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A map $\Phi :S_D\to \mathbb{R}$ is Hadamard differentiable at $s_0$ if there is a continuous linear functional ${\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\to 0$ with $t_j\ne 0$, $a_j\to a$ in $S$, and $s_0+t_j{a}_j\in S_D$,For tangential Hadamard differentiability, the limit directions $a$ are restricted to a specified tangent subspace and the derivative is continuous linear on that subspace. The domain condition is indispensable.

Assume $X_n\in S_D$ almost surely and use the usual measurable, separably supported setting for norm-valued convergence in distribution, with the limit in the tangent space. The functional delta method givesTo justify this by Skorokhod representation theorem, a sufficient precise version is: for weakly convergent random elements in a separable complete metric space, one can construct copies of the whole sequence and its limit on one probability space which converge almost surely in the metric. More generally the separable-limit-support version of the representation theorem can be used. Apply it to $Z_n=r_n(X_n-s_0)$, obtaining ${\tilde{Z}}_n\to \tilde{X}$ almost surely. The variables $s_0+r_n^{-1}{\tilde{Z}}_n$ have the same laws as $X_n$ and lie in $S_D$ on a common probability-one event. The definition of Hadamard differentiability, with $t_n=1/r_n$, gives convergence almost surely of the transformed quotients to ${\dot{\Phi }}_{s_0}(\tilde{X})$, and hence the claimed convergence in distribution. If $X$ is a Gaussian measure, the limit is Gaussian because the derivative is linear. Without membership of $X_n$ in $S_D$, the expression $\Phi (X_n)$ in the question is not defined.

The final function space is $C_b(\mathbb{R})$ with supremum norm; it is not separable as a whole. This does not affect the deterministic differentiability proof below. Any statistical application of the preceding representation argument must use its separability or tightness hypotheses, rather than assuming that every normed space automatically satisfies them.

Fix admissible $F,G$ and set $T(F,G)={\int }_0^{u}G(x)F^{\prime }(x)dx$. The candidate derivative, defined even for continuous directions $a,b$ without derivatives, isThis follows formally from the product rule and integration by parts. We now establish the uniform-direction limit required by Hadamard differentiability.

Let $F_j=F+t_j{a}_j$ and $G_j=G+t_j{b}_j$ belong to the stated domain, with $a_j\to a$, $b_j\to b$ in supremum norm. The domain gives $\int |F_j^{\prime }|\le 1$ and $\int |F^{\prime }|\le 1$, while $F_j\to F$ uniformly. We first claim that, for every continuous $b$ on $[0,u]$,For a continuously differentiable test function $v$, integration by parts gives convergence directly from uniform convergence of $F_j$: both boundary terms and $\int v^{\prime }{F}_j$ converge. By the Weierstrass approximation theorem, approximate $b$ uniformly on $[0,u]$ by a polynomial $v$. The difference between the two integrals introduced by this approximation is at most $2\Vert b-v{\Vert }_{\infty }$, by the uniform bounded variation bounds. Letting the approximation error decrease proves the claim. This is weak convergence of bounded-variation integrators.

Now decompose the quotient in the order that retains the bounded-variation bound:The first term differs from $\int b{F}_j^{\prime }$ by at most $\Vert b_j-b{\Vert }_{\infty }$, so it tends to $\int b{F}^{\prime }$. For the second term, integration by parts yieldswhich tends to the corresponding expression with $a$. Hence the quotient tends to ${\dot{T}}_{F,G}(a,b)$ for every admissible perturbation sequence. Finally,This proves that the derivative is a continuous linear functional on the product space. Therefore $T$ is Hadamard differentiable at every point of the stated domain. The bounded-variation restriction is what controls the apparently dangerous interaction between uniformly small perturbations and possibly large perturbation derivatives; no convergence of $a_j^{\prime }$ or $b_j^{\prime }$ has been assumed.

