Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 2

Let $L_{ik}=1$ when $k\le i$ and $L_{ik}=0$ otherwise. The matrix product has entriesso $A_n=L{L}^T$. Both $L$ and its matrix transpose are triangular matrices with diagonal entries one. Multiplicativity of the determinant therefore gives $\det A_n=(\det L{)}^2=1$ for every $n\ge 1$. This determinant of the minimum-index matrix also follows by subtracting each preceding row from the next, working from the bottom upwards: the resulting matrix is upper triangular with unit diagonal.

We prove Simplicity of the alternating group A5 by finding all its conjugacy classes. The alternating group has group order $|A_5|=5!/2=60$. Its even permutations have cycle types $1$, $(abc)$, $(ab)(cd)$ and $(abcde)$. There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles, $5\cdot 3=15$ products of disjoint transpositions, and $4!=24$ five-cycles.

For a three-cycle, a commuting permutation must rotate its three-element support and may interchange the two fixed points. Its centralizer in $S_5$ has order six, and only the three rotations are even. Thus its conjugacy class in $A_5$ has size $60/3=20$ and contains every three-cycle. The centralizer of $(12)(34)$ in $S_5$ has order eight: one may interchange the two pairs and independently interchange the elements within each pair. It contains the odd element $(12)$, so exactly half of these commuting elements are even. The conjugacy class in $A_5$ consequently has size $60/4=15$. A permutation commuting with a five-cycle is determined by the image of one point and must be a power of that cycle. All five powers are even, so each such conjugacy class has size $60/5=12$. The 24 five-cycles form exactly two classes.

A normal subgroup contains the identity and is a union of whole conjugacy classes. Its possible orders therefore lie inBy Lagrange's theorem, its order must divide 60. Only 1 and 60 in this list do so. Hence $A_5$ has no proper nontrivial normal subgroup and is a simple group.

Put ${\epsilon }_n={\sup }_{[-1,1]}|F_n-F|$. Uniform convergence means ${\epsilon }_n\to 0$, and continuity ensures that the functions have Riemann integrals. The integral triangle inequality givesThis proves that uniform convergence preserves integrals on a compact interval.

For the derivative assertion, take $F_n(x)=\sin (n^{2}x)/n$ and $F(x)=0$. These functions are differentiable, with $\sup |F_n|\le 1/n$, so their convergence is uniform. However, $F_n^{\prime }(0)=n$ whereas $F^{\prime }(0)=0$. The derivatives therefore need not converge even at one interior point.

Any two nonempty open sets in the cofinite topology on $\mathbb{N}$ intersect: their complements have finite union, which cannot exhaust $\mathbb{N}$. Thus this is a hyperconnected space. Suppose a continuous map $f$ took two distinct values $a$ and $b$. The usual real line is a Hausdorff space, so there are disjoint open intervals $U$ and $V$ containing $a$ and $b$. Their inverse images under the continuous map are disjoint, nonempty open sets, contradicting hyperconnectedness. Every such $f$ is constant.

There is a small defect in the printed definition: the listed family omits $\mathbb{R}$ and is not a topology, since ${\bigcup }_{n\ge 1}(-n,\infty )=\mathbb{R}$. We use the intended upper-ray topology on the real line, obtained by adjoining $\mathbb{R}$ to that family.

Take $f(n)=n$. For every real $a$,has finite complement and is therefore open in the cofinite topology. The inverse images of the empty set and the whole line are also open. Hence this nonconstant function is a continuous map for the corrected upper-ray topology on the real line. The same inverse-image calculation verifies the condition for every ray literally listed in the question.

The quotient $h=f/g$ is a holomorphic function because the denominator has no zeros, and $|h|=1$. Write $h=u+iv$. The Cauchy-Riemann equations say $u_y=-v_x$ and $v_y=u_x$. Differentiating $u^2+v^2=1$ in both real directions givesThe coefficient determinant is $-(u^2+v^2)=-1$, so $u_x=v_x=0$, and the Cauchy-Riemann equations also give $u_y=v_y=0$. Thus $h$ is locally constant. Since a domain is connected, it is constant throughout $\Omega$. Its constant value has modulus one and can be written $e^{i\alpha }$ with $\alpha \in \mathbb{R}$. Therefore $f=e^{i\alpha }g$ throughout $\Omega$.

