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A useful method of moments in probability states that if real random variables $Z_N$ have all moments, $\mathbb{E}{Z}_N^j\to m_j$ for every integer $j\ge 1$, and $(m_j)$ is the moment sequence of a moment-determinate probability distribution, then $Z_N$ converges to that distribution in the sense of convergence in distribution. In particular, convergence to the standard normal distribution follows from limiting odd moments zero and limiting even moments $(2k-1)!!$. The moment-generating function of a standard normal variable, finite in a neighbourhood of zero, ensures moment determinacy.

For the classical Erdős-Kac theorem, put $L=\log \log N$ and choose, for example, $\phi (N)=L^{1/4}$, with a harmless modification for small $N$. Truncate the prime omega function to primes $p\le y=N^{1/\phi (N)}$. The omitted number of distinct prime factors of any $n\le N$ is at most $\phi (N)$, since a product of $r$ omitted primes exceeds $N^{r/\phi (N)}$. Thus this truncation changes the normalized variable by $o(1)$ uniformly. Also, the Mertens theorem for reciprocal primes gives

Part (b) transfers every fixed normalized central moment of the truncated prime omega function to the independent Bernoulli random variables model. Indeed, its unnormalized error is at most $O_j(N^{2j/\phi (N)-1})$, using ${\sum }_{p\le y}1\le y$, and this tends to zero for each fixed $j$. In the independent expansion, any singleton index has zero expectation; the leading contributions are pairings, giving exactly the standard normal distribution moments, while blocks of size at least three are negligible after normalization. The method of moments in probability, followed by the uniformly negligible truncation and centering errors, proves the Erdős-Kac theorem.

The same strategy proves the bounded-prime-weight form in part (a). Choose $\phi (N)\to \infty$ slowly enough that $\phi (N)=o(B(N))$. The omitted weighted contribution is $O(\phi (N))$; its mean is likewise $O(\phi (N))$, and its independent variance is $O(\log \phi (N)+1)=o(B(N{)}^2)$ by the Mertens theorem for reciprocal primes. Hence the truncated model has the same normalization, and part (b) again transfers each fixed central moment.

Order configurations coordinatewise: $\omega \le {\omega }^{\prime }$ means ${\omega }_e\le {\omega }_e^{\prime }$ for every edge. An increasing event $A$ is a measurable set such that $\omega \in A$ and $\omega \le {\omega }^{\prime }$ imply ${\omega }^{\prime }\in A$. Opening additional edges cannot destroy an increasing event. The Harris-FKG inequality, the product measure version of the FKG inequality, statesfor two increasing events. Equivalently, bounded coordinatewise increasing random variables $f,g$ satisfy ${\mathbb{E}}_p(fg)\ge {\mathbb{E}}_{p}f{\mathbb{E}}_{p}g$. The event inequality also holds for two decreasing events, by taking complements.

For disjoint occurrence of increasing events, a finite open edge set $K$ witnesses $A$ in $\omega$ if every configuration agreeing with $\omega$ on $K$ belongs to $A$. Since $A$ is an increasing event and all edges of $K$ are open, this says that the configuration with exactly $K$ open already forces $A$. Define $A\square B$ to consist of configurations admitting disjoint finite witnesses $K,L$ for $A,B$. For events depending on finitely many edges, this is the usual disjoint-occurrence definition; it also applies to finite-connection events on the infinite graph. The van den Berg-Kesten inequality statesOn a countable graph, its finite-witness version follows by taking increasing unions over finite edge sets. The Harris-FKG inequality concerns simultaneous occurrence, whereas the van den Berg-Kesten inequality requires separate certificates using disjoint coordinates.

