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The geometric family law on the nonnegative integers has probability generating functionIts expected value is $G^{\prime }(1)=1$, so this is a geometric-offspring critical branching process. Starting from $G_0(s)=s$, verify the rational iterate by induction. If $G_n(s)=[n-(n-1)s]/[n+1-ns]$, thenwhich has the same form with $n$ replaced by $n+1$. Thus, for $n\ge 1$,The stated disk is the actual power-series convergence domain, not merely a place where the rational continuation can be evaluated. Its simple pole is at $s=1+1/n$.

The extinction probability at generation $n$ is the constant coefficient:For later use, expanding the remaining geometric series givesIn particular $P(X_n>0)=1/(n+1)$. At $n=0$ the initial population is one deterministically; its PGF is $s$, and the displayed radius involving $1/n$ is used only for $n\ge 1$.

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.

For each $n\ge 1$, choose the continuous probability density functionsThe second is at least $1/2$ and integrates to one. Their expected values differ byWriting $g_n/f=1+{\Delta }_n$ givesThe supplied product bound, also obtained from the chi-squared divergence of product measures, impliesFor the final elementary estimate, compare the exponential series with the geometric series: $e^x\le (1-x{)}^{-1}$ for $0\le x<1$. Thus the overlap of the product laws is at least $1/2$. Part (b) now gives the explicit mean-estimation minimax lower bound for continuous densitiesSo one may take $C=1/1152$, uniformly for all $n\ge 1$. The alternatives are allowed to depend on $n$, because the minimax risk takes a supremum over all continuous probability density functions separately at each sample size.

Write $e=y-g_{\theta }(x)$ and $h_{\theta }(x)=\partial g_{\theta }(x)/\partial \theta$. Holding the nuisance parameter $\eta$ fixed, the parametric score function is the derivative with respect to $\theta$ of the one-observation log-likelihood. The joint probability density function isBecause $\partial e/\partial \theta =-h_{\theta }(x)$, differentiating givesFor instance $E_{\eta }{h}_{\theta }(X{)}^2<\infty$ ensures a square-integrable score function. Its expected value is zero by independence and the centered normal distribution of the error. This is the Gaussian regression score.

Now $p_{\theta ,\eta }(x,y)=v(x)f(y-g_{\theta }(x))$. With $h_{\theta }={\partial }_{\theta }{g}_{\theta }$ and the location score $\rho (e)=-f^{\prime }(e)/f(e)$, differentiation of the log-likelihood givesTake the usual regularity conditions $h_{\theta }\in L^2(v)$ and $\rho \in L^2(f)$ so that this score function belongs to L2 space.

For the specified nuisance statistical path, differentiating $\log (1+t\gamma (e))$ at zero gives the score function $\gamma (e)$. Normalization requires $E_f\gamma =0$. There is a second constraint: the statistical path must preserve the model's zero error expected value. ThusThis constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric logistic distribution, for example, $\gamma (e)=\tanh (e)$ is bounded and has zero expected value, but $E[\epsilon \tanh (\epsilon )]>0$; it would not preserve the mean.

For any $\psi \in L^2(v)$, independence impliesAll these products are integrable by the Cauchy-Schwarz inequality, since $\epsilon \psi (X)$ is square-integrable. Thus every such error-density score function is orthogonal to every $e\psi (x)$. Covariate-density nuisance score functions $b(X)$ are also orthogonal to these functions, since $E\epsilon =0$. This is the mean-preserving error tangent space orthogonality used in the final calculation.

First use the additional admitted form of the efficient score. Write $\tilde{\ell }=e\zeta (X)$ and let ${\tau }^2=E_f{\epsilon }^2>0$. The difference $\dot{\ell }-\tilde{\ell }$ is the orthogonal projection onto the nuisance tangent space. Part (c) makes every $e\psi (X)$ orthogonal to that space. Therefore, for all $\psi \in L^2(v)$,The function in braces belongs to L2 space; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction isUnder the regular tail condition $sf(s)\to 0$ at both infinities, integration by parts gives $\int s{f}^{\prime }(s)ds=-1$, and the formula becomes $e{h}_{\theta }(x)/{\tau }^2$. This also confirms the sign.

