Past exam of the Probability, statistics and options II course of the University of Oxford

Source: Mathematics release, supplied bundle Part A 2007 11-1.pdf, pages 309–327. This nineteen-page Mathematics and Statistics paper is dated 30 June 2011. All 23 questions and all 86 printed leaf parts are addressed below, including a counterexample and explicit correction for the false algorithmic assertion in S2(a).

The department's answer-release policy excludes ordinary odd years. Bounded exact-paper checks found no matching advertisement of worked answers. No source part is skipped.

A splitting field of $f$ over $F$ is an extension field in which $f$ factors into linear factors and which is generated over $F$ by those roots. The generation requirement excludes unnecessary extra elements.

For existence, induct on the degree. An irreducible factor $p$ of $f$ gives a field $F_1=F[X]/(p)$, since an irreducible polynomial generates a maximal ideal in the polynomial ring over a field. The class $\alpha$ of $X$ is a root of $p$ and hence of $f$. By the factor theorem, $f=(X-\alpha )g$ over $F_1$, with $\mathrm{deg}g=\mathrm{deg}f-1$. Induction gives a splitting field of $g$ over $F_1$. It contains all roots of $f$; taking the subfield generated by these roots over $F$ gives exactly a splitting field of $f$. The degree-one case is immediate. Thus

Over ${\mathbb{F}}_2$, the cubic takes value one at both zero and one, so it has no root. A reducible cubic over a field has a linear factor and therefore a root; hence this cubic is irreducible. Adjoin a root:This extension has basis $1,\alpha ,{\alpha }^2$, degree three and eight elements. Squaring preserves polynomial equations with coefficients in ${\mathbb{F}}_2$, so ${\alpha }^2$ and ${\alpha }^4$ are roots too. The nonzero element $\alpha$ is not one; the multiplicative group has prime order seven, so $\alpha$ has order seven. The exponents $1,2,4$ are distinct modulo seven, making these three roots distinct. Thus they are all the roots and already lie in $K$. Therefore

All roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. We have $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

To show that adjoining $\sqrt{3}$ doubles the degree, suppose $\sqrt{3}=a+b\sqrt{2}$ with rational $a,b$. Squaring and using irrationality of $\sqrt{2}$ gives $ab=0$. If $b=0$, then $a^2=3$; if $a=0$, then $b^2=3/2$. Neither is a rational square, as the prime exponents in a rational square are even. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. Its quadratic polynomial remains irreducible over that field, and the tower law givesA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.

A composition series is a finite chain $1=G_0⊲G_1⊲\cdots ⊲G_r=G$ in which each inclusion is proper, each subgroup is normal in the next, and every quotient $G_i/G_{i-1}$ is a nontrivial simple group. These quotients are the composition factors.

For a finite group, use induction on its order. The trivial group has the empty series. Otherwise choose a maximal proper normal subgroup $N$; such a subgroup exists because there are finitely many subgroups and $1$ is proper. The correspondence theorem says normal subgroups of $G/N$ correspond to normal subgroups of $G$ containing $N$, so maximality makes $G/N$ simple. Inductively construct a composition series of the smaller group $N$ and append $G$. This proves existence.

A nonabelian example with abelian factors isBoth factors are simple because their orders are prime.

The cyclic group of order six has two different composition series:Their factor lists are $C_2,C_3$ and $C_3,C_2$ respectively; normality follows from commutativity.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same multiset of composition factors up to isomorphism. Equivalently, the factors of one series can be reordered to match those of the other by group isomorphisms. The actual intermediate subgroups need not be the same.

Let $C_n=⟨g⟩$ and take $H=⟨g^p⟩$. Since $n=mp$, the order of $g^p$ is $m$, so $H\cong C_m$. All subgroups of a cyclic group are normal. The quotient is generated by $gH$ and has order $n/m=p$, givingA group of prime order has no nontrivial proper subgroup. By the subgroup correspondence theorem, no normal subgroup lies strictly between $H$ and $C_n$. Thus $H$ is a maximal proper normal subgroup.

