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Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.

Use the inverse change of variables for probability densitiesIts absolute Jacobian determinant isThereforeSince ${\int }_0^{\infty }w{e}^{-w}dw=1$, its marginal densities are $f_V(v)=1_{(0,1)}(v)$ and $f_W(w)=w{e}^{-w}{1}_{(0,\infty )}(w)$. The joint density factorizes as $f_V{f}_W$, so $V$ and $W$ are independent random variables, with $V$ uniform on $[0,1]$ and $W\sim \mathrm{Gamma}(2,1)$ in the shape-rate convention. This is beta-gamma independence with both shape parameters equal to one.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.

If $B$ and $U$ are independent random variables, conditioning on the Bernoulli random variable $B$ giveswith the continuous value $M_Z(0)=1$. Indeed $Z=0$ when $B=0$, while the conditional probability distribution of $Z$ given $B=1$ is the original uniform distribution.

Under the literal assumptions alone, the law of total expectation gives onlyThis is not determined by $M_U$ and $p$. For an explicit counterexample, put $p=1/2$ and $B=1_{\{U>0\}}$. Then $B$ has the required Bernoulli distribution, butwhich differs from the independent-case formula.

First take integers $1\le k<n$. For independent copies $X_1,\dots ,X_n$ of $X$, defineThe two blocks involve disjoint families of independent random variables, so $A$ and $B$ are independent. The assumed strictly stable distribution gives $A\stackrel{d}{=}X$, $B\stackrel{d}{=}X$, andTherefore, on any fixed probability space carrying $X$ and an independent copy $X^{\prime }$, their joint probability distribution givesNow fix an arbitrary $0<c<1$ and choose rationals $r_m=k_m/n_m\in (0,1)$ tending to $c^{\alpha }$. Continuity of the real powers gives, on that fixed probability space,The left sides all have the probability distribution of $X$. Since almost sure convergence implies convergence in distribution, their limit has that same probability distribution. HenceThis proves strict stability implies self-decomposability. The limit argument uses no moment-generating function or finite moments, which is essential for general stable distributions.

For $B_n=\{X_n>a_n\}$, the definition of the common cumulative distribution function givesThe strict exceedance matters here: an atom of a measure at $a_n$ belongs to $\{X_n\le a_n\}$, so it is already included in $F(a_n)$. The events $B_n$ are mutually independent events, since each depends on a separate member of the independent random variables $X_n$.

If ${\sum }_n(1-F(a_n))<\infty$, countable subadditivity gives, for every $N$,Continuity of probability then gives $\mathbb{P}({\mathrm{limsup}}_n{B}_n)=0$. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and $1-u\le e^{-u}$ give, for every fixed $N$,Taking $M\to \infty$ shows that the probability of no exceedance after $N$ is zero. A countable union over $N$ shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves

The filtration is generated by the successive independent random variables $X_k$. Each $S_n$ is adapted, and $|S_n|\le n$ makes it an integrable random variable. Since independence gives $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=0$,Thus $(S_n)$ is a martingale.

Write the accumulated number of visits to zero asIt is adapted, and $0\le L_n\le n$. Thus $M_n=|S_n|-L_n$ is also adapted and an integrable random variable. For every integer $s$,Conditioning on $S_n$ and using independence of the next step therefore givesThe drift of the absolute value is cancelled exactly by the visit count, so $\mathbb{E}[M_{n+1}-M_n\mid {\mathcal{F}}_n]=0$. Hence both $(S_n)$ and $(M_n)$ are martingales. The latter is the absolute value martingale of a simple symmetric random walk, a discrete Tanaka formula in which visits to zero supply the correction at the corner of the absolute-value function.

For $m\ge 1$, fix signs $x_1,\dots ,x_m$. The constraints $X_j=x_j$ leave exactly two choices for the $(m+1)$ signs $Y_0,\dots ,Y_m$: choose $Y_0$ freely, then set $Y_1=Y_0{x}_1$ and $Y_j=x_{j-1}{x}_j$ for $2\le j\le m$. Each input tuple has probability $2^{-(m+1)}$ by independence. ConsequentlyEvery finite subfamily is a marginal distribution of one such initial segment, so the entire sequence is independent, with fair sign probability distributions.

