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Under fit.2, observations in each period are independent Poisson random variables with a common mean, so each sample variance should fluctuate around its sample mean. The plot instead has variances roughly 130–300 for means roughly 80–120, well above the line variance equals mean. This is overdispersion.

Day-to-day traffic variation can produce a Poisson mixture: a random daily intensity increases the marginal variance and can correlate observations from the same day. A useful replacement is a negative binomial regression, with variance $\mu +{\mu }^2/\kappa$, or a Poisson generalized linear mixed model with a day random intercept. A day-factor effect can also be included if inference is restricted to the observed days. A Quasi-Poisson regression estimates a dispersion multiplier for uncertainty, but does not itself explain the shared-day dependence and has no ordinary likelihood-based AIC. The means can retain the period-factor or selected jump structure.

Statement and definitions. For disjoint nonempty sets $A,B$ of vertices, their edge density of a bipartite graph is $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair of vertex sets, or an $\epsilon$-uniform pair, ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$ and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma states that for every $\epsilon >0$ and positive integer $m_0$, there are integers $M,n_0$ such that every graph of order $n\ge n_0$ has a set partitionwith $m_0\le k\le M$, $|V_0|\le \epsilon n$, equal nonzero sizes $|V_1|=\cdots =|V_k|$, and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $1\le i<j\le k$, that fail to be $\epsilon$-uniform. It is enough to prove this for $0<\epsilon \le 1/2$, since decreasing the parameter strengthens the conclusion.

Energy increment. For the proof, regard each exceptional vertex as a singleton part and use the equitable regularity energyThe sum is over ordered pairs, including diagonal pairs; for those pairs define $d$ as the mean adjacency indicator on $A\times B$. Thus $0\le q\le 1$. Under a refinement, the subrectangle densities have the original density as their weighted mean. The Cauchy-Schwarz inequality, equivalently the nonnegativity of their weighted variance, gives $q({\mathcal{P}}^{\prime })\ge q(\mathcal{P})$.

Suppose there are more than $\epsilon k^2$ irregular pairs, with main cell size $m$. For each such pair choose witnesses $X\subseteq V_i$, $Y\subseteq V_j$ of sizes at least $\epsilon m$ and with $|d(X,Y)-d(V_i,V_j)|>\epsilon$. Refine every main cell by all its witness subsets. Each splits into at most $2^{k-1}$ atoms, and we may use the larger bound $2^k$.

For an ordered irregular rectangle, write its refined densities as $d_{ab}$ and relative areas as $w_{ab}$. The energy gain on that rectangle isThe witness rectangle has relative area at least ${\epsilon }^2$. Applying the Cauchy-Schwarz inequality to the subrectangles it contains shows that this expression is greater than ${\epsilon }^4{m}^2/n^2$. There are two orientations of each irregular pair. Since $km=n-|V_0|\ge n/2$, the total gain is greater than

Restoring equal sizes without losing energy. Put $\ell =\lfloor m/4^k\rfloor$, assuming $m\ge 2\cdot 4^k$. Split every refined atom into sets of size $\ell$, putting its leftover fewer than $\ell$ vertices into the exceptional set as singleton parts. This is a further refinement, so the energy cannot decrease. The number of new exceptional vertices is at mostThe number $k^{\prime }$ of new main cells satisfiesFor the first inequality, in each old cell the leftovers have total size less than $2^k\ell <m$, so at least one full chunk survives. All previous exceptional singleton parts are kept. Treating leftovers as singleton parts is what makes the energy increment survive equitabilization exactly.

Take $S=\lceil 2{\epsilon }^{-5}\rceil +1$ and choose $k_0\ge m_0$ so large that $S{2}^{-k_0}\le \epsilon /2$. Start with $k_0$ equal main cells, with fewer than $k_0$ exceptional vertices. Define a finite bound $M$ by iterating $x\mapsto 2x{4}^x$ a total of $S$ times, starting from $k_0$. Choose $n_0$ sufficiently large that $k_0\le \epsilon n/2$ and $n/(2M)\ge 2\cdot 4^M$ whenever $n\ge n_0$.

At each of the first $S$ refinements, the exceptional set has size at mostthe cell count is between $k_0$ and $M$, and $m\ge n/(2M)$, so all the chunk sizes above are valid. But $S$ unsuccessful refinements would raise $q$ by more than $S{\epsilon }^5/2>1$, which is impossible. Thus the procedure terminates at a partition satisfying the Szemerédi regularity lemma.