Given a prime factorization $n=p_1\cdots p_r$, repeatedly apply part (c) to the cyclic group of order $n$: it has a chain of subgroups of successive ordersRead in increasing order, this is a composition series with factors isomorphic to the cyclic groups of orders $p_1,\dots ,p_r$. Any second prime factorization of the same integer produces another composition series of the same group $C_n$.

The Jordan–Hölder theorem equates the two multisets of simple factors. Since $C_p\cong C_q$ for primes $p,q$ implies $p=q$ by equality of group orders, the two prime multisets agree. HenceFor $n=1$ both factor lists are empty, corresponding to the trivial group.

A square residue modulo $p$ is a congruence class represented by $x^2$ for some integer $x$. For the final all-primes solvability statement, zero is included as a square. The Legendre symbol distinguishes the zero class from the nonzero square and nonsquare classes:When the term quadratic residue is reserved for nonzero classes, the zero case must be stated separately, as in this definition.

Euler's criterion states that, for an odd prime $p$ and $p\nmid a$,Thus this power is congruent to one for a nonzero square and to minus one for a nonsquare. The congruence also holds when $p\mid a$ if both sides are taken to be zero.

For distinct odd primes $p,q$, quadratic reciprocity isThe usual supplementary laws areThe first supplement also follows immediately by inserting $a=-1$ into Euler's criterion.

If $p$ divides $a$ or $b$, it divides their product and both sides of the proposed identity are zero. Otherwise use Euler's criterion and the ordinary exponent rule:Both sides are in $\{1,-1\}$, and these two values are distinct modulo the odd prime $p$. Hence the congruence is an equality:

For odd $m,n$,The last term is even, so induction over the prime factors, including repetitions, gives $\frac{b-1}{2}\equiv {\sum }_i\frac{p_i-1}{2}(\mathrm{mod}2)$. Using the first supplementary law in each Legendre factor,For $b=1$, the product is empty and both sides equal one.

If $b,c$ have a common prime factor, both Jacobi symbols are zero, and the identity holds. Otherwise write $b={\prod }_i{p}_i$, $c={\prod }_j{q}_j$ with multiplicity. All cross-pairs of primes are distinct. Multiplying quadratic reciprocity over these pairs givesThe exponent factors as a product of two sums. The parity identity from part (i) makes it congruent modulo two to $[(b-1)/2][(c-1)/2]$. ThereforeThe same formula includes a denominator equal to one by the empty-product convention.

Because $a=-b^2$ is odd, $b$ is odd. Choose $k=4b^2-1$, a positive integer congruent to three modulo four. Also $\gcd (k,b)=1$, so the square $b^2$ has Jacobi symbol one modulo $k$. Multiplicativity in the numerator follows from the corresponding Legendre property in each prime factor. ThusThis gives an explicit positive odd denominator with the required obstruction.

Choose a nonzero quadratic nonresidue $c$ modulo $q$. By the Chinese remainder theorem, choose a positive integer $k$ satisfyingThe moduli are pairwise coprime, and $k$ is coprime to $qb$. Jacobi reciprocity, with $k\equiv 1(\mathrm{mod}4)$, givesSince $t$ is odd, multiplying these symbols yieldsWhen $b=1$ its congruence imposes no restriction and its Jacobi factor equals one.

An integer square is a square modulo every prime, with the zero class allowed at primes dividing the integer. Conversely, suppose an odd integer $a$ is not an integer square. If every prime exponent in $|a|$ is even, the nonsquare condition forces $a=-b^2$, handled by part (i). Otherwise some odd prime has an odd exponent, so $a=\pm q^{t}b$ as in part (ii). In either case there is a positive odd $k$ with $(a/k)=-1$.

Factor $k$ into odd primes. A product of Legendre symbols equalling minus one has at least one factor equal to minus one. Thus $a$ fails to be a square modulo one of those primes, contradicting the hypothesis. HenceThe solvability formulation makes explicit the zero-residue convention needed by the printed final assertion. The prime two causes no extra issue because every odd integer is one modulo two.