The admitted product form is an extra restriction; it does not follow for every independent-error regression model. To locate the restriction precisely, suppose the error-density nuisance statistical paths preserve both normalization and mean to first order. Their closed mean-preserving error tangent space is $\{\gamma :E_f\gamma =E_f(\epsilon \gamma )=0\}$. The full nuisance tangent space is the orthogonal sum of this space and the centered functions of $X$.

For completeness, bounded functions satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its expected value, and then subtract a multiple of a fixed bounded centered function $q$ with $E_f(\epsilon q)\ne 0$. Such a $q$ exists by truncating $\epsilon$, since ${\tau }^2>0$. The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.

Let $m=E_v{h}_{\theta }(X)$. With $E_f\rho =0$ and $E_f(\epsilon \rho )=1$, the orthogonal projection of $h_{\theta }(X)\rho (\epsilon )$ onto the error nuisance space is $m\{\rho (\epsilon )-\epsilon /{\tau }^2\}$; its projection onto the covariate nuisance space is zero. Thus the general efficient score in independent-error regression isThe orthogonality of the two terms gives the displayed efficient information. For a normal distribution of the error, $\rho (e)=e/{\tau }^2$, so this reduces to the admitted formula. It also does so when $h_{\theta }$ is constant.

A concrete counterexample to the generality of the admission is $g_{\theta }(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard logistic distribution error. Here $m=0$ and $\rho (e)=\tanh (e/2)$, so the actual efficient score is $x\tanh (e/2)$. It cannot equal $e\zeta (x)$ because $\tanh (e/2)/e$ is not constant. This score function is already orthogonal to every nuisance score function: its factor $X$ is centered against error-only directions, and its factor $\rho (\epsilon )$ is centered against covariate-only directions. The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.

For the exponential distribution of mean $\mu$, $M_X(t)=(1-\mu t{)}^{-1}$. Substitution into the geometric-sum moment-generating function givesThis is the moment-generating function of an exponential distribution with rate $p/\mu$. By the uniqueness theorem for moment-generating functions, the geometric sum of exponential variables therefore has probability density functionIts expected value is $\mu /p$, agreeing with $\mathbb{E}[N]\mathbb{E}[X]=\mu /p$ in the random sum of independent claims formula.

Use the shape-rate convention for the gamma distribution. A shape $r$ and rate $\beta$ give expected value $r/\beta$ and variance $r/{\beta }^2$. The two moments here force $r=2$ and $\beta =2/\mu$. Thus $M_X(t)={\beta }^2/(\beta -t{)}^2$, and the geometric-sum moment-generating function becomesPut $r_0=\sqrt{1-p}$, $a=\beta (1-r_0)$ and $b=\beta (1+r_0)$. Then $0<a<b$, $ab=p{\beta }^2$, andConsequently the geometric sum of shape-two gamma variables has the same probability distribution as the sum of two independent exponential distributions with rates $a,b$. Their convolution of independent random variables yieldsThe probability density function is zero for $s<0$. It is nonnegative because $a<b$, and its integral isThis verifies normalization directly. An alternative check uses the conditional gamma distribution with shape $2n$:which is exactly the same density. The expected value $1/a+1/b=\mu /p$ supplies a further check.

Set $\overline{F}(M)=1-F_X(M)$. The compound Poisson distribution variance identity and the per-claim excess of loss reinsurance payouts giveApply differentiation under the integral sign. The endpoint terms $M^2{f}_X(M)$ in the first two terms cancel, while the last integrand is zero at its lower endpoint. Equivalently, differentiate the two payout squares inside their expected values; the finite second moment supplies a dominating integrable function. This gives the total variance stationary condition for excess of lossHence the specified equality makes $g^{\prime }(M)=0$. When $\overline{F}(M)>0$, its interpretation is $M=m_X(M)$, where the mean residual life is $m_X(M)=\mathbb{E}[X-M\mid X>M]$.

For the exponential distribution of mean $\mu$, $\overline{F}(M)=e^{-M/\mu }$ andThus $g^{\prime }(M)=2\lambda e^{-M/\mu }(M-\mu )$ is negative below $\mu$ and positive above it. The variance-minimizing exponential retention is the unique global minimizerFor an explicit value, put $m=M/\mu$. The capped claim moments and the excess-claim second moment giveThe sign argument establishes global minimality, rather than just stationarity.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

A credibility estimate blends the risk's own sample mean with a population or prior expected value. In the basic equal-exposure form it is $Z\overline{X}+(1-Z)\mu$, where the credibility factor $Z\in [0,1]$ is the weight attached to the individual experience. More data ordinarily increase that weight; greater within-risk variation relative to between-risk variation reduces it.

