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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.

Finiteness of the moment-generating function on an open neighborhood of zero gives its differentiability there. Using the permitted interchange of differentiation and expectation,Hence $M_X(0)=1$, $M_X^{\prime }(0)=\mu$ and $M_X^{\prime \prime }(0)=s^2$. The second-order Taylor expansion with its Peano remainder givesHere $s^2$ is the raw second moment, not the variance: $\mathrm{Var}(X)=s^2-{\mu }^2$. This distinction matters before specializing to a centered variable.

Put $\eta =\mathrm{Im}z>0$. For a real eigenvalue $\lambda$,The spectral theorem for real symmetric matrices therefore gives $\mathrm{Im}g>0$ and $\mathrm{Im}{q}_i\ge 0$, since $q_i$ is a sum of $(\lambda -z{)}^{-1}$ with nonnegative coefficients given by squared eigenvector coordinates of $x_i$. This is the positive imaginary part of a Stieltjes matrix resolvent. HenceApply the triangle inequality to the identity above, and add and subtract the minor trace:The allowed principal minor resolvent trace bound, with the printed $N$ normalization, bounds the second average by $c/(\eta N)$. Thus the explicit conclusion isIn particular, the final term is $O((\mathrm{Im}z{)}^{-3}{N}^{-1})$ with a constant independent of $N$. The upper-half-plane restriction gives the displayed positive denominators; in the lower half-plane the analogous bound uses $|\mathrm{Im}z|$.

The chosen centering also has a probabilistic meaning. The column vector $x_i$ is independent of the minor, has mean zero and coordinate variance $1/N$, so its conditional expectation satisfiesThis explains why the bound isolates fluctuations of a quadratic form around the minor trace. Mean centering alone is not a concentration estimate; no unrequested limiting assertion is being assumed.

An effective bandwidth converts a traffic distribution and a quality-of-service requirement into a capacity requirement, penalizing bursts as well as mean load. Let $X\ge 0$ be the traffic generated in one observation interval and assume its moment-generating function is finite for some positive parameters. Write $\Lambda (s)=\log \mathbb{E}[e^{sX}]$. The Chernoff bound follows immediately from Markov's inequality applied to $e^{sX}$:Optimizing gives $\boxed{\mathbb{P}(X\ge b)\le \exp [-\underset{s>0}{\sup }\{sb-\Lambda (s)\}]}$, where the supremum is restricted to finite exponential moments. This exhibits the effective bandwidth as the load appropriate to exponential tail control, not an extra physical stream of traffic.

For the mean and peak limits of effective bandwidth, if exponential moments exist near zero, the cumulant-generating function expansion isFor small $s$, $\boxed{\alpha (s)\to \mathbb{E}X}$, with the variance giving the first burstiness correction. convexity of $\Lambda$ and $\Lambda (0)=0$ show that its secant slope $\alpha (s)$ is nondecreasing; Jensen's inequality also gives $\alpha (s)\ge \mathbb{E}X$.

If $X$ is bounded with essential supremum $M$, then $\alpha (s)\le M$. For any $a<M$, the probability $p_a=\mathbb{P}(X>a)$ is positive andLet $s\to \infty$ and then $a\uparrow M$. For large $s$, $\boxed{\alpha (s)\to \mathrm{esssup}X}$, so very stringent exponential tail control approaches peak provisioning. If traffic is unbounded and every positive exponential moment is finite, the same lower-bound argument gives $\alpha (s)\to \infty$. If exponential moments cease to exist beyond a finite parameter, $\alpha$ is infinite there; one cannot assign a finite large-$s$ peak interpretation to that traffic.

For example, deterministic traffic $X=d$ has $\alpha (s)=d$. A burst of size $b$ with Bernoulli distribution probability $p$ giveswhich moves from $pb$ to $b$. For Poisson packet count of mean $\lambda$ and packet size $b$, $\alpha (s)=\lambda (e^{sb}-1)/s$, showing a mean limit $\lambda b$ but no finite peak limit.

For independent traffic sources, logarithmic generating functions add, so effective bandwidths add. For arrivals $A_r(t)$ in a window of length $t$, use the rate-valued definition ${\alpha }_r(s,t)=\log \mathbb{E}[e^{s{A}_r(t)}]/(st)$. A resource with rate capacity $C$ then satisfiesThus a sufficient admission condition for a target probability $\epsilon$ is ${\sum }_r{\alpha }_r(s,t)\le C-\log (1/\epsilon )/(st)$ for some $s>0$. Independence is needed for this additive form; correlated sources require the joint exponential moment.