A non-degenerate probability distribution is not concentrated at a single point. In extreme value theory, a non-degenerate distribution function $F$ is max-stable if, for each integer $n\ge 1$, constants $a_n>0,b_n\in \mathbb{R}$ satisfy $F(a_{n}x+b_n{)}^n=F(x)$ for every $x$. This says that a normalized sample maximum has the same law as one observation. A distribution function $F$ belongs to the maximum domain of attraction of a non-degenerate $G$ if there are $a_n>0,b_n$ such thatat every continuity point of $G$. Equivalently, $(M_n-b_n)/a_n$ has convergence in distribution to $G$.

The extremal types theorem says that any such non-degenerate limit is max-stable and, up to a positive affine change of variable, is exactly one of the following:These are respectively the Gumbel distribution, Fréchet distribution, and negative Weibull distribution. The theorem is also known as the Fisher–Tippett–Gnedenko theorem.

Useful sufficient conditions can be expressed using the survival function $\overline{F}=1-F$ and the right endpoint of a distribution $x_F$. An infinite endpoint with $\overline{F}(tx)/\overline{F}(t)\to x^{-\alpha }$ for every $x>0$ gives the Fréchet distribution domain. A finite endpoint with $\overline{F}(x_F-sx)/\overline{F}(x_F-s)\to x^{\alpha }$ as $s\downarrow 0$ gives the negative Weibull distribution domain. For the Gumbel distribution, it suffices that a positive Gumbel auxiliary function $a(t)$ gives $\overline{F}(t+a(t)x)/\overline{F}(t)\to e^{-x}$ for every real $x$ as $t\uparrow x_F$. These are regular variation and exponential tail-ratio conditions; they need no proof here.

In case (i), ${\overline{F}}_1(x)=(1-x{)}^2$ on $0<x<1$, so the finite-endpoint ratio is exactly $x^2$. Therefore the domain is negative Weibull with shape $2$. With $a_n=n^{-1/2}$ and $b_n=1$,and the limit is one for $x\ge 0$.

In case (ii), ${\overline{F}}_2(t)=e^{-\lambda t}$ has exponential tail ratio with the constant Gumbel auxiliary function $1/\lambda$. Therefore the domain is Gumbel. Choosing $a_n=1/\lambda$ and $b_n=\log n/\lambda$ givesIn case (iii), ${\overline{F}}_3(t)=(1+t{)}^{-\lambda }$ has regular variation of index $-\lambda$, so the domain is Fréchet with shape $\lambda$. The convenient choices $a_n=n^{1/\lambda }$ and $b_n=-1$ givewith limit zero for $x\le 0$.

Finally, define the empirical distribution function $F_n(x)=n^{-1}{\sum }_{i=1}^n{1}_{\{X_i\le x\}}$. On a sample with positive threshold $t=X_{(n-k)}$ and exactly $k$ observations strictly above $t$, its survival function has $1-F_n(t)=k/n$. Using the Tonelli theorem to integrate the finite nonnegative sum givesThis is the Hill estimator as an empirical plug-in version of the tail-integral limit. It does not prove consistency for fixed $k$; that is a separate issue.

There is a finite-sample qualification because the question permits ties and does not assume positive observations. The formula requires $1\le k<n$ and $t>0$. If $r=\#\{i:X_i>t\}<k$ because the threshold is tied, direct substitution giveswhere terms equal to the threshold contribute zero. This is generally $k/r$ times the displayed Hill estimator, rather than the same estimator. If $r=0$, the empirical plug-in denominator is zero. For a continuous probability distribution the no-tie condition holds almost surely; for a Fréchet distribution domain and fixed $k$, the threshold is positive with probability tending to one. Thus the stated plug-in identity holds at a positive untied threshold, and the literal claim for all samples from an arbitrary distribution function needs this qualification.

First, convergence in probability at $t=0$ gives $X_0=0$ almost surely, since every $X_0^n=0$. For any finite set of times, the corresponding vectors converge in probability: the union bound controls the probability that any coordinate differs by more than a fixed tolerance. The same holds for their increment vectors, hence also for their convergence in distribution.