Write $m(\theta )=\mu (\theta )$ for the conditional claim expected value, keeping $\mu$ for the prior-level parameter. Normalization of the probability density function givesAt interior points of its finite domain, differentiation under the integral sign yieldsThis is an instance of the exponential-family derivative identities for natural parameter $-\theta$. The minus sign is present in the PDF and is lost in the converted TeX. The conditional variance is likewise $(\log q{)}^{\prime \prime }(\theta )$.

Differentiate the log of the natural conjugate prior:Integrate on a compact subinterval and then let its endpoints approach those of the parameter interval. The prescribed vanishing of the prior density eliminates the boundary term. Since $m(\theta )\ge 0$, monotone convergence theorem justifies the limit of its expectation, and the identity givesIn particular the assumed proper prior and boundary conditions themselves establish this first-moment integrability.

For $T={\sum }_{i=1}^n{x}_i$ and $\overline{x}=T/n$, conditional independence makes the likelihood function proportional to $q(\theta {)}^{-n}{e}^{-\theta T}$. Multiplication by the prior distribution gives the Bayesian posteriorThus this is a conjugate prior family, with updated parameters

The integration-by-parts calculation for the posterior mean also needs vanishing posterior boundary values. Here that property can be checked, rather than assumed. Let $a,b$ be the essential lower and upper endpoints of the claim support. Every nondegenerate tilted claim law has $a<m(\theta )<b$, so the established prior mean satisfies $a<\mu <b$. Supported observations have $\overline{x}\in [a,b]$; hence $a<{\mu }_n<b$. At a finite parameter endpoint the prior's vanishing forces $q(\theta )\to \infty$, because its exponential factor has a finite positive limit; the increased power $q^{-k_n}$ makes the posterior vanish there too. At $\theta \to +\infty$, choose $a<d<{\mu }_n$ with positive base-measure mass below $d$. Then $q(\theta )\ge C{e}^{-d\theta }$ and the unnormalized posterior is at most $C^{-k_n}{e}^{-k_n({\mu }_n-d)\theta }\to 0$. At $\theta \to -\infty$, choose ${\mu }_n<d<b$ with positive mass above $d$ to obtain the corresponding bound $C^{-k_n}{e}^{k_n(d-{\mu }_n)\theta }\to 0$. These bounds also ensure posterior propriety at infinite endpoints. This is the endpoint control for a Laplace-family conjugate posterior.

Apply the previous score integration to the Bayesian posterior. The natural conjugate credibility identity isThus $Z=n/(n+k)$ is the credibility factor. This is an exact posterior mean, rather than an affine approximation to one.

For the specific shape-two gamma distribution, takeThen $m(\theta )=2/\theta$ and $\mathrm{Var}(X\mid \theta )=2/{\theta }^2$. The gamma rate gamma conjugacy prior isthat is, $\Theta \sim \mathrm{Gamma}(2k+1,\text{rate }k\mu )$. Its normalizing constant in the notation of the question is $c(\mu ,k)=\Gamma (2k+1)/(k\mu {)}^{2k+1}$. The gamma distribution density vanishes at both zero and infinity for $k>0$. The posterior has shape $2k+2n+1$ and rate $k\mu +T$, so its conditional-mean posterior mean is again $(k\mu +T)/(k+n)$.

For the last calculation, gamma distribution integration givesThe second identity requires the printed $k>1/2$. Therefore the expected process variance and variance of hypothetical means, denoted by $a,b$ in this question, areConsequently $a/b=k$, and the Bühlmann credibility factor agrees with the exact Bayesian weight:This is the exact Bühlmann credibility for gamma claims, expressed using the reciprocal rate as the random scale. The symbols $a,b$ here are the process and between-risk variances, respectively; their roles should not be interchanged when comparing with other notation for the Bühlmann model.

