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If nonnegative integer-valued random variables $X_n$ have all fixed factorial moments converging to ${\lambda }^k$, then their convergence in distribution is to the Poisson distribution with mean $\lambda$. The first-moment bound gives uniform tightness, and bounds on higher moments give uniform integrability of each fixed lower moment. Every subsequential limit therefore has the indicated factorial moments. For such a limit $X$ and $t\ge 0$, the nonnegative binomial expansion and the monotone convergence theorem give $\mathbb{E}(1+t{)}^X={\sum }_{k\ge 0}{t}^k{\lambda }^k/k!=e^{\lambda t}$. This identifies its moment-generating function as that of the Poisson distribution, uniquely determining the limit.

First prove the concentration statement with its correct scale. Let $U_i=R_i^2$, the $i$th uniform order statistic. The number of sample points at most $z$ is $B_z\sim \mathrm{Bin}(n,z)$. Therefore $U_i>z$ means $B_z<i$, and $U_i<z$ means $B_z\ge i$, up to probability-zero ties. Take ${\epsilon }_n=1/\log n$ and $M=\lceil n^{1/10}\rceil$. Applying Chernoff bounds at $z=(1\pm {\epsilon }_n{)}^{2}i/n$ gives, for an absolute $c_0>0$ and sufficiently large $n$,When the upper endpoint exceeds one its upper-tail event is empty. A union bound makes the total failure probability for all $M\le i\le n$ at most $2n\exp (-c_{0}M/(\log n{)}^2)=o(1)$. On this event,where integral comparison gives ${\sum }_{i=M}^n{i}^{-1/2}=2\sqrt{n}+O(\sqrt{M})$.

The first vertex block has right endpoint $R_1$ and always contains $L_1$, so its initial chord contributes vertex degree two. All its other endpoints are those $L_i$ with $i\ge 2$ that fall below $R_1$. Conditional on the right endpoints, these events are independent with probabilities $R_1/R_i$. ThusThe fewer than $M$ early summands contribute at most $M=o(\sqrt{n})$ to this conditional mean. For the rest, the uniform estimate and uniform tightness of $\sqrt{n}{R}_1$ giveTo control the actual fluctuations without assuming unconditional independence, restrict to the concentration event and $\sqrt{n}{R}_1\le A$. Then $\mu (R)\le M+2+CA\sqrt{n}$. Conditional Chebyshev inequality bounds the probability of a deviation exceeding $\eta \sqrt{n}$ by $(M+2+CA\sqrt{n})/({\eta }^{2}n)=o(1)$. The discarded endpoint event has limiting probability $e^{-A^2}$; let $A\to \infty$. ThereforeSince $W$ has a continuous distribution, the corrected first-vertex degree in the LCD model tail isThis disproves the printed exponent $-y^2/8$ for $G_n^{(1)}$. The hint's factor is independently contradicted at $i=n$: its proposed upper bound tends to $1/\sqrt{2}$, whereasThus the printed concentration event has probability tending to zero, not one. Replacing its $2n$ by $n$ leads to the proven exponent $-y^2/4$. The exponent $-y^2/8$ would instead arise for $d_1(2n)/\sqrt{n}$, but then the graph has $2n$ vertices and the part (b) endpoint limit at $x/\sqrt{n}$ becomes $e^{-2x^2}$. It cannot repair all the printed claims with one convention. The literal part (c) is false for the named model; the full corrected limit and the source of the discrepancy are established above.

On the real line, Prohorov's theorem states that a sequence of probability measures is uniformly tight if and only if every subsequence has a further subsequence converging weakly to a probability measure. Here uniformly tight means that, for every $\epsilon >0$, some compact interval contains mass at least $1-\epsilon$ for every measure in the sequence. This is relative sequential compactness for weak convergence of probability measures, not merely the existence of one convergent subsequence.

We first prove the characteristic-function tail bound. Put $c_0=1-\sin 1>0$. For $y\ne 0$,At $y=0$ the expression is zero by continuity. If $|y|\ge \lambda$, the supplied elementary bound, with $t=|y|/\lambda$, makes it at least $c_0$. The integrand is nonnegative, so Tonelli's theorem yieldsThus one valid universal constant isThe estimate controls spatial tails using only the characteristic function near zero.

Apply it to each ${\mu }_n$. For fixed $\lambda$, pointwise convergence of the characteristic functions and the bound $|{\varphi }_n|\le 1$ allow dominated convergence on the finite integration interval, givingContinuity of a characteristic function and $\varphi (0)=1$ make the last expression tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ making it less than $\epsilon /2$; then for all sufficiently large $n$, the preceding tail is less than $\epsilon$. Enlarge the compact interval to handle the finitely many remaining probability measures. This proves uniform tightness of the whole sequence.

By Prohorov's theorem, every subsequence has a further weakly convergent subsequence, say with limit $\nu$. The real and imaginary parts of $e^{iuy}$ are bounded continuous functions, so along that subsequenceThe uniqueness theorem for characteristic functions implies $\nu =\mu$. All subsequential weak limits are therefore $\mu$. If the original sequence failed to converge weakly to $\mu$, a bounded continuous test function and a subsequence would keep its integrals a fixed distance from the limiting integral; a further weakly convergent subsequence would contradict that. HenceThis supplies the required proof of the Lévy continuity theorem in the stated case.