For $0=t_0<t_1<\cdots <t_m$, set $D_j^n=X_{t_j}^n-X_{t_{j-1}}^n$ and $D_j=X_{t_j}-X_{t_{j-1}}$. The characteristic function of a random vector factors for the independent increments of each $X^n$:Pass to the limit using convergence in distribution and bounded continuous functions. The resulting factorization of the characteristic function of a random vector, with the uniqueness theorem for characteristic functions, proves independence of the $D_j$. Similarly $X_{s+t}^n-X_s^n\stackrel{d}{=}{X}_t^n$ passes to the limit, proving stationary increments for $X$.

It remains to prove stochastic continuity. For $\epsilon >0$, the triangle inequality and the union bound imply, for every fixed $n$,The second term vanishes by stochastic continuity of $X^n$. Now let $n\to \infty$ and use the additional near-zero approximation hypothesis. We obtain $X_t\to 0$ in probability as $t\downarrow 0$. The stationary increments transfer this to every time: both $X_{s+h}-X_s$ and $X_s-X_{s-h}$ have the law of $X_h$ for $h>0$ when defined. Thus $X$ has all the intrinsic Lévy process properties, proving closure of Lévy processes under locally controlled convergence in probability.

If one requires the supplied process itself to be càdlàg, the hypotheses justify a càdlàg modification, rather than that stronger pathwise assertion. To see the distinction, take $X^n\equiv 0$, let $U$ have uniform distribution on $(0,1)$, and put $X_t=1_{\{t=U\}}$. At each fixed time $t$, $X_t=0$ almost surely, so both approximation hypotheses hold with zero error probability. Nevertheless every path has an isolated spike at $U$ and is not right-continuous there. Its identically zero modification of a stochastic process is a Lévy process with càdlàg paths. The conclusion is exact under the intrinsic definition, and exact up to modification under the convention requiring càdlàg paths.

One general form of the Erdős-Kac theorem is the following. Let $f$ be a real strongly additive arithmetic function, with $|f(p)|\le C$ for all primes, and defineIf $B(x)\to \infty$, then under the discrete uniform distribution on $[N]$,Here the arrow denotes convergence in distribution and $\mathcal{N}(0,1)$ is the standard normal distribution. Equivalently, the probability of being at most $z$ tends to the standard normal distribution function $\Phi (z)$ for every real $z$. Replacing $B(x{)}^2$ by ${\sum }_{p\le x}f(p{)}^2(1-1/p)/p$ is equivalent, because the difference is bounded by $C^2{\sum }_p{p}^{-2}$.

This is a general bounded-prime-weight form, not just the special case $f=\omega$. In that case the Mertens theorem for reciprocal primes gives $A(N)=\log \log N+O(1)$ and $B(N{)}^2=\log \log N+O(1)$, so the classical prime-factor central limit theorem is

A useful method of moments in probability states that if real random variables $Z_N$ have all moments, $\mathbb{E}{Z}_N^j\to m_j$ for every integer $j\ge 1$, and $(m_j)$ is the moment sequence of a moment-determinate probability distribution, then $Z_N$ converges to that distribution in the sense of convergence in distribution. In particular, convergence to the standard normal distribution follows from limiting odd moments zero and limiting even moments $(2k-1)!!$. The moment-generating function of a standard normal variable, finite in a neighbourhood of zero, ensures moment determinacy.

For the classical Erdős-Kac theorem, put $L=\log \log N$ and choose, for example, $\phi (N)=L^{1/4}$, with a harmless modification for small $N$. Truncate the prime omega function to primes $p\le y=N^{1/\phi (N)}$. The omitted number of distinct prime factors of any $n\le N$ is at most $\phi (N)$, since a product of $r$ omitted primes exceeds $N^{r/\phi (N)}$. Thus this truncation changes the normalized variable by $o(1)$ uniformly. Also, the Mertens theorem for reciprocal primes gives

Part (b) transfers every fixed normalized central moment of the truncated prime omega function to the independent Bernoulli random variables model. Indeed, its unnormalized error is at most $O_j(N^{2j/\phi (N)-1})$, using ${\sum }_{p\le y}1\le y$, and this tends to zero for each fixed $j$. In the independent expansion, any singleton index has zero expectation; the leading contributions are pairings, giving exactly the standard normal distribution moments, while blocks of size at least three are negligible after normalization. The method of moments in probability, followed by the uniformly negligible truncation and centering errors, proves the Erdős-Kac theorem.

