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If two parameters are separated by at least $2r$ in a metric, any estimator gives a test by deciding whether its distance from the first parameter is below $r$. For $0<\eta <1$, the event $|Z-1|\le \eta$ and Markov inequality give the displayed minimax risk bound, where $Z$ is the product likelihood ratio. The unrestricted claim for every positive $\eta$ is false when both factors on the right are negative.

Nonparametric statistics studies statistical inference without restricting the unknown probability distribution or target function to a fixed finite-dimensional parametric statistical model. In density estimation and nonparametric regression, the number of features that an estimator can represent may grow with the sample size. Rank tests offer another approach: the probability distribution of their test statistic under the null hypothesis can be obtained without specifying the common continuous probability distribution.

"Nonparametric" does not mean free of assumptions or tuning choices. A kernel density estimator still requires a kernel for density estimation and a smoothing bandwidth, and its interpretation depends on assumptions about the statistical sample. The freedom is useful when a prescribed shape, such as a single normal distribution, would hide features that the observations might reveal.

Use the first order statistic and correct its bias:This is a measurable function of the sample. Under $U_{\theta }^n$, write $Z=X_{(1)}-\theta$. It lies in $[0,1]$, and independence givesFor a bounded nonnegative random variable, integrating the identities $Z={\int }_0^1{1}_{\{Z>u\}}du$ and $Z^2={\int }_0^{1}2u{1}_{\{Z>u\}}du$ proves the tail integral formula for moments by Tonelli's theorem. HenceThus $\hat{\theta }$ is an unbiased estimator. Its quadratic risk is the variance of $Z$, independent of $\theta$:Taking the supremum over all real $\theta$ and then the infimum over estimators givesThe final inequality follows from $n^3\le (n+1{)}^2(n+2)$. Together with part (i), this proves that the minimax risk has order $n^{-2}$ for all $n\ge 1$. In the last line of the source, the request for the upper-bound constant calls it $C$; here it is denoted $C^{\prime }$ consistently with the upper-bound inequality.

Use two translated uniform distributions, with parameters $0$ and $a=1/(2n)$. Their distance in the usual metric is $a$, so take the separation radius $\delta =a/2=1/(4n)$ and the loss function $\Phi (r)=r^2$. The supplied testing reduction, equivalently the Le Cam two-point lemma, and part (a)(i) implyTo see the reduction directly, turn any estimator into a binary hypothesis test by choosing the closer of $0$ and $a$. If the test is wrong, the absolute estimation error is at least $a/2$. The worst quadratic risk is at least its average over these two parameters, giving precisely this bound.

Under Lebesgue measure on ${\mathbb{R}}^n$, the two sample probability density functions are the indicators of the cubes $[0,1{]}^n$ and $[a,a+1{]}^n$. Their intersection is $[a,1{]}^n$, of volume $(1-a{)}^n$. Thereforewhere the inequality is Bernoulli's inequality. ConsequentlyThis also fits the Hellinger distance approach: the two one-sample affinities give $H^2(U_0,U_a)=a$, while equal-height indicator densities imply $V(U_0^n,U_a^n)=H^2(U_0^n,U_a^n)$. Part (a)(ii) thus yields the same bound $V\le na=1/2$. The change of support produces Squared Hellinger distance proportional to $a$, explaining the parameter scale $a\asymp n^{-1}$ and quadratic risk scale $n^{-2}$.

Let $Y_i=\theta (X_i)f(X_i)/g(X_i)$. The independent summands have expectation $\mu$ and second momentWhen this integral is finite, independence givesIf the second-moment integral diverges, the estimator can still be unbiased and consistent, but its variance is infinite; the finite-variance assumption cannot be omitted.

Write $f=\tilde{f}/Z_f$ and $g=\tilde{g}/Z_g$. If the constants needed to evaluate $f/g$ are unavailable, use the self-normalized importance sampling estimatorThe unknown common factor $Z_g/Z_f$ cancels. Under full target probability support and integrability, the numerator and denominator sample means converge to $\int \tilde{f}\theta /Z_g$ and $\int \tilde{f}/Z_g$, respectively, whose ratio is $\mu$. The same construction applies if only one normalization is unknown by taking the other probability density function already normalized. Sampling from $g$ itself is assumed available, even when its normalization is not explicitly known.