Put $d=np=\gamma \log n$ and $\mu =n^{1-\gamma }$. The hypothesis implies $\mu \ge e^{\omega (n)}\to \infty$. For the number $I$ of isolated vertices, the isolated vertices in the Erdős-Rényi model formulas giveThus the Chebyshev inequality gives $I=(1+o(1))\mu$ with high probability, uniformly over the specified $\gamma$ range.

The Cayley formula supplies a spanning tree on any connected vertex set. Consequently the expected value $C_j$ of the number of graph components of order $j\ge 2$ is bounded byChoose a fixed $\eta \in (0,1/2)$ with ${\gamma }_0(1-\eta )(k+1)>1$. For $k+1\le j\le \eta n$, with $j\ge 2$, the binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le (en/j{)}^j$ givesHere $ed{e}^{-d(1-\eta )}\le e\log n{n}^{-{\gamma }_0(1-\eta )}\to 0$. The geometric series and a union bound therefore exclude this entire size range, since $n[e\log n{n}^{-{\gamma }_0(1-\eta )}{]}^{k+1}=o(1)$. For $\eta n\le j\le n/2$, an empty graph cut would be necessary. Its total probability is at mostThere are consequently no graph components of orders between $k+1$ and $n/2$, with high probability.

For each fixed $2\le j\le k$, a graph component containing a graph cycle has a spanning tree and at least one extra edge. Counting that extra edge gives an expected value $O_k(d^j{e}^{-dj})=o(1)$; hence every small graph component is a tree component. The total number $R_2$ of vertices in tree components of orders $2,\dots ,k$ satisfiesThe Markov inequality gives $R_2=o(\mu )$ with high probability. When $k=1$, this sum is empty and $R_2=0$. Since $\mu =o(n)$, the remaining vertices must lie in one graph component larger than $n/2$. It is unique, and its order is $n-(1+o(1))n^{1-\gamma }$. Every other graph component is a tree with at most $k$ vertices. This is the logarithmic-regime giant component mechanism: almost all missing vertices are isolated.

Write $\mu =1+\epsilon$, $p=\mu /n$, $q=1-p$, and let $T$ be the total progeny of a binomial branching process. The preceding binomial branching survival correction gives $\rho =(2+o(1))\epsilon$. For a branching process conditioned on extinction, the offspring probability generating function is $f((1-\rho )s)/(1-\rho )$, with meanfor sufficiently large $n$. Summing expected generation sizes therefore gives $\mathbb{E}[T;T<\infty ]=(1-\rho )/(1-f^{\prime }(1-\rho ))=O(1/\epsilon )$.

Put $K=\lceil {\epsilon }^{-3}\rceil$ and let $N_{\ge K}$ count vertices in graph components of order at least $K$. A breadth-first exploration of a binomial random graph is dominated by $T$. The Markov inequality applied to finite total progeny givesAlways $L_1\le K+N_{\ge K}$. Since $n{\epsilon }^4\ge n^{1/3}\to \infty$, we have $K=o(\epsilon n)$ and $n/(\epsilon K)=O(n{\epsilon }^2)=o(\epsilon n)$. Hence $\mathbb{E}{L}_1\le (2+o(1))\epsilon n$. This bound controls the expected value directly, including rare large graph components.

For the lower bound, use the breadth-first exploration of a binomial random graph with $U_t$ unseen vertices, $A_t$ active vertices and $t$ explored vertices, starting from $U_0=n$, $A_0=0$. Let $b_t=1_{\{A_{t-1}=0\}}$ indicate a new root. Conditional on the past, the newly discovered count $Z_t$ has a binomial distribution $\mathrm{Bin}(U_{t-1}-b_t,p)$. Put $D_t=Z_t-p(U_{t-1}-b_t)$. The unseen-count recursion givesHere $M_t$ is a martingale. Up to $T_0=\lceil 3\epsilon n\rceil$, martingale-difference orthogonality gives $\mathbb{E}{M}_{T_0}^2\le C\epsilon n$, since each conditional variance is at most $np<2$ and $q^{-2T_0}$ is bounded.