The same strategy proves the bounded-prime-weight form in part (a). Choose $\phi (N)\to \infty$ slowly enough that $\phi (N)=o(B(N))$. The omitted weighted contribution is $O(\phi (N))$; its mean is likewise $O(\phi (N))$, and its independent variance is $O(\log \phi (N)+1)=o(B(N{)}^2)$ by the Mertens theorem for reciprocal primes. Hence the truncated model has the same normalization, and part (b) again transfers each fixed central moment.

Fix $u$. Put $m=\phi (u)$ and form the centered real vectorsThey are bounded independent and identically distributed random variables, so the multivariate central limit theorem applies to $n^{-1/2}{\sum }_j{Y}_j=(\mathrm{Re}{\mathcal{C}}_n(u),\mathrm{Im}{\mathcal{C}}_n(u){)}^T$.

To identify its limiting covariance matrix, put $Z=e^{iu{X}_1}-m$. Its complex covariance and pseudo-covariance are respectivelyFor $Z=A+iB$, $|Z{|}^2=A^2+B^2$ and $Z^2=A^2-B^2+2iAB$. ThusThe two conditions on $\Gamma (u)$ identify exactly this real covariance matrix: the second is ${\mathrm{Cov}}_{\mathbb{C}}(\Gamma (u),\overline{\Gamma (u)})=\mathbb{E}\Gamma (u{)}^2$, with a conjugate in its second argument. A centered multivariate normal distribution is determined by its covariance matrix, including when that matrix is singular. ThereforeThis proves convergence in distribution at each fixed $u$; it does not assert convergence of the entire empirical characteristic process.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.

The symmetric group has $n!$ elements, not the printed order $n$. For $n\ge 2$ the identity and the $n-1$ adjacent transpositions have probabilities $1/n$, so the specified transition matrix is normalized. Its stationary distribution is uniform and it satisfies detailed balance, since each transposition is its own inverse. Adjacent transpositions generate the symmetric group, making this an irreducible Markov chain; the positive holding probability makes it an aperiodic Markov chain.

For a fixed label $i$, right multiplication swaps positions, so $f(\sigma )={\sigma }^{-1}(i)$ is the label's position. Under the uniform distribution on a finite set it is uniform on $\{1,\dots ,n\}$. ThereforeFor each generator $s_j=(j,j+1)$, the squared change $[f(\sigma s_j)-f(\sigma ){]}^2$ is one if the label occupies either of the two positions, and zero otherwise. Its stationary expected value is $2/n$, givingThus the tagged-label Rayleigh quotient for adjacent transpositions gives the corrected bound

The printed factor six cannot be obtained: the hint's variance of a uniform distribution is wrong. Directly, $\mathrm{Var}(U)={\int }_0^1{u}^{2}du-({\int }_0^{1}udu{)}^2=1/12$. Bounded convergence in distribution of $U_n$ does imply convergence of its first two moments, hence convergence of its variance, but the limit is $1/12$.

More strongly, the claimed asymptotic bound itself is false for this kernel. The Aldous spectral gap theorem states that the interchange process on a finite connected weighted graph has the same spectral gap as the single-label random walk with identical edge rates; see Theorem 1.1 and its proof. Here $P-I$ is that interchange process generator on a path graph with edge rates $1/n$. For the tagged-label generator, direct substitution of $g_{\ell }(k)=\cos (\ell \pi (k-1/2)/n)$ gives eigenvalues $-2(1-\cos (\ell \pi /n))/n$, $0\le \ell <n$, including the reflecting endpoint equations. These $n$ distinct eigenvalues exhaust the tagged-label function space. Consequently the theorem yieldswhich exceeds $6/n^3$ asymptotically. This use of a substantial external theorem is only to certify the printed claim's failure; the corrected factor-twelve upper bound above follows directly from the requested test function and the variational formula.