This ratio estimator is generally biased at finite $n$, unlike the ordinary normalized-weight estimator. Also, its variance is not the formula in the next part: that formula concerns the ordinary unbiased estimator. Full probability support for $f$ is needed here because the denominator estimates the entire target normalizing integral, not just the probability support of $f\theta$.

Assume independent samples $X_i\sim g$, $\int f|\theta |<\infty$, and $g>0$ wherever $f\theta$ can contribute. The importance weight is $w(x)=f(x)/g(x)$, and the ordinary importance sampling estimator isIts expectation is $\int \theta (x)f(x)dx=\mu$, by cancellation of $g$ in the expectation integral. The strong law of large numbers gives almost-sure convergence to $\mu$. Common probability support is sufficient; the precise support condition for importance sampling only requires covering the nonzero integrand for this particular expectation.

Put $A=\int |\theta (x)|f(x)dx$, with $0<A<\infty$. The Cauchy-Schwarz inequality applied to $|\theta |f/\sqrt{g}$ and $\sqrt{g}$ givesEquality holds exactly when $|\theta |f/\sqrt{g}$ is proportional to $\sqrt{g}$ on the contributing probability support. Therefore the minimum-variance importance distribution isIf $\theta$ has one sign, $A=|\mu |$ and the ideal proposal gives zero variance. Its normalization or exact sampling may nevertheless be impractical. If $A=0$, the integrand vanishes almost everywhere and the estimator is identically zero.

The minimizing probability density function may vanish where $\theta =0$ even though $f>0$ there. This is valid for the ordinary estimator, which needs probability support only for its integrand. If one insists on exactly the same positive probability support as $f$, the bound is instead an infimum whenever this vanishing occurs, approached by $g_{\delta }=(1-\delta )g_0+\delta f$ as $\delta \downarrow 0$: because $g_{\delta }\ge (1-\delta )g_0$, the second moment lies between $A^2$ and $A^2/(1-\delta )$, proving convergence to the bound.

For $x\ge k$, the proposal CDF is $G(x)=1-k/x$. Therefore inverse transform sampling with $V\sim U(0,1)$ givesEquivalently, put $U=1-V$, which is also uniform, and use $X=k/U$. This equivalent parametrization matches the uniforms used in the final estimator.

For a weakly stationary process with finite second moments, let $\mu =\mathbb{E}{X}_t$ and $\gamma (h)=\mathrm{Cov}(X_{t+h},X_t)$, independent of $t$. If $\gamma (0)>0$, its autocorrelation function is $\rho (h)=\gamma (h)/\gamma (0)$. Given observations $x_1,\dots ,x_N$, one usual sample autocovariance function isThe correlogram plots $\hat{\rho }(h)$ against lag $h$. The partial autocorrelation function at lag $h$ is the correlation between $X_t$ and $X_{t-h}$ after linearly projecting each onto the intervening observations. Its sample version can be calculated by solving the Yule-Walker regression equationsthen taking the last coefficient ${\hat{\varphi }}_{h,h}$. Equivalently, use the corresponding residual correlation from the sample regressions, with a consistent covariance convention.

For a minimal causal autoregressive model of order $p$, the population partial autocorrelation function is zero after lag $p$, while the autocorrelation function follows an exponentially decaying or damped oscillatory tail. For a moving-average model of order $q$, its autocovariance and hence autocorrelation are zero after lag $q$, because observations farther apart share no driving-noise terms; its partial correlations usually tail off. These order identification by autocorrelation cutoffs patterns suggest AR or MA candidate orders. Sample cutoffs are approximate and should be judged with sampling uncertainty and residual diagnostics; the familiar white-noise bands near $\pm 1.96/\sqrt{N}$ are a null approximation, not universal confidence limits for a dependent series.