Use the probability space $([0,1{]}^{{\mathbb{Z}}^d},\mathcal{F},\mathbb{P})$ with the product measure of the uniform distribution on $[0,1]$. Its coordinate random variables $U_v$ are independent and identically distributed random variables. Define simultaneously for every parameterFor $r\le s$, the indicator function inequality $X_r(v)\le X_s(v)$ holds pointwise at every graph vertex. For fixed $p$, each coordinate has the Bernoulli distribution with parameter $p$, and the coordinates are independent because they are functions of distinct independent $U_v$. Thus every marginal configuration is exactly the required site percolation model. This is a continuum-indexed family of processes, all on one probability space, and is the standard monotone coupling of Bernoulli percolation.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

Choose any strictly positive pricing kernel $Z\in \mathcal{Z}$. The assumed componentwise expectations and the linearity of expectation giveThe integrand is nonnegative, so its expectation is also nonnegative and hence zero. A nonnegative random variable with zero expectation vanishes almost surely. Since $Z>0$ almost surely, this forces $H\cdot P=0$ almost surely. Both conclusions follow:The strict positivity of the pricing kernel is essential: a kernel allowed to vanish could miss a positive payoff on its zero set. No normalization $\mathbb{E}Z=1$ is assumed in this abstract market; that normalization follows only if a unit-priced unit-payoff cash asset is included.

Apply Markov's inequality to the nonnegative random variable $e^{\lambda X}$, since $X>t$ implies $e^{\lambda X}>e^{\lambda t}$:This exponential Markov bound remains true, though uninformative, when the expected value is infinite. For a standard normal distribution, completion of the square gives its moment-generating functionMinimize ${\lambda }^2/2-\lambda t$ over positive $\lambda$; the choice $\lambda =t$ gives the Chernoff bound

The conditional density givesusing either integration by parts or the expected value of an exponential distribution. HenceThis is a finite value for almost every realization of $X$, but not an integrable random variable: $\mathbb{E}[1/X]={\int }_0^{\infty }{e}^{-x}/xdx=\infty$. Since $Y\ge 0$, its conditional expectation remains well-defined in the nonnegative, extended-expectation sense; the tower property of conditional expectation then gives $\mathbb{E}Y=\infty$.

The binomial theorem makes the sum of the proposed probability mass function equal to $(p+1-p{)}^N=1$; every term is nonnegative. Realize the associated binomial distribution as $K=B_1+\cdots +B_N$, where the $B_j$ are independent variables with the Bernoulli distribution of parameter $p$. Each has expected value $p$ and variance $p(1-p)$. Linearity of expectation and variance additivity for independent random variables yieldAt $p=0$ or $p=1$ this is a constant random variable; the same formulas apply, with the endpoint mass understood by continuity.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.

Specify an initial value or initial probability distribution. A strong solution of a stochastic differential equation uses a prescribed Brownian motion on a prescribed filtered probability space. It is continuous, adapted to the completed filtration generated by that Brownian motion and the initial value, and satisfiesalmost surely for every $t$, with ${\int }_0^t(|b(X_s)|+\sigma (X_s{)}^2)ds<\infty$ almost surely. The given initial variable is independent of future Brownian motion increments. Some courses allow a larger prescribed filtration in their definition of a strong solution of a stochastic differential equation; the Brownian-generated convention states explicitly what absence of extra randomness means.

A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion relative to its filtration, and a continuous adapted process $X$ satisfying the same integrability and integral equation. Here the space and noise are part of the unknown; the filtration may contain randomness beyond the initial variable and the driving path.

Pathwise uniqueness means that on any common filtered probability space, any two solutions with the same driving Brownian motion and the same initial value are indistinguishable. Uniqueness in law means that all weak solutions with the same initial probability distribution have the same law as path-valued random variables, even on different spaces. Pathwise uniqueness compares shared-noise paths; uniqueness in law compares path distributions. Neither uniqueness notion by itself asserts existence.

A coupling of probability distributions is a joint probability distribution of random variables $(X,Y)$ with the specified marginal distributions: $\mathbb{P}(X\in A)=\mu (A)$ and $\mathbb{P}(Y\in A)=\nu (A)$. No independence is required.

For every set $A$, the difference of the two indicator functions vanishes when $X=Y$. ConsequentlyMaximizing over $A$ gives the coupling inequality for total variation:

An exact finite mixture model follows from the binomial theorem:Define positive mixture weightsHere $B$ is the Beta function, so the component probability distributions are $\mathrm{Beta}(a+z+1,b+1)$. All coefficients are computable from the observed integer counts.