False. Take $k(x)=1/(1+|x|)$. This is bounded and measurable on $\mathbb{R}$, and tends to zero as $x\to \infty$, butThus decay to zero alone does not imply Lebesgue integrability.

False. Let $k$ be the indicator of the positive integers. It is measurable, bounded by one and zero almost everywhere, so its Lebesgue integral on $(0,\infty )$ is zero. Yet $k(n)=1$ for every positive integer $n$, so it does not tend to zero at infinity. Lebesgue integrability is unchanged by modifications on a null set, whereas a pointwise limit can change.

Let $H_0={\int }_0^{\pi }|h(u)|du<\infty$. On each period, substitute $x=j\pi +u$ and use periodicity:Summing over $j\ge 0$ gives the convergent geometric boundMeasurability is preserved under multiplication by the exponential, so this proves Lebesgue integrability.

Use the nonnegative parts $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$, so $h=h^{+}-h^{-}$. The zero-mean assumption gives $H:={\int }_0^{\pi }{h}^{+}={\int }_0^{\pi }{h}^{-}<\infty$. On the $k$-th period, each of the positive and negative weighted integrals lies between $H/((k+1)\pi )$ and $H/(k\pi )$. Their difference therefore satisfies the stronger estimateThe sum of the absolute values of these block integrals converges, proving the existence of the finite limit of the integrals from $\pi$ to $n\pi$.

For genuine Lebesgue integrability we instead need the integral of the absolute value. Periodicity givesIf $H>0$, summation diverges by the harmonic series. If $H=0$, then $h=0$ almost everywhere and the function is integrable. ThusThe finite signed block limit therefore need not be a Lebesgue integral.

The monotone convergence theorem states that for measurable functions $0\le f_1\le f_2\le \cdots$ converging pointwise to $f$, the limit is measurable andBoth sides may be infinite. Monotonicity and nonnegativity replace any need for an integrable dominating function.

The partial sums $S_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase pointwise to the extended-valued sum $S={\sum }_{k=1}^{\infty }{a}_k$. The monotone convergence theorem and finite linearity giveA nonnegative measurable function with finite integral is finite almost everywhere; if it were infinite on a positive-measure set, its integral would be infinite. Define it arbitrarily on the null exceptional set if a real-valued representative is desired. Hence the series converges almost everywhere to an integrable function and

Substitute $x=e^{-t}$ to getwhere the second substitution $u=(k+1)t$ reduces the integral to the supplied gamma integral.

On $(0,1)$, $-x\log x\ge 0$. Expand $x^{-x}=e^{-x\log x}={\sum }_{j=0}^{\infty }(-x\log x{)}^j/j!$. The partial sums increase, so the monotone convergence theorem permits termwise integration. The integral of the $j$-th term is $1/(j+1{)}^{j+1}$, including $j=0$. Reindexing givesThis is finite, for instance because the summands with $k\ge 2$ are bounded by $2^{-k}$.

Expand $x^x=e^{x\log x}={\sum }_{j=0}^{\infty }(x\log x{)}^j/j!$. The sum of the integrals of the absolute values of its terms is ${\sum }_{k=1}^{\infty }{k}^{-k}<\infty$ by part (c). Absolute summability therefore permits interchange of the series and the integral. Applying the moment formula from part (c),Equivalently, the absolute exponential series is $x^{-x}$, an integrable majorant, so dominated convergence gives the same interchange.

A connected space has no separation into two disjoint nonempty open subsets whose union is the space. A continuous surjection to the discrete space $\{0,1\}$ produces such a separation by the two point preimages. Conversely, from a separation define a function equal to zero on one open piece and one on the other; it is continuous and surjective. Thus these formulations of connectedness are equivalent.

An interval in $\mathbb{R}$ is connected: if a continuous map from it to $\{0,1\}$ took both values, the intermediate value theorem would force value $1/2$, impossible. Conversely, a subset of $\mathbb{R}$ that is not an interval has points $a<b$ in it and a point $c\in (a,b)$ outside it. Intersecting it with $(-\infty ,c)$ and $(c,\infty )$ separates it. The empty set is connected, and singleton intervals are included.

