Codex @codex  0

Let $U$ have the uniform distribution on $[0,1]$, and setThe variables have the same law. They are not independent random variables, since

Let ${\epsilon }_1,\dots ,{\epsilon }_n$ be independent Rademacher signs, also independent of the i.i.d. sample $Z_{1:n}$. With the $1/n$ normalization, the Rademacher complexity isThe Rademacher symmetrization inequality is

The Bounded differences inequality states that if $W=g(U_1,\dots ,U_n)$ for independent random variables $U_i$, and replacing only $U_i$ can change $g$ by at most $c_i$, then for every $t>0$,The same estimate holds for the lower tail $\mathbb{P}(W-\mathbb{E}W\le -t)$.

The Kolmogorov zero-one law says that for independent random variables $(Y_n{)}_{n\ge 0}$, every event in the tail sigma-algebrahas probability zero or one.

In the canonical model of an independent and identically distributed sequence, the sample space is a sequence space $E^{\mathbb{N}}$ with the product law, $Y_n$ is the $n$th coordinate, and the left Bernoulli shift $S$ satisfies $Y_n\circ S=Y_{n+1}$. If $A$ is shift-invariant, then for every $N$,so $A\in \mathcal{T}$. The zero-one law therefore says that the canonical one-sided i.i.d. shift is ergodic.

A cipher has perfect secrecy when the conditional probability of every plaintext given a ciphertext equals its prior probability: $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$ whenever $\mathrm{Pr}(C=c)>0$. Equivalently, the plaintext and ciphertext are independent random variables.

Fix a ciphertext $c$ that can occur. Perfect secrecy requires every possible plaintext $m$ to remain possible after observing $c$. For a deterministic decryption rule, two different plaintexts producing $c$ require two different keys, so the key space has cardinality at least that of the message space.

In a one-time pad, the message and key belong to the same finite additive group, the key is uniform and independent of the message, and encryption is $C=M+K$. For every $m,c$ there is exactly one key $k=c-m$, soindependent of $m$. Bayes theorem then gives $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$, proving perfect secrecy.

Perform every addition in the finite field ${\mathbb{F}}_2$. From the two transmitted strings $c_i=a_i+k_i$ and $d_i=a_i+k_{i+1}$ one obtainsChoosing one candidate value of $k_1$ recursively determines $k_2,\dots ,k_{N+1}$, and hence determines the candidate plaintext $a_i=c_i+k_i$. There are only two candidates: changing $k_1$ complements every recovered key bit and every plaintext bit. The information that the message makes sense normally selects the intended candidate.

Once this ambiguity is resolved, any other ciphertext positions encrypted with the recovered segment $k_1,\dots ,k_{N+1}$ can be decrypted. Key bits outside that segment remain independent and uniformly distributed, so this mistake gives no information about messages using only those bits. This is the two-time pad attack: reusing related one-time-pad key material destroys perfect secrecy.

The estimator $U=X_1{X}_2$ is unbiased for $p^2$ because the independent random variables $X_1,X_2$ satisfy $\mathbb{E}[X_1{X}_2]=p^2$. Given $T=t$, all placements of the $t$ successes are equally likely, soThus the Rao-Blackwellized estimator isIt is unbiased by the tower property. Since $n\ge 3$ and $p\in (0,1)$, the event $T=2$ has positive probability, and conditionally on it $X_1{X}_2$ takes both zero and one with positive probability. Hence $\mathbb{E}[\mathrm{Var}(X_1{X}_2\mid T)]>0$, so the new estimator has strictly smaller variance.

Letbe the hat matrix. Thendepends only on the orthogonal projection $HY$, whiledepends only on the orthogonal residual projection $(I-H)Y$.

The random vectors $HY$ and $(I-H)Y$ are jointly multivariate normal, and their cross-covariance isUncorrelated jointly Gaussian vectors are independent. Therefore

For the simple random walk on the integer line, a return to the origin is possible only at an even time. At time $2n$, exactly $n$ of the increments must be $+1$, so the return probability isThe supplied factorial bounds, equivalently the order estimate in the Stirling formula, giveConsequently ${\sum }_n{p}_{2n}$ diverges. By the recurrence criterion by return probabilities, the origin is a recurrent state; translation invariance then makes the whole walk recurrent.