Using the usual everywhere sign-valued representatives, for every $n\ge 1$ we haveThe right side is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. Therefore $Y_0$ is measurable with respect to their intersection $\mathcal{G}$.

Write $\mathcal{T}={\bigcap }_n{\mathcal{T}}_n$. The Kolmogorov zero-one law applies to the independent random variables $(X_n)$, so every event $C\in \mathcal{T}$ has probability zero or one. Also $Y_0$ is independent of $\mathcal{Y}$: this follows first for finite intersections of sign events from the given independence, then for their generated sigma-algebra by part (a).

The sets $D\cap C$ with $D\in \mathcal{Y}$ and $C\in \mathcal{T}$ form a pi-system generating $\mathcal{H}$. If $\mathbb{P}(C)=0$, the event $D\cap C$ has probability zero and is independent of every event in $\sigma (Y_0)$. If $\mathbb{P}(C)=1$, it differs from $D$ by a null set, so the same independence follows from that of $Y_0$ and $\mathcal{Y}$. Another application of part (a) provesIn particular $\mathcal{H}$ cannot contain the nontrivial event $\{Y_0=1\}$, whereas $\mathcal{G}$ does. Thus adjoining a sigma-algebra need not commute with a decreasing intersection.

There is a small version qualification if the assumptions specify only the probability distributions, allowing exceptional values other than signs. Put $q_n=X_n/(Y_1\cdots Y_n)$ when the denominator is nonzero, and $q_n=0$ otherwise. For $k\ge n$, $q_k$ is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. The limit superior of $(q_k)$, after scalar clipping to $[-1,1]$, is therefore $\mathcal{G}$-measurable and equals $Y_0$ almost surely. Literal measurability of an arbitrary version of $Y_0$ need not follow without choosing sign-valued representatives or adjoining the ambient null sets. All the independence conclusions are unaffected.

Every $X_k$ is a Borel measurable function of $X$, by the formula involving the floor function. ThusConversely, part (i) gives the everywhere identity $X={\lim }_{k\to \infty }{X}_k$. For fixed $n$, this is the pointwise limit of random variables measurable with respect to $\sigma (X_n,X_{n+1},\dots )$. Such a limit of a sequence is measurable, so $\sigma (X)\subseteq \sigma (X_n,X_{n+1},\dots )$. ConsequentlyThese are increasing-precision observations even when the first finitely many are discarded; they do not behave like the independent random variables whose tail sigma-algebra appears in Question 1(b).

Taking the transpose gives $M^T=(X^T+X)/\sqrt{2}=M$, so $M$ is a symmetric matrix. For $i<j$, $M_{ij}=(X_{ij}+X_{ji})/\sqrt{2}$ has the standard normal distribution, whereas $M_{ii}=\sqrt{2}{X}_{ii}$ has the normal distribution $N(0,2)$. Distinct upper-triangular entries depend on disjoint sets of the original independent random variables, so they remain independent.

Consequently the joint probability density function, relative to ${\prod }_{i\le j}d{m}_{ij}$, isHere $\mathrm{tr}(m^2)={\sum }_i{m}_{ii}^2+2{\sum }_{i<j}{m}_{ij}^2$ because $m$ is a symmetric matrix. This is the Gaussian orthogonal ensemble, with off-diagonal variance one and diagonal variance two; the rescaled matrix $M/\sqrt{n}$ has the corresponding $1/n$ off-diagonal variance.

We will also use its entry covariance. Expanding in the independent entries of $X$, for which $\mathbb{E}[X_{ij}{X}_{kl}]={\delta }_{ik}{\delta }_{jl}$, givesThe symbols ${\delta }_{ab}$ are Kronecker deltas. This formula includes the diagonal case, where the two terms both contribute.

Write $s={\sigma }^2\ge 1$. The two multivariate normal distribution probability density functions have ratioHere $\Vert x\Vert$ is the Euclidean norm. Consequently the smallest rejection sampling envelope constant, and its acceptance probability, areThe supremum is attained at $x=0$; hence no smaller constant can dominate the ratio. Draw $Y\sim N(0,s{I}_d)$ and an independent $U$ from the uniform distribution on $(0,1)$. Accept $Y$ whenotherwise repeat with fresh independent random variables. One can generate $Y=\sqrt{s}G$ using a random vector $G$ with independent standard normal distribution coordinates.

