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Put ${\delta }_i=t_i-t_{i-1}$, $\Delta ={\max }_i{\delta }_i$, $v_i={\int }_{t_{i-1}}^{t_i}\sigma (s{)}^{2}ds$ and $S=\Vert {\sigma }^4{\Vert }_{\infty }$. The deterministic integrand in the Itô integral implies that the increments $Z_i=X_{t_i}-X_{t_{i-1}}$ are independent random variables with normal distributions $N(0,v_i)$. The Gaussian fourth moment givesThese identities include $v_i=0$. Since the summands of $M_n$ are centered and independent, all cross terms in its second moment vanish. Consequently,Thus $D=2$ works for every observation partition. The last step uses the total interval length, rather than assuming equally spaced observations.

For each ball, the event of landing in $S_j$ is a Bernoulli distribution trial. Under the null hypothesis its probability is $\epsilon$. Under the selected alternative $Q_j$, its probability is $p_1=\pi +(1-\pi )\epsilon =\epsilon +\pi (1-\epsilon )$. The placements are independent random variables conditional on that selected index, so the count has the binomial distribution:The second assertion concerns $Q_j$, rather than the unconditional mixture model $P_1$.

The Hoeffding lemma for a Bernoulli distribution variable gives $\mathbb{E}{e}^{s(B-p)}\le e^{s^2/8}$. Multiplication of the moment-generating functions of the independent random variables gives $\mathbb{E}{e}^{s(X-np)}\le e^{n{s}^2/8}$. Thus the centered count is a sub-Gaussian random variable with variance proxy $n/4$. The Chernoff bound, optimized at $s=4t/n$, givesThe same estimate holds for the lower tail by applying the Chernoff bound to $-(X-np)$. No independence between different bin counts is needed.

Because the scalar signal and the noise vector are independent random variables with normal distributions, their sum is a Gaussian random vector. Its expected value is zero, and the covariance matrix isThus this is a rank-one covariance spike. Each summand $X_i{X}_i^{\top }$ has expected value ${\Sigma }_S$, so $\mathbb{E}\hat{\Sigma }={\Sigma }_S$. Here the empirical matrix is an uncentered empirical second-moment matrix; because the population expected value is known to be zero, it is an unbiased estimator of the covariance matrix. The TeX's apparent factorial denominator is a transcription error.

Use the convention $\mathbb{E}{e}^{sX}\le e^{{\sigma }^2{s}^2/2}$ for the centered sub-Gaussian random variable. Applying the two-sided Chernoff bound to $X$ gives $\mathbb{P}(|X|>x)\le 2e^{-x^2/(2{\sigma }^2)}$. Since $\mathbb{E}{X}^2\ge 0$, this already supplies the requested bound:For averages of independent random variables, the stronger useful statement is that the centered square of a sub-Gaussian random variable is a sub-exponential random variable. The Bernstein bound for independent sub-exponential variables then give $\mathbb{P}(n^{-1}{\sum }_i(X_i^2-\mathbb{E}{X}_i^2)>t)\le 2\exp [-cn\min \{t^2/{\sigma }^4,t/{\sigma }^2\}]$ for a universal positive $c$. These are the concentration theorems used for quadratic scan statistics.

Uniform expectations require nonempty finite sets; assume this throughout the analytic argument. For real-valued functions on $X\times Y$, define the box norm byAll four variables are sampled as independent random variables with replacement, so repeated coordinates are included. The defining fourth power equalsIf it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give ${\mathbb{E}}_{x}f(x,y{)}^2=0$ for each $y$, so $f=0$. Absolute homogeneity of a norm follows directly: $\Vert af{\Vert }_{\square }=|a|\Vert f{\Vert }_{\square }$.

For the triangle inequality, prove the mixed box Cauchy-Schwarz inequality. WriteSeparating the two $y$ averages and applying the Cauchy-Schwarz inequality in $(x_0,x_1)$ givesExpanding the square and reversing the order of finite sums,by a second Cauchy-Schwarz inequality. Thus $|\Lambda |\le {\prod }_{i,j}\Vert f_{ij}{\Vert }_{\square }$. Expand $\Lambda (f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\Vert f{\Vert }_{\square }+\Vert g{\Vert }_{\square }{)}^4$. Taking fourth roots yields the triangle inequality. Together with definiteness of a norm and absolute homogeneity of a norm, this proves the box norm is a norm. Complex functions require conjugates and are outside the printed real-valued convention.

