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Let $W_n$ be the interpolation in part (a). The operationis continuous, since $|I(f)-I(g)|\le \Vert f-g{\Vert }_{\infty }$. By the Donsker invariance principle and the continuous mapping theorem,The exact trapezoidal integral of the linear interpolation isTherefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using independence, zero means, and unit variances givesThis error tends to zero in $L^2$, hence in probability. The Slutsky theorem now proves the integrated random-walk limit:The limiting law can also be made explicit. The time integral is a Gaussian random variable, as a mean-square limit of linear combinations of a Gaussian process. It is centered, and the covariance identity $\mathbb{E}(B_s{B}_t)=\min (s,t)$ givesThus the terminal value of integrated Brownian motion here has law $N(0,1/3)$.

For $0\le s<t$, linearity of the isonormal Gaussian process givesThe squared $L^2$ norm of this indicator is $t-s$, so part (a) yieldsAlso, the covariance formula gives $\mathrm{Cov}(W_s,W_t)=s\wedge t$, the Brownian covariance kernel. The Gaussian statement concerns the signed increment.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.

The structural parameters remain $m_0=2$, $v=2$ and $a=1/3$, so the credibility factor after $n$ years is $Z_n=n/(n+6)$. Writing $C_n={\sum }_{i=1}^n{x}_i$, the Bühlmann credibility estimate isThe target is $\mathbb{E}(X_{n+1}\mid \lambda )=\lambda$. It is also the best affine predictor of the next year's count, because its extra conditional noise has zero covariance with past observations.

For a real weakly stationary process, extend its autocovariance by ${\gamma }_{-k}={\gamma }_k$. The exact existence of a time-series spectral density condition is that its spectral measure of a stationary time series be absolutely continuous with respect to Lebesgue measure. Absolute summability ${\gamma }_0+2{\sum }_{k\ge 1}|{\gamma }_k|<\infty$ is a useful sufficient condition, not a necessary one.

We use the conventional angular-frequency density on $[-\pi ,\pi ]$, restricted to $[0,\pi ]$ by symmetry. Under the absolute-summability condition, the Fourier series and its inverse relation areThus $2{\int }_0^{\pi }f={\gamma }_0$. If a one-sided density is instead normalized to integrate to the full variance, use $g=2f$ and omit the factor two in the inverse formula. This is the positive-frequency spectral normalization convention difference. With merely an integrable spectral density, the inverse relation remains valid; one must not assume pointwise convergence of the unweighted Fourier series. Its Fejér sums recover the density in $L^1$:

If $X$ and $Y$ are independent stationary processes, the cross covariances vanish, soLinearity of the inverse Fourier series relation, or addition of the spectral measures of a stationary time series, gives $f_Z=f_X+f_Y$. Independence can in fact be weakened to zero cross-covariances at every lag.

For the ARMA$(1,1)$ representation, away from an uncancelled unit root the transfer function givesThe usual causal case has $|\varphi |<1$; the same expression holds for the two-sided stationary solution when $|\varphi |>1$. We work in the nondegenerate case $v,w>0$ and $\varphi \ne \pm 1$; cancellations and zero-noise cases are obtained by the appropriate reduced representation or limits. Since independent white noise of variance $w$ contributes $w/(2\pi )$, putThenThe white-noise addition to an ARMA(1,1) process problem is therefore the factorizationDefineBoth are positive under the stated nondegeneracy conditions. An invertible choice isThese satisfy $|\alpha |<1$, $\lambda (1+\alpha {)}^2=P$, and $\lambda (1-\alpha {)}^2=Q$. HenceIn particular, it is the autoregressive coefficient $\varphi$ that enters the terms from the added observation noise.

To obtain a representation of the actual $Z$, defineThe stable inverse exists because $|\alpha |<1$, and spectral filtering gives $f_{\xi }=\lambda /(2\pi )$. Thus $\xi$ is weak white noise, andIts variance can equivalently be writtenWhen $\alpha \ne 0$, also $\lambda =(v\theta -w\varphi )/\alpha$; when $\alpha =0$, necessarily $C=0$ and $\lambda =A$, so the latter quotient should not be used. The model may reduce in order through cancellation. Boundary limits with $P=0$ or $Q=0$ give $|\alpha |=1$ and need not be stably invertible; the displayed stable reconstruction applies to the nondegenerate case above.

Because both vectors are linear transformations of the same multivariate normal response, $(\hat{Y},e)$ is jointly multivariate normal. Its cross-covariance matrix isZero cross-covariance implies independence for jointly multivariate normal vectors, including singular ones. Therefore the fitted values and the entire vector of regression residuals are independent. The fitted-residual orthogonality identity gives the zero covariance; the normal distribution assumption is what upgrades it to independence.

Group means are independent random variables with normal distributionsThe resulting univariate sampling distributions areThe same variance calculation holds for every $i\ge 2$, since the difference involves two independent group means. Therefore the standard errors satisfyReplacing $\sigma$ by a common estimated error standard deviation preserves this ratio. Although the individual group means are independent, the contrasts ${\hat{\alpha }}_i$ share the baseline mean and have covariance ${\sigma }^2/J$ for distinct $i\ge 2$.