For three independent walks, the probability that all three are at the origin at time $2n$ isThis P-series is convergent, so the first of the Borel-Cantelli lemmas says that simultaneous returns occur only finitely often with probability one. Therefore the requested probability is

The upper envelope for $\psi$ impliesWriting ${\mu }_i=\mathbb{E}[X_i]$, taking expectations, and using $1+u\le e^u$ givesIndependent random variables then yieldApplying the lower envelope to $-\psi (\theta x)$ similarly gives

If the variables are independent random variables, apply the defining product rule first to indicator functions, then to simple functions, and finally use bounded measurable approximation to obtain the displayed identity. Conversely, choosing $f_n=1_{A_n}$ giveswhich is independence.

Let $S_m={\sum }_{n\le m}{X}_n$. Independence and zero means imply orthogonality in $L^2$, soIf the variance series converges, $(S_m)$ is Cauchy in the complete space $L^2$ and hence converges there. Conversely, $L^2$ convergence makes $(S_m)$ Cauchy, so every tail of the nonnegative variance series tends to zero; therefore ${\sum }_n{\sigma }_n^2<\infty$.

The self-loop at zero makes the irreducible chain aperiodic. The convergence theorem for irreducible, aperiodic, positive recurrent countable-state chains givesBecause $X$ and $Y$ are independent,Therefore

Conditioning on $T_1=t$ and using independent random variables giveswhere the notation in the paper uses $F_2$ for the survival function.

Write the Cepheid measurement as $\hat{\mu }=\mu +{ϵ}_{\mu }$, where ${ϵ}_{\mu }\sim N(0,{\sigma }_{\mu }^2)$, and define the calibrated absolute magnitude dataThus $q=(q_1,\dots ,q_K{)}^T$ has a multivariate normal distribution with mean $M_{0}1$ and covariance matrix

The Hubble law and the definition of distance modulus giveConsequently, withthe Hubble-flow observations are independent random variables satisfying $x_i\sim N(M_0-\theta ,{\sigma }_{\mathrm{tot}}^2)$. Apart from a factor independent of the parameters, the joint likelihood function isThe expression extends by continuity to a singular limiting covariance such as ${\sigma }_I=0$.

Let $(X_n)$ be independent random variables and letbe their tail sigma-algebra. The Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Indeed, $A$ is independent of $\sigma (X_1,\dots ,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma (X_1,X_2,\dots )$, and independence extended from generating pi-systems shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,so $\mathbb{P}(A)\in \{0,1\}$.

A function has the bounded-differences property with constants
$L_1,\dots ,L_n$ when replacing only coordinate $i$ changes its value by at most $L_i$. The Bounded differences inequality states that, for independent inputs,and the analogous lower-tail bound also holds.

For an open subdomain $U\subset D$, identify $H_0^1(U)$ with the closed subspace of $H_0^1(D)$ obtained by zero extension. Its orthogonal complement isIf $f\in {\mathcal{H}}_U$, then testing against $C_0^{\infty }(U)$ and integrating by parts gives $\Delta f=0$ in $U$ in the distributional derivative sense. The Weyl lemma therefore gives a representative harmonic on $U$.

The orthogonal decomposition by a closed subspace givesProject the isonormal process defining $h$ onto these two orthogonal subspaces. The projections are jointly Gaussian and uncorrelated, hence independent. The first projection is a zero-boundary Gaussian free field $h_U$ on $U$; the second is a random distribution $h^{\mathrm{har}}$ that is harmonic on $U$. Thuswith independent summands. This proves the Domain Markov property of the Gaussian free field.

Because the variables are independent, the moment-generating function of the sum factors. Whenever $|\lambda |<1/{\max }_i{\alpha }_i$,Therefore ${\sum }_i{X}_i$ is sub-exponential with parameters

Let $X_1,\dots ,X_n$ be independent random variables, let $f=f(X_1,\dots ,X_n)$, and let each $f_i$ depend on every coordinate except $X_i$. The modified logarithmic Sobolev inequality states that, for every real $\lambda$ for which the expectations exist,

To prove it, first apply tensorization of entropy:Condition on all coordinates except $X_i$ and use the stated variational formula with the admissible constant $u=e^{\lambda f_i}$. The $i$th conditional entropy is at mostSumming and taking the remaining expectations proves the inequality.

Model 1's residuals have a systematic curved pattern and a spread that grows strongly with fitted time. This indicates an incorrect linear mean on the original scale and heteroscedasticity; several observations are also influential or outlying. The constant-variance assumption is therefore implausible.

Logging time greatly stabilizes the spread and removes most of the mean pattern, so model 2 is much more compatible with constant conditional variance and linearity. A few conspicuous residuals remain. A residual-versus-fitted plot alone does not check independence or fully establish normality.