For the trend problem, write a symmetric estimator asReproducing arbitrary constant and quadratic trends requires $2a+2b+c=1$ and $8a+2b=0$; the linear-trend condition holds automatically by symmetry. These constraints follow by expanding $T_{t+j}=T_t+j(b_0+2c_{0}t)+c_0{j}^2$ for a trend $T_t=a_0+b_{0}t+c_0{t}^2$. Thus $b=-4a$, $c=1+6a$. Since the errors are uncorrelated and the estimator is unbiased, its mean squared error isThe strictly convex quadratic is minimized at $a=-3/35$, giving the five-point symmetric quadratic trend smootherThis weighted moving average allows negative coefficients; a convex average could not cancel the quadratic moment.

For three points, symmetry gives weights $(a,b,a)$. Unbiasedness for every quadratic requires $2a+b=1$ and $2a=0$, so the only solution is $(0,1,0)$. Thus there is no nontrivial three-point quadratic-preserving smoother. The literal claim that there is no unbiased estimator at all is false, since ${\hat{T}}_t=X_t$ is unbiased with error variance ${\sigma }^2$. This is the three-point quadratic reproduction forces the identity filter distinction.

For seasonal extraction by a centred moving average, use the additive decomposition $X_t=T_t+S_t+{\epsilon }_t$, where $S_t$ has period $m$ and its values over one period sum to zero. An average covering a full period removes the seasonal component and approximately retains a slowly changing trend. If $m$ is odd, take the centred length-$m$ average; if $m$ is even, centre it by averaging two adjacent length-$m$ averages, giving half weights at the two endpoints. Subtract this trend estimate from the data, average residuals within each seasonal phase over different cycles, and centre the resulting phase estimates to sum to zero. For a multiplicative decomposition, use ratios and normalize the phase estimates to average one.

Successive moving averages compose by convolution of their weights. Even an initially uncorrelated noise series becomes correlated: for weights $w_j$, the filtered noise has covariance ${\sigma }^2{\sum }_j{w}_j{w}_{j+h}$. Further smoothing changes this covariance again and can turn random variation into apparently persistent patterns while obscuring genuine short-scale features. Spectrally, repeated use of a transfer function $H$ multiplies the spectrum by $|H{|}^{2r}$ after $r$ applications, so repeated smoothing is not harmless repetition of the same estimate.

In simple exponential smoothing, choose $0<\alpha <1$ and update ${\ell }_t=\alpha X_t+(1-\alpha ){\ell }_{t-1}$. Iteration gives ${\ell }_t=\alpha {\sum }_{j=0}^{t-1}(1-\alpha {)}^j{X}_{t-j}+(1-\alpha {)}^t{\ell }_0$, so older observations receive exponentially decreasing weights. Forecast ${\hat{X}}_{t+h\mid t}={\ell }_t$ for every $h\ge 1$. This is appropriate for a locally stable level with no persistent trend or seasonality.

For a local linear trend, double exponential smoothing supplies level and slope. Holt's linear trend method usesAnother standard convention is Brown's double exponential smoothing: smooth twice with the same gain, $s_t^{(1)}=\alpha X_t+(1-\alpha )s_{t-1}^{(1)}$ and $s_t^{(2)}=\alpha s_t^{(1)}+(1-\alpha )s_{t-1}^{(2)}$, then put $a_t=2s_t^{(1)}-s_t^{(2)}$, $b_t=\alpha (s_t^{(1)}-s_t^{(2)})/(1-\alpha )$ and forecast $a_t+h{b}_t$. These variants handle a nonseasonal trend; seasonal effects need separate modelling, and a long-horizon linear extrapolation assumes the current local slope persists.

The empirical distribution gives each observed value mass $1/n$, so its empirical distribution function isFor each of $B$ independent replications, draw $n$ values with replacement from this empirical distribution, recompute the same estimator ${\hat{\theta }}^{\ast (b)}$, and use their sample standard deviation:This estimates the conditional bootstrap standard error; dividing it by $\sqrt{B}$ would instead estimate simulation uncertainty in the bootstrap mean and is not the requested sampling standard error.