For subject $i=1,\dots ,100$ and $t_j\in \{0,2,4,6,8,10\}$, the random-intercept linear mixed model isAll subject random effects and measurement errors are mutually independent. The fixed effects ${\beta }_0,{\beta }_1$ describe the population mean baseline and weekly gain. Conditional on $b_i$, a person's observations have independent normal distributions; after integrating out $b_i$, their covariance is ${\tau }^2$ at distinct weeks and their common variance is ${\tau }^2+{\sigma }^2$. Therefore their correlation is ${\tau }^2/({\tau }^2+{\sigma }^2)$.

This Gaussian linear mixed model permits correlated repeated readings while keeping different subjects independent. The fitted standard deviations are $\hat{\tau }=21.67593$ kg and $\hat{\sigma }=14.23496$ kg, giving within-person correlation about $0.699$. All persons still have the same latent weekly slope in this model.

Allow each subject to have a separate baseline and slope through a correlated random-intercept and random-slope model:The subject random effects are independent of all measurement errors, and subjects are independent. The within-person covariance between times $s,t$ is ${\tau }_0^2+(s+t)\rho {\tau }_0{\tau }_1+st{\tau }_1^2$, with an additional ${\sigma }^2$ when the same reading is used twice.

Prefer the model with a random slope. The maximum likelihood estimation refits improve twice the log-likelihood by $227.9604$ while adding two covariance statistical parameters. Both the Akaike information criterion ($5165.779$ to $4941.819$) and Bayesian information criterion ($5183.367$ to $4968.200$) strongly favour it. The printed nominal likelihood-ratio test is overwhelming. The usual reference chi-squared distribution with two statistical degrees of freedom is not an exact regular calibration, since zero slope variance is a boundary and the intercept–slope correlation is then unidentified; a design-specific parametric bootstrap could calibrate it. This qualification does not undermine the substantial descriptive improvement shown by both information criteria.

The preferred fit estimates population mean baseline strength ${\hat{\beta }}_0=61.09286$ kg and weekly gain ${\hat{\beta }}_1=2.83143$ kg/week. The between-person baseline standard deviation is ${\hat{\tau }}_0=15.785582$ kg, and the between-person slope standard deviation is ${\hat{\tau }}_1=2.738666$ kg/week. Their estimated correlation $\hat{\rho }=0.117$ is weakly positive, giving random-effect covariance about $5.058$ kg${}^2$/week; its uncertainty is not supplied. The measurement-error standard deviation is $\hat{\sigma }=9.923280$ kg. The printed $-0.038$ instead describes the correlation between the estimated fixed effects, not between the subject random effects. These statistical parameter estimates come from restricted maximum likelihood; the model comparison uses the separate maximum likelihood estimation fits.

The population mean fitted trajectory is $61.09286+2.83143t$ kg. HenceUsing the reported slope standard error, an approximate 95% confidence interval for the population mean gain is $28.3143\pm 1.96(2.984464)=(22.46,34.16)$ kg; a Student t confidence interval with 499 statistical degrees of freedom is almost identical.

Successful individuals can improve far more than the average. Their latent ten-week gains have fitted normal distributionA clear estimate for an upper-performing group uses a between-person slope quantile: the 95th percentile is $28.3143+1.645(27.38666)\simeq 73.4$ kg, and the 97.5th percentile is about $82.0$ kg. The upper 5% of fitted underlying gains begin around 73 kg. These are person-to-person performance quantiles, not confidence intervals for the population mean. Identifying the best observed trainee would require that person's data or fitted subject random effects; the aggregate output cannot identify a literal maximum. If performance means the observed difference of endpoint readings, add the measurement-error variance $2{\sigma }^2$ to the latent gain variance.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).

Write $s=\sigma /\sqrt{n}$, $\mu =\delta /s$, and let $Z_1,Z_2$ be independent standard normal random variables obtained by centering and scaling the separate stage means. ThenThe second statistic uses all patients, so the statistics are correlated even though the stages' new observations are independent. Their covariance is $1/\sqrt{2}$ and each variance is one. This gives the exact bivariate normal distributionUnder the null hypothesis both means are zero. At $\delta ={\delta }^{\ast }$ replace $\mu$ by $\sqrt{n}{\delta }^{\ast }/\sigma$; the mean vector is $(\sqrt{n}{\delta }^{\ast }/\sigma ,\sqrt{2n}{\delta }^{\ast }/\sigma {)}^T$. In this group sequential design the full-sample statistic may be viewed as a potential statistic from the underlying sequence of outcomes, even on paths where recruitment stops.