Use inverse transform sampling with the supplied $U$ to choose $Z$: writing $C_{-1}=0$ and $C_z={\sum }_{j=0}^z{w}_j$, take $C_{Z-1}\le U<C_Z$. Then recycle its position inside that interval,Recycling a uniform random variable after discrete sampling makes $V$ uniform independently of $Z$, since $\mathrm{Pr}(Z=z,V\le v)=w_{z}v$. It is also independent of all supplied random variables with gamma distributions. Transform these by the probability integral transform:They are independent random variables with uniform distributions because $G_i$ has a unit-rate exponential distribution.

Conditional on $Z=z$, put $k=a+z+b+1$ and $r=a+z+1$. Form the $k$ values $V,W_1,\dots ,W_{k-1}$ and return their $r$th order statistic. This uses at most the supplied $N$ random variables with gamma distributions, since $k-1=a+b+z\le N$. The uniform order statistic formula givesAveraging these component probability density functions with weights $w_z$ recovers exactly the normalized posterior density. Unused inputs can be discarded. When $N=0$, $Z=0$, $k=r=1$, and the output is $U$, as required. Interval endpoints and ties have probability zero; any consistent convention there is harmless. Exactness here concerns the mathematical algorithm; finite-precision arithmetic can be stabilized with logarithmic mixture weights.

Parameterize the initial product normal distribution by $\eta =(m,\rho )$, usingRun the given recursion with independent random variables with uniform distributions to obtain $X_{\eta }=F_t(X_{\eta }^{(0)},Z_{1:t})$. Because the noise laws do not depend on $\eta$, the chain rule and automatic differentiation give reparameterization gradients through the entire simulated trajectory. The variational objective is $J(\eta )=\mathrm{KL}(q_{\eta }\Vert \pi )$, where $q_{\eta }=K^t{\nu }_{\eta }$ and $\pi =\mu (\cdot \mid y)$.

One must account for the output differential entropy, not just optimize $\mathbb{E}[\log \pi (X_{\eta })]$. Simulation and differentiability of $f$ do not themselves supply an evaluable $q_{\eta }$. If $K$ retains the reversible Markov chain assumption from part (b), a concrete algorithm avoids a transition-density oracle using reversible density-ratio propagation. Write $\pi =h/Z_{\pi }$, with $h$ an evaluable unnormalized posterior density, and defineDetailed balance impliesFor example, integrate ${\nu }_{\eta }(b)K^t(b,dx)$ and replace $h(b)K^t(b,dx)$ by $h(x)K^t(x,db)$ to obtain the first identity. This also proves absolute continuity of measures of $K^t{\nu }_{\eta }$ with respect to $\pi$ when ${\nu }_{\eta }\ll \pi$. An everywhere finite differentiable log-posterior has positive density, so an ordinary product normal distribution has this domination.

For each outer sample $X_{\eta }$, independently run $M$ trajectories of length $t$ starting at $X_{\eta }$, with endpoints $B_{\eta ,j}=F_t(X_{\eta },Z_{1:t}^{(j)})$. These are reverse trajectories because of detailed balance; the same recursion simulates them. Estimate $A_{\eta }(X_{\eta })$ byAverage $\log \hat{A}$ over an outer batch. Compute it stably by a logarithmic sum of exponentials. Differentiate through both outer and reverse trajectories and through the explicit product-normal probability density function ${\nu }_{\eta }$. Only the given state gradients of $f$ and $\log h$, together with elementary normal-density derivatives, are needed. Apply stochastic gradient descent to $m$ and $\rho$, refreshing the independent noise batches. The logarithmic scale parameters keep all variances positive.

This is nested Monte Carlo optimization of the exact objective in the limit of adequately increasing inner and outer sample sizes. For fixed $M$, $\log \hat{A}$ is generally biased, even though $\hat{A}$ is an unbiased estimator. Finite logarithmic moments, suitable uniform integrability, and justified differentiation under the integral sign are needed for consistency and gradient interchange. Under those conditions increasing $M$ and controlling the optimization error yields a justified numerical approximation; nonconvexity prevents a generic guarantee of a global optimum.