The Lévy continuity theorem states that if characteristic functions ${\phi }_n$ of probability distributions ${\mu }_n$ converge pointwise on $\mathbb{R}$ to $\phi$, and $\phi$ is continuous at zero, then $\phi$ is the characteristic function of a unique probability distribution $\mu$ andHere the arrow denotes weak convergence of probability measures, equivalently convergence in distribution of associated random variables. Conversely, weak convergence of probability measures implies pointwise convergence of their characteristic functions to the characteristic function of the limit. The continuity-at-zero condition is essential to ensure that the pointwise limit is a characteristic function.

Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.

Let $F_n$ be the distribution function of $X_n$. The limiting constant zero has distribution function $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$. For every $\epsilon >0$, both $-\epsilon$ and $\epsilon$ are continuity points of $F$. Thus convergence in distribution implies $F_n(-\epsilon )\to 0$ and $F_n(\epsilon )\to 1$. ThereforeThis provesThe same proof after subtracting a fixed real $c$ proves convergence in distribution to a constant implies convergence in probability. A deterministic limit fixes the coupling automatically; the counterexample in (c) exploits a nonconstant limit whose probability distribution alone does not fix that coupling.

For independent and identically distributed random variables with distribution function $F$, let $M_n={\max }_{1\le i\le n}{X}_i$ be the sample maximum. In extreme value theory, $F$ belongs to the maximum domain of attraction $D(G)$ of a nondegenerate distribution function $G$ when there exist $a_n>0$ and $b_n\in \mathbb{R}$ such thatat every continuity point of $G$. Thus the normalized sample maxima have convergence in distribution to $G$.

A positive measurable function $q$ has regular variation at infinity with index $\rho$, written $q\in {\mathrm{RV}}_{\rho }$, ifThe case $\rho =0$ defines a slowly varying function; equivalently a regularly varying function is $q(t)=t^{\rho }\ell (t)$ with $\ell$ a slowly varying function.

Write $\overline{F}=1-F$ for the survival function and $x_F=\sup \{x:F(x)<1\}$ for the right endpoint of a distribution. For $\alpha >0$, use the following standard extreme value theory normalization conventions:They are respectively the Fréchet distribution, the negative Weibull distribution, and the Gumbel distribution for maxima. The negative Weibull distribution is supported to the left of its finite endpoint; it is not the usual positive Weibull distribution.

The necessary and sufficient maximum domain of attraction criteria are:

For the corresponding sufficient hazard function conditions, assume $F$ is absolutely continuous near $x_F$, its survival function tends to zero there, and its probability density function $f$ is positive there. Write $\eta (t)=f(t)/\overline{F}(t)$. The Von Mises conditions for extreme values giveThe last condition additionally assumes that the reciprocal hazard function is continuously differentiable near the endpoint. It concerns the derivative of the reciprocal hazard function, not the derivative of the hazard function itself. Equivalently, when $f$ is differentiable, it is $\overline{F}(t)f^{\prime }(t)/f(t{)}^2\to -1$.

To see the connection, taking an integral of the hazard function givesThe first two limits make the logarithmic tail ratios over multiplicative or endpoint-distance intervals tend to $-\alpha \log u$ or $\alpha \log u$, giving the required regular variation. For the Gumbel distribution, set $a(t)=1/\eta (t)$. The derivative condition makes $a(t+a(t)v)/a(t)\to 1$ uniformly on bounded $v$-intervals, soThe endpoint assumptions ensure these shifts are admissible: at an infinite endpoint $a(t)/t\to 0$, and at a finite endpoint $a(t)/(x_F-t)\to 0$. These are sufficient smooth-tail conditions; the tail-ratio characterizations above are necessary and sufficient without differentiability assumptions. For a source using the same reciprocal-hazard convention, see Smith and Weissman's university lecture notes, equations (2.18)–(2.24).