For the union assertion, any separation of $X$ must put each nonempty connected $A_i$ entirely on one side, since intersecting the separation with $A_i$ would otherwise disconnect it. Pairwise nonempty intersections force all the $A_i$ onto the same side. Their union is $X$, contradicting a nonempty second side. The empty-union case is already connected.

The plane is the union of all lines through the origin. Each line is homeomorphic to $\mathbb{R}$ and connected, and every pair meets at the origin, so ${\mathbb{R}}^2$ is connected. For the punctured plane, let $R_{+}$ be the positive horizontal ray, and for each $r>0$ let $C_r$ be the circle of radius $r$. Each circle is a continuous image of $[0,2\pi ]$ and hence connected; $C_r\cup R_{+}$ is connected because its two connected pieces meet at $(r,0)$. These unions all contain $R_{+}$ and cover the punctured plane, proving it connected as well.

Finally suppose a continuous injection $g:{\mathbb{R}}^2\to \mathbb{R}$ existed. Its image is a connected interval, since a continuous image of a connected space is connected. It contains at least three distinct values; choose $p$ whose value lies strictly between two others. The punctured plane ${\mathbb{R}}^2\setminus \{p\}$ is connected by translation of the preceding result. By injectivity its image is precisely $g({\mathbb{R}}^2)\setminus \{g(p)\}$, which is disconnected by the missing interior point. This contradicts connectedness of a continuous image. Therefore

A sequentially compact space is one in which every sequence has a subsequence converging to a point of the space. A limit point $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus \{\ell \}$.

In a compact metric space, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no limit point, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a limit point in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.

For the remaining assertions assume the compact space is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The extreme value theorem gives a minimum $\delta =q(p)$. If $\delta >0$, then $p\ne f(p)$, and strict decrease of distances impliescontradicting minimality. Thus $\delta =0$ and $p$ is a fixed point. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the fixed point is unique.

On the noncompact real line, takeIts derivative $e^x/(1+e^x)$ is strictly between zero and one. The mean value theorem therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log (1+e^{-x})>0$, so it has no fixed point. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.

The $j$-th partial derivative is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is $J_f(x)=({\partial }_j{f}_i(x){)}_{i,j}$. When the total derivative exists, all partial derivatives exist and $J_f(x)$ is its matrix in the standard bases; its $j$-th column is $Df(x)e_j$. Existence of partial derivatives alone does not guarantee total differentiability.

For a scalar-valued function, the gradient is the column vectorWhen $f$ is differentiable, its total derivative is the linear functionalThus the Jacobian is the transpose of the gradient in standard Euclidean coordinates.

The chain rule at the specified point isIn standard coordinates this becomesThe matrix dimensions are respectively $k\times n$, $k\times m$, and $m\times n$, matching the composition of the two linear derivative maps.

Put $Q=1+x^2+y^2>0$. Differentiation of $Q^{-1}$ givesThe positive denominator also shows that the function and these derivatives are smooth throughout ${\mathbb{R}}^2$.

At $(-1,0)$ the gradient is $(1/2,0{)}^T$. The total derivative is therefore the linear functional

The derivative of the curve isIts image lies on the unit circle, so $f(g(t))=1/2$ and $\boxed{D(f\circ g)(t)=0}$. Meanwhileand multiplication by $J_g(t)$ gives $-\sin t\cos t/2+\cos t\sin t/2=0$. This verifies the chain rule explicitly.

Because $2|x|\le 1+x^2\le Q$, we have $|f_x(x,y)|=2|x|/Q^2\le 1/Q\le 1$; similarly $|f_y|\le 1$. Apply the one-variable mean value theorem on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The triangle inequality givesThus the function is globally one-Lipschitz for the ${\ell }^1$ metric on the plane.

For a normal constrained extremal, introduce a constant Lagrange multiplier and use $L=F-\lambda G$. The Euler-Lagrange equation isWhen both integrands have no explicit $x$ dependence, the Beltrami identity gives $L-y^{\prime }{L}_{y^{\prime }}=\text{constant}$, as follows by differentiating this expression and using the Euler-Lagrange equation.