When the original observations are distinct, the first resampled index may be arbitrary. After $j$ distinct indices have been drawn, the next index avoids them with probability $1-j/n$. Multiplying conditional probabilities givesHere $P_{\ast }$ denotes probability conditional on the data.

For jackknife resampling, calculate the leave-one-out estimates ${\hat{\theta }}_{(-i)}$ by deleting observation $i$, and their average ${\overline{\theta }}_J=n^{-1}{\sum }_i{\hat{\theta }}_{(-i)}$. The jackknife variance estimator isThe factor $n-1$ compensates for the small differences between the overlapping samples. For sufficiently smooth statistics it estimates their sampling variance. The related estimated bias is $(n-1)({\overline{\theta }}_J-\hat{\theta })$, leading to the jackknife bias correction $n\hat{\theta }-(n-1){\overline{\theta }}_J$.

For the nonparametric bootstrap, replace the unknown distribution by its empirical distribution functionIndependently sample $n$ observations with replacement from the original observations, compute the estimator on that bootstrap sample, and repeat this for $B$ independently generated bootstrap samples. If the resulting values are ${\hat{\theta }}_1^{\ast },\dots ,{\hat{\theta }}_B^{\ast }$, estimate the conditional bootstrap variance byIts interpretation as an approximation to the original sampling variance relies on the bootstrap approximating the sampling law of the statistic; increasing $B$ only reduces simulation error and cannot repair a failure of that approximation.

Negative covariance can be exploited through antithetic variates, but it must belong to an actual joint sampling scheme. Two separately independent samples do not inherit negative covariance merely because one can imagine a coupling of their marginal distributions. Interpret the data as $h=\min (n,m)$ independent matched pairs $(X_i,Y_i)$ with covariance ${\sigma }_{XY}<0$, together with independent unmatched observations if the sample sizes differ. Other known pairing schemes can be handled by substituting their actual covariance of the two means.

Let $A=\overline{X}$, $B=\overline{Y}$, and putFor any fixed $w$, the combination $wA+(1-w)B$ is an unbiased estimator of the common mean. Its variance is $w^2{V}_X+(1-w{)}^2{V}_Y+2w(1-w)C$. Differentiation, or completing the square, gives the minimum-variance combination of unbiased sample means:With $C<0$, both weights are positive. This is efficiency among fixed linear unbiased combinations, not a universal claim about every possible estimator for arbitrary distributions $F,G$.

For an immediately implementable exactly unbiased estimator without known moments, use fixed $w=n/(n+m)$, the pooled sample mean. Its variance iswhich shows directly the gain from negative covariance. For any fixed weight, estimate $V_X,V_Y$ using $s_X^2/n,s_Y^2/m$, and $C$ using $h{s}_{XY}/(nm)$, with $s_{XY}$ the usual unbiased sample covariance of the matched pairs. Substitute them into the variance quadratic. At least two matched pairs are needed to estimate their covariance nonparametrically.

If the unknown optimal weights are desired while retaining exact unbiasedness, a cross-fitted combination of unbiased sample means gives an algorithm using the supplied data. Split the independent pairs and unmatched observations into two folds, keeping each pair together and retaining observations of both types in each fold. Estimate the moments from fold 2 to choose the weight for the fold-1 sample means, adjusting the variance terms for fold 1's sample sizes; do the converse for fold 2. Clip estimated weights to $(0\le w\le 1)$, to keep them bounded when training moment estimates are unstable. Average the two held-out combinations with deterministic weights, for example one half each for balanced folds.

Conditional on its training fold, each held-out pair of sample means has expectation $\theta$, and its weights sum to one. Each held-out combination, and hence their average, is therefore exactly an unbiased estimator. Under consistent moment estimation with increasing balanced folds, the weights approach $w_{\ast }$ and recover its first-order variance advantage. In contrast, estimating a weight from the same data to which it is applied need not yield an unbiased estimator.

