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For fixed-design nonparametric regression, let $x_i=i/n$ and $K_h(u)=h^{-1}K(u/h)$. The degree-$p$ local polynomial estimator takes the fitted intercept ${\hat{m}}_h(x;p)={\hat{\beta }}_0$, whereUse a nonnegative regression kernel and an invertible local polynomial Gram matrix to obtain a unique fit. For $p=0$, solving the single weighted least squares normal equation gives the local constant estimator:At the interior point $x_0$, use the standard unit-integral regression kernel convention ${\mu }_0(K)=1$. Otherwise the leading variance below is $R(K)/(nh{\mu }_0(K{)}^2)$; unlike a kernel density estimator, the local constant estimator itself is unchanged by rescaling $K$.

The local constant estimator is a linear estimator in nonparametric regression with weights $W_i=K_h(x_i-x_0)/(n{s}_{0,h})$. The independent unit-variance errors implyTo control the bias of an estimator, differentiability at this one point is enough. WriteSymmetry makes ${\mu }_1(K)=0$, and the supplied moment approximation gives $s_{1,h}=o(h)$. Therefore the linear contribution to the bias of an estimator is $m^{\prime }(x_0)s_{1,h}/s_{0,h}=o(h)$. The remaining contribution has absolute value at most $\eta (h)h$, because only $|x_i-x_0|\le h$ receive positive weight. This is the interior first-order bias cancellation for local constant regression, and gives squared bias of an estimator $o(h^2)$. Combining with the variance provesFor fixed $\epsilon >0$, take $h_{\epsilon }={\epsilon }^{-1/3}{n}^{-1/3}$. It satisfies both asymptotic smoothing bandwidth conditions, and the displayed remainder, multiplied by $n^{2/3}$, tends to zero for this fixed $\epsilon$. ThusLetting $\epsilon \downarrow 0$ proves the optimized scaled pointwise error tends to zero. The order of these limits matters: no remainder uniform in $\epsilon$ has been assumed.

For the lower bound over the whole function class, the absence of a uniform bound on derivatives permits a particularly direct argument. It also avoids assuming a normal distribution for errors when the model only supplies their first two moments. Fix $n$ and $x_0$. Choose a smooth bump function $b$ on the interval, with $b(x_0)=1$, whose support contains no design point except possibly $x_0$. Such a bump exists because the design is finite; at an endpoint use the restriction of a smooth bump on the real line. Every $m_a=ab$, $a\in \mathbb{R}$, belongs to the differentiability class.

If $x_0$ is not a design point, all observation means are zero for every $a$. The Le Cam two-point lemma applied to $a=\pm A$ has total variation distance zero and gives a lower bound tending to infinity with $A$. In fact, the supremum of the mean squared error is infinite at every such $x_0$.

If $x_0$ is a design point, only its response depends on $a$. The resulting location family is the single observation $Y=a+ϵ$; all remaining observations are independent of $a$ and give no extra information. The single-observation location minimax bound isHere is a proof covering discrete as well as continuous errors. Give $a$ the prior distribution $U[-A,A]$, independently of the noise. Put $p_M=\mathbb{P}(|ϵ|\le M)$ and $v_M=\mathrm{Var}(ϵ\mid |ϵ|\le M)$. Consider an easier experiment in which an oracle also reveals whether $|ϵ|\le M$, and reveals $a$ itself on the complementary event. On the event $|ϵ|\le M$ and $|Y|\le A-M$, every possible truncated noise value places $a=Y-ϵ$ within $[-A,A]$. The flat prior distribution then makes the Bayesian posterior noise law exactly its truncated original law. The smallest conditional squared-error loss is its conditional variance $v_M$, attained by the posterior mean.

The probability of that event is at least $p_M(1-2M/A)$ for $A>2M$, by restricting further to $|a|\le A-2M$. Thus every rule's supremum risk function is at least the original Bayes risk, which is at least the easier experiment's Bayes risk, and therefore at least $p_M(1-2M/A)v_M$. Let $A\to \infty$ and then $M\to \infty$. Finite second moment and zero mean give $p_M{v}_M\to 1$. The rule $T(Y)=Y$ has constant mean squared error one, proving the equality.