A queue involves a supremum over time windows, not merely one fixed window. For independent identically distributed discrete-time traffic increments $X_i$, constant service $C>\mathbb{E}X$, and stationary workloadthe workload Chernoff bound for independent increments follows from a union bound:, whenever $a(s)=e^{\Lambda (s)-sC}<1$,The requirement is precisely $\alpha (s)<C$. This explains why effective bandwidth is useful for buffer and delay design: the relevant time scale and tail parameter depend on the service rule and target overflow risk. A single-window bound cannot be silently substituted for an infinite-horizon workload bound.

Let $X_i$ be independent and identically distributed random variables with mean $0$ and variance ${\sigma }^2\in (0,\infty )$, and put $S_0=0$, $S_k={\sum }_{i=1}^k{X}_i$. Define the polygonal interpolationwith the endpoint $W_n(1)=S_n/(\sigma \sqrt{n})$. The Donsker invariance principle stateswhere $B$ is standard one-dimensional Brownian motion and the arrow denotes weak convergence of probability measures on this path space. Equivalently, the step interpolation converges in the Skorokhod space with its $J_1$ topology. The continuous interpolation version is sufficient for the maximum in part (c).

For the simple symmetric random walk, the increments have mean $0$ and variance $1$. Use the polygonal interpolation from (b). The maximum functional is Lipschitz in the supremum norm, sinceEach interpolating segment is linear, so its maximum occurs at a grid endpoint. The continuous mapping theorem applied to the Donsker invariance principle givesThe limiting maximum has no atom at $a>0$ by the Brownian reflection principle, so the closed-tail probabilities converge at this threshold. Including $j=0$ does not change the event because $a>0$. Therefore the random-walk maximum limit from Donsker invariance isHere $\Phi$ is the standard normal distribution function. The normalization and the positive-threshold condition are those in the PDF; the TeX aid's $\sqrt{t}$ is not the printed expression.

For a pathwise differentiable statistical functional, an influence-function representer is a mean-zero function $\phi \in L^2(P)$ such that every admissible score function $g$ satisfies $D{\psi }_P(g)=P(\phi g)$. The efficient influence function, also called the canonical gradient, is the unique such representer in the statistical tangent space. Equivalently, it is the orthogonal projection of any representer onto that statistical tangent space. The Pythagorean theorem in an inner-product space shows that it has the smallest squared L2 norm among all representers.

Here the statistical tangent space is all of $L_0^2(P_f)$. To verify the closure explicitly, take $h\in L_0^2(P_f)$, truncate it to $h_m=\max (-m,\min (h,m))$, and set $g_m=h_m-P_f{h}_m$. Then $g_m$ is bounded and centered, and $g_m\to h$ in $L^2(P_f)$, by dominated convergence and the Cauchy-Schwarz inequality. Part (c) supplies a representer already in this space. HenceIts variance is $P_f{a}^2-(P_{f}a{)}^2$.

The closure must be taken in the density-weighted space $L^2(P_f)$. An unweighted reading of $L^2[0,1]$ in the printed hint is false. For example, when $f(u)=2u$, the function $h(u)=u^{-3/4}-8/5$ has $P_{f}h=0$ and $P_f{h}^2=36/25$, so bounded centered truncations converge to it in $L^2(P_f)$; nevertheless $h\notin L^2([0,1],du)$. This illustrates density of bounded centered scores and fixes the measure in the closure statement.

Use the normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design matrix $X$, independent errors ${\epsilon }_i\sim N(0,{\sigma }^2)$, and ${\sigma }^2>0$. Its log-likelihood isFor fixed ${\sigma }^2$, differentiating with respect to $\beta$ gives $X^T(Y-X\hat{\beta })=0$. Since $X^{T}X$ is positive-definite, the unique least-squares estimate isLet $\mathrm{RSS}=\Vert Y-X\hat{\beta }{\Vert }^2$. The variance score equation is $-n/(2{\sigma }^2)+\mathrm{RSS}/(2{\sigma }^4)=0$, givingThis is the maximum-likelihood estimator, whereas the residual standard error reported by linear-regression software is $s=\sqrt{\mathrm{RSS}/(n-p)}$. The latter uses the unbiased residual estimate of Gaussian noise variance. These formulas assume $n>p$ and positive RSS; exact interpolation gives a variance-boundary degeneracy rather than an interior maximum.

A Poisson model assumes independent counts between rats conditional on time, a common mean at a given time, and conditional variance equal to that mean. Comparable observation units and consistent counting are needed; unexplained rat-to-rat heterogeneity is excluded by this simple specification. The quadratic Poisson regression isThe printed nested analysis of deviance tests $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$. Its likelihood-ratio statistic is $G^2=3.8548$, approximately ${\chi }_1^2$ under the null hypothesis and regular Poisson asymptotics. The reported $p=0.0496$ gives borderline rejection of the log-linear time effect at 5%, conditional on the Poisson assumptions. This is curvature of the log mean, rather than an ordinary quadratic model for the count itself.

There are only three distinct time values, so the intercept, linear term, and quadratic term span every possible set of three log means. This is Poisson mean saturation at three distinct times: the quadratic fit is saturated for the time-group means, which are $14/7=2$, $17/8=2.125$, and $47/7\simeq 6.7143$. A grouped deviance comparison against unrestricted time-group means has zero statistical degrees of freedom: the quadratic functional form cannot be tested at these three times. The individual-observation deviance 24.515 still describes within-group departure from the Poisson distribution, but small means and zeros make the naive ${\chi }_{19}^2$ goodness-of-fit calibration doubtful. Poisson variation itself can be checked using the replicates, preferably by a parametric bootstrap; it is not rendered untestable by the saturated mean specification.

Overdispersion means conditional variance exceeding the variance assumed by the Poisson mean model. For example, rats may have different latent susceptibility, so if $Y\mid \Lambda \sim \mathrm{Pois}(\Lambda )$, the law of total variance gives $\mathrm{Var}(Y)=\mathbb{E}\Lambda +\mathrm{Var}(\Lambda )>\mathbb{E}Y$. A quasi-Poisson model instead uses $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson-residual dispersion estimate is $\hat{\varphi }=1.254068$, using $n-p=22-3=19$.

For the conventional quasi-likelihood comparison with estimated dispersion, useIts upper-tail probability is about 0.096, so the evidence for curvature no longer reaches 5%. The F-distribution is an approximate calibration, not an exact small-sample result for arbitrary overdispersed counts; quasi-likelihood does not supply a full Poisson likelihood. With known dispersion one would instead compare $G^2/\varphi$ approximately with ${\chi }_1^2$. The displayed dispersion code uses the undefined name acf.glm2; its fitted values must be those of aom.glm2 for the stated calculation.

The moment-based approach specifies marginal means, variances, and within-student covariances, without supplying a joint probability distribution for each count profile. It is a quasi-likelihood or generalized estimating equation strategy. The parameter interpreted as a structural-zero fraction is not automatically an actual probability of a structural-zero component merely because it appears in the moment formulas. A valid positive-definite working covariance and parameter identifiability must also be checked.

The printed covariance specification is not admissible for all the stated parameter values. For example, take $\pi =1/2$, $\tau =1$, and all three component means equal to 10. It gives diagonal variance 80 and off-diagonal covariance 100, so $\mathrm{Var}(Y_{i1}-Y_{i2})=80+80-2(100)=-40$. A covariance matrix cannot have a negative variance in any direction. Thus the moment approach requires additional admissibility restrictions or a valid working covariance; the parameter ranges alone do not define a valid model. This does not alter the hierarchical model, whose covariance derived below is positive semidefinite.

The alternative gives a full hierarchical mixture distribution, specifically a shared zero-inflated Gamma-Poisson count model: a common student random effect is zero with probability $\pi$, and otherwise has a Gamma distribution with mean 1 and variance $\tau$. Conditional on this effect, the three counts are independent Poisson random variables. Integrating it out induces both excess zeros and positive within-student dependence. Zero inflation is shared for the whole student, rather than independently reselected in each term. This full distribution supports likelihood-based inference but is more sensitive to the distributional assumptions.

The marginal mean in the hierarchical model has the same form as the proposed moment mean. Its marginal covariance does not, in general, equal the printed working covariance; the law of total covariance calculation below makes the distinction explicit. At $\tau =0$ the Gamma component is interpreted as its point-mass-at-one limit. At $\pi =1$ all counts are zero, and the log-mean parameterization and regression effects are not identifiable.