Parametrize the surface by $(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The cross product of its two coordinate tangents has magnitude $f\sqrt{1+(f^{\prime }{)}^2}$, so integration in $\theta$ gives $A=2\pi {\int }_a^{b}f\sqrt{1+(f^{\prime }{)}^2}dz$. Integrating the cross-sectional disk areas gives $V=\pi {\int }_a^b{f}^{2}dz$.

Dividing the augmented functional $A-\lambda V$ by $\pi$, useThe Euler-Lagrange equation isThe Beltrami identity becomes $2f/\sqrt{1+(f^{\prime }{)}^2}-\lambda f^2=C$. At an endpoint $f\to 0$, the absolute value of the first term is at most $2f$, so $C=0$, even if the profile has an infinite endpoint slope. For a nondegenerate regular closed surface take $f>0$ on $(a,b)$. Then $\lambda >0$, and, with $R=2/\lambda$,Combining this identity with the Euler-Lagrange equation yields $f^{\prime \prime }=-R^2/f^3$. Consequently $(f^2{)}^{\prime \prime }=2((f^{\prime }{)}^2+f{f}^{\prime \prime })=-2$. The two zero endpoint values uniquely determineThe positive radius profile is therefore a semicircle and generates a sphere, centred at height $(a+b)/2$, with radius $R=(b-a)/2$. In particularThe graph is smooth in the interior; its vertical slopes at the two poles are compatible with a smooth spherical surface. Its volume is $4\pi R^3/3$, the volume associated with this stationary candidate.

For minimality, the three-dimensional isoperimetric inequality says that every closed surface enclosing volume $V$ has $A^3\ge 36\pi V^2$, with equality for a sphere. The displayed extremal attains equality, proving that it minimizes area at its fixed enclosed volume. The uniqueness above concerns the regular connected positive profile with the two stated endpoint heights.

For symmetric quadratic coefficient matrices, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$, where $T$ is the positive-definite mass matrix. Substitute a normal-mode oscillation $q=v\cos (\omega t+\varphi )$ to getThe positive roots ${\omega }^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.

The particle's kinetic energy is $m|\dot{q}{|}^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, Hooke's law gives tension $\lambda (r-a)/a$ and elastic energy $\lambda (r-a{)}^2/(2a)$. With $\lambda =mg$ and gravitational potential $mgz$, this gives exactlyThis is the taut-string regime containing the equilibrium. Differentiating the potential givesAt $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.

With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta +(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leavingThe mass matrix remains $mI$, and the stiffness matrix is $\mathrm{diag}(mg/(2a),mg/(2a),mg/a)$. ThereforeThe horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a normal mode. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.

The general time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$. For a spherically symmetric wave function, ${\nabla }^2\psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$; the Coulomb attraction between the proton and electron is $V=-q^2/(4\pi {ϵ}_{0}r)$. These supply respectively the kinetic and potential terms in the displayed radial equation. Rearranging identifiesas the Bohr radius.

Substitute $\psi =(A+Br)e^{-\lambda r}$ into the rearranged equation and put $ϵ=2mE/{\hslash }^2$. Equating the coefficients of $r,1,r^{-1}$ after cancelling the exponential givesA nonzero state has $A\ne 0$. If $B=0$, these identities give $\lambda =1/a$ and $ϵ=-1/a^2$. If $B\ne 0$, they give $\lambda =1/(2a)$, $B=-A/(2a)$, and $ϵ=-1/(4a^2)$. Thus the two independent eigenfunctions and their corresponding energies areThey are eigenfunctions at different energies, rather than two independent solutions at one fixed energy.

For the higher-energy state, radial probabilities include the spherical volume factor $4\pi r^2$. The normalization constant and $4\pi$ cancel in the ratio. Using the supplied integral at $b=a$ and at infinity,HenceThis calculation fixes the denominator as $8e$ in the source's unparenthesized expression; interpreting it as a product with $e$ would give a negative probability.