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This covariance is also zero, but symmetry is essential to the proof. Write ${\epsilon }_t={\eta }_t|{\epsilon }_t|$, where ${\eta }_t$ is an independent fair sign. The normal distribution makes that sign independent of its magnitude and of every other driving variable. Conditional on the magnitudes and all noise except this sign, changing ${\eta }_t$ flips $X_t$ and leaves $X_t^2$ unchanged. Every future conditional scale uses only squared past observations, so $X_{t+h}$ is unchanged for $h>0$.

Thus $f(X_{t+h})$ is independent of the remaining fair sign in $X_t$. Averaging that sign gives $\mathbb{E}[X_{t}f(X_{t+h})]=0$, and henceThe Cauchy-Schwarz inequality again guarantees integrability. For odd $h$, the parity decomposition of a lag-two ARCH process also gives independence directly. For even $h$, the sign symmetry of an ARCH process supplies the argument; the martingale difference sequence property alone would not justify this reversed covariance.

In the Black-Scholes model, absence of arbitrage and completeness give the price as the discounted expectation under the risk-neutral measure. Under that measure,The physical drift $\mu$ therefore does not enter the option price. PutFor the standard normal distribution function $\Phi$, $\mathbb{P}(S_T>K)=\Phi (d_2)$. Completing the square in the Gaussian density also gives $\mathbb{E}[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Expanding the call payoff into these two truncated moments yields

Weak convergence $X_k\Rightarrow X$ means that $\mathbb{E}f(X_k)\to \mathbb{E}f(X)$ for every bounded continuous real function $f$ on ${\mathbb{R}}^d$; it is a statement about their distributions, not their coupling.

The central limit theorem says that for independent identically distributed real variables of finite mean $\mu$ and variance ${\sigma }^2$, $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2)$. For $\sigma >0$ this is equivalent to convergence of $(\sum X_j-n\mu )/(\sigma \sqrt{n})$ to the standard normal distribution. For $\sigma =0$, every variable equals $\mu$ almost surely and the limit is the point mass at zero.

For the proof when $\sigma >0$, put $Y=(X-\mu )/\sigma$. Its characteristic function has expansion ${\phi }_Y(t)=1-t^2/2+o(t^2)$. To justify this using only a second moment, subtract the first two Taylor terms of $e^{itY}$ and use dominated convergence theorem, with the second-order remainder bounded by a constant times $t^2{Y}^2$. Independence gives the characteristic function of the normalized sum as $[{\phi }_Y(t/\sqrt{n}){]}^n\to e^{-t^2/2}$. The Lévy continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies weak convergence. Since $e^{-t^2/2}$ is the characteristic function of $N(0,1)$, the result follows.

This is a generalized linear model with independent Bernoulli distributions for the responses and the logit link. Under treatment coding with colour one as the reference level, the linear predictor isColour is a categorical variable, not a numerical trend. Width has the same slope for every colour, since no interaction term is fitted. The parameters are estimated by maximum likelihood estimation.

The approximate Wald confidence interval for the width coefficient is $\boxed{0.46\pm 1.96(0.106)}$, or approximately $(0.2522,0.6678)$. This uses the approximate normal distribution of the maximum-likelihood estimator, rather than a confidence interval for an individual binary response.

For the requested crab, the fitted linear predictor is $\hat{\eta }=-11.38-0.22+0.46(20)=-2.40$. Applying the inverse logit link gives $\boxed{\hat{p}=(1+e^{2.40}{)}^{-1}\simeq 0.0832}$. The rounded displayed coefficients need not reproduce the reported test statistics exactly.

Write $y=N^{1/\phi (N)}$, $L={\sum }_{p\le y}|g(p)|$, and $A={\sum }_{p\le y}g(p)/p$. Put $I_p(n)=1_{p\mid n}$ and $Y_p=I_p-1/p$. The strongly additive arithmetic function identity is $g-A={\sum }_{p\le y}g(p)Y_p$.