Three linearly large uniform sets. Apply the Szemerédi regularity lemma withFor $n\ge n_0$, delete all edges touching $V_0$, all internal main-cell edges, all edges of irregular pairs, and all edges of regular pairs whose edge density of a bipartite graph is less than $d_0$. The respective losses are bounded byThus the total loss is at most $0.00102n^2$, and the remaining graph has at leastedges. Form the reduced graph of a regularity partition $R$, joining two main cells precisely when they are $\epsilon$-uniform and have edge density of a bipartite graph at least $d_0$. If their common size is $m$, the remaining edge count is at most $e(R)m^2$. Since $km\le n$, we obtain $e(R)>k^2/4$. The Mantel theorem gives a triangle in a graph in $R$. Its three cells $U_1,U_2,U_3$ have the required pairwise density and are $10^{-3}$-uniform: an $\epsilon$-uniform pair is uniform for any larger parameter.

Each selected cell has size $m\ge (1-\epsilon )n/M$. ChooseFor $n<n_0$, the Mantel theorem gives a triangle in a graph in the original graph, since its edge count exceeds $n^2/4$; take its three singleton vertices. Their pairs have density one and are uniform for every positive parameter, and $1>cn$. Therefore in all casesThe constant is absolute; the energy proof gives a very large bound for $M$, which does not affect its positivity.

A standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ satisfying $B_0=0$ almost surely, having continuous paths almost surely, and having independent increments withHere $N(0,t-s)$ is the normal distribution with zero expected value and variance $t-s$. Thus the increments are also stationary increments.

If Brownian motion is defined relative to a specified filtration $({\mathcal{F}}_t)$, additionally require that it is an adapted process and that $B_t-B_s$ is independent of ${\mathcal{F}}_s$. For its natural filtration, this follows from independent increments. These are the defining properties of standard Brownian motion.

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

Multiplication by the integrating factor $e^{t/2}$ and the Itô product rule give the Ornstein-Uhlenbeck processThe two summands are independent centered normal random variables, because $X_0$ is independent of the Brownian motion and the integrand is deterministic. By the Itô isometry, their variances sum to $e^{-t}+{\int }_0^t{e}^{-(t-u)}du=1$. ThusFor $t\ge s$, split the Itô integral at $s$ to obtain $X_t=e^{-(t-s)/2}{X}_s+{\int }_s^t{e}^{-(t-u)/2}d{B}_u$, where the latter term is centered and independent of $X_s$. ConsequentlyIn particular this initialization makes the Gaussian process a stationary process.

The expression is a Test-function pairing with a Gaussian free field, rather than the pointwise integral of an ordinary random function. For a real test function $\varphi$, take the Dirichlet solutionusing the Green-kernel identity allowed in the original PDF. In particular, $-\Delta f_{\varphi }=2\pi \varphi$. Integration by parts gives, initially for smooth compactly supported $u$ and then by continuity in the Dirichlet energy space,The integral functional is continuous in the energy norm: pull back to the unit disc, where the transformed test function has compact support, and use the Poincare inequality. Hence $f_{\varphi }$ is its energy-space representer. DefineBy the isonormal Gaussian process definition, it has mean zero and varianceThe logarithmic singularity is locally integrable, so this variance is finite for smooth compactly supported test functions. ThusThe local TeX corrupted the double integral and the bracketed Green-kernel identity; the PDF supplies the identity above and does not assert an infinite variance.

Put $v_k=p_k(1-p_k)$ and assume $0<p_k<1$. For fixed $n$ and a fixed nonzero difference $\Delta =p_0-p_1$, the approximate statistical power of the Wald test increases as its noncentrality $|\Delta |/\sqrt{V}$ increases, whereDifferentiating with respect to the continuous allocation givesHence the unique minimum and its associated minimum variance areThis is Neyman allocation. Integer allocations use a feasible adjacent integer to the continuous optimum, comparing the resulting variances; $n_0,n_1\ge 1$ must still hold. It is a large-sample normal approximation argument, not an exact finite-sample power theorem for arbitrary response-adaptive randomization. Under the null $p_0=p_1$, the approximate power is the significance level for any allocation, although the same allocation still minimizes the stated variance.

The difference of the Bernoulli distribution variances factors asThe first factor is positive, whereas $p_0>1-p_1$ makes the second negative. Thus $v_0<v_1$ and $R^{\ast }=\sqrt{v_0/v_1}<1$. Therefore Neyman allocation assigns more patients to treatment 1, the inferior treatment. Minimizing the variance of a treatment comparison and maximizing the number of desirable responses are different objectives.

For the electronic-record observational study, three strengths are:

Three weaknesses are:

For two-sample Mendelian randomization, three strengths are:

Three weaknesses are:

Thus the agreement supports the hypothesis, but does not establish the effect of a dietary intervention.