Take $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$. Then $h\to 0$, $\epsilon =o(h)$ and $h^{2}n{\epsilon }^3\to \infty$. The Doob L2 maximal inequality impliesThe Taylor expansion, uniformly for $t\le T_0$, gives $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$. Thus $F(t)\ge h{\epsilon }^{2}n/3$ throughout the integer interval $[\lceil h\epsilon n\rceil ,\lfloor (2-h)\epsilon n\rfloor ]$ for large $n$. The new-root sum is nonnegative, so on the preceding event $A_t>0$ throughout this interval. No graph component finishes there: all these explored vertices belong to one graph component, of order at least $(2-2h)\epsilon n-O(1)$, with high probability. Its expected value is therefore at least $(2-o(1))\epsilon n$. Combining both inequalities provesThe barely-supercritical largest-component expectation uses both a positive exploration window and a finite-progeny bound; a bare convergence-in-probability assertion would not by itself justify this expectation.

The Kruskal-Katona theorem states that a colexicographic initial segment minimizes the lower shadow of a uniform set family of prescribed size. Its numerical form is as follows. For $m>0$, write the unique combinatorial number system expansionEvery $\mathcal{F}\subseteq [n{]}^{(r)}$ of size $m$ satisfiesThe empty family has empty lower shadow. Equality is attained by the first $m$ $r$-sets in colexicographic order, where the largest differing element belongs to the later set. Iterating the Kruskal-Katona theorem shows that the same colexicographic initial segment minimizes every iterated lower shadow.

For $1\le r<n/2$, the Erdős-Ko-Rado theorem givesfor an intersecting family $\mathcal{F}\subseteq [n{]}^{(r)}$. All $r$-sets containing one prescribed point show that the bound is sharp.

For the proof from the Kruskal-Katona theorem, put $k=n-r$ and form the complement family $\mathcal{G}=\{[n]\setminus A:A\in \mathcal{F}\}\subseteq [n{]}^{(k)}$. Let $\mathcal{S}$ be its rank-$r$ iterated lower shadow. No member $B\in \mathcal{F}$ belongs to $\mathcal{S}$: containment in $[n]\setminus A$ would give $A\cap B=\varnothing$, impossible for an intersecting family of nonempty sets. HenceSuppose $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The first $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$ members of the colexicographic order are all $k$-sets of $[n-1]$. Their rank-$r$ iterated lower shadow is all $r$-sets of $[n-1]$, since $r<k$. The next $k$-set contains $n$ and has an $r$-subset containing $n$, so taking even one more member strictly enlarges that iterated lower shadow. By the iterated Kruskal-Katona theorem, $|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. Together with $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ this contradicts Pascal's identity and the preceding inequality. This proves the Erdős-Ko-Rado theorem.

For the Katona circle method proof, fix a cyclic ordering of $[n]$. A cyclic interval of length $r$ is specified by its final position. If no such interval belongs to $\mathcal{F}$, the bound below is automatic. Otherwise rotate one selected interval so that it ends at position $n$, and thus occupies positions $n-r+1,\dots ,n$. Intervals ending at positions $r,\dots ,n-r$ are disjoint from it and cannot be selected. Among the remaining endpoints, pair $j$ with $n-r+j$ for $1\le j\le r-1$. The corresponding intervals are disjoint when $n\ge 2r$, so at most one interval per pair is selected. Together with the interval ending at $n$, this gives the cyclic interval intersection bound of $r$ selected intervals.

Choose a uniformly random permutation and read its positions cyclically. A fixed $r$-set is a cyclic interval with probability $n/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$: each cyclic position gives a uniformly distributed $r$-set, and for $r<n$ the $n$ intervals are distinct. Summing over $\mathcal{F}$, the expected value of the number of selected intervals is $n|\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$. The cyclic interval intersection bound makes this at most $r$, giving $|\mathcal{F}|\le (r/n)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$.

Finally apply the Katona circle method to an arbitrary antichain $\mathcal{A}$. If it contains $\varnothing$ or $[n]$, it has exactly one member and the LYM inequality holds with equality. Otherwise all its members have sizes $1,\dots ,n-1$. In a fixed cyclic ordering, the cyclic intervals sharing a final position form a nested chain as their lengths increase. At most one of them can belong to the antichain. Summing over the $n$ final positions proves the cyclic interval antichain bound of $n$ intervals. Taking the expected value over a uniformly random permutation givesAfter division by $n$, this is the LYM inequality proved by cyclic intervals. The case $n=0$ consists only of the empty set and is immediate.