For any tuple $p_1,\dots ,p_j$, expand ${\prod }_{r=1}^j(I_{p_r}-1/p_r)$ by choosing an indicator in some positions and the constant in the others. If the selected positions contain distinct primes whose product is $d$, thenFor the corresponding product of the independent Bernoulli random variables $X_p$, the expectation is exactly $1/d$. This includes repeated primes, since their indicator functions are idempotent, and the empty selection has zero error. Summing absolute values of all expansion coefficients bounds the difference bywhen $y\ge 2$. Expanding the $j$th power and summing the absolute coefficient weights gives

It remains to center at the actual expectation. Since ${\mathbb{E}}_N{I}_p=\lfloor N/p\rfloor /N$, we have $|{\mathbb{E}}_{N}g-A|\le L/N$. For each $n$, both $g(n)-A$ and $g(n)-{\mathbb{E}}_{N}g$ have absolute value at most $L$, because all their indicator coefficients have absolute value at most one. The identity $u^j-v^j=(u-v){\sum }_{r=0}^{j-1}{u}^{j-1-r}{v}^r$ therefore bounds the change in the central moment by $j{L}^j/N$.

Finally, independence factors each term of the independent central moment. If $q_1,\dots ,q_w$ are the distinct primes of the tuple and $a_i$ their multiplicities, the desired comparison becomesIf $y<2$, then $g=0$ and both sides vanish, so the formula remains valid. The argument also covers complex coefficients $g(p)$; a real-valued assumption is needed for the later normal distribution conclusion, not for this centered moment comparison for additive arithmetic functions.

A standard real Brownian motion relative to a filtration $({\mathcal{F}}_t)$ is an adapted stochastic process with $B_0=0$ almost surely, continuous paths almost surely, andEquivalently, it has independent increments and stationary increments, each with a normal distribution of mean zero and variance equal to the elapsed time, together with continuous paths and the stated initial value. The filtration may be its natural filtration. A Brownian motion started at $z$ is $z+W_t$ for a standard process $W$; multidimensional standard Brownian motion has independent standard real coordinate processes. The normalization is variance $t$ at time $t$.

Let $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$, $H=X(X^{T}X{)}^{-1}{X}^T$, ${\hat{y}}_i=x_i^T\hat{\beta }$, and $e_i=y_i-{\hat{y}}_i$. The horizontal axis of the residual-versus-fitted plot is ${\hat{y}}_i$, the fitted values, and the vertical axis is $e_i$, the regression residual, in millimetres. Under the normal linear model, $\mathrm{Var}(e_i)={\sigma }^2(1-h_{ii})$, where $h_{ii}$ is the regression leverage.

In the quantile-quantile plot, the vertical values are the ordered standardized regression residualsand the horizontal values are corresponding theoretical quantiles ${\Phi }^{-1}(p_k)$ of the standard normal distribution, with plotting positions such as $p_k=(k-1/2)/n$. The exact plotting-position convention has little practical effect here. Under Gaussian errors these points should approximately follow a straight line; they are not independent because fitting induces residual correlations.

The main visible concern is curvature in the conditional mean. The red smooth in the residual-versus-fitted plot descends from positive residuals at low fitted values, becomes negative in the middle, and rises again at high fitted values. This suggests the strictly linear time effects may be inadequate. There is no clear monotone widening of the residual scatter, so strong heteroscedasticity is not apparent. The quantile-quantile plot has modest tail deviations and a few labelled observations, but does not show a dramatic departure from normality. These plots cannot establish independence across days or laboratories; check residuals against day, laboratory, and sampling order as well. A large coefficient of determination does not remove the visible mean-model concern.

For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0{)}^T$. Its fitted mean isWrite $h_0=x_0^T(X^{T}X{)}^{-1}{x}_0$. The prediction interval in a normal linear model iswhere $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha /2}$ is the indicated Student's t-distribution quantile. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient standard errors alone: the coefficient covariances are also needed.