Estimate the variance of this fitted-weight procedure by a bootstrap that resamples matched pairs as blocks and unmatched observations separately, repeats the entire split/weight/combination procedure, and takes the sample variance of the resulting estimates. This preserves the negative covariance and includes variability of the fitted weights. Resampling the $x$ and $y$ values independently would erase the feature responsible for the improvement. The bootstrap estimate has its usual large-sample qualification; the fixed-weight plug-in variance formula above is the simpler alternative when weights are predetermined.

Use the product-limit estimate in (b). Between consecutive event times there are no mortality jumps, soThus the estimated relative survivor curve rises between events when background mortality is positive, and is constant if it is zero. Censoring changes the risk set and can change the slope, but supplies no downward event jump. At the next event time the curve is multiplied by $1-d_k/r_k$.

Provided the curve is still positive, the ratio of the post-event estimates at the two consecutive event times isFor $d_k<r_k$, the later value is larger precisely whenFor example, suppose there are 100 individuals in the risk set after the earlier event, no intervening censoring, a 20-day gap, and one death at its end. If every background hazard is $0.001$ per day, then the integrated background contribution is $0.02$ andThe individual background hazards are small, but the interval contains less observed mortality than expected from that background. This can occur by chance in a small event sample or when the cohort is healthier than the life-table reference on other dimensions.

A nonnegative true excess hazard makes $F_E$ nonincreasing, but this unconstrained estimator need not inherit that restriction. An increase is therefore not proof of a clinical survival benefit. If the exponential cumulative-hazard version is chosen instead, its corresponding condition is ${\int }_{t_j}^{t_k}{\overline{h}}_B>d_k/r_k$; the qualitative explanation remains the same. If the risk set is exhausted by deaths, the product-limit factor is zero and no later increase is possible.

An unordered bootstrap sample is determined by the multiplicities $N_i\ge 0$ of the $n$ distinct observations, with ${\sum }_i{N}_i=n$. Conversely each such vector specifies exactly one sample up to rearrangement. Place $n$ identical marks into $n$ boxes using $n-1$ separators. The stars and bars argument therefore givesThese bootstrap count vectors are not equally likely: the probability of a vector is $n!/(n^n{\prod }_i{N}_i!)$, counting its possible ordered sequences. The ordered index sequences themselves number $n^n$.

The first simulation uses independent standard Cauchy distributions and returnsThe tail probability and estimator moments areThe indicator count has a binomial distribution with parameters $n,\theta$; the infinite variance of a Cauchy observation is irrelevant because this estimator averages bounded indicators.

For the next code block, each row is an independent sample of size $n$ from the empirical distribution of the original $x_i$. Its entry in the vector $v$ is the corresponding bootstrap indicator mean ${\hat{\theta }}_b^{\ast }$. The last expression is the variance of these $B=199$ replicate estimates, with denominator $B-1$:Condition on the original data and write a star on bootstrap expectations. Every resampled indicator has a Bernoulli distribution with success probability $\hat{\theta }$, and the $n$ draws within a row are independent. Thus the conditional bootstrap variance of a sample mean givesIt estimates the sampling variance of the tail-probability estimator, not the variance of the raw Cauchy data. Its square root is an estimated standard error. Finite $B$ leaves bootstrap simulation noise. Averaging over the original data, its expectation is $(n-1)\theta (1-\theta )/n^2$, so it is not exactly unbiased for the true variance $\theta (1-\theta )/n$ at finite $n$.

In the importance-sampling block, the transformation $Y=2/(1-U)$ gives, for $y\ge 2$,The variable $Y$ is a proposal draw supported entirely in the integration region. The deterministic value computed from it is the importance sampling weightConsequentlyThe code uses ordinary importance sampling of a Cauchy tail, not self-normalization; $Y$ itself is neither an observation from the target Cauchy law nor the estimator's summand. The weight compensates for sampling from its proposal law.