For the final request, the printed $Q$ is not defined; the intended function is $\varphi$, the characteristic function of a summand. We prove that ${\varphi }^{\prime }(0)=ia$ without assuming integrability. Write $Y_n=S_n/n$. Independence gives $\mathbb{E}{e}^{iu{Y}_n}=\varphi (u/n{)}^n$. Moreover convergence in probability to $a$ implies uniform convergence of these characteristic functions on each bounded frequency interval: for $|u|\le T$ and $\delta >0$,First let $n\to \infty$, then $\delta \downarrow 0$.

The potential logarithm-branch ambiguity needs to be resolved, not discarded. By continuity of $\varphi$ near zero, its principal complex logarithm is defined at $u/n$ for all $|u|\le 1$ and all large $n$, with value zero at $u=0$. Alsouniformly on $[-1,1]$, so $\mathrm{Log}{\Psi }_n(u)\to 0$ uniformly there. The two continuous logarithms differ by an integer multiple of $2\pi i$:That integer is continuous in $u$ and zero at zero, hence identically zero. ConsequentlyFor arbitrary nonzero $h\to 0$, take $n=\lfloor 1/|h|\rfloor$ and $u=nh$. Eventually $1/2\le |u|\le 1$, so dividing the preceding error by $u$ shows $\mathrm{Log}\varphi (h)/h\to ia$. Exponentiating at zero proves the derivative forced by the weak law:This is the weak law forces a characteristic-function derivative argument. Its estimates use bounded exponentials, not $\mathbb{E}|X_1|$. If $X_1$ is integrable, dominated convergence permits differentiation inside the expectation and gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1$, so then $a=\mathbb{E}{X}_1$.

Absolute integrability is genuinely unnecessary. For an explicit weak law without an absolute first moment, take a symmetric variable withwith mass $1-1/e$ at zero and no mass with $0<|X|<e$. This is a valid law: the specified tail decreases from $1/e$ to zero, and its derivative defines the continuous part beyond $e$. Its absolute first moment is infinite because ${\int }_e^{\infty }dx/(x\log x)=\infty$. For independent copies, set $Y_{n,j}=X_j{1}_{\{|X_j|\le n\}}$. Symmetry gives $\mathbb{E}{Y}_{n,j}=0$, and integration of the tail givesFor the last estimate, split the integral at $\sqrt{n}$; the initial part is $O(\sqrt{n})$ and the remainder is at most $2n/\log n$, apart from a fixed initial constant. Chebyshev's inequality now gives $n^{-1}{\sum }_{j\le n}{Y}_{n,j}\to 0$ in probability. The probability that any truncation changes a summand is at most $n\mathbb{P}(|X|>n)=1/\log n\to 0$, so $S_n/n\to 0$ in probability although $\mathbb{E}|X|=\infty$. The derivative is zero, but the usual expectation of $X$ does not exist.

If either the positive or negative part is integrable, the finite weak-law limit does force full integrability. For example, if $\mathbb{E}{X}^{-}<\infty$, then $X\wedge A$ is integrable and its sample averages have the weak law of large numbers limit $\mathbb{E}(X\wedge A)$. Since they are bounded above by the original averages, $\mathbb{E}(X\wedge A)\le a$ for every $A$. Letting $A\to \infty$ shows $\mathbb{E}{X}^{+}<\infty$. The symmetric argument handles the other one-sided hypothesis. The nonintegrable example necessarily has both one-sided expectations infinite.

Enlarge the space if needed to choose $Y_0$ with density $p$, independent of the driving Brownian motion, and solve the same equation with the same driver as $X_0=x$. The second moment of $p$ makes the initial difference square integrable. By parts (a) and (b),while part (c) givesWe must not assume that the smooth bounded $f$ is globally Lipschitz. The hypotheses give $C_Y={\sup }_t\mathbb{E}{Y}_t^2<\infty$, hence uniform tightness of $(Y_t)$. For any $R$ and $0<\delta \le 1$, define the modulus of continuity of $f$ on $[-R-1,R+1]$ by ${\omega }_R(\delta )$. Splitting according to $|Y_t|\le R$ and $|X_t-Y_t|\le \delta$, and applying Markov inequality, givesFirst choose $R$ large, then $\delta$ small using uniform continuity on that compact interval, and finally let $t\to \infty$. The right side can be made arbitrarily small. Together with part (a), this provesThis is convergence by synchronous coupling for bounded continuous test functions. It only needs the constant expectation from part (b) and uniform second moments; it does not add an unstated global bound on $f^{\prime }$ or require a separate stationarity theorem. As explained in part (c), the literal bounded-drift plus strict-contraction assumptions have no global example; the calculation records the intended consequence under compatible dissipative-drift hypotheses as well.

A probability measure is tight if compact sets carry arbitrarily nearly all its mass. Every Borel probability measure on a Polish space is tight. The stronger family condition is uniform tightness.