To prove coverage, write $Y_0=x_0^T\beta +{\epsilon }_0$, with ${\epsilon }_0\sim N(0,{\sigma }^2)$ independent of the original sample. The error $Y_0-x_0^T\hat{\beta }$ has a normal distribution with mean zero and variance ${\sigma }^2(1+h_0)$. By the joint distribution of least-squares and variance estimators, $26s^2/{\sigma }^2\sim {\chi }_{26}^2$ and is independent of $\hat{\beta }$; it is also independent of ${\epsilon }_0$. Hence the pivotal quantityConsequently $\mathrm{Pr}(-t_{26,1-\alpha /2}\le T\le t_{26,1-\alpha /2})=1-\alpha$, which is exactly the claimed prediction interval coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean confidence interval.

The fitted formula and coefficient table describe a random-slope linear mixed model:with random effects independent of the errors. There is a common fixed intercept, a population slope ${\beta }_1$, and a laboratory random slope deviation $b_j$. In tumour1, the three laboratory slopes are unrelated unknown fixed coefficients. Here they are modelled as draws from a common normal distribution; estimated laboratory deviations are shrunk toward zero. After integrating out $b_j$, observations in laboratory $j$ have covariance ${\tau }^2{t}_{ij}{t}_{kj}+{\sigma }^2{1}_{\{i=k\}}$, so they are marginally dependent even when their conditional errors are independent.

The displayed command contains a source inconsistency: size ~ 1 + (0 + days | lab) would impose zero population slope and could not produce the displayed fixed days coefficient. The output corresponds to size ~ days + (0 + days | lab). Its estimates are ${\hat{\beta }}_0=0.6464$, ${\hat{\tau }}^2=17.08$ and ${\hat{\sigma }}^2=227.76$. The estimated daily increase averaged over the laboratory population isbecause $\mathbb{E}{b}_j=0$.

A random slope is appropriate if these laboratories can reasonably be viewed as exchangeable representatives of a larger population and the goal includes population inference or predictions for other laboratories. If these particular three laboratories are the whole set of interest, or have systematic differences incompatible with exchangeability, laboratory-specific fixed slopes are preferable. Only three groups give weak information about between-laboratory variance; the mixed model does not itself resolve possible nonlinear growth. Its population mean is distinct from a realized finite-group average ${\beta }_1+(b_A+b_B+b_C)/3$.

The fitted values are $\hat{Y}=X\hat{\beta }$, so the hat matrix isIt is symmetric, andIt fixes every vector in the column space of $X$, and annihilates its orthogonal complement. Thus $H$ is the orthogonal projection onto that column space, has matrix rank and trace $p$, and $I-H$ is the complementary orthogonal projection. ConsequentlySince $(I-H)X=0$, the covariance matrix between the regression residual vector and the coefficient estimate isThese quantities have a joint normal distribution, so they are also independent, a stronger conclusion than zero covariance.

Now $\hat{e}=(I-H)\epsilon$ has mean zero. Using the expectation of a quadratic form,Thus the maximum-likelihood estimator hasThe unbiased alternative isThe orthogonal projection of a Gaussian vector also yields $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$, explaining the residual degrees of freedom.

Check that the chosen mean structure adequately describes the transformed response. Examine the residual-versus-fitted plot and residuals versus each predictor: systematic curvature suggests omitted nonlinear effects or an interaction term. Plot residual spread against fitted values, for example using a scale-location plot, to assess homoscedasticity. Compare standardized residuals with a normal distribution using a quantile-quantile plot, especially for finite-sample Student's t-distribution and F-test inference.

Also check independence using the sampling design and residuals versus time, collection site, or other groups; spatially related photovoltaic systems may have correlated errors that are invisible in a residual-versus-fitted plot. Investigate large standardized regression residuals, high regression leverage, and influential observations using Cook's distance. Check that the design matrix has full rank and that severe multicollinearity is not making estimates unstable. Reassess whether the response transformation and retained predictors improve these diagnostics and prediction; a coefficient p-value alone cannot establish model adequacy. If observations are independent only conditional on site effects, use an appropriate dependence model rather than treating correlated systems as independent replicates.