For the bilinear correlation bound for the box norm, use the normalized L2 norm: let $B={\mathbb{E}}_{x,y}f(x,y)u(x)v(y)$ and $\Vert u{\Vert }_2^2={\mathbb{E}}_{x}u(x{)}^2$, $\Vert v{\Vert }_2^2={\mathbb{E}}_{y}v(y{)}^2$. The Cauchy-Schwarz inequality in $y$ givesThe remaining factor is ${\mathbb{E}}_{x,x^{\prime }}u(x)u(x^{\prime }){\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. By the Cauchy-Schwarz inequality in $(x,x^{\prime })$, its magnitude is at mostTaking square roots proves

For the tripartite graph, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency indicator functions. Let $f=g_{XY}-\alpha$, so $\Vert f{\Vert }_{\square }\le c$. The normalized triangle count isThe constant-degree hypothesis is exactly ${\mathbb{E}}_y{g}_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ isFor fixed $z$, apply the bilinear correlation bound for the box norm with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are indicator functions, $\Vert v_z{\Vert }_2^2=\beta$ and $\Vert u_z{\Vert }_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the Cauchy-Schwarz inequality in $z$ gives the sharper triangle counting with one box-uniform pair and constant opposite degree estimateMultiplying by $|X||Y||Z|$ yieldsThe middle bound also handles $\beta =0$ or $\gamma =0$, when the triangle count is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and ${\mathbb{E}}_y{f}_2(y,z)=0$ for every $z$. Consequently ${\mathbb{E}}_{x,y,z}{f}_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha \beta \gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the triangle count itself is zero, but the printed uniform expectations and edge density of a bipartite graph values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

The negative binomial regression assumes independent random variables conditional on their recorded days, withThe logarithmic link function applies to the conditional expectation, not to the observed count. The displayed maximum-likelihood estimates areand the dispersion parameter for the fitted family is one. Thus the fitted variance is ${\hat{\lambda }}_i+{\hat{\lambda }}_i^2/\hat{r}$, practically the Poisson distribution variance at these means.

Increasing the day by one multiplies the conditional expectation by $e^{{\hat{\beta }}_1}\simeq 0.996836$, a decrease of about $0.3164\%$. Comparing days 365 and 1 gives $e^{364{\hat{\beta }}_1}\simeq 0.3155$. The negative slope has Wald statistic approximately $-35.79$, so under the stated model assumptions there is very strong evidence of a decreasing conditional mean:This is an association with calendar day, not a causal effect established by the regression. The printed negative-binomial residual deviance is $7544.3$ on $1298$ residual degrees of freedom, which is a reason to investigate the adequacy of the assumed count model and dependence structure. In particular, the small reported slope standard error cannot make a misspecified model reliable.

For independent random variables $Y_i\sim \mathrm{Bernoulli}(p_i)$, the full logistic regression has linear predictorLayout A is the reference level in a regression factor. The smaller model sets ${\beta }_B={\beta }_C=0$, leaving the same logit link and the price predictor. In either caseThe saturated statistical model allows a separate probability per observation; its maximizing probabilities are $p_i=y_i$, with log-likelihood zero under $0\log 0=0$. Hence the binomial deviance for either fitted model isEquivalently, write the summands as $2[y_i\log (y_i/{\hat{p}}_i)+(1-y_i)\log ((1-y_i)/(1-{\hat{p}}_i))]$. The fitted probabilities differ between the two models, but the definition of binomial deviance is the same. A residual deviance for individual binary data need not itself have an accurate chi-squared distribution approximation; the nested-model difference used next has a different asymptotic justification.

The tail sigma-algebra isIts events depend only on arbitrarily late coordinates. The Kolmogorov zero-one law states that for a sequence of independent random variables, every tail event has probability zero or one:Identical probability distributions and integrability are not required for the Kolmogorov zero-one law.

Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ for $c>0$.

For an integer $n\ge 1$, set $T_0=0$ and successively hit the levels $a,2a,\dots ,na$. The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of $T_a$. Thus, for independent copies $T_a^{(j)}$,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.

Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ be the natural filtration. An integer-valued random variable $M$ is a stopping time with respect to $(X_i)$ whenThus whether one has stopped by time $n$ can be decided from the first $n$ observations, without seeing future variables. Equivalently, $\{M\ge i\}\in {\mathcal{F}}_{i-1}$.

Write the stopped sum asThe event $\{M\ge i\}$ depends only on $X_1,\dots ,X_{i-1}$, so its indicator and $X_i$ are independent random variables. Since $\mathbb{E}|X_1|<\infty$ and $\mathbb{E}M<\infty$, Tonelli theorem givesWe may consequently exchange expectation and summation. Using the tail-sum formula for $M$ proves Wald's equation:

For the renewal process, put $S_n=X_1+\cdots +X_n$ and $N_t=\max \{n:S_n\le t\}$. The random variable $N_t+1$ is a stopping time. Since $S_{N_t+1}>t$, Wald's equation givesand hence

For $c>0$, set $Y_i=X_i\wedge c$ and ${\mu }_c=\mathbb{E}{Y}_i$. Wald's equation still applies to the stopping time $N_t+1$ and the independent variables $Y_i$. Because $S_{N_t}\le t$,ThereforeBy the monotone convergence theorem, ${\mu }_c\uparrow \mu$ as $c\to \infty$. Combining the bounds proves the elementary renewal theorem

For a fixed word over $q$ equally likely symbols, the waiting time for a word in independent uniform symbols has mean $\sum q^k$, where $k$ runs through the lengths of the borders of a word, including the full word. The word lava' has no proper nonempty border, so $$ \boxed{\mathbb E\tau_{\mathtt{lava}}=100^4=100\,000\,000}. $$ The word aa' has borders of lengths $1$ and $2$, soFor a direct check, let $E_0$ be the expected remaining time with no trailing a' and $E_1$ with one trailing a'. Thenwhich gives $E_0=10100$.