For any measurable set $B$, the joint acceptance probability is ${\int }_{B}q(y)p(y)/(M_{\ast }q(y))dy=M_{\ast }^{-1}{\int }_{B}p(y)dy$. Dividing by $1/M_{\ast }$ proves that the returned random vector has probability density function $p$. The number of proposals has a geometric distribution on the positive integers, with expected value $M_{\ast }$. When $s=1$, every proposal is accepted.

Use several complementary Markov chain Monte Carlo convergence diagnostics, monitoring both $p$ and $\beta$ rather than judging the chain solely by its length:

A million iterations and stable-looking plots do not prove convergence. Agreement across dispersed chains, exploration of all visible modes and adequate effective sample size of a Markov chain provide stronger evidence. Estimate the Monte Carlo standard error using the chain's autocorrelation; the independent random variables formula from 3(b)(ii) does not apply directly to these dependent draws.

Integrating the envelope $p\le Mq$ shows $M\ge 1$. One trial of rejection sampling draws $Y\sim q$ and an independent $U$ from the uniform distribution on $(0,1)$, and accepts $Y$ if $U\le p(Y)/(Mq(Y))$. Repeat with fresh independent random variables after each rejection. Ratios need only be evaluated where $q>0$; the envelope makes the target zero wherever $q=0$.

For any measurable set $B$, integration of the conditional probability of acceptance givesTaking the whole domain gives acceptance probability $1/M$, and dividing gives $\mathbb{P}(Y\in B\mid \mathrm{accept})={\int }_{B}p(y)dy$. Fresh trials therefore supply independent and identically distributed random variables $X_1,\dots ,X_m$ with target probability density function $p$. The corresponding Monte Carlo estimator isThe proposal count $T$ up to and including one acceptance has the geometric distributionFor example, the tail-sum formula gives $\mathbb{E}T={\sum }_{t\ge 0}(1-1/M{)}^t=M$. Thus the expected cost is $M$ proposal draws per accepted target draw.

For a measurable set $C$, (i) gives the joint acceptance probabilityDivide by $Z_p/(M{Z}_q)$ from (ii). The resulting conditional probability isThus every accepted value has target probability density function $p$. Repeating rejected trials with fresh independent random variables does not change this law: the contribution from acceptance on trial $j$ is multiplied by $(1-Z_p/(M{Z}_q){)}^{j-1}$, and summing that geometric series gives the same conditional law. Independent repetitions produce independent target draws.

Continuity and strict increase make the quantile function $F^{-1}$ well-defined on $(0,1)$. The endpoints of the uniform distribution occur with probability zero and do not affect the construction. For every real $x$,Thus each $X_i$ has cumulative distribution function $F$. Applying a fixed measurable function separately to independent random variables preserves their independence; explicitly, joint lower-tail probabilities factor asConsequentlyThis is inverse transform sampling. Since $F^{-1}$ preserves order, it also sends the central order statistic to the central order statistic, justifying $M_X=F^{-1}(M_U)$.

Two assumptions of the usual normal linear model used for the reported coefficient Student t-tests are:

The second statement includes homoscedasticity. Normality concerns the conditional errors, rather than requiring the weights themselves to have a normal distribution.

By linearity of expectation, $\mathbb{E}(S_n)={\sum }_{i=1}^{n}p=np$. Part (ii) gives $\mathrm{Var}(X_i)=\mathbb{E}(X_i^2)-[\mathbb{E}(X_i){]}^2=p-p^2$. Independent random variables have zero pairwise covariance, so the variance of their sum is the sum of their variances. Therefore

For independent random variables, the probability generating function of an independent sum is the product of the individual probability generating functions. ThusExpanding this last expression as $e^{-(\lambda +\mu )}{\sum }_{r\ge 0}(\lambda +\mu {)}^r{s}^r/r!$ identifies each probability coefficient. Therefore

Because $T$ and $U$ are independent random variables, their product's expected value factors:The original independence assumption is essential to this factorization.

The independent random variables have a product joint probability mass function. For integers $r,t\ge 0$,It is zero for all other pairs. The factors are the respective Poisson distribution masses, with the product justified by independence.