The residual smooth rises from negative values, arches above zero, then falls at high fitted values. This suggests a nonlinear altitude effect on the log mean. Since the fitted log mean in the linear model increases with altitude, the horizontal fitted-value axis is also an increasing rescaling of altitude. A generalized additive model with Quasi-Poisson regression is therefore a reasonable candidate:The centering constraint identifies the intercept. Here bs="cr" uses a penalized cubic regression spline. The plotted count residuals are deviance residuals, with the Q-Q panel using leverage-standardized versions. The quantile-quantile plot also shows tail discrepancies, but normality of count residuals is not a distributional assumption of this model: its main requirements are an adequate mean, variance relation, and independence. Investigate the labelled islands and possible excess variation as well.

The estimated smooth in the second plot increases fairly rapidly at low altitude, flattens around altitude 20 to 25, and is nearly level at the upper end. The pointwise bands widen where there are fewer observations. A penalized spline can describe a rise followed by a plateau without forcing a global parabola. A quadratic log mean has slope ${\beta }_1+2{\beta }_{2}a$, so negative curvature eventually forces a decline and its curvature is constant everywhere. A cubic regression spline allows different curvature over different ranges, with a penalty controlling unnecessary wiggles and natural linear tails outside its boundary knots. The displayed smooth supports this greater flexibility, but it is not by itself a formal rejection of every quadratic model. Select complexity and compare predictive performance using cross-validation, then reassess residuals and the Pearson dispersion estimator.

Write $t_{ij}$ for stirring rate of observation $i$ in furnace $j$. The independent random-intercept and random-slope model isErrors and random effects are independent. The separate R terms (1 | furnace) and (0 + stir | furnace) impose independent random intercepts and random slopes; (1 + stir | furnace) would instead estimate their covariance as well. The estimates areThe corresponding estimated standard deviations are $6.9813$, $0.1147$, and $6.0249$; they are not additional parameters.

For a furnace with predictor vector $t_j$, the marginal formulation, obtained by integrating the Gaussian random effects, isindependently across furnaces. In particular,In stacked notation $Y=X\beta +Zb+\epsilon$, $b\sim N(0,D)$, giving the Gaussian linear mixed model marginal law $N(X\beta ,V)$ with $V=ZD{Z}^T+{\sigma }^{2}I$.

The residual-versus-fitted plot on page 17 has residuals of both signs over the fitted range, with no convincing smooth curvature or clear fan shape. Its quantile-quantile plot is roughly straight in the central region, with some tail departures and a few large negative residuals. There is no decisive visible violation, but investigate those observations. These plots mainly address conditional observation errors; they do not validate the distribution of furnace random effects or independence within furnaces. With only three furnaces, normality and the variance of the random effects are especially difficult to assess.

Use the usual clinical trial sampling assumption that all patients' responses are independent, with fixed positive arm sizes and positive known variances. Marginal normal distributions alone would not specify the variance of a treatment contrast without this assumption. Put ${\overline{Y}}_i=n_i^{-1}{\sum }_j{Y}_{ij}$, $v_i={\sigma }_i^2/n_i$, and $s_i=\sqrt{v_i+v_0}$. The sample mean contrast estimates ${\delta }_i$ and has standard error $s_i$. Thus the one-sided Wald statistics areLarge positive values of these Wald test statistics provide evidence against the respective null hypotheses.

The printed strong white noise assumption does not imply a normal distribution or even a finite fourth moment. Thus it does not, by itself, determine the covariance of a quadratic transform. We give the intended Gaussian calculation, and then the general finite-fourth-moment answer.

If $\eta$ is Gaussian, $X$ is a centered Gaussian process. Put $v={\gamma }_X(0)=136/15$ and $U_t=X_t/\sqrt{v}$. Since $H_1(u)=u$ and $H_2(u)=u^2-1$, the centered transform isThe supplied Hermite polynomial identity makes the cross terms vanish and givesEquivalently,The constant $a$ has no effect on covariance.