Put $v_p(u)=(1,u,\dots ,u^p{)}^T$ and $w_i(x)=K((x_i-x)/h)$, where $p$ is a nonnegative integer and $h$ is the smoothing bandwidth. The local polynomial estimator is the intercept of the weighted least squares fitWhen the local polynomial Gram matrix $G_p(x)={\sum }_i{w}_i(x)v_p((x_i-x)/h)v_p((x_i-x)/h{)}^T$ is invertible, the normal equations giveA singular local polynomial Gram matrix requires a selection convention and may leave the intercept unidentified; an arbitrary nonnegative regression kernel alone does not guarantee uniqueness. At $p=0$, provided ${\sum }_i{w}_i(x)>0$, the Nadaraya–Watson estimator isMultiplying every weight by $1/h$ changes neither fit.

For the uniform kernel, let $I_x=\{i:|i/n-x|\le h\}$ and $N_x=|I_x|$. Then ${\hat{m}}_h(x)=N_x^{-1}{\sum }_{i\in I_x}{Y}_i$. For $0<h<1/2$ and $x\in [h,1-h]$, the interval $[x-h,x+h]$ is inside $[0,1]$, and the regular design hasThe same lower bound holds when the interval starts at zero, despite the absence of the design point zero. In particular the Nadaraya–Watson estimator is defined throughout the integration interval.

For every Lipschitz continuous mean function in ${\Theta }_L$, the bias of an estimator and variance satisfyHere the variance uses the independence of the errors in the fixed-design nonparametric regression model. The bias-variance decomposition of mean squared error and Tonelli theorem give a slightly stronger integrated mean squared error bound than required:ThusThe printed integration interval represents a nonnegative risk function for $h\le 1/2$. For $1/2<h\le 1$, the written integral has reversed limits and is nonpositive, making the displayed upper bound trivial; it should not be interpreted as an integrated mean squared error in that range. For $h>1$, evaluation outside $[0,1]$ would additionally require a specified extension of the mean function. The subsequent optimization restricts to $h<1/3$.

Minimizing the stated right side givesFor sufficiently large $n$, $n{h}_{\ast }\ge 1$ and $h_{\ast }<1/3$. For example take $n_0=1+\lceil \max \{1,2L/(\sqrt{3}\sigma ),81{\sigma }^2/(4L^2)\}\rceil$. Therefore the upper bound holds with

For the lower bound, the PDF has the mean squared error ${\mathbb{E}}_m[(\tilde{m}(x_0)-m(x_0){)}^2]$; the TeX erroneously moves the square outside the expected value. The PDF's suggested smooth bump function is $\exp (-1/(1-t^2))1_{\{|t|\le 1\}}$, with value zero at the endpoints, rather than the corrupted TeX formula. A triangular alternative suffices because only Lipschitz continuity is needed.

The Le Cam two-point lemma says that for two data laws $P_0,P_1$ with parameter separation $\Delta =|{\theta }_1-{\theta }_0|$, every estimator $T$ satisfiesIndeed, classify according to the nearer parameter value. Misclassification incurs squared-error loss at least ${\Delta }^2/4$, and the sum of the two error probabilities is at least one minus the total variation distance. Averaging the two risk functions proves the stated form.

Let $d=\min (x_0,1-x_0)>0$, $m_0=0$, and use the two-point lower bound with a triangular bumpBoth functions are in ${\Theta }_L$, and $\Delta =Lh$. Assume first that $h\le d$ and $nh\ge 1$. At most $2nh+1$ design points meet the bump's support. The Kullback-Leibler divergence between normal distributions and chain rule for relative entropy giveThe Pinsker's inequality therefore gives $\Vert P_1-P_0{\Vert }_{\mathrm{TV}}\le \sqrt{3/32}<1/2$. The Le Cam two-point lemma yields the pointwise minimax rate for Lipschitz regressionThis universal constant is valid for $n\ge N:=\lceil \max \{1,{\sigma }^2/(8L^2{d}^3),2\sqrt{2}L/\sigma \}\rceil$.

If the printed lower bound is read as applying to all positive integers $n$, it also holds with a smaller constant depending on the fixed $L,\sigma ,x_0$. For $n<N$, use the constant mean functions $m_0=0$ and $m_1=\sigma /\sqrt{n}$. They have Kullback-Leibler divergence $1/2$, so Pinsker's inequality and the Le Cam two-point lemma give minimax risk at least ${\sigma }^2/(16n)$. Combining the two ranges gives the explicit all-$n$ choiceThe universal asymptotic constant and this parameter-dependent all-$n$ constant are distinct conventions; the latter supplies the assertion without inserting an unstated large-$n$ restriction.