This is the constant-modulus holomorphic function principle. Connectedness is essential to a single global phase: on two disjoint discs, taking $g=1$ and choosing $f=1$ on one disc and $f=-1$ on the other gives equal moduli without one common phase.

In polar coordinates, the Laplace equation isFor a single-valued, $2\pi$-periodic harmonic function on an annulus, separation of variables and its angular Fourier series giveIndeed a nonzero angular mode has radial equation $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, with solutions $r^{\pm n}$; the zero mode has solutions 1 and $\log r$. This expression describes the periodic annular problem here; angular terms not periodic, such as $\theta$, are excluded.

Decay at the origin leaves only positive powers in the interior, and decay at infinity leaves only negative powers in the exterior. Continuity on the unit circle matches the corresponding Fourier coefficients. If the common cosine coefficient is $K_n$, the interior and exterior terms are $K_n{r}^n\cos n\theta$ and $K_n{r}^{-n}\cos n\theta$. Their exterior-minus-interior radial derivative at the circle is $-2n{K}_n\cos n\theta$. Thus harmonic matching across a circle with a derivative jump gives $K_2=-1/4$, $K_4=-1/8$, and zero for every other sine or cosine mode. ConsequentlyEach branch satisfies the Laplace equation and its respective decay condition. Both give $-\cos 2\theta /4-\cos 4\theta /8$ at $r=1$, and direct differentiation gives the required jump $\cos 2\theta +\cos 4\theta$. The same mode calculation leaves no nonzero homogeneous solution in this class.

In vacuum magnetostatics, the two relevant Maxwell equations areOrient the unit normal $\mathbf{n}$ from the minus side to the plus side. Integrating Gauss's law for magnetism over a pillbox straddling the current sheet and shrinking its thickness gives $\mathbf{n}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})=0$.

To obtain the tangential condition, choose an arbitrary tangent unit vector $\mathbf{t}$ and an infinitesimal rectangle with its long sides of length $\ell$ parallel to $\mathbf{t}$ on the two sides of the sheet. Orient its surface normal as $\mathbf{n}\times \mathbf{t}$, so that its plus-side long edge is traversed along $\mathbf{t}$. The line integral in Ampère's law tends to $\ell \mathbf{t}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})$. The enclosed electric current tends to $\ell \mathbf{s}\cdot (\mathbf{n}\times \mathbf{t})$. HenceBecause this holds for every tangent direction and the normal jump is zero,Here $\mathbf{s}\cdot \mathbf{n}=0$, as required for a surface current density.

The force on a current sheet from the mean magnetic field isThe mean removes the sheet's equal and opposite local self-fields, leaving the field that acts on its current through the Lorentz force. Using a one-sided limiting magnetic field would incorrectly include the sheet's own force.

The two given energy eigenstates are orthonormal. Resolve the initial state as $\psi (0)=\cos \theta {\psi }_0-\sin \theta {\psi }_1$. The Schrödinger equation therefore givesThe second component, which is the probability amplitude for the requested final state, is $\sin \theta \cos \theta (e^{-i{E}_{0}t/\hslash }-e^{-i{E}_{1}t/\hslash })$. By the Born rule,This includes $P(t)=0$ when the energy eigenvalues coincide or when the initial state is itself an energy eigenstate.

Use the weighted inner product $⟨u,v⟩={\int }_0^{1}u(x)v(x)xdx$. A monic quadratic orthogonal polynomial ${\pi }_2(x)=x^2+ax+b$, orthogonal to 1 and $x$, satisfiesThus $a=-6/5$, $b=3/10$, and its two zeros are $x_{-}=(6-\sqrt{6})/10$ and $x_{+}=(6+\sqrt{6})/10$. Matching the first two weighted moments givesso $w_{-}=(9-\sqrt{6})/36$ and $w_{+}=(9+\sqrt{6})/36$. The two-node Gaussian quadrature with linear weight is thereforeFor completeness, a polynomial $P$ of degree at most three can be divided as $P=q{\pi }_2+r$, where both $q$ and $r$ have degree at most one. Orthogonality makes the weighted integral of $q{\pi }_2$ zero, and this term vanishes at both nodes. The two moment equations integrate $r$ exactly. Thus the Gaussian quadrature is exact for every cubic polynomial. No two-node rule can integrate every quartic: the square of its nodal polynomial has quadrature value zero and strictly positive weighted integral.

