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It suffices to prove the assertion after replacing $\omega$ by $\min \{\omega ,\sqrt{\log n}\}$: that replacement gives a tighter interval. Thus assume $\omega \to \infty$ and $\omega =o(\log n)$, and putAt $p_{-}$, let $J$ count isolated vertices in $L$. The same indicator random variable calculation as for all isolated vertices givesBy the Chebyshev inequality, $J>0$ with high probability. Since $\ell \ge 2$ eventually, the marked vertices are then not all connected.

At $p_{+}$, the logarithmic-regime giant component result applies with $k=2$ and any fixed ${\gamma }_0\in (1/3,1/2)$. Its unique giant component misses $(1+o(1))n^{1/2}{e}^{-\omega /4}$ vertices. On the event that it misses at most twice that number, exchangeability makes the missed vertex set uniform conditional on its size. A union bound therefore givesThus all of $L$ is connected at $p_{+}$ with high probability.

For the uniform random graph process, assign uniform labels to the edges of the complete graph as independent random variables and reveal them in label order. The binomial random graph $G(n,p)$ is the prefix containing the $M(p)$ labels at most $p$, where $M(p)$ has a binomial distribution with parameters $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and $p$. Its variance at $p_{\pm }$ is $O(n\log n)$, whereas the gap between $\mathbb{E}M(p_{\pm })$ and the relevant endpoint $m_{\pm }$ is $(1+o(1))\omega n/8$. The Chebyshev inequality gives $M(p_{-})\ge m_{-}$ and $M(p_{+})\le m_{+}$ with high probability, with harmless integer rounding. The monotone graph property that all marked vertices are connected now impliesThe marked-set connectivity threshold is lower than the threshold for connecting every vertex, because only about $\sqrt{n}$ specified vertices need to avoid the small graph components.

If some choice $\omega$ produced a graph with largest component of a graph smaller than $2n/3$, the balanced component cut result with packing parameter one would give an empty graph cut whose sides both have at least $n/3$ vertices. For any fixed such vertex set partition, a uniform edge of the complete graph crosses with probabilityTo avoid crossing in row $i$, at least one of its $k$ offered edges must be noncrossing. The row's probability is $1-q^k\le 1-(4/9{)}^k$. The rows consist of independent random variables, so at $m=cn$ a union bound over at most $2^n$ vertex set partitions givesFor example, the positive integermakes this bound at most $e^{-n}$. Therefore every choice has a component of order at least $2n/3$, with probability tending to one. The simultaneous giant for fixed random-edge choice estimate includes choices made after seeing the entire array; it does not assume a particular online selection rule.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

Use the normal linear modelHere $Y$ is the response random vector, $X$ is the known design matrix, $\beta \in {\mathbb{R}}^p$ contains the unknown regression coefficients, and ${\sigma }^2>0$ is the common error variance. Conditional on $X$, the errors have normal distributions and are independent random variables. Require $\mathrm{rank}X=p\le n$ for identifiability of $\beta$ and invertibility of $X^{T}X$. Usually $n>p$ is needed to estimate the error variance from the regression residuals. An intercept, when included, is represented by a column of ones in $X$.

Group means are independent random variables with normal distributionsThe resulting univariate sampling distributions areThe same variance calculation holds for every $i\ge 2$, since the difference involves two independent group means. Therefore the standard errors satisfyReplacing $\sigma$ by a common estimated error standard deviation preserves this ratio. Although the individual group means are independent, the contrasts ${\hat{\alpha }}_i$ share the baseline mean and have covariance ${\sigma }^2/J$ for distinct $i\ge 2$.

A loss network models calls that need several resources simultaneously and are rejected, rather than queued, when insufficient capacity remains. Take finite resource and route sets. Let $A_{jr}\ge 0$ be the integer amount of resource $j$ required by a route-$r$ call, $C_j$ its capacity, andUnder fixed routing, each arriving call has a predetermined resource requirement. Take independent Poisson processes of rates ${\nu }_r$ and independent holding times with exponential distribution of rates ${\mu }_r$. A feasible arrival changes $n$ to $n+e_r$ at rate ${\nu }_r$; a departure changes it to $n-e_r$ at rate ${\mu }_r{n}_r$. Put ${\rho }_r={\nu }_r/{\mu }_r$ and assume each route uses a finite positive-capacity resource, so the state space is finite.

The product-form stationary distribution of a loss network isFor any feasible adjacent pair,These detailed balance equations prove stationarity and make the process a reversible Markov chain. Equivalently, Independent random variables with Poisson distributions of means ${\rho }_r$ are conditioned on satisfying the joint capacity constraints. The conditioning makes resource occupancies dependent even though the unconstrained counts are independent.

By Poisson arrivals see time averages, a route-$r$ arrival sees acceptance probability $G(C-A_r)/G(C)$, interpreting the numerator as zero for a negative capacity. Hence its blocking probability is $1-G(C-A_r)/G(C)$ and the expected value of its number in service is ${\rho }_{r}G(C-A_r)/G(C)$. This connects a stationary occupancy law to observable rejection and carried traffic.

The insensitivity of loss networks extends this occupancy formula to independent general holding-time distributions with the same expected values, under the usual fixed resource requirements and admission rule. Counts alone then need not be a Markov chain; residual holding times belong in a Markov description. The invariant occupancy formula survives. This is useful because detailed call-duration distributions can be difficult to estimate, while their means are much easier to measure.

In the slotted ALOHA model, a station retains one packet until it successfully transmits. Conditional on $(S_t,N_t)=(S,N)$, take fresh independent trials with a Bernoulli distribution of parameter $1/S$ for the $N$ backlogged stations. Let arrivals be independent across slots and independent of these trials, with Poisson distribution of mean $\nu$. New arrivals join the next slot's backlog. An idle slot serves nobody, exactly one attempt serves one packet, and a collision serves nobody. Consequently the backlog changes by arrivals minus the success indicator.

Under these independence assumptions, $(S_t,N_t)$ is a time-homogeneous Markov chain: the feedback probabilities and the independent arrival law depend only on its current state. Merely specifying Poisson arrival marginals would not suffice. For example, take $S_t=1$, $a=b=c=0$, $N_0=0$, and arrivals $Y_0=H$, $Y_1=K$, $Y_2=H$ for independent random variables $H,K$ with Poisson distributions. Histories with $(N_1,N_2)=(0,2)$ and $(2,2)$ both have current backlog two, but the next backlog is respectively two and four. The usual model therefore includes fresh arrivals as an assumption.

Write $p_0=(1-1/S{)}^N$ and $p_1=(N/S)(1-1/S{)}^{N-1}$. Set $p_1=0$ for $N=0$, and use their exact binomial distribution probabilities at $S=1$. Away from the clipping boundary, the conditional drifts areNear $S=1$, the latter must use $\max (1,S+d)-S$ for each update $d$, rather than $d$ itself.

When both coordinates are large with $N/S\to \kappa$, the Poisson limit theorem approximates the attempt count, which has a binomial distribution, by a Poisson distribution of mean $\kappa$. Thus $p_0\to e^{-\kappa }$ and $p_1\to \kappa e^{-\kappa }$. Rescaling state by $K$ and slot time by $K$ motivates the fluid approximation of slotted ALOHAThis is an interior fluid approximation of slotted ALOHA, not the exact conditional drift at a clipped boundary.

Here is one explicit set of sufficient conditions, independent of the arrival rate within the subcritical range:Put $c_0=2/(e-2)$. ThenHence $g$ is negative below one and positive above one: the controller decreases the attempt denominator when offered contention is too small, and increases it when contention is too large.

Use the ratio time change for a homogeneous fluid model, $du/dt=1/s$. The equations becomeFor $\kappa \ge 1$,since $g(\kappa )\ge 0$ and $A(1+c_0)>1$. Also $h(1)=\nu -e^{-1}<0$. Therefore $h$ is negative on $[1,\infty )$, and by continuity on $[\bar{\kappa },\infty )$ for some $\bar{\kappa }<1$. Its value at zero is $\nu \ge 0$, so nonnegative backlog is preserved.

The ratio remains bounded by $K=\max (\kappa (0),1)$ and eventually enters $[0,\bar{\kappa }]$: on the compact interval $[\bar{\kappa },K]$, its derivative is bounded above by a strictly negative number. Once inside it cannot cross upward. The bounded ratio and smooth coefficients make the transformed equations exist for all $u\ge 0$. In this region $g(\kappa )\le g(\bar{\kappa })<0$, so $s(u)$ decreases at least exponentially in $u$. The original time satisfies $t(u)={\int }_0^{u}s(v)dv$ and has a finite limit $T$; boundedness of $\kappa$ gives $n(u)=\kappa (u)s(u)\to 0$ as well. Thus every nonnegative interior fluid trajectory drains to the origin in finite fluid time. At the origin the ratio equation is undefined; the usual stopped fluid trajectory is held there afterward. A state with $s=0<n$ enters the interior under the continuous limiting boundary drift $\dot{s}=c>0$; the same argument then applies.

This proof supplies sufficient conditions, not a characterization of every stabilizing triplet. It also does not assert the positive recurrent Markov chain property of the original stochastic chain solely from the heuristic ODE.

If $\nu >e^{-1}$, the ALOHA throughput bound givesNo choice of the three feedback increments can make this fluid backlog drain, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$. For the displayed controller, the supercritical fluid trajectory is global and grows instead of draining. At $\nu =e^{-1}$ the ray $n=s>0$ consists of stationary fluid states, explaining why the strict load inequality matters.

Write $p=p(n)$ and let $T$ be the triangle count in a binomial random graph. For every three-element set $S$ of vertices, let $I_S$ be the indicator random variable that its three edges are present. ThenFor two distinct triples $S,R$, their indicator random variables are independent random variables unless they share an edge. Sharing just one vertex still leaves the sets of relevant edges disjoint. If they share an edge, their union has five edges, and thereforeThere are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)$ unordered pairs of distinct triangles in a graph sharing an edge: choose that edge, and then choose their two additional vertices. The variance of a sum consequently gives the exact formulaFor sufficiently large $n$, the hypothesis $np\to \infty$ ensures $p>0$, so $\mu >0$. Dropping negative terms and simplifying the binomial coefficients,Here $n^{2}p=n(np)\to \infty$ as well. The event $T\ge 2\mu$ implies $|T-\mu |\ge \mu$. Applying the Chebyshev inequality from part (i) gives the requested conclusion:Thus the triangle count is concentrated around its expectation throughout this range of the binomial random graph.

Write $m_X=\mathbb{E}{X}_1$ and $v_X=\mathrm{Var}(X_1)$. For the random sum of independent claims, conditioning on $N$ givesThe law of total expectation and the law of total variance therefore give the aggregate momentsThe first term in the variance measures variation of the individual claims at a fixed count; the second measures variation of the count itself. These formulas require the indicated moments to be finite.

For the moment-generating function, independent random variables givewhere $G_N(z)=\mathbb{E}[z^N]$ is the probability generating function. This identity holds wherever the expectations are finite; in particular a moment-generating function need not exist for positive $t$ for an arbitrary positive claim distribution. The empty sum for $N=0$ is zero.

The Hoeffding inequality states that if $X_1,\dots ,X_n$ are independent random variables with $a_i\le X_i\le b_i$ almost surely, then for $t>0$The same bound holds for the lower tail. Consequently the two-sided tail is at most twice this bound. When all ranges have length zero, the sum is deterministic and every positive tail probability is zero.

We construct an exponential packing of a Hamming cube. Choose an inclusion-maximal separated set $S$ in $\{-1,1{\}}^n$ with separation at least $r=n/8$ for the Hamming distance. It exists by a finite greedy procedure: keep adding any vertex at distance at least $r$ from all selected vertices until none remains. By maximality, every vertex is within distance strictly less than $r$ of some member of $S$. This is the principle that maximal separated sets give covers.

For a uniformly random vertex $U$ and any fixed vertex $s$, the mismatch indicators in different coordinates are independent Bernoulli random variables with success probability $1/2$. Hence $\rho (U,s)$ has the binomial distribution with parameters $n,1/2$. The lower-tail Hoeffding inequality givesThus the Hoeffding lower-tail bound for Hamming balls shows that every strict-radius ball $B_{<r}(s)$ has at most $2^n{e}^{-9n/32}$ vertices. Using a strict ball handles noninteger $n/8$ without changing the required separation.

The strict balls centered at $S$ cover the entire Hamming cube. Counting their union by the sum of their sizes yieldsThereforeIn particular this proves the assertion for every $n\ge 8$. Maximality here means that no further vertex can be added; finding a largest separated set is unnecessary.

A stationary causal time series exists. The infinite moving-average representationconverges in the sense of mean-square convergence because the driving variables are independent random variables and ${\sum }_j{4}^{-j}<\infty$. Shifting the series gives $X_t=\frac{1}{2}{X}_{t-1}+{\epsilon }_t$. Its expected value is zero and its autocovariance isThus it is a weakly stationary process; since the driving variables have a normal distribution, it is also a Gaussian process and a strictly stationary process. Throughout the stationarity discussion, take ${\sigma }^2>0$; degenerate zero noise permits trivial constant solutions.

The Box-Muller transform uses independent uniforms to setThe radius has the unit Rayleigh distribution, with density $r{e}^{-r^2/2}$ for $r>0$, and the angle is independently uniform on $[0,2\pi )$. Their joint density is $(2\pi {)}^{-1}r{e}^{-r^2/2}$. The Cartesian change of variables has absolute Jacobian determinant $r$, so the joint density of $(X_1,X_2)$ isThe factorization proves that the outputs are independent random variables, each with a standard normal distribution.

The moment-generating function of a random variable $Z$ is $m_Z(\theta )=\mathbb{E}[e^{\theta Z}]$, at the real values of $\theta$ for which that expectation is finite. For a standard normal distribution variable $X$, combining the exponentials in its probability density function yieldsThe last equality follows by scaling the Gaussian integral. The variables $X_i^2$ are independent random variables, so the expectation of their product factorizes. ThereforeThis is the moment-generating function of a chi-squared distribution with $n$ degrees of freedom. For positive $n$ the defining integral diverges at and above $\theta =1/2$.

Interpret $Y_n$ as the length of a shortest Euclidean travelling salesman tour, and take the points to be independent random variables with the uniform distribution on the square. We assume $n\ge 2$; a closed tour on one point has length zero, so the printed lower bound would be false for $n=1$. For two points the closed tour traverses the joining segment twice.

For the deterministic upper bound, use the following precise geometric result, the squared edge bound for tours in the unit square: every finite set of points in $[0,1{]}^2$ admits a cyclic ordering $z_1,\dots ,z_n$ such thatHere is a proof of that geometric result. First consider a right triangle with hypotenuse endpoints $a,b$, hypotenuse length $c$, and right-angle vertex $v$. The quadratic path bound in a right triangle supplies a path from $a$ to $b$ through any finite set inside the triangle whose sum of squared edge lengths is at most $c^2$. For an empty set use the edge $ab$. For a single point $z$, use $a,z,b$: the triangle lies in the disk with diameter $ab$, so $\angle azb\ge \pi /2$ and the law of cosines gives $\Vert a-z{\Vert }^2+\Vert z-b{\Vert }^2\le c^2$.

For several points, draw the altitude from $v$ to the hypotenuse. It splits the triangle into two smaller right triangles with hypotenuses $av$ and $vb$. Their squared hypotenuse lengths sum to $c^2$ by the Pythagorean theorem. Repeated altitude subdivision eventually makes every cell small enough to contain at most one of the given distinct points: each child is similar to the original triangle and its diameter contracts by at most the fixed factor $\max (\Vert a-v\Vert ,\Vert b-v\Vert )/c<1$. Construct the base-case paths and combine them from the bottom of this finite subdivision tree. The two child paths concatenate at $v$. If $v$ is an auxiliary point, delete it by joining its two neighbours directly. Both neighbours lie in the right-angle sector of the parent triangle at $v$, so their angle at $v$ is at most $\pi /2$. The law of cosines now shows that this shortcut does not increase the sum of squared edge lengths. The combined path still has endpoints $a,b$ and squared cost at most $c^2$.

One can first put the given points in generic positions, avoiding subdivision boundaries, and then pass to a limit. There are only finitely many possible visiting orders, so a subsequence keeps the same order and its squared cost converges; this also handles coincident locations. No assumption about a minimum separation remains in the result.

Split the square along a diagonal $ab$. Apply the triangle result in each half, following one path from $a$ to $b$ and the other back to $a$. Each half contributes at most $\Vert a-b{\Vert }^2=2$. Remove either diagonal endpoint if it was auxiliary. At a square corner all neighbours lie in a sector of angle $\pi /2$, so the same squared-cost shortcut applies. This gives the asserted cyclic ordering with total squared cost at most four. It does not assert that the tour minimizing ordinary length itself has this squared-edge property. The Cauchy-Schwarz inequality applied to the tour supplied by the theorem provesThus the upper bound is deterministic, independently of the sampling law.

For the lower bound, put $R_i={\min }_{j\ne i}\Vert X_i-X_j\Vert$, the nearest neighbour distance at $X_i$. Both edges incident to $X_i$ in any closed tour have length at least $R_i$. Summing over vertices and then dividing by two gives $Y_n\ge {\sum }_i{R}_i$. Conditional on $X_i=x$, the union bound and the uniform distribution giveBoundary clipping only makes the ball smaller. Apply the layer-cake representation of the expected value, restricting the integral to $0\le t\le 1/(2\sqrt{n})$:Linearity of the expected value now yieldsThe unspecified sampling phrase in the paper must therefore be understood as independent uniform sampling: an arbitrary distribution concentrated in a tiny region cannot satisfy this lower bound.

