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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 ${\beta }_1+(b_A+b_B+b_C)/3$.

Retain the random-slope linear mixed model of part (e), but replace the within-laboratory independent error covariance by continuous-time autoregressive residual correlation:Errors from different laboratories remain independent and are independent of the random slopes. The resulting marginal covariance within laboratory $j$ isThe exponential decay applies to the conditional errors, not the entire marginal correlation, which also contains the shared random slope.

Use nlme, since lme4::lmer does not fit this general residual correlation structure. With a data frame containing the observed variables, suitable code isHere corCAR1 estimates $\rho =e^{-\kappa }\in (0,1)$, the error correlation one day apart; correlation at separation $\Delta$ is ${\rho }^{|\Delta |}$. This works for noninteger or unequally spaced days. At equally spaced integer days, corAR1(form = ~ days | lab) gives the corresponding discrete-time construction when its parameter is positive. Distinct times within each error series are required: separate tumours with repeated day values should have separate series identifiers or an additional independent-error component, rather than being assigned perfectly correlated errors. The continuous-time autoregressive residual correlation documentation describes the grouping and continuous-time conventions: stat.ethz.ch/R-manual/R-patched/library/nlme/html/corCAR1.html. The fixed days term here follows the fitted output rather than the inconsistent command in part (e).

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.

The three independent sample means satisfy ${\overline{Y}}_i\sim \mathcal{N}({\mu }_i,v_i)$. A linear transformation of their joint multivariate normal distribution is again normal. The two treatment contrasts have covarianceWriting $m_i={\delta }_i/s_i$ and $\rho =v_0/(s_1{s}_2)$, their joint distribution isThe positive correlation coefficient arises from the shared control arm. Under positive variances, $0<\rho <1$.

Use the original-noise coefficients ${\psi }_0=1$ and ${\psi }_j=-11\cdot 4^{-j}$ for $j\ge 1$. The autocovariance function of a causal linear process is ${\gamma }_X(h)={\sum }_{j\ge 0}{\psi }_j{\psi }_{j+|h|}$, since the original noise has variance $1$. ThusFor $h\ge 1$,By covariance symmetry,In particular, all nonzero-lag covariances are negative, despite the positive autoregressive coefficient.

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.

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 $\mu$.

With angular frequency $\lambda$, the periodogram isAt Fourier frequencies $2\pi j/T$ it is the squared modulus of the normalized discrete Fourier transform. If a nonzero mean is unknown, subtract the sample mean first.

For a zero-mean stationary process, expanding the square givesUnder absolute summability of the covariances this converges to the spectral density of a stationary process. Thus the periodogram is generally biased at finite $T$, but asymptotically unbiased under this short-memory time series condition.

It is nevertheless not a pointwise estimator with statistical consistency unless it is smoothed. For Gaussian white noise, at a nonzero Fourier frequency other than the Nyquist frequency, the real and imaginary Fourier components are independent normal variables. Exactly,where $f={\sigma }^2/(2\pi )$. The variance does not decrease with $T$. Under usual short-memory time series assumptions the same exponential limit is asymptotic for general processes. Averaging nearby frequencies or using a lag-window estimator reduces variance; a frequency bandwidth tending to zero while $T$ times that bandwidth tends to infinity can give statistical consistency. The raw plot remains useful for detecting strong periodic peaks, but increasing the record length alone does not remove its pointwise noise.

A nonstationary process has a statistical law that changes with the time origin. In the weak sense, this includes a time-dependent mean or variance, or a covariance that depends on the two times separately rather than only on their lag. A strictly stationary process requires invariance of every finite-dimensional distribution under a common time shift.

In the preceding plots, series $1$ suggests a stochastic trend or changing level, and series $4$ suggests a deterministic seasonal mean. These are the intended nonstationary examples. A plot alone cannot prove either conclusion: in particular, a stationary random-phase sinusoid can have a seasonal-looking trace and an oscillatory ACF. The interpretation of series $4$ as nonstationary concerns a seasonal mean tied to calendar time.

