Let $M_j=s^{\prime \prime }(j)$. A natural cubic spline has $M_0=M_2=0$. The interior spline equation on equal unit intervals isso $M_1=6$. The standard second-derivative interpolation formula, or direct integration of the linear second derivative, givesThe two pieces have equal value, first derivative, and second derivative at one, and $s^{\prime \prime }$ vanishes at zero and two.

Adding the linear function $x$ changes the nodal data by $(0,1,2)$ without affecting second derivatives. By uniqueness, the second requested natural cubic spline is $s(x)+x$, with pieces $x^3$ and $-x^3+6x^2-6x+2$ respectively.

Use Gaussian elimination with partial pivoting, choosing the largest absolute available entry in the current pivot column. The first pivot is 2, with multipliers $1/2,1/2,1/5$. After eliminating that column, the augmented system isThe second pivot is $-1$, already largest in its remaining column. Its multipliers are zero and $2/5$, giving row four $(0,0,3,-2\mid 0)$. In column three, partial pivoting now requires exchanging rows three and four, since $3>3/4$. The final elimination multiplier is $-1/4$. ThusBack substitution gives $\boxed{(w,x,y,z)=(1,-1,0,0)}$.

The permutation and recorded multipliers, with the earlier multiplier rows swapped at the third pivot, areDirect multiplication verifies $PA=LU$. Since $U$ is invertible and $A=P^{T}LU$,The two matrices are similar, and similar matrices have the same characteristic polynomial and eigenvalues.

The Neyman-Pearson lemma says that for testing a simple null density $f_0$ against a simple alternative $f_1$, a size-$\alpha$ test that rejects for the largest values of $f_1/f_0$ is most powerful. Randomization on a likelihood-ratio boundary can be used to achieve the exact size.

The source writes the same ${\lambda }_0$ in both hypotheses but immediately introduces ${\lambda }_1>{\lambda }_0$. Use the intended alternative $\lambda ={\lambda }_1$. If read literally as two identical hypotheses, there is no such distinct alternative to test.

For Weibull shape two, the sample likelihood is proportional toThuswhich is increasing in $S={\sum }_i{X}_i^2$. The most powerful critical region is thereforeIndeed, $2S/{\lambda }_0^2\sim {\chi }_{2n}^2$, so $C=({\lambda }_0^2/2)q_{2n,1-\alpha }$. The critical value depends only on the null and size, not on the chosen larger alternative. Hence the same test is uniformly most powerful for $\lambda >{\lambda }_0$, and in particular for the printed restricted range $\lambda >{\lambda }_1$.

Differentiating the least-squares residual norm gives the normal equationsIf $X$ has full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$, and thereforeHere $\bar{x}=0$ and $\bar{z}=2$, so the design matrix and its Gram matrix areThe centered predictor columns are perpendicular, givingBoth estimates are linear combinations of the jointly normal errors, hence jointly normal. Zero covariance therefore also gives independence. The normality assumption is what permits this last step; orthogonality alone would only prove uncorrelatedness.

A Poisson process of rate $\lambda >0$ starts from zero, has independent stationary increments, has unit jumps, and satisfies $N_t-N_s\sim \mathrm{Poisson}(\lambda (t-s))$ for $t>s$. Its inter-arrival times are independent with the exponential distribution of rate $\lambda$: any finite joint density is ${\prod }_i\lambda e^{-\lambda x_i}$ on positive coordinates.

The second arrival is a sum of two such variables. Convolution givesThis is a gamma distribution of shape two and rate $\lambda$. If $S$ is the first arrival of the independent rate-$\mu$ process, exponential independence gives

The Markov property says that, conditional on the present state, the next state is independent of the earlier history:Two states communicate if each is reachable from the other with positive probability in finitely many steps. The equivalence classes are communicating classes; a class is closed if no positive-probability transition leaves it.

The displayed probabilities require $0\le p\le 1/2$; negative values allowed by the printed one-sided inequality are not transition probabilities. For $p=0$, every state is an absorbing singleton, so all singleton classes are closed. For $0<p\le 1/2$, neighboring states communicate in both directions, so there is one closed irreducible class $S$ and no open classes.

For $m=3$, the detailed balance relations give $2{\pi }_0={\pi }_1={\pi }_2=2{\pi }_3$. Normalization yieldsFor $0<p<1/2$, positive holding probabilities make the finite irreducible chain aperiodic, so its transition probabilities converge to ${\pi }_j$ from every initial state. At $p=1/2$, every step changes parity and the period is two; convergence from a fixed initial state fails. A stationary initial mixture still has distribution $\pi$ at every time, but this does not give convergence of the transition probabilities asked for.

The inverse change of variables is $x=2u+v$, $y=v$, with absolute Jacobian determinant two. Positivity of $x,y$ givesThe change of variables for probability densities givesFor fixed $u$, integrate over $v>\max (0,-2u)$:Thus $\boxed{f_U(u)=\frac{1}{2}{e}^{-|u|}}$, the standard Laplace distribution.