There is a real assumption issue: part (c) redefines $K$ without explicitly restating reversibility. The algorithm above uses the reversible interpretation inherited from part (b). For an arbitrary differentiable recursion, one additionally needs access to output densities or a simulatable reverse kernel and regularity of the objective. Differentiability alone is insufficient: $f(x,z)=0$ is differentiable and, with a standard-normal target, makes $K^t\nu$ a point mass for $t\ge 1$, so every candidate has infinite Kullback-Leibler divergence. Thus a universal finite-objective algorithm cannot be justified solely from the literal differentiability assumptions. An invertible recursion with computable density transformations offers a different sufficient route, but invertibility is not stated here.

Let $k=m/2$. After offset subtraction, each summed image has mean $N$ analogue-to-digital units per detector pixel. For a constant detector conversion gain $g$ and independent Poisson distribution detected counts, $\mathrm{Var}(A)=\mathrm{Var}(B)=N/g$: summing $k$ independent frames adds both their means and variances. Here $N$ is the mean of the sum, not one frame.

Write $A=N+a$, $B=N+b$. At high counts, the delta method gives $A/B\simeq 1+a/N-b/N$. Independence therefore givesThus the formula is a high-signal gain estimation from flat-field ratios, not an exact identity for a ratio of random variables. A Poisson distribution denominator can even be zero at low counts. The exposures must be linear and unsaturated, offsets removed, and read noise negligible. Being close to saturation increases the signal but does not establish linearity; clipping or charge-induced correlations biases a variance estimate. A large ensemble of suitable detector pixels supplies the measured standard deviation. In normal photodetector terminology $g$ is collected Electrons per ADU, consistent with detected photons here only under the stated unit-yield convention; see the instrument gain definition.

For a discrete random variable with probabilities $p_i$, its Shannon entropy in bits is $H(X)=-{\sum }_i{p}_i{\log }_2{p}_i$, with $0{\log }_{2}0=0$. Gibbs inequality states thatwith equality exactly when $p=q$. If $q_i=0<p_i$, the left side is infinite. Otherwise apply $-\ln u\ge 1-u$ to $u=q_i/p_i$ for $p_i>0$:Equality requires $q_i=p_i$ on the support and no remaining mass outside it. In particular, comparison with the uniform probability distribution on two outcomes gives binary Shannon entropy at most $1$.

Write $q=1-p_1$. For $q>0$, grouping the last two outcomes givesEquality holds if $p_2=p_3$; if $q=0$ it also holds, and then both are zero. Thus equality holds exactly when $p_2=p_3$.

For the discrete memoryless channel, assume $\alpha ,\beta \ge 0$ and $\alpha +\beta \le 1$, and set $w=1-\alpha -\beta$. Both conditional output probability distributions have Shannon entropy $H(w,\alpha ,\beta )$. The third output has probability $\beta$ regardless of the input; the first two have total probability $1-\beta$. Their Shannon entropy is maximized by making them equal, which a uniform input achieves. Hence the channel capacity isand thereforeThe usual zero-times-log convention includes $\beta =1$, when $C=0$. If $w=\alpha$, the rows coincide and every input achieves zero mutual information; otherwise the uniform input is the unique maximizing input.

For $c>0$, Brownian scaling and the bijection $t\mapsto c^{2}t$ of $[0,\infty )$ give $Z\stackrel{d}{=}{c}^{-1}Z$, or $\boxed{cZ\stackrel{d}{=}Z}$. The supremum is a measurable extended random variable because continuity allows the supremum over rational times, and $Z\ge 0$.

For $z>0$, the function $F(z)=\mathbb{P}(Z\le z)$ must satisfy $F(z)=F(z/c)$ for every $c>0$, so it is constant on $(0,\infty )$. Letting $z\downarrow 0$ and $z\uparrow \infty$ shows $\mathbb{P}(Z=0)=\mathbb{P}(Z<\infty )$, hence $\boxed{Z\in \{0,+\infty \}\text{ almost surely}}$. This scaling argument alone does not decide the mass at zero; part (d) does.

Pointwise, an integer-valued nonnegative random variable has the representationwhen $T=k$, exactly the terms $n=0,\dots ,k-1$ equal one. Applying the Tonelli theorem, or the monotone convergence theorem to the finite sums of indicator random variables, gives the tail-sum formula for expectationBoth sides may be infinite; no integrability assumption is necessary.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers: