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A Gibbs sampler updates one coordinate by drawing from its full conditional distribution, leaving the others unchanged and using their current values. Each coordinate kernel $K_i$ preserves the target joint law $P$: integrate first over the unchanged coordinates, then over the old coordinate, and finally draw the new coordinate from the same conditional distribution. The resulting joint law is again $P$. In the discrete notation, if two states share the remaining coordinates,which is symmetric in $x_i,x_i^{\prime }$. Thus a coordinate update satisfies detailed balance. A systematic full sweep has kernel $K_1\cdots K_k$ and remains invariant because each factor preserves $P$. A fixed-probability random-scan mixture also preserves $P$; the mixture is reversible, whereas a systematic composition need not be.

In the autologistic binary-image model, count each neighboring pair once. Changing one pixel from zero to one raises $n_1$ by one and $n_2$ by the number $z_{ij}$ of neighboring ones. The ratio of its two conditional weights is $\exp (\alpha +\beta z_{ij})$. ThereforeFor single-site Gibbs sampling, select a pixel, compute its current neighbor sum and replace it by a draw from a Bernoulli distribution with this probability. One may use random scans or complete ordered sweeps. The normalizing constant is not needed. For finite $\alpha ,\beta$, every conditional probability lies strictly between zero and one; the finite chain is irreducible and has self-transitions, so its stationary law is unique and iterating converges to it.

For checkerboard Gibbs sampling, no edge joins two black pixels or two white pixels. Conditional on the whites, the black-pixel law factorizes:Draw all black pixels independently with these probabilities, then draw all white pixels independently conditional on the newly drawn blacks. Each color block is a full conditional update, so the two-block sweep preserves the joint target. Pixels of the same color can be updated together because their conditional distribution factorizes.

For the noisy image, the log likelihood contributes $-\frac{1}{2}{\sum }_v(y_v-x_v{)}^2$. Since $x_v^2=x_v$, dropping terms independent of $x$ givesThe Gaussian-noise posterior for an autologistic image therefore has conditional probabilitiesUse this probability in both single-pixel and sequential color-block schemes. The independent observation factors change only the local field, so conditional independence within each color is retained. The same invariance proof then targets the posterior distribution rather than the prior.

The expected check loss is finite because ${\rho }_{\tau }(u)\le |u|$. It is also Lipschitz in its location argument, with constant at most one. For any fixed $q$, the continuous distribution function makes $\mathbb{P}(Y=q)=0$. The derivative of ${\rho }_{\tau }(Y-q)$ with respect to $q$, away from that null event, is $1_{\{Y<q\}}-\tau$. Difference quotients are bounded by one, so the dominated convergence theorem givesThe continuous, strictly increasing distribution function has limits zero and one. Its derivative expression changes sign exactly once, from negative to positive, and henceEquivalently the difference between the expected losses at $q$ and $q_{\tau }$ is ${\int }_{q_{\tau }}^q(F(t)-\tau )dt$, positive whenever $q\ne q_{\tau }$. This proves population quantiles minimize check loss directly.

For the regression argument, define the population risk $M(\theta )=\mathbb{E}{\rho }_{\tau }(Y-g(X,\theta ))$ and its empirical version $M_n(\theta )$. Boundedness of $g$ and integrability of the error give integrability of $Y$ and of all these losses. Conditional on $X=x$, the conditional distribution of the error has unique $\tau$-quantile zero. Applying the preceding calculation conditionally shows that its expected loss is uniquely minimized when the fitted displacement $g(x,\theta )-g(x,{\theta }_0)$ equals zero. Thuswith strict inequality whenever that displacement is nonzero on an event of positive probability. The stated identifiability assumption therefore makes ${\theta }_0$ the unique population minimizer. This is conditional quantile identification.

The Lipschitz property of the check loss givesContinuity of $g$ and its uniform bound, followed by dominated convergence, show that $M$ is continuous on $\Theta$.

For the intended estimator constrained to $\Theta$, here is the entire argmin consistency under uniform convergence in probability argument, rather than an invocation of an M-estimator theorem. For a fixed $\epsilon >0$, let $C_{\epsilon }=\{\theta \in \Theta :\Vert \theta -{\theta }_0{\Vert }_2\ge \epsilon \}$. If this set is empty, there is nothing to prove. Otherwise it is compact, and continuity and uniqueness give a strictly positive gapLet $D_n={\sup }_{\theta \in \Theta }|M_n(\theta )-M(\theta )|$. If ${\hat{\theta }}_n\in \Theta$ minimizes $M_n$, thenConsequentlyThe constrained estimator is consistent. Existence of a constrained minimum follows from continuity of the sample criterion and compactness. The argument applies to any measurable choice of minimizer.

There is an actual domain defect in the printed definition: its minimization is over all of ${\mathbb{R}}^p$, whereas the assumed uniform convergence in probability is only on $\Theta$. That literal unconstrained consistency claim is false. The preceding conclusion needs minimization over $\Theta$, or another assumption that confines the selected minimizers to a set where the convergence and separation arguments apply. No change of question heading is needed to make this qualification explicit.

Here is a counterexample satisfying even identifiability over the whole line. Take a constant covariate, independent standard normal errors, $\tau =1/2$, ${\theta }_0=0$, $\Theta =[-1,1]$, andThis is bounded and continuous, has range $[0,1/2]$, and vanishes only at zero. The conditional error distribution function is continuous and strictly increasing with median zero. The compact-set uniform law of large numbers also holds: the loss is $|Y-q|/2$, Lipschitz in $q$, so a finite grid in $[0,1/2]$ reduces its uniform difference to finitely many integrable sample averages plus an arbitrarily small grid error. The weak law of large numbers then gives the asserted convergence.

Let $m_n$ be a sample median, and put $c_n=\min (1/2,\max (0,m_n))$. Absolute loss over the range $[0,1/2]$ is minimized at $c_n$. Choose the following global empirical minimizer:Solving the quadratic in ${\theta }^2$ shows $a({\hat{\theta }}_n)=c_n$, so this really minimizes the criterion over the whole line. For odd $n$, symmetry and continuity give $\mathbb{P}(m_n>0)=1/2$. Moreover $m_n\to 0$ in probability: for each fixed positive $\delta$, the proportion of observations below $\delta$ converges to a number greater than one half, and the analogous proportion below $-\delta$ converges to a number less than one half. Thereforealong odd $n$. The selected unconstrained estimator is not consistent. This illustrates why a global argmin may escape a compact convergence set.

Work conditionally on a positive-probability atom of ${\mathcal{F}}_{T-1}$. Let $Z$ have the conditional distribution of $S_T$ there, and let $Z^{\prime }$ be an independent copy of that conditional law. The finite sample space makes all expectations finite. ThenStrict increase of $g$ makes the integrand nonnegative, and strictly positive whenever $Z\ne Z^{\prime }$. The conditional variance is positive by part (c), so the conditional distribution is nondegenerate and $\mathbb{P}(Z\ne Z^{\prime })>0$. Thus the numerator of the hedge ratio is strictly positive on every such atom. The denominator is also positive, givingThis is strict positive covariance with an increasing payoff. The independent copy is taken from the conditional distribution, not from an unrelated unconditional distribution.

Multiple imputation by chained equations specifies one conditional distribution for each incomplete variable. In general, let $V_1,\dots ,V_q$ be the incomplete variables, $Z$ the fully observed predictors, and ${\varphi }_j$ the parameters of a model $p_j(V_j\mid V_{-j},Z;{\varphi }_j)$. The notation $V_{-j}$ means all variables except $V_j$. The models should respect each variable's support and contain the terms needed by the planned analysis.

First choose suitable models and initialise only the missing entries with admissible provisional values. One sweep updates $j=1,\dots ,q$ in turn. At update $j$, fit its model to records whose $V_j$ is observed, using the latest filled-in predictors. DrawThe first draw represents parameter uncertainty through a Bayesian posterior; the second represents residual uncertainty through a posterior predictive distribution. Replace the missing entries with these draws, leaving every genuinely observed entry fixed. Repeat whole sweeps until the chain has settled, assessing traces, mixing and plausibility. Obtain $M$ completed datasets from independently initialised chains, or sufficiently separated draws after convergence. Both parameter and missing-value uncertainty must be propagated; repeated deterministic prediction is not multiple imputation.

Here use a logistic regression for $C$ conditional on $P$, age and sex. For $P$, use an appropriate discrete count model conditional on $C$, age and sex, such as a suitably checked negative binomial regression, allowing enough flexibility for skewness and nonlinear effects. Include interactions or transformations needed by the analysis in the imputation models. If condom use is structurally undefined for a record, preserve that applicability rule rather than treating it as ordinary missing data.

Impute the eligible survey information before discarding records solely because the observed $P$ is absent. In each completed dataset select its own $P>1$ records and fit the requested logistic regression. This allows a record with missing $P$ to contribute to the target group in some imputations and not others. Models must respect any known eligibility restrictions; if eligibility was independently known, use that information.

For a scalar regression coefficient, let ${\hat{Q}}_m$ be its estimate and $U_m$ its estimated variance in completed dataset $m$. Rubin's rules giveThe standard error is $\sqrt{T}$. Use the corresponding multiple-imputation confidence interval, rather than treating one completed dataset as fully observed. For vector coefficients replace the squared differences by outer products and $U_m$ by the covariance matrix. This combines within-imputation and between-imputation uncertainty.

Ordinary multiple imputation by chained equations relies on missing at random, conditional on the variables included in its models, together with adequate model specification and suitable conditional distributions. For Bayesian inference that ignores the response mechanism, the usual sufficient condition also includes distinct parameters for data and missingness models, with an appropriately separate prior.

The suspected dependence on withheld responses makes the usual missing-at-random assumption doubtful. An accurately fitted observed-data model cannot, on its own, establish what the nonresponders would have reported. Use substantive judgement and sensitivity analysis under plausible missing not at random departures, for example a pattern-mixture sensitivity analysis shifting the imputed condom log odds or partner-count distribution relative to the observed-data prediction. Report how the final association changes. Additional informative measurements may make missing at random more credible, but cannot verify it.

Treat the extra measurements as auxiliary variables for multiple imputation. Include alcohol consumption and disease history as predictors in both the $C$ and $P$ conditional distributions, together with age, sex and the other incomplete response. If these auxiliary variables themselves have missing entries, give them appropriate models and update them within the same multiple imputation by chained equations sweeps. Preserve observed entries throughout.

Useful auxiliary measurements can improve predictions, precision and the plausibility of missing at random. Their association with the missing responses reduces uncertainty in the imputed values; association with nonresponse may explain some selection that otherwise depends on unobserved values. This can reduce selection bias under the enlarged missing at random model and reduce between-imputation variance. It does not guarantee elimination of missing not at random dependence. They need not be added to the final substantive logistic regression merely because they help imputation: the final analysis should still estimate its intended association.

Interpret the contests as standard all-pay auctions: the highest effort among entrants wins, ties are shared uniformly, and an unentered contest awards no prize. The printed question does not specify a prize allocation rule; the highest-effort convention is the one used for standard all-pay contests in the course author's 2014 lecture slides. Under this convention, we can characterize the unique symmetric participation probabilities and bid marginals. Uniqueness of the entire joint mixed strategy requires a further restriction on dependence, as explained below.

Let $q_j$ be the probability that a player omits contest $j$. Since every player enters exactly two contests, $q_1+q_2+q_3=1$. Let $G_j(b)$ be the probability that a rival is absent from contest $j$ or enters it with effort at most $b$. Rivals' strategy draws are independent between players. For a positive bid $b$ outside a measure atom, the expected payoff from this contest isAn all-pay auction cannot have a positive-effort measure atom in a symmetric equilibrium: slightly overbidding that measure atom gives a positive discrete increase in the winning probability at an arbitrarily small extra cost. Nor can active bids have a measure atom at zero when entry has positive probability, because a small positive bid beats tied zero bids. Gaps inside the active effort support are impossible: moving a bid from the top of a gap to just above its bottom preserves its winning probability and lowers its cost. The effort support starts at zero, because lowering its positive lower endpoint would preserve the chance that every rival is absent.

The all-pay indifference equation with random entry consequently gives the maximal per-contest payoffEvery $q_j$ lies strictly between zero and one. First, if $q_j=1$, the other two contests have certain entry and zero per-contest payoff, while deviating into the unused contest wins a positive prize. This contradicts equilibrium. Next, if $q_j=0$, the remaining omission probabilities sum to one and neither can equal one, so both are positive. Contest $j$ has zero per-contest payoff, whereas both other contests have strictly positive payoffs. Every pair containing $j$ is then worse than omitting it and entering the other two, contradicting certain entry in $j$.

Every omission therefore occurs with positive probability. The three entered pairs must give the same maximal payoff, which forces $u_1=u_2=u_3=u>0$. Normalizing the omission probabilities gives the two-of-three all-pay participation equilibrium:Conditional on entering contest $j$, the effort has distribution functionextended by zero below this interval and one above it. The inequalities $q_1\le q_2\le q_3$ show that the larger prizes are entered more often.

To construct an equilibrium, omit $j$ with probability $q_j$, then draw the two active efforts independently with their respective conditional distributions $H_k$. Every bid in a contest's effort support earns $u$; a bid above $w_j-u$ earns at most $w_j-b<u$. Thus no effort deviation or choice of a different pair improves on total payoff $2u$. This proves existence and verifies the Nash equilibrium without relying only on the indifference equations. The arguments above also prove uniqueness of the omission probabilities and the per-contest marginal distributions.

For the full joint strategy law, however, the printed uniqueness claim is too strong. Given an entered pair $k,\ell$, either use two independent uniform random variables $U,V$ and bids $(H_k^{-1}(U),H_{\ell }^{-1}(V))$, or use a single uniform $U$ and bids $(H_k^{-1}(U),H_{\ell }^{-1}(U))$. These are different copulas with the same conditional marginals. A fixed deviation's expected additive payoff only uses the rivals' per-contest marginal distributions, so both constructions remain Nash equilibria. This is the marginal-equivalent equilibria in additive contests phenomenon. The participation probabilities and bid marginals are unique; the full joint mixed strategy is not unique unless a dependence convention is imposed. Independent conditional sampling gives one canonical representative.