For the Gamma distribution with integer shape $m$ and rate one, let ${\overline{F}}_m(t)={\int }_t^{\infty }{u}^{m-1}{e}^{-u}du/(m-1)!$. The base case is ${\overline{F}}_1(t)=e^{-t}$, and integration by parts givesInduction provesThis survival function has the asymptotic form ${\overline{F}}_m(t)\sim e^{-t}{t}^{m-1}/(m-1)!$; for $m=1$ the relation is exact. Put $t_n={\beta }_n+x$. For every fixed real $x$, $t_n\to \infty$ and $t_n/\log n\to 1$. The proposed centering satisfiessoOnly the highest-degree term contributes to the limit, with the same reasoning covering $m=1$. The Taylor expansion $\log (1-u)=-u+O(u^2)$ now givesbecause $n{\overline{F}}_m({\beta }_n+x{)}^2\to 0$. Therefore the Gumbel limit for gamma maxima isEquivalently, $M_n-{\beta }_n$ has convergence in distribution to the standard Gumbel distribution, with scale $a_n=1$.

Suppose first thatThis is convergence in L2, hence convergence in probability, which implies convergence in distribution. By part (i), the respective laws are ${\mu }_n$ and $\nu$, so ${\mu }_n$ converges weakly to $\nu$. Part (a)(i) also gives

Conversely, suppose that ${\mu }_n$ converges weakly to $\nu$ and the displayed second moments converge. By the stated quantile coupling for convergence in distribution,Part (i) identifies their second moments with those of the measures, so part (a)(ii) applies and gives convergence in $L^2$. ThereforeThis is the one-dimensional second-Wasserstein convergence criterion realized by the common-quantile coupling.

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

Convergence in distribution $W_n\Rightarrow W$ means that the cumulative distribution functions satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The Lévy continuity theorem states that this is equivalent to pointwise convergence of characteristic functions to the characteristic function of $W$, provided the limiting function is continuous at zero.

For fixed $t$, independence and a Gaussian integral giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

For $n>m$, independence and $\mathbb{E}{X}_k=0$ giveThus $(S_n)$ is a Cauchy sequence in the complete L2 space and converges in $L^2$ to some random variable $S$. The Cauchy-Schwarz inequality gives $\Vert Z{\Vert }_1\le \Vert Z{\Vert }_2$ on a probability space, so the convergence also holds in $L^1$. Finally, convergence in $L^2$ implies convergence in probability, which implies convergence in distribution.

WriteExpanding $R_i^2$ gives the leading term ${\epsilon }_i^2{\xi }_i^2$ and terms of the forms treated in parts b and c, together with their versions obtained by interchanging $(\epsilon ,F)$ and $(\xi ,G)$. For example, the pure error terms are ${\epsilon }_i^2{G}_i^2$, ${\xi }_i^2{F}_i^2$, and $F_i^2{G}_i^2$, and each cross term is controlled by Cauchy-Schwarz from these. HenceAssuming this limit is positive, the continuous mapping theorem yields ${\tau }_D\stackrel{p}{\to }\{\mathbb{E}({\epsilon }_1^2{\xi }_1^2){\}}^{1/2}$. Combining this with the assumed convergence in distribution of $\sqrt{n}{\tau }_N$ and applying the Slutsky theorem givesThis is the studentization of the generalized covariance measure statistic.

LetHere $S_n$ has the binomial distribution with parameters $(n,\lambda /n)$. Part a, now with independent coordinates, givesBy Pinsker's inequality, the probability mass functions therefore converge in total variation, and in particular $S_n$ converges in distribution to $Z\sim \mathrm{Poisson}(\lambda )$.

The joint probability of the observed row depends only on $S_n$:Taking logarithms in any fixed base and choosing $c_n=\log n$ giveswhereThe convergence lemma supplied in the question now yieldsThis sparse triangular array therefore has a random limiting normalized self-information rather than the constant limit in the usual asymptotic equipartition property.