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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.

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