Let $\alpha$ be a root in a splitting field. The preceding characteristic-$p$ identity gives $X^p-a=(X-\alpha {)}^p$, so there is only one distinct root. If the polynomial were reducible over $F$, choose a monic proper factor $g\in F[X]$ of degree $d$, with $1\le d<p$. In the splitting field it must be $g=(X-\alpha {)}^d$. Its coefficient of $X^{d-1}$ is $-d\alpha$, which belongs to $F$. Since $d$ is nonzero and thus invertible in the field $F$, this forces $\alpha \in F$, contradicting the absence of a root there. Therefore $X^p-a$ is irreducible over $F$.

In field characteristic $p$, the binomial theorem gives $(X-b{)}^p=X^p-b^p$, since the intermediate binomial coefficients are divisible by the prime number $p$. If $b\in F$ is a root of $X^p-a$, then $b^p=a$, and hence

Full column rank gives the least-squares estimatorWrite $A=(X^{T}X{)}^{-1}{X}^T$. Since $AX=I$ and $\mathrm{Cov}(ϵ)={\sigma }^{2}I$, we have $\hat{\theta }=\theta +Aϵ$ andThis calculation needs the stated uncorrelated equal-variance errors, but not normality.

Set $s=S_D/\sqrt{17}$. Symmetry of $t_{16}$ gives $P(T>-1.746)=0.95$, soThe requested upper confidence bound is thereforeThere is a direction inconsistency in connecting this particular interval to the stated treatment alternative. Since $D$ is before minus after, a blood-pressure reduction means $\mu >0$. An upper interval $(-\infty ,b)$ excludes the null value zero only when $b<0$, which instead supports $\mu <0$.

For the requested reduction alternative, use the lower 95% intervalIt excludes zero in the appropriate direction exactly whenThis is the correct level-5% paired one-sided $t$ test of no mean change against a reduction. The requested upper bound is valid as a confidence statement, but its ordinary exclusion-of-zero test has the opposite direction.

The sign follows the duality between confidence intervals and hypothesis tests for the paired t-test.

Each before/after pair concerns the same individual, so subject-specific baseline blood pressure can affect both measurements. Analyzing the within-person difference retains this pairing and removes a common baseline contribution. Under the stated independent normal-difference model,The normal-sample variance identity givesMoreover $\overline{D}$ and $S_D^2$ are independent, because the sample-mean projection and centered residual projection of a Gaussian vector are orthogonal and hence independent. Thus the standardized normal variable $N=\sqrt{n}(\overline{D}-\mu )/\sigma$ is independent of $U=(n-1)S_D^2/{\sigma }^2\sim {\chi }_{n-1}^2$. ConsequentlyThis distribution is free of both unknown parameters.

The Markov property and conditioning on the intermediate state give, for the full history through time $2n$,Condition this identity again on the smaller history $(Y_0,\dots ,Y_n)$. Its right-hand side is already determined by $Y_n=X_{2n}$, so the tower property givesThus the sampled sequence is a time-homogeneous Markov chain with transition matrix $P^2$. Checking only its one-time marginal distributions would not by itself establish the Markov property.

It is the every-second-step case of a subsampled Markov chain.

Write the stationary probabilities as $({\pi }_1,{\pi }_2)$. The equation $\pi P=\pi$ reduces to $a{\pi }_1=b{\pi }_2$. Together with ${\pi }_1+{\pi }_2=1$, this givesThe positive transition probabilities make the two states communicate, so the stationary distribution is unique.

Rewrite the function asThese are equal mixture weights on the moment-generating functions of exponential distributions of rates $1$ and $2$. Moment-generating functions finite near zero determine the distribution, so $Y$ has that mixture law. Consequently

For rate $\lambda >0$,This function is finite near zero, so its third derivative there equals the third moment. Therefore

For each fixed $(x,y)$, the function $z\mapsto z^3+z+xy-1$ is continuous, strictly increasing because its derivative is $3z^2+1>0$, and tends to opposite infinities at the two ends of the real line. The intermediate value theorem therefore gives exactly one zero.

For $F(x,y,z)=z^3+z+xy-1$, its partial derivative $F_z=3z^2+1$ never vanishes. The implicit function theorem gives a local $C^1$ solution near every point of the solution set. Uniqueness makes these local solutions agree, producing a global $g\in C^1({\mathbb{R}}^2)$. Implicit differentiation givesAt $(1,1)$ the root is $g(1,1)=0$, so