,Since $n/(\lfloor nx\rfloor +1)\to 1/x$ and $\log (1-u)=-u+o(u)$ as $u\to 0$, taking logarithms gives the limit $e^{-1/x}$. For $x\le 0$ the probability is zero. ThereforeThis is the Fréchet limit for a reciprocal integer tail, with a Fréchet distribution of shape one. Its cumulative distribution function is continuous even at zero, so the pointwise calculation establishes convergence in distribution.

The sample maximum has cumulative distribution functionThe product $x^n$ follows from independence of random variables and the continuous uniform distribution of each observation. For every $x<1$, this cumulative distribution function tends to zero; for $x>1$ it equals one. These are exactly the values of the cumulative distribution function of the constant random variable $1$ at its continuity points. Therefore, by the definition of convergence in distribution,At $x=1$ the limiting cumulative distribution function is discontinuous, so the definition does not require checking that point, although $F_{M_n}(1)=1$ also holds.

The cumulative distribution function of the uniform distribution on $[0,1/n]$ isThe constant zero has cumulative distribution function $F(t)=1_{[0,\infty )}(t)$. Its continuity points are precisely $t\ne 0$. For fixed $t<0$, $F_n(t)=0=F(t)$; for fixed $t>0$, $F_n(t)=1=F(t)$ eventually. This proves convergence in distribution to zero directly from the definition.

Also, $\mathbb{P}(0\le U_n\le 1/n)=1$. For every $\epsilon >0$ and $n>1/\epsilon$, $\mathbb{P}(|U_n|>\epsilon )=0$. HenceThe convergence in probability conclusion holds for any realization of these random variables on a common probability space; no independence of random variables is needed.

Use $X_n\sim \mathrm{Unif}[0,1/n]$ from part (a). We already proved convergence in distribution to zero. For any prescribed $p\in [0,1]$, put $c_n=p/n$. Then $c_n\to 0$, and the cumulative distribution function satisfiesThis includes $p=0$ and $p=1$, since the uniform distribution assigns zero probability to each endpoint. The limit cumulative distribution function is discontinuous at zero, explaining why the conclusion of (ii) does not apply there.

The definition of convergence in distribution gives $F_n(t)\to F(t)$ at each continuity point $t$ of $F$. We must handle a moving argument rather than assume this convergence is uniform.

A cumulative distribution function has at most countably many discontinuities: each is a jump of positive size, and there are at most $r$ jumps of size at least $1/r$, since their sizes sum to at most one. Thus we can choose continuity points $u_j<c<v_j$ with $u_j\uparrow c$ and $v_j\downarrow c$. For fixed $j$, eventually $u_j<c_n<v_j$. The monotonicity of each cumulative distribution function impliesFirst let $n\to \infty$, using the definition of convergence in distribution at $u_j,v_j$. Then let $j\to \infty$, using continuity at $c$:No result relating different convergence of random variables has been assumed.

Separate a long future count from the early Poisson arrival time. Let ${\mathcal{F}}_t=\sigma (N_s:0\le s\le t)$ be the natural filtration, and defineThe truncated Poisson arrival time in $B_n$ is ${\mathcal{F}}_{n^2}$-measurable. The independent increments property makes $A_n$ independent of ${\mathcal{F}}_{n^2}$, and hence of $B_n$.

The numerator of $A_n$ is a centered Poisson random variable of parameter $n^4-n^2$. Apply the central limit theorem to the $n^4-n^2$ unit-time increments, then use $\sqrt{n^4-n^2}/n^2\to 1$ and Slutsky's theorem. This proves $A_n\stackrel{\mathrm{d}}{\to }N(0,1)$. Also, Markov's inequality and $\mathbb{E}{T}_n=n$ giveThus $B_n$ equals $(T_n-n)/\sqrt{n}$ outside an event of vanishing probability, so part (i) and Slutsky's theorem give $B_n\stackrel{\mathrm{d}}{\to }N(0,1)$. Exact independence of random variables for each $n$ now gives the joint limit $(A_n,B_n)\stackrel{\mathrm{d}}{\to }(Z_1,Z_2)$ with independent standard normal random variables.

Finally, $\mathbb{E}{C}_n=0$ and $\mathrm{Var}(C_n)=n^{-2}$, so Chebyshev's inequality gives $C_n\to 0$ in probability. The first coordinate in the question is $A_n+C_n$. The second differs from $B_n$ only on $\{T_n>n^2\}$, and so its difference tends to zero in probability. The vector version of Slutsky's theorem proves the required joint convergence in distribution. The boundary of $(-\infty ,x]\times [y,\infty )$ has zero limiting probability, since both marginal distributions have continuous cumulative distribution functions. The Portmanteau theorem therefore yieldsThis is $\mathbb{P}(Z_1\le x,Z_2\ge y)$ by independence of random variables.

Let $d=\gcd (a,b)>1$ and put $a^{\prime }=a/d$, $b^{\prime }=b/d$. Then $a^{\prime },b^{\prime }$ are coprime integers, and $p=b^{\prime }/(a^{\prime }+b^{\prime })$. The six requested revisions are as follows.

(b)(i): communicating classes. Every increment is divisible by $d$, so the residue class modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, Bézout's identity applied to $a^{\prime },b^{\prime }$ gives a path just as in (b)(i). Thus the original Markov chain is reducible on $\mathbb{Z}$, with exactly the $d$ closed communicating classesThe specified walk starting at zero stays in $d\mathbb{Z}$.

(b)(ii): period of a state in a Markov chain. The return equation $au=bv$ reduces to $a^{\prime }u=b^{\prime }v$. Hence $u=b^{\prime }k$, $v=a^{\prime }k$. For every state, the period of a state in a Markov chain is thereforeThis is still at least two, so no state is aperiodic.

