A natural cubic spline with the given knots is a $C^2$ function that is polynomial of degree at most three on each interval between consecutive knots and has $S^{\prime \prime }(x_0)=S^{\prime \prime }(x_4)=0$. With the usual whole-line convention, it extends affinely outside the extreme knots. The word natural refers to these endpoint second-derivative conditions; the interpolation values must be specified separately.

For $H_k(x)=(x-x_k{)}_{+}^3$, direct differentiation on either side of the knot givesBoth one-sided limits of the function and its first two derivatives at $x_k$ are zero, so these formulas are continuous there andIts third derivative jumps from zero to six, which is permitted for a cubic spline.

Subtract the two finite sums. In the shifted sum replace the summation index $k$ by $k-1$. For every interior index the new coefficient isby Pascal's identity; the two endpoint coefficients agree as well. ThereforeFor a degree-$d$ polynomial with leading coefficient $c\ne 0$, the backward difference has leading termSince $h>0$, this coefficient is nonzero for $d\ge 1$. Starting with $Q_{(0)}=P$, iteration gives degree $n-r$ and leading coefficient $c{h}^{r}n!/(n-r)!$ for $0\le r\le n$. The next difference of the resulting constant is zero:

Write $J_1,J_2,J_3$ for the actual successive rotations $J(1,2)$, $J(1,3)$, and $J(2,3)$, each selected using the currently transformed matrix. The first eliminates position $(2,1)$; the second eliminates $(3,1)$ while leaving row two unchanged. After these, the first column is $(r_{11},0,0{)}^T$. The final rotation of rows two and three preserves both zeros in that column and eliminates $(3,2)$. HenceA lower-triangular starting matrix does not force any further zeros above the diagonal. For example, the lower-triangular matrix with all entries on and below its diagonal equal to one gives, with the nonnegative pivot convention,which has all upper-triangular entries nonzero.

Inverting the product of rotations yields the QR factorizationThe order reverses when taking the transpose. If $A$ is orthogonal, then $R$ is both orthogonal and upper triangular. Its first column forces $|r_{11}|=1$ and orthogonality with the other columns forces $r_{12}=r_{13}=0$. Applying the same argument to the remaining block makes $R$ diagonal with entries of modulus one.

Our elimination rule makes $r_{11}$ and $r_{22}$ positive when nonzero, so they are both one. Every rotation has determinant one; thus $\det R=\det A$ andWithout a fixed pivot-sign convention one could only specify a diagonal sign matrix. The convention chosen in part (c) determines the precise signs above.

The columns outside indices $i,j$ are unchanged coordinate vectors. The remaining two columns have nonzero entries only in rows $i,j$, with the $2\times 2$ block $M(\theta )$. They have squared norm ${\cos }^2\theta +{\sin }^2\theta =1$ and mutual inner product zero; they are also orthogonal to all the unchanged columns. Hence $J^{T}J=I$.

Let $r=\sqrt{x_i^2+x_j^2}$. If $r>0$, choose $\theta \in [-\pi ,\pi ]$ with $\cos \theta =x_i/r$ and $\sin \theta =x_j/r$. ThenEvery other coordinate is unchanged. If $r=0$, choose $\theta =0$; the same conclusions hold. Therefore the procedure can be fixed with the nonnegative pivot conventionThis is a Givens rotation.

Using orthogonality,Both norms are nonnegative, so taking square roots provesThus an orthogonal matrix is a Euclidean isometry fixing the origin.

A real square matrix $Q$ is orthogonal when $Q^{T}Q=I$, equivalently $Q^{-1}=Q^T$. For $c=\cos \theta$ and $s=\sin \theta$,Thus $M(\theta )$ is orthogonal for every allowed angle, and indeed for every real angle.

The standard normal distribution function is strictly increasing. Thus $p>\alpha$ impliesRearranging givesThis is the confidence-interval/test duality for the matching one-sided procedures. The strict inequalities agree with the printed open interval.

For the observed mean, the one-sided $p$-value is the null probability of a sample mean at least as large:Standardization under the null gives

For every particular alternative mean ${\mu }_1>{\mu }_0$, the likelihood ratio is increasing in the same statistic $\overline{x}$. Its size-$\alpha$ threshold is fixed solely by the null distribution, so it is always the same $c_{\alpha }$ found in part (c), independent of ${\mu }_1$.

The single rejection region is therefore most powerful against every member of the composite alternative. Hence the test is uniformly most powerful for $H_1:\mu >{\mu }_0$.

The likelihood ratio between the two normal means isSince ${\mu }_1>{\mu }_0$, it is strictly increasing in $\overline{x}$. The Neyman–Pearson test therefore rejects for large sample means. Under the null, $\sqrt{n}(\overline{X}-{\mu }_0)\sim N(0,1)$, so choosing a right-tail probability $\alpha$ givesThis has size exactly $\alpha$ because the sampling distribution is continuous.

The Neyman-Pearson lemma concerns testing two simple hypotheses with densities $f_0$ and $f_1$ relative to a common measure. A test that rejects when $f_1(x)>k{f}_0(x)$, accepts when $f_1(x)<k{f}_0(x)$, and randomizes on equality if necessary to achieve size $\alpha$, is most powerful among all tests with size at most $\alpha$:Here $\phi$ is the described likelihood-ratio test and test functions take values in $[0,1]$. In the continuous normal setting below the boundary has probability zero, so randomization is unnecessary.

Independence and normality give $\overline{X}\sim N(\mu ,1/n)$. Consequently $Z=\sqrt{n}(\overline{X}-\mu )$ is standard normal, andRearranging the event gives $\mu >\overline{X}-z_{\alpha }/\sqrt{n}$. Thus the random interval $(\overline{X}-z_{\alpha }/\sqrt{n},\infty )$ covers the fixed parameter with probability $1-\alpha$. After observing the sample, its realized confidence interval isThe coverage refers to repeated samples, not to a posterior probability for the fixed parameter.

The sample mean has expectation $\theta$ and variance ${\theta }^2/n$. For $g(z)=\log z$, $g^{\prime }(\theta )=1/\theta$. The first-order delta-method expansion isTaking its mean and variance givesThese are large-sample approximations. Retaining the quadratic Taylor term would add the leading mean correction $-1/(2n)$, but it does not change the requested first-order expression.

After the third arrival, the future interarrival times are independent of the first three interarrival times and again generate a rate-$\lambda$ Poisson process. Conditional on $T_3=t<2$, exactly three arrivals have occurred by time $t$, and the number of further arrivals before two is Poisson with mean $\lambda (2-t)$. Consequently, in the regular conditional-distribution sense at $t=1$,Explicitly its conditional mass is zero for $k<3$ and isThis conditions on the third arrival time itself, rather than merely on the event that the third arrival has happened by time one.

Write $N_3=N_2+(N_3-N_2)$. Independent increments make $N_2$ independent of the second term. HenceThe overlapping time interval contributes the covariance; the extra unit of time contributes none.

Successive interarrival times $W_1=T_1$ and $W_2=T_2-T_1$ are independent exponential variables of rate $\lambda$. Their joint density is ${\lambda }^2{e}^{-\lambda (w_1+w_2)}$ for $w_1,w_2>0$. The transformation $t_1=w_1$, $t_2=w_1+w_2$ has Jacobian determinant one, givingIntegrating over $0<t_1<t_2<\infty$ gives one, as a joint density should.