Under missing completely at random, the observed second-year outcomes form a random subsample, so the direct pooled complete-case analysis estimate isIt is valid under missing completely at random because dropout no longer changes the marginal distribution of second-year use. Pooling is simpler than reporting separate conditional probabilities, but MCAR alone does not make this estimate more efficient than the estimate in the previous part. The fully observed first-year variable carries useful outcome information and can improve precision even when dropout is completely random.

To make the qualification explicit, write $X=Y_1$, $m(X)=P(Y_2=1\mid X)$, $\theta =E\{m(X)\}$ and $\rho =P(R=1)$ for the constant response probability. For $N$ patients the leading variance of the pooled estimator is $\mathrm{Var}(Y_2)/(N\rho )$. The standardization estimator in the preceding part has leading varianceIndeed its first-order centered contribution is $m(X)-\theta +R\{Y_2-m(X)\}/\rho$: the two terms have zero covariance, and their variances give that expression. The law of total variance therefore yieldsThe inequality is strict when there is dropout and first-year use predicts second-year use, as the distinct conditional rates suggest here. Thus the requested universal efficiency claim needs qualification. In the unrestricted joint binary-outcome model, maximizing the observed-data likelihood under either missing at random or missing completely at random gives the same standardization estimate $0.384889$: the all-patient first-year proportion and the two observed conditional second-year proportions maximize its factored outcome likelihood. Restricting the distinct missingness parameters to a common response probability affects their factor, not this estimate. The pooled value is the simple valid MCAR answer; retaining the first-year information gives the efficient MCAR answer and does not require replacing the previous estimate.

Both fits use the same Gamma distribution mean model and inverse link function. Their coefficient estimates, fitted means and hence their weight matrices coincide: the common dispersion parameter only multiplies the likelihood score by a scalar and therefore does not change its zero. The displayed calls likewise show the same family and formula; fixing dispersion in a summary changes the uncertainty calculation, not these coefficient estimates.

For $V(\mu )={\mu }^2$ and $g^{\prime }(\mu )=-{\mu }^{-2}$, the Fisher information weights areThus the common matrix $(X^{T}WX{)}^{-1}$ is multiplied by dispersion $1$ in the exponential case and by $\hat{\varphi }=0.3103711$ in the fitted gamma case. Taking square roots of the diagonal covariances givesFor instance $0.03607\times 0.5571\approx 0.02009$ for the intercept, and the same factor applies to all the coefficients. The standard errors are smaller because the fitted dispersion parameter is below one, not because a different mean function was fitted.

Series 1 wanders over a changing level rather than fluctuating around a stable local mean. Its sample autocorrelation function is strongly positive and decreases very slowly. This is the usual diagnostic evidence for an ordinary unit root: an autoregressive polynomial containing $1-z$, with a zero at $z=1$, and a stationary model after first differencing. The plots support an integrated model, rather than specifying the number of its remaining stationary autoregressive or moving-average terms.

Series 2 has a pronounced oscillation with period about six observations. Its sample autocorrelation alternates between large positive and negative values with little damping: approximately positive at multiples of six and negative halfway between. Together with the changing amplitude, this suggests a conjugate pair of unit-circle zeros nearThe associated real autoregressive factor is $1-2\cos (\pi /3)z+z^2=1-z+z^2$. A targeted filter $1-B+B^2$ removes this pair; the broader seasonal difference operator $1-B^6$ also contains it but introduces additional differencing factors. This is the oscillatory unit-root diagnosis from an undamped sample autocorrelation.

Thus Series 1 suggests a zero at 1; Series 2 suggests a conjugate pair on the unit circle at a seasonal frequency. These are model diagnoses, not deductions of exact roots from a finite sample. A stationary model very close to a unit root can look similar, and an undamped periodic covariance can also arise from a stationary random sinusoid. The figure does not identify exact orders or prove nonstationarity by itself.

Write $X_t={\sum }_{j\ge 0}{\psi }_j{Z}_{t-j}$, with innovation variance four. The generating function of this linear process isEquating coefficients gives ${\psi }_0=1$, ${\psi }_1=0.2$ and ${\psi }_j=-0.3{\psi }_{j-1}+0.28{\psi }_{j-2}$ for $j\ge 2$. Thus the first five coefficients, including lag zero, areFor all remaining lags, partial fractions giveThe coefficients are absolutely summable, so the series converges in mean square and defines the causal moving-average expansion. In the unit-variance convention of part (c), $X_t={\sum }_{j\ge 0}{c}_j{W}_{t-j}$ with $c_j=2{\psi }_j$; its first five coefficients are $(2,0.4,0.44,-0.02,0.1292)$. This is the two-geometric-coefficient expansion of a causal ARMA(2,1) process.

White noise orthogonality gives, for any integer $h$,To see this directly, expand the covariance of the two convergent series. Only matching noise indices contribute. Cauchy-Schwarz inequality makes the coefficient-product sum finite, justifying the covariance limit.

There is also a closed expression. Set $a=9/11$, $d=2/11$, $r=0.4$, $s=-0.7$. For $h\ge 0$,and negative lags follow by symmetry. Dividing by the expression at zero gives the same autocorrelation function.