The covariance matrix isIf $\mathrm{Cov}(X_1,X_2)=0$, the moment-generating function in part (ii) factors into the product of the two marginal moment-generating functions. Thus $X_1$ and $X_2$ are independent random variables; this is the independence of uncorrelated jointly normal variables.

By bilinearity of covariance,ChooseThen $(X_1,Y)$ is bivariate normal by part (i) and has zero covariance, so $Y$ is a normal random variable independent of $X_1$. Hence $X_2=a{X}_1+Y$.

For this model, the Gauss-Markov theorem states that $\hat{\beta }$ is the unique minimum-variance linear unbiased estimator of $\beta$: ifis unbiased for every $\beta$, then $\mathrm{var}(\tilde{\beta })\ge \mathrm{var}(\hat{\beta })$.

Indeed,so unbiasedness is equivalent to ${\sum }_i{a}_i{x}_i=1$. WriteThe constraint becomes ${\sum }_i{c}_i{x}_i=0$. Since the errors are independent with common variance ${\sigma }^2$,Equality holds exactly when every $c_i=0$, which gives $a_i=x_i/S_{xx}$ and $\tilde{\beta }=\hat{\beta }$.

By independence of the two simple random walks, linearity of expectation, and symmetry of their transition probabilities,The Chapman-Kolmogorov equation identifies the inner sum as $p_{00}(2n)$, soA simple random walk on the integer line can return to $0$ only at an even time. Since it is recurrent at $0$, part (a) givesTherefore $\boxed{\mathbb{E}Z=\infty }$.

Assume first that $A\subseteq {\mathbb{F}}_3^n$ is a cap set. Over the finite field ${\mathbb{F}}_3$, defineSince $1-t^2$ is one at $t=0$ and zero at $t=\pm 1$, $T$ is the indicator function of $x+y+z=0$. If $x,y,z\in A$ satisfy this equation, then either they are all equal or they are three distinct points; the latter is excluded. Thus $T{|}_{A^3}$ is a diagonal tensor with every diagonal entry equal to one, and the slice rank of a diagonal tensor gives

Expand $T$ as a polynomial. Every variable has exponent at most two, and every monomial has total degree at most $2n$. Splitting a monomial's degree among its $x$-, $y$-, and $z$-blocks, at least one block has degree at most $2n/3$. Assign each monomial to one such block and group together terms with the same low-degree block monomial. Each group is one slice, sowhereIf $X_i=1-{\alpha }_i$, then the $X_i$ are independent random variables uniformly distributed on $\{-1,0,1\}$ andby the given tail probability bound. Consequently every cap set satisfies

Put $C_0=3e^{-1/24}<3$. For $n\ge 27$, the preceding bound is at most $C_0^n$. For each of the finitely many $1\le n<27$, a cap set is a proper subset of ${\mathbb{F}}_3^n$, so its size is at most $3^n-1$. We may therefore chooseEvery cap set then has size strictly below $C^n$. Equivalently,

Let $Y_{ij}$ be the number of ice creams in price category $i\in \{\mathrm{high},\mathrm{low},\mathrm{medium}\}$ and score category $j\in \{\mathrm{excellent},\mathrm{good},\mathrm{poor}\}$. The fitted Poisson regression assumes that the nine counts are independent random variables withwhere the logarithmic link function is used and excellent is the reference level in a regression factor, so ${\beta }_{\mathrm{excellent}}=0$. Equivalently,This is the independence log-linear model for a two-way contingency table: price and score have main effects but no interaction.

Saying that $U$ has law $\mu$ means that $\mu$ is the pushforward measureFor a real random variable, $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

The pair $U,V$ consists of independent random variables whenfor every pair of Borel sets $A,B$; equivalently, their joint law is the product of their marginal laws.

At $p=1/2$, the output of the binary symmetric channel is independent of its input. It conveys no mutual information, and therefore

Conditional on the observed colleges, the stochastic block model assumes that the friendship indicators are independent random variables. For arbitrary linear predictors ${\theta }_{ij}$, their Bernoulli distribution likelihood isThus the three requested likelihoods are

For every Borel set $A\subseteq {\mathbb{R}}^d$, letbe its random point count. A Non-homogeneous Poisson point process with intensity function of a point process $\lambda$ is characterized by:

Because $\Lambda$ has a density with respect to Lebesgue measure, it is atomless, so the process is simple: almost surely no location occurs more than once.