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Because $X$ and $Y$ are independent random variables, conditioning on $Y$ merely translates $X$. The conditional entropy under a deterministic change of variables and the fact that conditioning reduces entropy therefore giveInterchanging $X$ and $Y$ similarly gives $H(X+Y)\ge H(Y)$, and henceThe proof applies to countable alphabets whenever the displayed information entropies are finite.

Because a strictly monotone map fixing both endpoints is increasing, the change of variables formula $t=h(s)$ givesAssuming $\mathbb{E}\Vert X{\Vert }^2>0$, defineThe Hilbert-space variance identity then gives

Under the regularity needed to interchange expectation and differentiation, $\mathbb{E}{h}^{\prime }(s)=1$ because $\mathbb{E}h(s)=s$. HenceThus $A=1$ exactly when the integrated covariance in the numerator vanishes. In particular, this holds when the amplitude $X$ and the time warp $h$ are independent random variables.

A one-time pad over the finite additive group ${\mathbb{Z}}_n$ uses a key $K$ that is uniform on ${\mathbb{Z}}_n$, independent of the plaintext $X$, as long as the plaintext, and never reused. Encryption and decryption are

For independent random variables $X,Y$, conditioning reduces entropy and translation by a known element of the finite additive group preserves entropy, soInterchanging $X$ and $Y$ gives $H(X+Y)\ge H(Y)$, henceThis is the entropy of a sum of independent finite-group variables.

The result explains why adding independent pad symbols cannot reduce uncertainty. For a uniform pad, $X+K$ is itself uniform and is independent of $X$, which is the perfect-secrecy property. Independence is necessary: if $X$ is nonconstant and $Y=-X(\mathrm{mod}n)$, then $X+Y=0$ is constant, so

The commands choose $n=100$ observations and $p=2$ predictors; generate a $100\times 2$ matrix $X$ of independent standard-normal values; generate independent binary random variables $Y_i$ with success probability $1/2$; and report the observed number of successes, ${\sum }_i{Y}_i=48$. The first glm call then fits the Bernoulli logistic-regression modelThe prediction command returns the fitted response probabilities ${\hat{p}}_i$ and sums them.

The likelihood isAt the maximum-likelihood estimator, the intercept component of the score function isThereforeexactly, by the fitted-mean balance for logistic regression with an intercept.

The second fit uses probit regression,where $\Phi$ is the standard normal distribution function. Its intercept score isa weighted residual equation. It does not imply ${\sum }_i{\hat{p}}_i=48$, so exact equality is not expected. The output should nevertheless be close to $48$: the data were generated with constant success probability $1/2$ independently of $X$, so the fitted slopes should be small, the linear predictors should cluster near a common intercept, and the score weights should be nearly constant.

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

Let $Y_i$ be the first team's goal count in match $i$, let $x_i^T$ be row $i$ of the design matrix, and let $\beta$ be the regression coefficient vector. The fitted Poisson regression assumes that the $Y_i$ are independent random variables withThe likelihood function maximized by glm is therefore

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

A Non-homogeneous Poisson point process with intensity function $\lambda$ is a random counting measure $N$ such that, for every finite family of pairwise disjoint bounded Borel sets $A_1,\dots ,A_k$, the counts $N(A_1),\dots ,N(A_k)$ are independent random variables andThe assumed local integrability makes each Poisson parameter finite. This count law and independence characterize the process.

They are not independent. The Itô formula gives ${\beta }_t^2=t+2{\int }_0^t{\beta }_{s}d{\beta }_s$, and the bilinear Itô isometry therefore givesIf $B_t$ and ${\beta }_t$ were independent random variables, then $B_t$ would also be independent of the measurable function ${\beta }_t^2$, and centeredness would instead give $\mathbb{E}[B_t{\beta }_t^2]=\mathbb{E}[B_t]\mathbb{E}[{\beta }_t^2]=0$. This contradiction disproves independence.

Write the molecular copy numbers as $\mathbf{x}=(x_1,x_2,x_3,x_4)$ and use the power-law reaction propensity convention of the paper. The seven propensities and stoichiometric vectors areA channel is assigned zero propensity whenever its update would leave the nonnegative integer lattice.

The Gillespie algorithm starts from $\mathbf{x}=(5,0,0,0)$ and $t=0$. At the current state compute $a_0={\sum }_{r=1}^7{a}_r$. If $a_0=0$, terminate the path. Otherwise draw independent random variables $U_1,U_2$ from the uniform distribution on $(0,1)$, set the next waiting time toand choose the least index $r$ satisfyingThen update $t\leftarrow t+\tau$ and $\mathbf{x}\leftarrow \mathbf{x}+{\nu }_r$, and repeat. The minimum of the seven competing reaction clocks has an exponential distribution of rate $a_0$, while reaction $r$ wins with probability $a_r/a_0$.