For the first sampling design, treat the fixed groups of 60 and 90 people as two independent, representative samples from the respective populations. Within each sample, assume independent individuals with common eye-colour probabilities and mutually exclusive, exhaustive categories. This gives two independent multinomial distributions. The null hypothesis is that the two populations have the same eye-colour probabilities; its estimated common probabilities are $40/150$, $70/150$ and $40/150$.

The expected counts under the null hypothesis areThe Pearson chi-squared test of homogeneity usesThere are $2(3-1)=4$ unrestricted category parameters and $3-1=2$ under the null hypothesis, so the asymptotic chi-squared distribution has two degrees of freedom. All expected counts exceed five, making the usual large-sample approximation reasonable. Reject equality at the 5% level, but not at the 1% level: $5.99<T<9.21$. The approximate p-value is $\mathbb{P}({\chi }_2^2\ge T)=e^{-25/7}\simeq 0.0281$. Independence and representative sampling, rather than merely the count sizes, are essential to population inference; the result does not establish a causal explanation.

In the second design, the total sample size is fixed but both row and column classifications are observed for each sampled person. Assume an independent, representative sample with a common joint distribution of handedness and eye colour. The six cell counts then have one multinomial distribution. The null hypothesis is independence of the two classifications, so joint probabilities factor into row and column probabilities. Their maximum-likelihood estimates are the observed marginal proportions, again giving exactly $E$ above.

The unrestricted joint distribution has five free parameters; independence leaves $(2-1)+(3-1)=3$. The Pearson chi-squared test of independence again has two degrees of freedom, the same statistic, and the same significance conclusion. The numerical analysis is unchanged, but the sampling model changes from two fixed-row samples to one joint sample. If handedness groups had instead been deliberately sampled with fixed row sizes, the first, homogeneity formulation would apply to those groups too.

For a Markov chain started in its stationary distribution, reversibility means that for every $n$, the joint distribution of $(X_0,\dots ,X_n)$ equals that of $(X_n,\dots ,X_0)$. More generally, stationarity then makes every finite segment invariant under reversal of its time order.

If the Markov chain is reversible, take $n=1$. Comparing the events $(X_0,X_1)=(i,j)$ and $(j,i)$ gives the detailed balance equationsConversely, an irreducible Markov chain with an invariant probability distribution has ${\pi }_i>0$ for every state. Under detailed balance, a transition is positive exactly when its reverse is positive. For a path with positive probability, substitute $P_{ij}={\pi }_j{P}_{ji}/{\pi }_i$ in each factor to obtainThe factors of $\pi$ telescope. If a path has a zero transition, its reverse does too, so the equality still holds. These are the respective forward and reversed path probabilities. Summing over any collection of paths proves equality of the joint distributions, hence a reversible Markov chain. The first factor in the PDF's balance equation is ${\pi }_i$; the converted TeX incorrectly changes it to ${\pi }_j$.

Associate to $Q$ its symmetric bilinear form by the polarization identity,We first construct a diagonal basis by induction on the dimension. A zero quadratic form is already diagonal. Otherwise choose $v$ with $Q(v)\ne 0$. Every $x$ has the decompositionwhose second term lies in $W=\{w:B(w,v)=0\}$. Also $\mathbb{R}v\cap W=\{0\}$, so $V=\mathbb{R}v\oplus W$ and $\dim W=n-1$. Apply the induction hypothesis to the restricted quadratic form on $W$. The cross terms with $v$ vanish, giving a diagonal basis on $V$. Rescale each vector with diagonal coefficient $a\ne 0$ by $1/\sqrt{|a|}$, and reorder positive, negative and zero coefficients. This yieldsNo nondegeneracy assumption was used.

