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Use the printed parameterization $\lambda ={\sigma }^2$, $v={\sigma }_0^2$, and $a=\sqrt{\rho }$. Let $T=365$, $X$ be the design matrix with rows $x_i^T$, and $r=\theta -X\beta$. Conditional on $\theta$ and the parameters, $Y$ contributes no extra dependence on $Z$ by conditional independence.

The stationary autoregressive normal distribution has prior density proportional toIts Gaussian autoregressive conditional precision is the tridiagonal matrix $J$ withCombining this prior with $r\mid Z\sim N(\lambda Z,vI)$ and completing the square givesThe precision matrix $Q$ is positive definite. Compute its lower Cholesky decomposition $Q=L{L}^T$, solve $Qm=h$, draw $\xi \sim N(0,I_T)$, and solve $L^T\eta =\xi$. Then return $Z=m+\eta$. This Gaussian sampling from a precision Cholesky factor is exact because $\mathrm{Cov}(\eta )=L^{-T}{L}^{-1}=Q^{-1}$. Tridiagonal structure gives linear cost in $T$ and preserves the conditional correlation between days. Notice that the solve uses $L^T$, not $L$.

These conditionals are normalized for every fixed positive ${\sigma }^2,{\sigma }_0^2$, despite the joint posterior impropriety explained in part (a). If the intended model uses $\sigma Z_i$ instead, replace the observation coefficient $\lambda$ by $\sigma$ throughout; the information and precision terms then use $\sigma /v$ and ${\sigma }^2/v$.

A procedure controls the familywise error rate at level $\alpha$ when, for every configuration of true nulls $I$,Let $i_{\ast }=\min I$. The step-down procedure can reject any true null only if it reaches and rejects $H_{i_{\ast }}$, which requires $p_{i_{\ast }}\le \alpha$. Since a valid p-value under its null satisfies $\mathbb{P}(p_{i_{\ast }}\le \alpha )\le \alpha$, the procedure controls FWER without any assumption on dependence among the p-values.

A distribution $P$ has the global Markov property for a directed acyclic graph $G$ when every D-separation statement $A{\perp }_{G}B\mid C$ implies the corresponding conditional independence $Z_A\perp \perp Z_B\mid Z_C$ under $P$.

If $G\in \mathcal{S}$ has nonadjacent vertices $j,k$, choose a topological ordering. Suppose $j$ precedes $k$. Then $j$ is a non-descendant and non-parent of $k$, so the directed local Markov property givesThe reversed ordering case is analogous. HenceUse the intersection-union test: reject $H_0$ exactly when $p_S\le \alpha$ for every $S$. Under $H_0$, at least one $H_{S_{\ast }}$ is true, soThis is nontrivial because it rejects whenever every tested conditional independence is rejected.

Conditional independence givesFor each satellite, Bayes' theorem gives $P(d_s\mid m)\propto P(m\mid d_s)/P(m)$. ThereforeIf ${\hat{p}}_0(m)$ is a kernel density estimator for the simulated marginal masses and ${\hat{p}}_s(m)$ estimates the posterior based on satellite $s$, thenThe one-dimensional integrals can be evaluated by numerical integration on a common mass grid.

For one object, the probabilistic graphical model factorization isEach factor is the normal distribution density specified by the model. This factorization displays the conditional independences of the latent variables ${\xi }_i,{\eta }_i$ and the noisy observations $x_i,y_i$.

A distribution is faithful to a Directed acyclic graph $G$ when every conditional independence in the distribution is implied by D-separation in $G$. Together with the graphical Markov property, faithfulness makes conditional independence equivalent to D-separation and rules out independences caused only by exact parameter cancellation.

Under $H_0$, conditional independence givesMultiplication by the measurable sign $s(z_1)$ and the law of total expectation prove the first identity.

WriteThenConditional orthogonality under $H_0$ and the variance bounds in (ii) giveand similarly for the term containing ${\xi }_i{a}_i$. The Cauchy-Schwarz inequality bounds the last term bywhich is $o_p(n^{-1/2})$ by $n{\mathrm{MSPE}}_f{\mathrm{MSPE}}_g\to 0$.

The leading summands ${ϵ}_i{\xi }_{i}s(z_i)$ are independent, centered, and have variance $\mathrm{Var}({ϵ}_1{\xi }_1)$ because $s^2=1$. The central limit theorem and consistency of ${\tau }_D$ therefore give, by the Slutsky theorem,This is the generalized covariance measure statistic.

Without the null, condition first on $(x_1,z_1)$. Sincethe term involving $\mathbb{E}(x_1\mid z_1)$ vanishes, givingFor the specified alternative, independence and $\mathbb{E}{x}_1=0$ make the right sideThe constant choice $s=1$ gives zero because $\mathbb{E}h(z_1)=0$, so no first-order power is expected. Takinginstead gives $\mathbb{E}(x_1^2)\mathbb{E}|h(z_1)|>0$, producing asymptotic power.

Under the null, ${\epsilon }_1$ and ${\xi }_1$ have conditional independence given $z_1$. The law of total expectation gives

Conditional on the covariates and all responses used to construct $\hat{g}$, the residual ${\epsilon }_i$ has conditional second moment at most $C$; conditional independence under the null is what permits this conditioning. ConsequentlyThe Markov inequality proves

Next, the Cauchy-Schwarz inequality givesThe first factor converges to zero in probability. The weak law of large numbers and part a make the second $O_p(1)$, so the product converges to zero in probability.

Conditional independence $A\perp B\mid D$ means that, for almost every $d$,equivalently $p(a\mid b,d)=p(a\mid d)$ wherever the conditional probabilities are defined.

Now use the chain rule and both assumed independences:This is exactly $A\perp (B,C)\mid D$, proving the contraction axiom for conditional independence.

Let $K={\Sigma }^{-1}$ be the precision matrix, let $R=\{1,\dots ,p\}\setminus \{j,k\}$, and condition on $V_R=v_R$. In the Gaussian exponent, all terms depending jointly on $v_j$ and $v_k$ are contained inIf $K_{jk}=0$, this conditional density is a product of one function of $v_j$ and one function of $v_k$, so the conditional independence of $V_j$ and $V_k$ given $V_R$ holds.

Conversely, conditional independence makes this everywhere-positive conditional density factorize. Its mixed second derivative must therefore vanish:Hence the conditional independence holds exactly when $({\Sigma }^{-1}{)}_{jk}=0$.

The PC algorithm starts with a candidate undirected graph, removes edges using conditional independence tests over increasing conditioning-set sizes, records separating sets, then orients unshielded colliders and propagates orientations. Under oracle tests and faithfulness of a directed acyclic graph, it estimates the graph's Markov equivalence class.

With an exact conditional independence oracle and faithfulness of a directed acyclic graph, the skeleton phase of the PC algorithm deletes precisely the nonedges. A true edge cannot be D-separated by a set excluding its endpoints. For nonadjacent vertices, choose the later vertex $v$ in a topological ordering; its parents separate it from the earlier nonparent. True parent edges survive throughout, so this separator remains available among the algorithm's candidate neighbour sets. For an unshielded triple $u-v-w$, any recorded separator of $u,w$ excludes $v$ exactly when the triple is an unshielded collider. The skeleton and collider characterization of Markov equivalence therefore identifies the correct equivalence class. Directions compelled throughout that class give its completed partially directed acyclic graph.

An unshielded collider in a Directed acyclic graph is a triple $a\to b\leftarrow c$ with $a$ and $c$ nonadjacent. The absence of an edge between its outer vertices makes this orientation detectable from conditional independence information under faithfulness of a directed acyclic graph.