(b)(iii): stationary distribution. The conclusion is unchanged: there is no stationary distribution on $\mathbb{Z}$. The maximum-mass proof in (b)(iii) applies verbatim, without using coprimality. Equivalently, each closed communicating class, after translating and dividing its state labels by $d$, is the same centered random walk with step sizes $a^{\prime },-b^{\prime }$ and has no stationary distribution.

(b)(iv): occupation of zero. On $d\mathbb{Z}$, the rescaled walk $S_n/d$ has exactly the transition probabilities of the coprime walk with parameters $a^{\prime },b^{\prime }$. The return-probability argument in (b)(iv) therefore makes this class null recurrent. The renewal process proof of the occupation limit applies unchanged:Starting in another residue class would make every visit to zero impossible, also giving occupation proportion zero.

(c)(i): modified communicating classes. These are unchanged: $C=\{-c-b,\dots ,c+a\}$ is one finite closed communicating class, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used coprimality. Resetting to the $b$ consecutive integers $c-b+1,\dots ,c$ connects different residue classes, so it would be incorrect to split $C$ according to residue classes modulo $d$.

(c)(ii): limiting distribution. The column-sum calculation, the irreducible Markov chain property on a finite class, positive self-transition probabilities, and geometric entrance argument also required no coprimality. Thus for every initial state the same convergence in distribution holds:In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.

The uniqueness theorem for moment-generating functions states that if $M_X$ and $M_Y$ are finite and equal on an open real interval containing zero, thenFor the convergence theorem for moment-generating functions, suppose that the moment-generating functions $M_{X_n}$ are finite on a common open real interval $(-a,a)$, $a>0$, and $M_{X_n}(t)\to M_X(t)$ for every $|t|<a$, where $M_X$ is finite there. ThenOne may equivalently assume convergence to a finite function $M$ on that interval: the theorem then asserts existence of a unique limiting probability distribution with moment-generating function $M$ as well as convergence in distribution.

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.

Interpret the source's missing variable in the limit arrow as $t\to \infty$. We prove convergence in distribution by showing that the angle of the Brownian motion becomes uniform on the circle group.

Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular Fourier coefficients satisfyThese are precisely the Fourier coefficients of the uniform distribution on the unit circle. Thus expectations of every trigonometric polynomial converge to its uniform angular average. By the Stone-Weierstrass theorem, trigonometric polynomials approximate continuous functions on the unit circle uniformly. Given an approximation error $\epsilon$, the discrepancy of expectations is at most the polynomial discrepancy plus $2\epsilon$. Taking $t\to \infty$ and then $\epsilon \to 0$ proves weak convergence of probability measures to uniform angular measure.

The function $z\mapsto (\mathrm{Im}z{)}^2$ is continuous on the unit circle, so the continuous mapping theorem and part (a)(ii) implyEquivalently, the limiting arcsine distribution has cumulative distribution function $2\arcsin \sqrt{x}/\pi$ for $0\le x\le 1$, and probability density function $1/(\pi \sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting mean.

For a partition $0=t_0<\cdots <t_n=1$, approximate the integral by $X_{\pi }={\sum }_j{t}_j(B_{t_{j+1}}-B_{t_j})$, using left endpoints. The defining independent Brownian increment of Brownian motion make this a centered Gaussian variable with variance ${\sum }_j{t}_j^2(t_{j+1}-t_j)$. As the mesh tends to zero, that variance tends to ${\int }_0^1{t}^{2}dt=1/3$.

The stochastic integral is the mean-square convergence of these sums. mean-square convergence implies convergence in distribution, while the characteristic function converge to $e^{-u^2/6}$. HenceThe probability is exactly one half because this centered nondegenerate normal distribution is symmetric and has no atom at zero.

The long-time Spitzer winding theorem states that, for the continuous unwrapped angle ${\Theta }_t$ of planar Brownian motion started at $1$,Here $C$ has the standard symmetric Cauchy distribution. The notation means convergence in distribution: for every real $x$,Equivalently, expectations converge for every bounded continuous test function. This is a statement about the laws at large deterministic times, without asserting almost-sure convergence of sample paths. The angle is unwrapped; a principal argument is bounded and would instead give zero after division by $\log t$.

Write $M_n={\max }_i{E}_i$. Each exponential distribution has cumulative distribution function $1-e^{-x}$ for $x\ge 0$. Independence therefore gives the maximum of independent exponential random variables formulaFor any fixed $y$, the threshold $y+\ln n$ is nonnegative for all sufficiently large $n$. ThenThis is convergence in distribution to the standard Gumbel distribution.

In a uniform balls-in-bins allocation with $m\to \infty$ and $n/m\to \lambda \in (0,\infty )$, a fixed bin count has convergence in distribution to a Poisson distribution variable of mean $\lambda$ by the Poisson limit theorem. If $N_j$ counts bins containing exactly $j$ balls, symmetry and the linearity of expectation give $\mathbb{E}[N_j]/m=\mathbb{P}(B_1=j)\to e^{-\lambda }{\lambda }^j/j!$ for each fixed $j$. In particular the expected empty fraction tends to $e^{-\lambda }$, and the expected fraction with at least two balls tends to $1-(1+\lambda )e^{-\lambda }$. These are statements about expectations; convergence of the random fractions requires an additional argument.

For a random walk with centered independent identically distributed increments of variance one, the maximum of its polygonal interpolation is the maximum at its grid vertices. The maximum functional is Lipschitz for the supremum norm, so the continuous mapping theorem applied to the Donsker invariance principle proves the displayed convergence in distribution. The Brownian reflection principle then gives upper-tail limits $2(1-\Phi (a))$ at all $a>0$, since the limiting maximum has no atom there.