For uniqueness, let $V_{+}$ be the positive coordinate subspace and $V_{\le 0}$ the span of all negative and zero coordinate vectors. Their dimensions are $p$ and $n-p$. Any subspace $W$ on which $Q$ has positive definiteness intersects $V_{\le 0}$ only in zero. The dimension formula for a sum of subspaces therefore gives $\dim W\le p$. Conversely $V_{+}$ itself has dimension $p$ and positive restriction. Thus $p$ is the largest possible dimension of a positive subspace, an intrinsic description independent of the basis. Applying the same argument to $-Q$ shows that $q$ is the largest dimension of a negative subspace. This proves Sylvester's law of inertia, including uniqueness of the zero count $n-p-q$.

The extremal subspaces themselves need not be unique. For $Q(x,y)=x^2-y^2$, the coordinate axes give one positive and one negative line. For any $t\ne 0$, putThen $Q(u)=1$, $Q(v)=-1$, $B(u,v)=0$, and the determinant of the change of basis is one. The distinct lines $\mathbb{R}u$ and $\mathbb{R}v$ give another positive/negative decomposition, with the same intrinsic counts $p=q=1$.

Suppose $K=\ker \phi$ for a group homomorphism $\phi :G\to H$. Its identity belongs to $K$, and for $k,l\in K$ we have $\phi (k{l}^{-1})=1$, so it is a subgroup. Moreover, for $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$; applying this also to $g^{-1}$ gives equality, so $K$ is a normal subgroup.

Conversely, if $K$ is a normal subgroup, multiplication of cosets in the quotient group $G/K$ is well-defined. The map $g\mapsto gK$ is a group homomorphism and has kernel of a group homomorphism exactly $K$. This proves both directions without requiring $\phi$ to be injective or surjective.

For the matrix application, reduce all entries modulo $p$:Reduction respects matrix multiplication and the determinant, so this is a group homomorphism. Its kernel of a group homomorphism consists exactly of matrices congruent to the identity modulo $p$, which are the given matrices. Therefore $K=\ker \rho$ is a normal subgroup, the level-$p$ principal congruence subgroup. Surjectivity of reduction is not needed for this argument.

The complete metric space $\mathbb{R}$ supplies a counterexample: $A_n=[n,\infty )$ are descending nonempty closed sets, but no real number belongs to them all.

Now assume shrinking diameters. Choose $x_n\in A_n$. Given $\epsilon >0$, choose $N$ so that $\mathrm{diam}(A_N)<\epsilon$. Whenever $m,n\ge N$, both chosen points lie in $A_N$, so $d(x_m,x_n)<\epsilon$. Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x\in X$. For each fixed $N$, the tail lies in the closed set $A_N$, so its limit lies there too. Hence $x\in {\bigcap }_n{A}_n$. Any two points of this intersection have distance at most every $\mathrm{diam}(A_n)$, so their distance is zero. The shrinking intersection contains exactly one point. This is the Cantor intersection theorem.

For the dense-open conclusion, begin inside an arbitrary nonempty open set $O\subseteq X$; taking $O=X$ gives the requested existence. Since $U_1$ is a dense subset, choose $x_1\in O\cap U_1$ and $0<r_1\le 1/2$ small enough that the closed ball $A_1=\overline{B}(x_1,r_1)$ is contained in $O\cap U_1$. Here $\overline{B}$ denotes the closed ball, not an assumption about the closure of the open ball.

Inductively, the open ball $B(x_n,r_n)$ meets the dense open set $U_{n+1}$. Choose $x_{n+1}$ in that intersection and a positive radius $r_{n+1}\le 2^{-(n+1)}$ so small thatSuch a radius exists by openness and the triangle inequality. These nonempty closed sets descend and have diameters at most $2r_n\to 0$. The preceding result gives a point in every $A_n$, hence in $O$ and every $U_n$. Therefore ${\bigcap }_n{U}_n$ is nonempty; in fact it is dense in $X$. This proves the relevant Baire category theorem directly, without assuming closed balls are compact.

For a sequence $ϵ$, let $C_n(ϵ)$ be its prefix cylinder set, consisting of sequences agreeing with $ϵ$ in the first $n$ coordinates. The specified collection consists exactly of unions of such cylinder sets.

The empty set meets the defining condition vacuously, and $X$ meets it because every cylinder set is contained in $X$. For a union ${\bigcup }_{\lambda }{Y}_{\lambda }$, a point lies in one $Y_{\lambda }$ and has a prefix cylinder set contained in that set, hence contained in the union. For a finite intersection, a point has a cylinder neighbourhood in every constituent. Taking the largest of the finitely many prefix lengths gives one cylinder set contained in all of them. Therefore the collection contains the empty and whole sets, is closed under arbitrary unions and finite intersections, and is a topology on $X$, the prefix topology of Cantor space.

The weighted binary-sequence metric is finite because $0\le d(ϵ,\eta )\le {\sum }_{n\ge 1}{2}^{-n}=1$. It is symmetric and nonnegative. If $d(ϵ,\eta )=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|{ϵ}_n-{\zeta }_n|\le |{ϵ}_n-{\eta }_n|+|{\eta }_n-{\zeta }_n|$ in each coordinate and sum. This proves the triangle inequality, so $d$ is a metric.

If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the open ball $B_d(\eta ,2^{-n})$ is contained in $C_n(\eta )$. For every $\eta \in C_n(ϵ)$, the latter cylinder equals $C_n(\eta )$, so each cylinder set is open in the metric topology. Every set in the specified topology, being a union of cylinder sets, is therefore metric-open.

Conversely, agreement in the first $n$ coordinates givesGiven a metric-open set containing $ϵ$, choose a ball $B_d(ϵ,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(ϵ)\subseteq B_d(ϵ,r)$, establishing the required prefix-neighbourhood condition. The two topologies coincide. The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.

Define a candidate antiderivative by integration along the straight segment from zero:For $z$ in the open unit disc and sufficiently small $h$, the triangle with vertices $0,z,z+h$ lies within the disc, since the disc is convex. By the Cauchy integral theorem, its boundary contour integral vanishes. ConsequentlyFor $h\ne 0$, subtract $f(z)$ after division by $h$. The absolute value of the remainder is at most ${\sup }_{0\le t\le 1}|f(z+th)-f(z)|$, which tends to zero by continuity of the holomorphic function $f$ at $z$. Hence the complex derivative exists and $g^{\prime }(z)=f(z)$ for every point of the disc. This proves that $g$ is a holomorphic function and supplies the requested primitive directly from the Cauchy integral theorem.

Use coordinates $\mathbf{r}$ measured from the center of mass $G$, so $\int \mathbf{r}dm=0$. The displacement from $P$ is $\mathbf{r}-\mathbf{X}$. Expand the defining inertia tensor:All terms linear in $\mathbf{r}$ integrate to zero. The remaining constant terms give the tensor parallel-axis theorem,For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal inertia tensor entries zero. Direct integration gives $\int x_i^{2}dm=M{a}^2/3$ for each coordinate, and therefore $I(G)=(2M{a}^2/3)I_3$. With $\mathbf{X}=a(1,1,1{)}^T$, the tensor parallel-axis theorem givesThe unit eigenvector $u=(1,1,1{)}^T/\sqrt{3}$ has eigenvalue $2M{a}^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $11^T$ and hence has eigenvalue $11M{a}^2/3$. One orthonormal basis of that eigenspace isThe principal moments are $2M{a}^2/3$ along the body diagonal and $11M{a}^2/3$ twice in its perpendicular plane. Any orthonormal pair in that plane supplies the remaining principal axes.

Put $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$. The magnetic field is the curl of the magnetic vector potential. Since $\mathbf{J}({\mathbf{r}}^{\prime })$ does not depend on the differentiated variable $\mathbf{r}$,Taking the curl under the source integral consequently gives the Biot-Savart law,Outside the compact source this differentiation is ordinary differentiation of a smooth kernel. For regular volume currents, the locally integrable $R^{-2}$ derivative kernel gives the same formula at interior points.

For the loop, choose positive current circulating counterclockwise as viewed from the positive $z$-axis. Parametrize its source position and directed line element byAt $(0,0,z)$ the denominator is the constant $(a^2+z^2{)}^{3/2}$, whileThe transverse terms integrate to zero around the loop. The Biot-Savart law therefore yieldsThe magnetic field is axial, with the stated magnitude for $I>0$; reversing the electric current reverses its direction.

Take the momentum operator $p=-i\hslash d/dx$ and sufficiently regular real $f$. Acting on a test wave function, the commutator is $[p,f]=-i\hslash f^{\prime }$. Expanding the factorized quantum Hamiltonian givesso $V(x)=(f(x{)}^2-\hslash f^{\prime }(x))/(2m)$, which is real.

Set $Q=p-if$. Its zero-mode equation iswith solutionIt satisfies $H{\psi }_0=Q^{†}Q{\psi }_0/(2m)=0$ as a differential equation. It is an actual quantum state only when it is square-integrable. If $f(x)=c{x}^n+\cdots$, $c\ne 0$, its antiderivative has leading term $c{x}^{n+1}/(n+1)$. For even $n$, this leading term has opposite signs at the two ends of the line, so the wave function grows at one end and cannot be normalized. For odd $n$, it has the same sign at both ends. Hence normalizability requires odd $n$, and in fact holds exactly when $n$ is odd and $c>0$. A negative leading coefficient makes it grow at both ends; $f=0$ gives a nonnormalizable constant. These are the normalizable zero mode of a polynomial factorized Hamiltonian conditions.

For a normalized state in the operator domain, integration by parts, with the appropriate vanishing boundary term, givesThus every energy eigenvalue is nonnegative. Equality forces $Q\psi =0$ almost everywhere, whose first-order equation has the one-dimensional solution space above. Whenever its zero mode is normalizable, every energy eigenstate not proportional to ${\psi }_0$ has strictly positive energy. If the formal zero mode is not normalizable, there is no zero-energy eigenstate; the positivity statement still applies to every normalizable eigenstate. This argument concerns eigenvalues and does not assert a strictly positive bound on an arbitrary continuous spectrum.

For the quantum harmonic oscillator, take $f(x)=m\omega x$ with $m,\omega >0$. ThenIts zero mode is normalizable and has no competitor of lower energy. Using the Gaussian integral for its normalization gives the concise answerup to an arbitrary constant phase.

Let $r=Ax-b$ and consider an arbitrary variation $x+th$. The objective for the linear least-squares problem isAt a minimizer the derivative at $t=0$ must vanish for every $h$, so $A^{T}r=0$. Conversely, when this holds, the same expansion at $t=1$ gives $\Vert A(x+h)-b{\Vert }^2=\Vert r{\Vert }^2+\Vert Ah{\Vert }^2\ge \Vert r{\Vert }^2$ for every $h$. Thus $x$ minimizes the residual norm if and only if $A^T(Ax-b)=0$. These least-squares normal equations do not require $A$ to have full rank; without full rank the minimizing vector need not be unique.

The original PDF has $A_{22}=1$, not the 2 in the converted TeX. Using the PDF's matrix givesThe least-squares normal equations consequently have the unique solutionAs a direct check, $Ax-b=(-279,-95,238,-61{)}^T/123$, whose inner product with either column of $A$ is zero. The positive determinant verifies full column rank for this particular problem.

A simple hypothesis specifies one probability distribution. Let the null hypothesis and alternative have probability density functions $f_0,f_1$ with respect to a common measure. A possibly randomized statistical test is a measurable function $\phi (x)\in [0,1]$, its conditional probability of rejecting the null hypothesis. Its size of a statistical test is ${\mathbb{E}}_0\phi$, and its statistical power at the alternative is ${\mathbb{E}}_1\phi$. A most powerful test at level $\alpha$ maximizes power among tests with size at most $\alpha$.

The Neyman-Pearson lemma says to choose $k\ge 0$ and a test ${\phi }_{\ast }$ of size $\alpha$ with ${\phi }_{\ast }=1$ where $f_1>k{f}_0$ and ${\phi }_{\ast }=0$ where $f_1<k{f}_0$, choosing values on the equality set to reach the specified size. Equivalently, threshold the likelihood ratio, treating positive alternative density with zero null density as an infinite ratio. Randomization on a threshold atom ensures exact size when needed. If $k=0$, all remaining null-only mass can be used to reach size without changing power.

For any test $\phi$ of size at most $\alpha$, the signs of the two factors givepointwise. Integrating provesThis is the required optimality proof. If $k>0$ and another test is equally powerful, both inequalities must be equalities. It must have size $\alpha$ and agree with ${\phi }_{\ast }$ wherever $f_1\ne k{f}_0$, apart from null sets for the relevant densities. Thus freedom is confined to the equality set and sets null under both hypotheses; at $k=0$, additional size need not improve power. This also explains the nonuniqueness below.

For the unit-scale Laplace distribution, put $\delta ={\theta }_1-{\theta }_0>0$. Its likelihood ratio isIt is nondecreasing, so an upper-tail critical region has the Neyman-Pearson lemma form, even if the cutoff lies inside a constant-ratio tail. Solving ${\mathbb{P}}_{{\theta }_0}(X>c_{\alpha })=\alpha$ gives the most powerful test for a Laplace location shift,Continuity of the distribution makes this nonrandomized test exactly size $\alpha$. It need not be the unique most powerful test: within a flat extreme likelihood ratio region, other selections having the same null probability are equally good.

Its statistical power isHere $\beta$ denotes power, not the probability of a type-II error. For $\alpha =0.05$, $c_{\alpha }={\theta }_0+\log 10$. Thus the power is $0.05e^{\delta }$ for $\delta \le \log 10$, and $1-5e^{-\delta }$ otherwise. The first branch is at most $1/2$. On the second branch, power at least $0.95$ is equivalent to $5e^{-\delta }\le 0.05$. Therefore the required pairs are exactly ${\theta }_1-{\theta }_0\ge \log 100$, with no restriction on the common location.

Read the arrows in the original diagram as follows: $0$ moves to $1^{\prime }$ with probability one; from an unprimed $n\ge 1$, probability $q$ leads to $n-1$ and probability $p$ leads to $(n+1{)}^{\prime }$; from $n^{\prime }$ the corresponding destinations are $n$ and $(n+1{)}^{\prime }$. This is the two-lane Markov chain with a deterministic restart. It is irreducible: every state can reach zero by a finite path of $q$ transitions, and zero can reach every primed state by outward $p$ transitions and then every unprimed state by a $q$ transition. Thus all states have the same recurrence classification.

Let $a_n$ and $b_n$ be the hitting probabilities of zero starting from $n$ and $n^{\prime }$, with $a_0=1$. First-step conditioning givesSubtracting yields $b_n=(1+q)a_n-q{a}_{n-1}$. Eliminating $b_{n+1}$ givesIts characteristic roots are $1$ and $r=q/[p(1+q)]=q/(1-q^2)$. Put $q_c=(\sqrt{5}-1)/2$, so $r=1$ precisely when $q=q_c$.

For $q>q_c$, the general solution is $a_n=A+B{r}^n$. The bound $0\le a_n\le 1$ and $r>1$ force $B=0$, and $a_0=1$ then gives $a_n=b_n=1$. For $q=q_c$, the general solution is $a_n=A+Bn$; boundedness again gives $a_n=b_n=1$. In both cases, the return probability from zero is $b_1=1$, so the Markov chain is recurrent.

For $q<q_c$, $0<r<1$, and the pairis a nonnegative bounded solution of the first-step equations. To identify it as the actual hitting probability, use that hitting probability is the minimal nonnegative harmonic extension. This fact follows directly by iteration: a nonnegative solution with boundary value one at zero dominates the probability of hitting zero within $N$ steps, for every $N$. The actual bounded solution has form $a_n=A+(1-A)r^n$, where nonnegativity as $n\to \infty$ requires $A\ge 0$. Minimality bounds it above by $r^n$, while this form bounds it below by $r^n$. Hence it is exactly the displayed solution. Starting at zero, the first return time has finite-return probability $b_1=q^2/p<1$. The Markov chain is therefore transient.

It remains to distinguish positive and null recurrence. Write $u_n={\pi }_n$ and $v_n={\pi }_{n^{\prime }}$ for invariant probabilities. Balance at zero and at $1^{\prime }$ gives ${\pi }_0=q{u}_1$ and $v_1={\pi }_0$. For the finite cut consisting of zero and both lanes up to level $n$, invariance equates its outward and inward probability fluxes:Balance at the unprimed state $n$ also gives $u_n=q(u_{n+1}+v_n)$. Combining these identities yieldsTherefore every invariant probability distribution would necessarily have $u_n=({\pi }_0/q)t^{n-1}$ and $v_n={\pi }_0{t}^{n-1}$. Conversely, these weights satisfy all individual balance equations, including $v_{n+1}=p(u_n+v_n)$, so their summability is the only remaining condition.