Let ${\xi }_{1,n},{\xi }_{2,n}$ be independent random variables with the standard normal distribution. The unconstrained Milstein method step isThe $X$ noise is additive, so its Milstein correction vanishes. The diagonal, independent noise fields commute, so no cross iterated stochastic integral is required.

Implement each reflecting boundary condition for a diffusion by folding the proposed coordinate back into $[-1,1]$. One formula that also handles multiple overshoots isThe complete update isUnder the usual smoothness assumptions, the Milstein discretization has strong order of convergence one and weak order of convergence one. The reflection enforces the boundary pathwise.

Because $X,Y,Z$ are independent random variables, adding $Y$ is an independent noise channel, sois a Markov chain. The data processing inequality for mutual information givesTranslation in the finite additive group preserves conditional entropy, and independence therefore givesandSubstitution proves the required entropy submodularity for three independent sums.

Letwhere $\phi$ is the golden ratio. If the union-closed family consists of one nonempty set, any element of that set has frequency one, so assume its cardinality exceeds one. Choose independent uniform members $A,B$ and let $X,Y\in \{0,1{\}}^n$ be their characteristic vectors of sets. Then $H(X)=H(Y)=\log |\mathcal{A}|>0$.

Suppose for a contradiction that every element has frequency $p_i<c$. Put $q_i=1-p_i>1/\phi$, and let $Z=X\mathrm{OR}Y$, the characteristic vector of a set of $A\cup B$. The chain rule for information entropy and conditioning reduces entropy givebecause $Z_{<i}$ is a function of $(X_{<i},Y_{<i})$.

Fix the two prefixes and setThe two conditioned bits are independent random variables, and $Z_i=0$ exactly when both are zero. The supplied binary entropy product inequality therefore givesAveraging over the independent prefixes yieldswhenever either conditional entropy is positive. Summing and using $H(X)>0$ gives $H(Z)>H(X)$.

But set union keeps $A\cup B$ inside the union-closed family, so $Z$ is supported on $\mathcal{A}$. The maximum entropy distribution on a finite set gives $H(Z)\le \log |\mathcal{A}|=H(X)$, a contradiction. Some element must therefore occur in at least $c|\mathcal{A}|$ members, proving the entropy bound for a union-closed family.

Write $\delta =d_R(X,Y)$, $A=U_1+U_2$, $B=V_1+V_2$, and $S=A+B$. Distances depend only on distributions, so all variables used in any one application may be realized as independent random variables.

First, the entropy submodularity for three independent sums impliesand analogously for $V$. The Entropic Ruzsa triangle inequality gives $d_R(U,U)\le 2d_R(U,X)$ and $d_R(V,V)\le 2d_R(V,Y)$. Consequently the relevance of independent self-sums gives

Apply the Conditioned entropic Ruzsa distance of a summand first to $(A,B,X)$ and then to $(B,A,Y)$:Three applications of the Entropic Ruzsa triangle inequality giveAdding all these bounds yieldsThus the required absolute constants may be taken as $a=4$ and $b=2$.

The relevant invariance principle for a low-degree multilinear polynomial is the following. Let $(X_i)$ and $(Y_i)$ be sequences of independent random variables satisfyingIf $f$ is multilinear of degree at most $k$ and $\Vert {\psi }^{(4)}{\Vert }_{\infty }\le M$, then

For the proof, use the Lindeberg replacement method. Replace $X_i$ by $Y_i$ one coordinate at a time and write $f=u_i+z_i{v}_i$, where $u_i=E_{i}f$ and $v_i=D_{i}f$ depend only on the other coordinates. A third-order Taylor expansion of $\psi (u_i+z_i{v}_i)$ has identical expected terms through order three for $z_i=X_i$ and $z_i=Y_i$, because the first three moments match. Each fourth-order remainder is bounded by $M|z_i{v}_i{|}^4/24$, so the $i$th replacement costs at mostPart (i), applied in the hybrid product space, bounds each of the two fourth-moment terms by $9^k{\mathrm{Inf}}_i(f{)}^2$. Thus the cost is at most $(M/12)9^k{\mathrm{Inf}}_i(f{)}^2$. The triangle inequality and summation over $i$ prove the result.

For every $\theta \ge 0$, the exponential Markov bound and independence giveTaking the infimum over $\theta \ge 0$ yieldsFor $x\ge m$, convexity makes this supremum equal to ${\psi }^{\ast }(x)$, proving the upper bound in the stated tail form of Cramér theorem.