The first four jobs all have deadlines at most ten, but their total processing time is $2+4+3+2=11$. They cannot all be completed by time ten on one machine, so at least one is late in every schedule.

Moore's algorithm processes jobs in nondecreasing deadline order. Whenever the electric current on-time list first misses the new job's deadline, remove from that list a job with the largest processing time and put it on the late list. Continue, then append the rejected jobs after the on-time list. The theorem for this algorithm states that it minimizes the number of late jobs when all jobs are initially available.

The source order is already deadline order. The steps areAfter the first removal the cumulative time is seven; after the second it is eleven. Append jobs two and five. One optimal schedule isIts completion times are $2,5,7,11,15,20$. Comparing with the corresponding deadlines verifies that precisely jobs two and five are late. Moore's theorem certifies that no schedule can improve on that number.

Richardson extrapolation combines approximations with a known leading error power to cancel that error. If $A(h)=I+c{h}^p+O(h^{p+q})$, then $(2^{p}A(h)-A(2h))/(2^p-1)$ removes $c{h}^p$.

Let $x_j=a+jh$ for an even number $n$ of equal subintervals, with $x_n=b$. The composite trapezoidal rule isThe Simpson's rule isAssume an error expansion $T_h=I+c_2{h}^2+O(h^4)$, with the same coefficient $c_2$ on refined uniform meshes; $f\in C^4([a,b])$ is sufficient for this form. The coarser rule is $T_{2h}=h[f_0+f_n+2{\sum }_{\mathrm{even}}{f}_j]$. Direct coefficient collection givesThe leading $h^2$ errors cancel, leaving $S_h=I+O(h^4)$. The identity of the numerical integration weights is exact; the error-order conclusion additionally uses the smoothness and even-power error assumptions.

A Givens rotation acts as the identity except on coordinates $i,j$, where its block can be writtenThe block has orthonormal columns because ${\cos }^2\theta +{\sin }^2\theta =1$ and their inner product is zero, so $J^{T}J=I$. For $r=\sqrt{v_i^2+v_j^2}>0$, choose $\cos \theta =v_i/r$, $\sin \theta =v_j/r$. The new coordinates are $(r,0)$, eliminating the $j$th entry. If both entries are zero, the identity already works.

For the specified vector, one suitable matrix isIts rows are orthonormal, so it is orthogonal; in fact its determinant is one. The positive nonzero output component must be five because an orthogonal matrix preserves Euclidean norm.

The matrix is not unique. Left multiplication by any orthogonal matrix fixing the second coordinate vector preserves this output. For example arbitrary rotations of the first and third coordinates provide infinitely many alternatives.

Minimizing the squared residual norm gives the normal equationsFull column rank makes $X^{T}X$ invertible. With columns corresponding to $1,x,x^2-4$, the design matrix isSymmetry gives zero cross products between the odd column and either even column, and $\sum (x^2-4)=28-28=0$ gives the remaining orthogonality. Their squared norms are seven, 28 and 84, so $X^{T}X=\mathrm{diag}(7,28,84)$. ThereforeSince the error covariance is ${\sigma }^{2}I$, least squares has covariance ${\sigma }^2(X^{T}X{)}^{-1}$. ThusThe estimators have zero covariances. independence would require an additional distributional assumption such as Gaussian errors; the stated mean and variance assumptions alone give only the covariance calculation.

The statistical power of a test $\phi$ is $\beta (\theta )={\mathbb{E}}_{\theta }\phi (X)$, its rejection probability at parameter $\theta$. Its size is ${\sup }_{\theta \in H_0}\beta (\theta )$; for a simple null this is just $\beta ({\theta }_0)$.

The Neyman-Pearson lemma states that for testing a simple null probability density function $f_0$ against a simple alternative probability density function $f_1$, a test rejecting where $f_1/f_0>k$, accepting where it is below $k$, and if needed randomizing on equality to attain the desired size is most powerful among all tests of that size. It follows by integrating $(\phi -{\phi }_{\ast })(f_1-k{f}_0)\le 0$ and using the size constraint.

The proposed probability density function is nonnegative, andFor an independent sample, cancellation of the factors $(x_i+1)$ in the likelihood ratio givesFor ${\theta }_1<{\theta }_0$ this is increasing in $S={\sum }_i{X}_i$. The most powerful test therefore has rejection regionThe sum has a continuous positive probability density function on $(0,\infty )$, so no boundary randomization is required. The threshold depends on the null and chosen size, but not on which ${\theta }_1<{\theta }_0$ was specified. The same test is therefore a uniformly most powerful test for the composite alternative $0<\theta <{\theta }_0$.

A confidence interval is a pair of statistics $L(X),U(X)$ such that, for each parameter value $\mu$,In repeated sampling, at least the stated fraction of intervals constructed by this rule cover the fixed unknown parameter. It is not a posterior probability statement about the parameter after one interval has been observed. Discrete exact constructions can have coverage greater than the nominal value.

For independent observations from a Poisson distribution, $S={\sum }_i{X}_i$ has a Poisson distribution with mean $\lambda =n\mu$. Invert equal-tail tests of that total mean. At an observed integer $s>0$, the lower endpoint solvesand the upper endpoint solvesThen divide both endpoints by $n$. If $s=0$, use lower endpoint zero. Each inversion allocates at most $\alpha /2$ noncoverage probability to one side, giving at least $1-\alpha$ total coverage.

Here the single total count is 39. The supplied values with the correct adjacent count indices are $F_{27.7}(38)=0.975$ for the lower endpoint and $F_{53.3}(39)=0.025$ for the upper one. ThusThe values 28.6 and 52.2 use the other count index and are not the endpoints of this central interval. This is adjacent count tails in an exact Poisson confidence interval.

The moment-generating function is $M_X(t)=\mathbb{E}{e}^{tX}$ on the real values where this expectation is finite, normally including a neighborhood of zero. Direct substitution giveswhenever the right side is finite.

For the given Gamma distribution and $t<\lambda$,using the Gamma function after the scaling $u=(\lambda -t)x$. At $t\ge \lambda$ the positive integral diverges.

independence multiplies moment-generating functions. ThusUniqueness of a moment-generating function on a neighborhood of zero identifieswhere $\lambda$ is the rate parameter, as in the printed probability density function.