Write $V=\{1,\dots ,p\}$. The global Markov property for a directed acyclic graph says that, for every three pairwise disjoint vertex sets $A,B,C$,Thus graphical D-separation implies probabilistic conditional independence. Causal minimality means that $P$ has this global Markov property for a directed acyclic graph, but has it for no proper subgraph of $\mathcal{G}$ obtained by deleting edges while keeping all vertices. Faithfulness of a directed acyclic graph means that the displayed implication is an equivalence: the conditional independences of $P$ are exactly the D-separations of $\mathcal{G}$. In particular it includes the global Markov property for a directed acyclic graph.

A causally minimal distribution can nevertheless have an extra independence. Here is a causally minimal nonfaithful Gaussian model. Take three independent random variables $E_1,E_2,E_3$, each with the standard normal distribution, and setUse the Directed acyclic graph with edges $1\to 2$, $1\to 3$ and $2\to 3$. The equations give a directed acyclic graph factorization, so the global Markov property for a directed acyclic graph holds. Alternatively this complete acyclic graph imposes no nontrivial D-separations. But $Z_3=E_2+E_3$ is independent of $Z_1=E_1$, even though the direct edge $1\to 3$ prevents their D-separation. Hence faithfulness of a directed acyclic graph fails.

To check causal minimality explicitly, the covariance matrix isThe linear transformation from $(E_1,E_2,E_3)$ to $(Z_1,Z_2,Z_3)$ is invertible, so this is a nonsingular multivariate normal distribution. Deleting $1\to 2$ would require $Z_1\perp Z_2$, whereas $\mathrm{Cov}(Z_1,Z_2)=1$. Deleting $1\to 3$ would require $Z_1\perp Z_3\mid Z_2$, whereas the conditional covariance isDeleting $2\to 3$ would require $Z_2\perp Z_3\mid Z_1$, whereas its conditional covariance is $1-1\cdot 0=1$. By Gaussian conditional independence, each nonzero conditional covariance rules out the required independence. Every proper subgraph is contained in a graph obtained by a single edge deletion, and further deletion can only add Markov restrictions. Thus no proper subgraph is Markov for $P$, proving causal minimality.

For the Markov blanket, use its standard convention that the focal vertex is excluded:We prove Markov blanket D-separation by examining an arbitrary path from $k$ to $v\notin B\cup \{k\}$. If its first edge points into $k$, its next vertex is a parent of $k$, hence a conditioned noncollider in $B$, and the path is blocked. Otherwise the path starts $k\to h$ at a child $h\in B$. If the next edge points out of $h$, then $h$ is again a conditioned noncollider. If it points into $h$, the path begins $k\to h\leftarrow u$. Although conditioning on the collider $h$ opens that local segment, $u$ is another parent of $h$, hence belongs to $B$. The outgoing arrow $u\to h$ makes $u$ a noncollider on the path, so conditioning on $u$ blocks it. Since $v$ is outside $B$, none of these blocking vertices can be the endpoint.

Every path is therefore blocked by $B$. Applying the global Markov property for a directed acyclic graph givesThis proves the requested screening property without any positivity assumption on $P$.

Finally assume faithfulness of a directed acyclic graph and let $A$ have the stated screening property. Then $k$ and every vertex outside $A\cup \{k\}$ must be D-separated by $A$. If a parent or child $h$ of $k$ were missing from $A$, the one-edge path between them would be active, a contradiction. Thus every parent and child is in $A$. If another parent $u$ of a child $h$ were missing from $A$, the path $k\to h\leftarrow u$ would be active: its only interior vertex is the collider $h$, already in $A$. This is again a contradiction. ConsequentlyUnder faithfulness the graphical blanket is contained in every screening set. This is the minimal Markov blanket under faithfulness; the preceding argument also shows that the blanket itself is a screening set.

Draw independent $U_1,U_2$ uniformly on $(0,1)$, and setThis is the Box-Muller transform. The endpoint $U_1=0$ has probability zero; an implementation should avoid evaluating its logarithm.