Let $S={\sum }_i{Y}_i$ and $V_H={\sum }_i(b_i-a_i{)}^2$. To prove the Hoeffding lemma, take a centered random variable $Y\in [a,b]$ and its cumulant-generating function $\psi (t)=\log \mathbb{E}{e}^{tY}$. Under exponential tilting, ${\psi }^{\prime \prime }(t)$ is the variance of $Y$. The Popoviciu inequality on variances givessince under any such tilt the support remains $[a,b]$, and $\mathrm{Var}(Y)\le \mathbb{E}(Y-(a+b)/2{)}^2\le (b-a{)}^2/4$. Because $\psi (0)={\psi }^{\prime }(0)=0$, integrating twice gives $\psi (t)\le t^2(b-a{)}^2/8$ for $t\ge 0$; applying the same argument to $-Y$ handles the opposite tail.

The independence of the $Y_i$ gives $\mathbb{E}{e}^{tS}\le \exp (t^2{V}_H/8)$. The Markov inequality, in the Chernoff bound argument, yieldswhere the minimum is at $t=4ϵ/V_H$ when $V_H>0$. Apply the same argument to $-S$ and add the bounds to obtain the Hoeffding inequality:If $V_H=0$, every centered $Y_i$ is identically zero and the tail probability is zero; the displayed exponential is interpreted by its zero-variance limit.

For the Bennett inequality, write $v=n{\sigma }^2$, and first assume $M>0$ and $v>0$. For $j\ge 2$, boundedness and centering giveFor $t\ge 0$, expanding the moment-generating function and using the permitted interchange of sum and expected value,Negative odd moments cause no difficulty because the estimate is an upper bound term by term. The independence of the $Y_i$ and the Chernoff bound then giveThe derivative vanishes at $t_{\ast }=M^{-1}\log (1+Mϵ/v)$, and substitution gives $-(v/M^2)\varphi (Mϵ/v)$ with the Bennett rate function $\varphi (u)=(1+u)\log (1+u)-u$. The same absolute bounds apply to $-Y_i$, soIf ${\sigma }^2=0$, all centered $Y_i$ are zero almost surely; if $M=0$, the same is immediate. Handle those cases directly rather than divide by zero. Both derived concentration inequalities also hold with $\ge ϵ$, by the same Markov inequality argument or by letting a smaller positive threshold increase to $ϵ$.

For the binomial distribution, write $X={\sum }_{i=1}^n{Z}_i$ with independent and identically distributed random variables $Z_i$ having the Bernoulli distribution with parameter $p=p_n$. Then $Y_i=Z_i-p$ has range $[-p,1-p]$, width one, and variance $p(1-p)$. For large $n$, $p<1/2$, so $M=1-p$ and $v=np(1-p)$. Set $ϵ=C\sqrt{v}$.

The Hoeffding inequality givesAfter truncation at one, this is asymptotically the trivial probability bound. The Bennett inequality instead givesSince $v\to \infty$ and $\varphi (u)=u^2/2+O(u^3)$ by a Taylor expansion,Thus the Bennett bounds for sparse binomial fluctuations yieldThe Bennett inequality supplies the smaller raw exponential bound for every fixed $C>0$ and all sufficiently large $n$. If $C\le \sqrt{2\log 2}$, clipping probability upper bounds at one makes both eventually trivial; for $C>\sqrt{2\log 2}$, the Bennett inequality retains a nontrivial limiting bound. This distinction avoids claiming a probability improvement after clipping when both raw bounds exceed one.

Taking expectations in the single-step discrete Tanaka formula, the martingale transform term vanishes because $f$ is bounded and ${\mathcal{F}}_T$-measurable. HenceAt a state with $p_K=\mathbb{P}(S_T=K)>0$, the tower property of conditional expectation gives $\mathbb{E}[S_{T+1}\mid S_T=K]=K$, so the conditional second moment of the increment is ${\sigma }^2(T,K)$. Part (a) gives $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)$. Combining these identities yields the discrete Dupire equationThe conditional variance needs a positive-probability conditioning state. The bound $|K-S_0|\le T$ does not guarantee $p_K>0$. If $p_K=0$, both the call-price curvature and the time increment above are zero; the equation can be extended by assigning an arbitrary finite value to ${\sigma }^2(T,K)$ at such a state, but its conditional variance is not determined there.

Let $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)>0$ and $D_K=C(T+1,K)-C(T,K)$. Given $S_T=K$, the next value is one of $K-1,K,K+1$. The martingale property makes its upward and downward conditional probabilities equal; their sum is the conditional variance of the increment, ${\sigma }^2(T,K)=2D_K/p_K$. ThusThe discrete Dupire equation therefore recovers all positive-probability one-step conditional transitions from the European call option price surface. No Markov property is required: these are conditional probabilities given the current value, not necessarily given the whole past.

Recovery is impossible at a zero-probability state. For example, $S_t\equiv S_0$ satisfies every stated assumption, but at $T\ge 1$ a different integer $K$ can satisfy $|K-S_0|\le T$ while $p_K=0$. Its conditional transitions can be specified arbitrarily without altering $C$. Thus the literal all-states claim needs the positive-probability qualification.