Consequently, at every $x_0\in [0,1]$, the unrestricted differentiability class satisfies the stronger conclusionThus $c=1$ works for every $n\ge 1$. There is no conflict with the preceding fixed-function limit: the worst functions, including arbitrarily narrow bumps, can vary with $n$. This is the pointwise versus uniform risk distinction. A smoothness ball with a common derivative bound would pose a different minimax risk problem.

At zero a centered normal distribution with standard deviation $\sigma$ has density $1/(\sqrt{2\pi }\sigma )$. The two prior standard deviations are $1/\sqrt{n_0}$ and $c/\sqrt{n_0}$, givingThe wider prior distribution has a lower density at its center.

Write $\overline{Y}=\beta +\epsilon$, with independent $\epsilon \sim N(0,1/n)$ and $\beta \mid H_i\sim N(0,1/q_i)$. The convolution of independent random variables is again a normal distribution, so the prior predictive laws areThese are predictive distributions before observing $y$, hence the Bayesian model evidence for the observed mean. The residual information in the original observations is common to both models and cancels in their Bayes factor.

Divide the two normal distribution predictive densities to obtain the Bayes factorAt $y=0$ this becomesThe wider prior distribution spreads its predictive mass over more possible means, giving the narrower model more Bayesian model evidence for observations very near zero. Away from zero, the exponential term opposes that factor.

Under the point hypothesis $\beta =0$, the test statistic $\sqrt{n}\overline{Y}$ has a standard normal distribution. The observed statistic is three, giving a two-sided p-value $2[1-\Phi (3)]\simeq 0.00270$. This is conventionally strong evidence against that point hypothesis. The narrow model $H_0$ is a continuous prior distribution around zero, rather than a point hypothesis.

For $n_0/n=100$ and $c=100$, Normal-normal conjugacy givesThe narrow-model posterior mean is strongly pulled toward zero and its standard deviation is about $0.0995/\sqrt{n}$. The wide-model posterior mean is about $2.9703/\sqrt{n}$, very close to the observation, with standard deviation about $0.9950/\sqrt{n}$. Each model produces a markedly different posterior, so choosing between them requires their predictive evidence.

With equal model prior probabilities, Bayes factor updating gives $w_0=B_{01}/(1+B_{01})$ and $w_1=1/(1+B_{01})$. The Bayesian model averaging posterior density is thereforeThis is a two-component mixture model of the normal distributions already derived. For the numerical observation above, $w_0\simeq 0.10829$ and $w_1\simeq 0.89171$. Consequently it is predominantly the wide-model posterior, not predominantly the narrow one.

In a normal linear model, attach a continuous spike-and-slab prior to each regression coefficient. With suitably scaled predictors, introduce indicators ${\gamma }_j\sim \mathrm{Bernoulli}(\pi )$ and setThe narrow component describes practically negligible effects; the wide component permits substantial ones. Fit the joint Bayesian posterior of coefficients, indicators and any unknown residual variance. Bayesian model averaging gives shrinkage toward zero for poorly supported effects, while $\mathbb{P}({\gamma }_j=1\mid \mathcal{D})$ quantifies wide-component support. A shared Beta distribution prior on $\pi$ can represent uncertainty about how many effects are substantial.

Select on scientifically meaningful effect size, for example a high $\mathbb{P}(|{\beta }_j|>{\delta }_j\mid \mathcal{D})$ for a prechosen threshold in meaningful predictor units. Membership in the wide component alone does not imply a large realized effect: its normal distribution still permits values near zero. Correlated predictors also require interpretation of the joint Bayesian posterior, rather than treating each coefficient as an isolated test.

Use a proper normal distribution prior centered at zero on the log odds ratio. For example,puts approximately 95 percent of its mass between $-2$ and $2$, giving odds ratios roughly between $e^{-2}$ and $e^2$, close to $1/8$ and $8$. Centering at zero treats reciprocal odds ratios symmetrically. If the desired central 95 percent interval is exactly $(1/8,8)$, use standard deviation $\log 8/{\Phi }^{-1}(0.975)\simeq 1.0610$. A soft prior is appropriate for implausibility, whereas a bounded uniform prior would declare effects outside the limits impossible.

The overall mean $\mu$ and the between-study heterogeneity $\tau$ encode different information. A proper broad normal distribution prior such as $\mu \sim N(0,2^2)$ is one possible weak prior for the mean log odds ratio; information about the spread of trials alone does not determine its center.