The final block bootstraps these same proposal observations $Y_i$ and recomputes the weight average in each row. Since $W$ is a deterministic function of $Y$, this is equivalent to resampling the observed weights $W_i$. If ${\tilde{\theta }}_b^{\ast }$ denotes a replicate, the final output is its sample variance over the $B$ replicates. Its conditional expectation isIts expectation over the original proposal sample is $(n-1)\mathrm{Var}(W)/n^2$.

The variance reduction can be calculated, rather than inferred just from the appearance of the weights. Write $A=\arctan (1/2)$. Thenbecause an antiderivative of the integrand without its prefactor is $\frac{1}{2}(\arctan y-y/(1+y^2))$. HenceBy comparison, $\theta (1-\theta )\approx 0.125803$. For $n=100$ the true estimator variances are approximately $9.55253\times 10^{-7}$ and $1.25803\times 10^{-3}$ respectively. The importance estimator has about 1/1,317 of the indicator estimator's variance, because all proposals land in the tail and the weights remain in the narrow interval $[2/(5\pi ),1/(2\pi ))$.

Thus the last bootstrap output should usually be much smaller than the first, and its expected value is smaller by the same variance ratio. This is not a deterministic ordering of two finite-run outputs: if all the original Cauchy draws happen to be at most two, the first bootstrap variance is zero, while unequal importance weights can still give a positive second variance.

The Metropolis–Hastings algorithm proposes $y\sim q(\cdot \mid x)$ and moves from the current state $x$ with acceptance probabilityOn rejection it retains $x$. The target need only be known up to its normalizing constant, but it must define a proper probability distribution. The accepted off-diagonal probability flow iswhich is symmetric in $x,y$ and proves detailed balance. This permits considerable flexibility in proposals, including joint moves when a posterior distribution has strongly correlated parameters, but requires proposal tuning and can waste computation on rejections.

A Gibbs sampler instead updates a coordinate or block from its exact full conditional distribution, holding the other coordinates fixed. This is a special Metropolis–Hastings algorithm update with acceptance probability one: the joint density factors into the unchanged marginal density and the proposed conditional density, which cancel in the acceptance ratio. The coordinate update preserves the target; composing such updates therefore preserves it too, although an entire deterministic sweep need not itself be reversible.

Use Gibbs sampling when the conditional laws are easy to simulate and suitably chosen blocks mix well. Scalar updates can mix poorly under strong posterior distribution correlation, whereas a blocked Gibbs sampler can alleviate that dependence. Prefer a flexible Metropolis proposal when conditional draws are difficult or a well-tuned joint proposal is more effective. Variations include random or systematic coordinate scans, random-walk Metropolis algorithms, independence proposals, and using Metropolis updates for difficult blocks inside a Gibbs scheme. Both methods generate dependent draws; usable long-run estimates require an appropriate irreducible, aperiodic chain and assessment of mixing, not merely a formal update rule.

For the change-point calculation, put $v={\sigma }^2$ andMultiplication of the Gaussian likelihood function by the specified prior distribution gives the formal posterior distribution kernel, with respect to $d\theta d\varphi dv$ and counting measure on $m$,

There is an empty component under an improper prior at $m=n$. The sum $S_n$ is independent of $\varphi$, whose prior distribution is flat on all of $\mathbb{R}$. For every fixed finite $\theta$ and $v>0$, the kernel is a strictly positive constant in $\varphi$, henceThe configuration has positive prior probability $1/n$. Integrating over, for example, a bounded interval of $\theta$ and $1\le v\le 2$ already makes the joint normalization infinite. Thus the printed prior distribution does not produce a posterior probability distribution. This failure is present for every data set, not just for a special arrangement of observations.