For a general iid noise with $E{\eta }^4<\infty$, put $m_3=E{\eta }^3$ and ${\kappa }_4=E{\eta }^4-3$, its fourth cumulant. Independence and expansion of third and fourth moments give, for $k=|h|$,whereSumming the geometric series explicitly givesThe covariance of quadratic transforms of a linear process follows from a fourth-moment expansion consisting of the three Isserlis theorem pairings, plus the fourth-cumulant contribution when all four noise indices coincide. The third-moment contribution similarly requires three coincident indices. This proves the general formula without assuming a normal distribution. For Gaussian white noise $m_3={\kappa }_4=0$, recovering the simpler answer. If $c\ne 0$ and the noise has infinite fourth moment, $Z_t$ need not have finite variance, so an autocovariance function may not exist.

For a causal autoregressive model of order $p$, the population partial autocorrelation function cuts off after lag $p$, whereas the autocorrelation function generally decays. For an invertible moving-average model of order $q$, the autocorrelation function cuts off after lag $q$, whereas the partial correlation coefficients generally decay. For a mixed autoregressive moving-average model, both generally decay. Exponential decay may alternate in sign or show damped oscillations. Sample correlation coefficients only approximate these patterns, so isolated crossings of the significance bands are not exact order tests.

In PDF Figure 1, the autocorrelation function alternates sign, with a large negative lag-one correlation coefficient and geometrically decreasing magnitude. The partial autocorrelation function has essentially one substantial spike, negative at lag one and about $-0.8$; later values mostly lie within the displayed bands. Thus a stationary AR(1) with a negative coefficient is the natural first model, with $\hat{\varphi }$ initially near $-0.8$. The trace fluctuates around an approximately constant mean and shows no evident deterministic trend.

Fit that model, compare nearby low-order alternatives using likelihood function and an Akaike information criterion or Bayesian information criterion, and inspect the residual autocorrelation function and residual variance. A few small later PACF spikes are expected in a plot with many lags. The roughly $\pm 1.96/\sqrt{T}$ white noise bands are a guide, rather than simultaneous guarantees for every lag.

The sign alternation also suggests concentration of spectral power near the high-frequency end: the AR(1) denominator $1+{\varphi }^2-2\varphi \cos \lambda$ is smallest near $\lambda =\pi$ when $\varphi <0$. From estimated $\gamma (0)$ and lag-one correlation coefficient one can estimate innovation variance as ${\hat{\sigma }}_{\epsilon }^2\approx \hat{\gamma }(0)(1-{\hat{\varphi }}^2)$. The persistence parameter controls the decay rate and long-run variance of a stationary process. These plots do not establish a normal distribution, independence, or the absence of nonlinear dependence; correlation coefficients of squared residuals can provide a separate variance diagnostic.

Put ${\delta }_i=t_i-t_{i-1}$, $\Delta ={\max }_i{\delta }_i$, $v_i={\int }_{t_{i-1}}^{t_i}\sigma (s{)}^{2}ds$ and $S=\Vert {\sigma }^4{\Vert }_{\infty }$. The deterministic integrand in the Itô integral implies that the increments $Z_i=X_{t_i}-X_{t_{i-1}}$ are independent random variables with normal distributions $N(0,v_i)$. The Gaussian fourth moment givesThese identities include $v_i=0$. Since the summands of $M_n$ are centered and independent, all cross terms in its second moment vanish. Consequently,Thus $D=2$ works for every observation partition. The last step uses the total interval length, rather than assuming equally spaced observations.

Because the scalar signal and the noise vector are independent random variables with normal distributions, their sum is a Gaussian random vector. Its expected value is zero, and the covariance matrix isThus this is a rank-one covariance spike. Each summand $X_i{X}_i^{\top }$ has expected value ${\Sigma }_S$, so $\mathbb{E}\hat{\Sigma }={\Sigma }_S$. Here the empirical matrix is an uncentered empirical second-moment matrix; because the population expected value is known to be zero, it is an unbiased estimator of the covariance matrix. The TeX's apparent factorial denominator is a transcription error.