A concrete prior calibration for normal random-effect range can make the factor-of-50 statement simultaneous across all six trials. Conditional on $\tau$, each pair difference has normal distribution ${\beta }_j-{\beta }_k\sim N(0,2{\tau }^2)$. Set $R=\log 50$, $m=\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$, $z={\Phi }^{-1}(1-0.05/(2m))$ andFor every $\tau \le A$, each pair exceeds $R$ in absolute value with probability at most $0.05/m$. The union bound therefore gives $\mathbb{P}({\max }_j{\beta }_j-{\min }_j{\beta }_j>R)\le 0.05$, and integrating over the uniform prior preserves that bound. This is one explicit interpretation of “very unlikely”; a different elicited probability would change the bound. A smoother proper scale prior could be calibrated similarly.

The proposed $1/\tau$ improper prior is unsuitable. The observed-data likelihood function approaches the positive common-effect likelihood as $\tau \downarrow 0$. After restricting $\mu$ and the intercepts to a compact interior region, it is bounded below there by a positive constant. HenceThis is an improper posterior from a log-uniform random-effect scale prior. Proper conditional sampling distributions do not repair the improper joint posterior, and an arbitrary tiny cutoff would make inference depend on that cutoff.

Conditional on $\mu ,\psi >0$, let $\lambda \sim {\chi }_4^2$ and independently $Z\sim N(0,1)$. The normal distribution with the stated precision parameter can be generated asThereforeby the defining normal distribution and chi-squared distribution representation of Student's t-distribution. Its density isThus $\mu$ is the location and $\psi$ is the scale, not the standard deviation: $\mathrm{Var}({\beta }_j\mid \mu ,\psi )=2{\psi }^2$. Each study gets its own independent chi-squared draw in this Student t random-effect model.

The Student t random-effect model is useful when most studies are comparable but occasional genuine departures are more frequent than a normal distribution hierarchy allows. Its heavier tails permit a study effect far from $\mu$ without forcing a large common between-study heterogeneity scale on every study. In the Gaussian scale mixture representation, a small study-specific ${\lambda }_j$ lowers its precision parameter and weakens its shrinkage.

Use this as robust partial pooling when occasional atypical effects are plausible. Known systematic population or design differences should still be modeled explicitly; a heavy tail cannot identify or correct within-study bias by itself.

Fit both the normal and Student t random-effect model with comparable proper prior distributions. Compare priors on the same spread measure: a normal distribution scale $\tau$ is a standard deviation, whereas the $t_4$ standard deviation is $\sqrt{2}\psi$.

Use a posterior predictive check: draw study effects and binomial counts from each fitted hierarchy and compare replicated dispersion and extreme study contrasts with the observations. For predicting a new study, generate a new effect from the hierarchy rather than reusing an existing fitted effect. A Leave-one-out cross-validation with entire studies held out can compare integrated predictive probabilities for both arms of each omitted trial, averaging over hyperparameters and its unobserved study effect. Leave-one-study-out influence analysis also reveals whether the difference is driven by a single trial.

Prefer the heavier-tailed hierarchy if it improves the relevant predictive checks and held-out study predictions robustly to reasonable prior choices. The deviance information criterion can supplement the comparison, but its effective parameter count can depend on the latent-variable representation, and six studies give limited information about tail shape. A small numerical criterion difference alone is insufficient evidence.

Multiply the linear stochastic differential equation by $e^{-at}$. The Itô product rule givesThe integrand is deterministic, so the Itô integral has a centered normal distribution. Its variance, by the Itô isometry, isHenceAt $a=0$ the correct continuous-limit formula is $N(z,b^{2}t)$. A zero variance, for example when $b=0$, denotes the deterministic distribution. For $a<0$ the variance is still positive because both numerator and denominator in its quotient are negative. This is the explicit Ornstein-Uhlenbeck solution, allowing either sign of the linear drift coefficient.