First, the absolute values in the printed limit are an error: the correct long-run variance of a stationary process is the signed sum of autocovariances. Directly,soEach coefficient tends to $1$ and has absolute value at most $1$. Absolute summability and the dominated convergence theorem therefore giveThe sum of absolute values is an upper bound, not the general limit. For a concrete counterexample to the printed assertion, take $X_t={\eta }_t-{\eta }_{t-1}$ with unit-variance iid noise. Then $\gamma (0)=2$, $\gamma (\pm 1)=-1$, and other covariances vanish. The absolute sum is $4$, but ${\overline{X}}_T=({\eta }_T-{\eta }_0)/T$, so $T\mathrm{Var}({\overline{X}}_T)=2/T\to 0$.

For correlated data the central limit theorem can still hold under suitable strong mixing of a stationary process and moment conditions, but the limiting variance is the long-run variance of a stationary process, rather than the one-observation variance. When it is positive,Positive serial dependence usually increases the standard error, while negative dependence can reduce it. The approximate effective sample size of a stationary sample is $T{\gamma }_X(0)/{\sum }_h{\gamma }_X(h)$ when that denominator is positive. A long-memory time series can require a different normalization or a different limit law. Absolute covariance summability alone does not prove a CLT: if $X_t=V{\eta }_t$ with a common independent random scale $V$ taking values $1$ and $2$ with equal probabilities and iid ${\eta }_t$ with the standard normal distribution, the off-diagonal covariances are zero, yet $\sqrt{T}{\overline{X}}_T$ has the nonnormal scale-mixture law $VN(0,1)$. The example is not an ergodic stationary process. If the long-run variance of a stationary process is zero, the usual nondegenerate square-root-$T$ CLT is unavailable.

For the causal AR(1), $|\varphi |<1$ andThe conditional distribution follows because the current innovation is independent of the past.

For a fixed known $\varphi$, condition on the observed $X_1$ and use the $n=T-1$ transitions $t=2,\dots ,T$. Their conditional maximum likelihood criterion is, up to constants,Differentiating in $\mu$ yieldsSince $X_t-\varphi X_{t-1}=(1-\varphi )\mu +{\epsilon }_t$,Thus it is an unbiased estimator, even conditionally on $X_1$, andIt has statistical consistency with mean-square convergence, and the iid-noise strong law of large numbers also gives almost-sure statistical consistency. Its conditional distribution is exactly normal with the displayed mean and variance. If $X_0$ is also observed and all $T$ transitions are used, replace $T-1$ by $T$.

The fixed-$\varphi$ qualification is necessary for the exact finite-sample claims. If $\varphi$ is jointly estimated, conditional likelihood function is linear regression with an intercept: writing ${\bar{X}}_{-}=n^{-1}{\sum }_{t=2}^T{X}_{t-1}$ and ${\bar{X}}_{+}=n^{-1}{\sum }_{t=2}^T{X}_t$, the unconstrained estimators areThis ratio is not generally an unbiased estimator and does not have the preceding finite-sample variance. Under the usual stationary regression conditions it has statistical consistency; its asymptotic variance is ${\sigma }_{\epsilon }^2/(1-\varphi {)}^2$. Indeed, it differs from ${\bar{X}}_{+}$ by $\hat{\varphi }(X_T-X_1)/(n(1-\hat{\varphi }))$, an asymptotically negligible endpoint term. Profiling an unknown innovation variance does not change the fixed-$\varphi$ estimate of $\mu$.

Write $z=(Z_{s_1},\dots ,Z_{s_n}{)}^T$, let ${\Sigma }_{ij}=C(\Vert s_i-s_j\Vert )$, let $c_i=C(\Vert s_i-s_0\Vert )$, and put $v_0=C(0)$. Assume the covariance matrix $\Sigma$ is invertible. Because the mean is known to be zero, every linear predictor $w^{T}z$ is unbiased. Its mean squared prediction error isCompleting the square yieldsThe positive-definite matrix property makes the last term nonnegative, proving the simple kriging formulasThe residual is uncorrelated with $z$. Since the joint law is multivariate normal, it is independent of $z$, so the same predictor and variance describe the conditional distribution. Consequently a 95-percent prediction interval isThere is no constraint that the weights sum to one: that is required for an unknown constant mean in ordinary kriging, not for the stated known-zero-mean problem. If $\Sigma$ is singular, use a Moore-Penrose inverse with the corresponding covariance compatibility condition. At an already observed location prediction of that very same field value has zero error; predicting an independent noisy replicate is a different target. Estimated covariance parameters make this a plug-in interval and generally add uncertainty that the formula does not include.