Three standard responses are to remove a fitted deterministic trend or seasonal mean; to use regular or seasonal differencing for an appropriate trend or seasonal component; and to stabilize a changing variance by a transformation or explicit seasonal scale model. For example, logarithms or a Box–Cox transformation can address level-dependent variance. The operation should match the source of nonstationarity; differencing a varying variance does not generally make it stationary.

Use the usual normalization that ${\epsilon }_t$ is mean-zero unit-variance white noise orthogonal to the past, and extend $\varphi (\nu ),\sigma (\nu )$ periodically in $\nu$. A nondegenerate causal periodic autoregressive model of order one requires $|{\prod }_{\nu =1}^T\varphi (\nu )|<1$. Write $V_{\nu }={\gamma }_{\nu }(0)$, with season indices understood modulo $T$.

For $h\ge 1$, multiply the recursion by $X_{nT+\nu -h}$ and use orthogonality of the current innovation to the past. The periodic Yule-Walker equations areAt lag zero, expansion of the squared recursion givesIn particular, ${\gamma }_{\nu }(1)=\varphi (\nu )V_{\nu -1}$, so an equivalent variance equation is $V_{\nu }=\varphi (\nu ){\gamma }_{\nu }(1)+\sigma (\nu {)}^2$. Negative lags are obtained from the given symmetry ${\gamma }_{\nu }(-h)={\gamma }_{\nu +h}(h)$.

Replacing covariances by their sample versions givesTake the nonnegative scale and require a positive estimated predecessor variance. Using matched seasonal sample covariance matrices makes the quantity under the square root nonnegative by Cauchy-Schwarz inequality; with unmatched covariance estimates one can instead fit seasonal ordinary least squares and use the nonnegative residual variance. Statistical consistency of the covariance estimates gives statistical consistency of these plug-in estimators. If the driving noise variance were not normalized to $1$, the equations would identify $\sigma (\nu {)}^2\mathrm{Var}({\epsilon }_t)$, not the two scales separately.

Put $c_t=\cos (2\pi t/S)$ and assume ${\sigma }^2>0$. ThenThus $Y$ generally has a periodic mean, and $Z$ generally has a periodic variance. The edge cases allowed by $S\in {\mathbb{N}}^{\ast }$ matter: $Y$ is stationary when $S=1$, because the mean is then constant, and $Z$ is stationary for $S=1$ or $S=2$. In the latter case multiplying iid centered Gaussian white noise by $(-1{)}^t$ leaves its iid distribution unchanged. For every $S\ge 3$, $c_0^2=1$ but $c_1^2<1$, so $Z$ is nonstationary. For $S>1$, $Y$ is nonstationary because $c_t$ is not constant.

Use the seasonal difference operator ${\Delta }_S=1-B^S$. Since $c_t=c_{t-S}$,This is a stationary moving-average model of iid Gaussian white noise. Its covariance is $2{\sigma }^2$ at lag $0$, $-{\sigma }^2$ at lags $\pm S$, and zero otherwise.

For the variance-modulated process the same operation givesIts variance is $2{\sigma }^2{c}_t^2$ and its covariance at lag $S$ is $-{\sigma }^2{c}_t^2$. Therefore it remains nonstationary for $S\ge 3$, and the operation doubles the marginal variance at nonzero seasons. Seasonal differencing does not remove periodic variance: it removes a periodic mean, but the periodic variance generally remains. Practical alternatives are a periodic model or seasonal variance standardization; at seasons with $c_t=0$ the observations are deterministically zero, so division by $c_t$ is not possible there. For the already stationary special cases $S=1,2$, filtering preserves stationarity.

A periodically correlated process with period $S$ has its mean and two-time covariance unchanged when both times are shifted by $S$. For $Z$, the mean is zero andHence $Z$ is a periodically correlated process. Its covariance period can be smaller than $S$: when $S$ is even, $c_{t+S/2}^2=c_t^2$, so $S/2$ is already a covariance period.