For $r\ge 0$, $P(R\le r)=1-e^{-r^2/2}$, so $R$ has density $r{e}^{-r^2/2}$. It is independent of the angle, which has a uniform distribution on $[0,2\pi )$, and their joint density is $(2\pi {)}^{-1}r{e}^{-r^2/2}$. Transforming to Cartesian coordinates divides by the polar-coordinate Jacobian $r$, yieldingThus the outputs are independent random variables with the standard normal distribution, proving the algorithm.

Let $p=\Phi (1)-\Phi (0)$, and write $\phi$ for the standard normal distribution density. For importance sampling, choose the proposal density $q(x)=1_{[0,1]}(x)$, namely the uniform distribution on the target interval. Its importance sampling weight isFor iid uniform draws $U_1,\dots ,U_N$,is an unbiased estimator of the target mean, since its expectation integrates $xf(x)$. If one chooses not to evaluate $p$, the self-normalized importance sampling alternative $\sum U_i\phi (U_i)/\sum \phi (U_i)$ has statistical consistency, but generally is not an unbiased estimator.

Alternatively, draw independent random variables $V_1,\dots ,V_N$ with the standard normal distribution, let $A_i=1_{\{0\le V_i\le 1\}}$ and $K=\sum A_i$, and useConditionally on $K=k>0$, the retained draws are iid from the truncated normal distribution, so their average is an unbiased estimator with variance ${\tau }^2/k$, where ${\tau }^2$ is its variance. If $K=0$ the ratio is undefined; one can continue drawing until a prescribed number of retained observations is obtained.

Here importance sampling uses every draw inside the interval, whereas only $p\approx 0.3413$ of the normal proposals are retained. The advantage can also be quantified. Integration using ${\phi }^{\prime }(x)=-x\phi (x)$ givesFor uniform importance sampling,For the retained-draw ratio with a fixed total proposal budget, its asymptotic variance coefficient is ${\tau }^2/p\approx 0.233347$. Thus this importance choice has substantially smaller error for the same proposal count; it also avoids wasting about two thirds of the standard normal distribution draws. This comparison is specific to this well-matched importance sampling proposal; arbitrary proposal choices do not share this guarantee.

On $x>0$, the exponential proposal is the proper density $g(x)=e^{-x}{1}_{\{x>0\}}$; its integral is $1$. The word “improper” in the question is therefore not applicable on its stated positive support.

The target-to-proposal ratio isIts maximum occurs at $x=1$, so the smallest valid envelope constant isOne implementation draws $Z=-\log U_1$ and accepts it when $U_2\le e^{-(Z-1{)}^2/2}$, for independent random variables with the uniform distribution on $(0,1)$. Accepted observations have the positive half-normal density by the usual rejection sampling calculation.

Use the ergodic theorem for a positive Harris recurrent Markov chain: for an invariant distribution $\pi$ and an integrable function $h$, the time average converges almost surely to $\int hd\pi$ under the usual ergodicity conditions. This is the relevant theorem for correlated Metropolis–Hastings algorithm output, rather than the strong law of large numbers. Here $h(x)=x_1$ is integrable. Indeed, puttingwe have $V(x)\ge (x_1-4{)}^2+(x_2-4{)}^2-32$, which bounds the unnormalized density by a constant times an integrable Gaussian density and also gives finite first moments.

For a proposal scale $s>0$, initialize $X_0$ anywhere in ${\mathbb{R}}^2$. At each step draw independent random variables $Z_1,Z_2$ with the standard normal distribution and an independent uniform $U$, propose $Y=X_t+s(Z_1,Z_2)$, and accept whenOtherwise set $X_{t+1}=X_t$. The normal proposal is symmetric, so its density cancels; the target normalizing constant $c$ also cancels. Include every state in the average, including repeated states after rejection. The ergodic theorem for a positive Harris recurrent Markov chain then givesA fixed discarded initial segment does not change this limit.

Small proposal variance gives high acceptance but tiny moves and strong serial dependence. Large variance gives more ambitious moves, but many proposals enter very low-density regions and are rejected, creating long runs at one state. An intermediate scale should be assessed by exploration and effective sample size of a Markov chain, not acceptance rate alone.

This density has two modes, at $(2+\sqrt{3},2-\sqrt{3})$ and their interchange. To see this, stationary points satisfy $x_1(1+x_2^2)=x_2(1+x_1^2)=4$, implying either $x_1=x_2$ or $x_1{x}_2=1$. The unequal solutions have $x_1+x_2=4$ and are local minima of $V$; the equal solution is a saddle. Mode switching is therefore a material part of proposal-scale selection: a chain confined to one mode can have high acceptance and a misleading finite-run estimate.