Choose the market price of risk ${\lambda }_t=(\mu -r{S}_t)/\sigma$. The Girsanov theorem, with the hypotheses allowed in the question, gives an equivalent martingale measure $\mathbb{Q}$ under whichThus the stock under the risk-neutral measure is a linear Gaussian diffusion. In particular, for $\tau =T-t$ its conditional mean and standard deviation arePut $\varphi (x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ and let $\Phi$ be the standard normal distribution function. For $N\sim N(0,1)$,since ${\int }_{-d}^{\infty }x\varphi (x)dx=\varphi (d)$. Discounting this expectation gives the call price in an arithmetic stock model with interest. A particularly convenient expression isAt maturity define $C(T,s)=(s-K{)}^{+}$. This value is nonnegative because it is a discounted expectation of a nonnegative payoff.

For $t<T$, hold ${\pi }_t=C_s(t,S_t)$ shares and hold ${\beta }_t=[C(t,S_t)-{\pi }_t{S}_t]/B_t$ units of the continuous-time bank account. The pricing function solvesThe Itô formula under the physical measure therefore givesThis proves self-financing and terminal replication, with wealth always $C(t,S_t)\ge 0$. The coefficients are locally smooth before maturity, and the strategy extends to maturity through its continuous wealth limit and the square-integrable discounted payoff representation.

To see minimality, any other nonnegative self-financing portfolio replicating the payoff has discounted wealth a nonnegative local martingale under $\mathbb{Q}$, hence a supermartingale. Its initial capital $x$ must satisfy $x\ge {\mathbb{E}}^{\mathbb{Q}}[e^{-rT}(S_T-K{)}^{+}]=C(0,S_0)$. The strategy constructed above attains equality. ThusThe additive physical diffusion may take negative stock values; the formula and nonnegative replicating wealth remain valid. Replacing it by a multiplicative Black–Scholes diffusion would give the wrong price and hedge.

For fixed rational $0\le t_1<t_2<t_3<t_4$, writeBy independent increments, $Z$ is independent of $(V,W)$ and has normal distribution $N(0,t_3-t_2)$. Conditional on $(V,W)$, the displayed difference has a continuous probability density function; in particular it equals zero with probability zero. Taking the countable intersection over all such rational quadruples proves that, on one event of probability one, the maxima over any two separated rational closed intervals are distinct.

Suppose a local maximum at $t$ were not a strict local maximum. There is a neighbourhood in which every value is at most $B_t$, and, arbitrarily close to $t$, some other time $u$ strictly inside that neighbourhood has $B_u=B_t$. Choose separated rational closed intervals containing $t$ and $u$, both lying inside that neighbourhood. When one time is zero, use a first interval with left endpoint zero. Both interval maxima equal $B_t$, contrary to the preceding event.

Therefore every local maximum of Brownian motion is a strict local maximum almost surely, simultaneously over all times. This countable-interval argument avoids an invalid intersection of probability-one events over uncountably many candidate times.

Choose independent random variables ${\xi }_i\sim N(0,1)$ on a countable product probability space. For each fixed $t$, the diagonal assumption givesThe partial sums ${\sum }_{i\le n}{k}_i(t){\xi }_i$ are therefore Cauchy in $L^2$. DefineChoose a representative of this limit for each $t$. No path continuity or simultaneous series convergence over all uncountably many times is being asserted.

For any finite list $t_1,\dots ,t_m$ and real coefficients $a_j$, the linear combination ${\sum }_j{a}_j{X}_{t_j}$ is the $L^2$ limit of centered Gaussian variablesTheir variances converge, so their characteristic functions converge to that of a centered normal distribution. This proves that every finite-dimensional vector is Gaussian and hence that $X$ is a Gaussian process. Taking $L^2$ limits also givesThis is the Gaussian process construction from square-summable features.

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.

Two useful features of autoregressive conditional heteroscedasticity are persistent changes in conditional scale and excess unconditional kurtosis. Its time-varying conditional variance can explain volatility clustering, where large absolute returns occur in groups even when signed returns have little autocorrelation. A homoscedastic autoregressive moving-average model has a fixed innovation variance.

Also, a conditional normal distribution with a random scale is a Gaussian scale mixture. Its unconditional kurtosis can exceed $3$, or its fourth moment can be infinite. A Gaussian autoregressive moving-average model remains jointly Gaussian and cannot reproduce this effect. In the particular lag-two ARCH process, persistence of the squared scale occurs within each parity subsequence; the two parity subsequences are independent.