The formal full conditionals expose exactly where the requested sampler breaks. For $m<n$, completing the two mean squares givesThe variance kernel is that of $\mathrm{IG}(n/2,S_m/2)$ when $S_m>0$, and the discrete conditional probabilities are proportional to $e^{-S_m/(2v)}$. At $m=n$ the conditional kernel for $\varphi$ is constant and cannot be normalized; it is not a normal distribution with an infinite variance. Therefore no valid Gibbs sampler, and no value of $P(m=n\mid x)$, exists for the model as printed. A chain that simply substitutes an arbitrary distribution for this missing conditional would be sampling a different model.

Here is a fully specified proper-prior repair retaining the no-change configuration. Choose fixed hyperparametersindependently, with $t_{\theta },t_{\varphi },a_0,b_0>0$. Its Gaussian change-point posterior with proper priors isIt is normalizable even for degenerate data: the likelihood function is at most $(2\pi {)}^{-n/2}{v}^{-n/2}$, and its integral against the variance prior distribution is bounded byThe normal mean priors integrate to one, and the normalizing constant is positive. This proves posterior propriety for the repaired model.

Completing the conditional squares gives the Gaussian change-point Gibbs updates. DefineThe four updates areAt $m=n$, $M_{\varphi }={\mu }_{\varphi },V_{\varphi }=t_{\varphi }$, giving its proper prior distribution unchanged. Starting from any finite means, positive variance and allowed $m$, cycle through these four updates using the latest values. The sum-of-squares terms can be evaluated from cumulative data sums; normalize the discrete log weights after subtracting their maximum to avoid numerical underflow. The proper positive conditionals allow movement among all split values and throughout the continuous parameter support.

Under this specified repair, $P(m=n\mid x)$ is the posterior probability of no change within the observed sequence: all $n$ data points belong to the first mean regime. For $N$ retained draws after initialization has ceased to affect the estimates, its Markov chain Monte Carlo estimator isAlternatively average the displayed conditional probability of $m=n$ at the sampled continuous parameters, giving an estimate by Rao-Blackwellization. Autocorrelation must be accounted for in its Monte Carlo uncertainty. These are probabilities and estimators for the explicitly repaired model; the original unnormalizable kernel supplies neither.

Assume $n>p\ge 1$, the ordinary nonsaturated Gaussian linear model. The log likelihood is, up to a constant,The normal equations and full column rank give the least-squares estimator $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Write $H=X(X^{T}X{)}^{-1}{X}^T$ and $Q=I-H$, an orthogonal projection of rank $n-p$. Maximizing in ${\sigma }^2$ yieldsSince $QX=0$, the residual sum of squares is ${\epsilon }^{T}Q\epsilon$. Expanding its expectation uses only $\mathbb{E}{\epsilon }_i{\epsilon }_j={\sigma }^2{\delta }_{ij}$:This does not use Cochran's theorem. It proves bias $-p{\sigma }^2/n$ and givesIn the saturated case $n=p$, residuals vanish identically, the likelihood is unbounded as the positive variance tends to zero, and the unbiased residual estimator is unavailable. Thus the usual variance-estimation request requires $n>p$.

Fix any estimator $\hat{\eta }$ and put $A(x)=d(\hat{\eta }(x),f)$, $B(x)=d(\hat{\eta }(x),g)$ and $D=d(f,g)$. The triangle inequality for the metric gives $A+B\ge D$. HenceThe larger of the two risk functions dominates their average. Using nonnegativity to replace both probability density functions by their minimum,The right side is independent of the estimator, so taking the infimum provesThis metric squared-loss two-point bound uses overlap of the observation laws to quantify how hard it is to distinguish the two separated parameters.

Apply the preceding argument to the two product observation laws and the real-valued estimator $T={\hat{\psi }}_n$. Alternatively, directly useThe average of the two mean squared errors is at least $(\theta -\tau {)}^2/4$ times the overlap of the two joint probability density functions. Their overlap equalsTaking the infimum over estimators therefore yieldsHere $R_n^{\ast }$ is the minimax risk for this functional. The factor $\frac{1}{2}\Vert P_f^{(n)}-P_g^{(n)}{\Vert }_1$ is the total variation distance, so the same metric squared-loss two-point bound applies to a functional even when distinct density parameters have the same functional value.