Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 (AS2, XMST 4306)

Source: Mathematics release, supplied bundle Part A 2007 11-1.pdf. This nineteen-page Mathematics and Statistics paper is dated 25 June 2009. Complete source statements and numerical tables are retained in the private editable transcription.

Bounded exact-paper checks found no matching advertisement of worked answers. All printed tasks and surrounding requests are answered, including explicit distinctions between the full-data and grouped-data likelihoods in O5.

The characteristic is the least positive integer $r$ with $r{1}_K=0$, if it exists, and zero otherwise. The prime subfield is the smallest subfield containing $1_K$; in characteristic $p$ it is $F=\{0,1,\dots ,(p-1)1_K\}\cong {\mathbb{F}}_p$. Every field automorphism fixes $1_K$ and addition, so it fixes every element of this prime subfield.

The additive sequence $0,1,2\cdot 1,\dots$ must repeat because $K$ is finite, so the characteristic is positive. If its least positive value were composite, $r=ab$ with $1<a,b<r$, then $(a{1}_K)(b{1}_K)=r{1}_K=0$ while neither factor would be zero. A field has no zero divisors, giving a contradiction. Hence $\boxed{\mathrm{char}K=p\text{ is prime}}$.

The finite field $K$ is a finite-dimensional vector space over its prime subfield $F$ of size $p$. A basis has some positive length $n$, and every element is uniquely specified by $n$ coefficients in $F$. There are $p$ choices per coefficient, so $\boxed{|K|=p^n}$ and $[K:F]=n$.

Since $K=F(\lambda )$ and the extension is finite, the element $\lambda$ is algebraic. Its minimal polynomial is irreducible, and evaluation gives an isomorphism $F[X]/(m_{\lambda })\cong F[\lambda ]=F(\lambda )=K$. If its degree is $d$, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis. Comparing with $[K:F]=n$ gives $\boxed{\mathrm{deg}{m}_{\lambda }=n}$.

Define $\vartheta (x)=x^p$ on all of $K$. The binomial coefficients between the endpoints are divisible by $p$, so $(x+y{)}^p=x^p+y^p$; also $(xy{)}^p=x^p{y}^p$ and $1^p=1$. Thus this is a field homomorphism. Its kernel is zero, and an injective self-map of the finite set $K$ is surjective. Hence $\boxed{\vartheta \text{ is an automorphism}}$, the Frobenius automorphism, with the required image of $\lambda$. It fixes $F$ by the preceding prime-subfield argument.

Take $0<m<n$, the intended positive-exponent range. If ${\vartheta }^m(\lambda )=\lambda$, then ${\vartheta }^m$ fixes both $F$ and its field generator, so it fixes every element of $K=F(\lambda )$. Every element would then be a root of $X^{p^m}-X$. But that polynomial has degree $p^m<p^n=|K|$, impossible for a nonzero polynomial over a field. Thus $\boxed{{\vartheta }^m(\lambda )\ne \lambda (0<m<n)}$. The source writes only $m<n$; zero must be excluded, since ${\vartheta }^0(\lambda )=\lambda$.

For nonzero $x\in K$, Lagrange's theorem in $K^{\times }$ gives $x^{p^n-1}=1$; zero satisfies $x^{p^n}=x$ too. Thus ${\vartheta }^n$ is the identity, and (iii) excludes every smaller positive order. Hence $\boxed{|⟨\vartheta ⟩|=n}$.

Any field automorphism of $K$ fixes $F$ and is determined by its value on $\lambda$. That value must be a root of the degree-$n$ minimal polynomial, so there are at most $n$ automorphisms. The $n$ powers of $\vartheta$ already attain this bound. Consequently $\boxed{\mathrm{Aut}(K)=⟨\vartheta ⟩\cong C_n}$.

A left action is a map $(g,\omega )\mapsto g\omega$ satisfying $1\omega =\omega$ and $(gh)\omega =g(h\omega )$. The orbit of $\omega$ is $G\omega =\{g\omega :g\in G\}$. The action is transitive when there is exactly one orbit on the nonempty acted-on set, equivalently when any point can be moved to any other by a group element. An action is equivalently a homomorphism from $G$ to the permutation group of the set.

Act by left multiplication on the five cosets of $H$. The kernel is the core ${\bigcap }_{g\in G}gH{g}^{-1}$, a normal subgroup contained in $H$, so the hypothesis makes it trivial. Thus $G$ embeds in $S_5$, and its image has order $60$ and index two. An index-two subgroup is normal, so the quotient is of order two. Every three-cycle maps trivially to that quotient, since its order divides three. Three-cycles generate $A_5$, so the image contains $A_5$; their equal order gives $\boxed{G\cong A_5}$.

For completeness, prove simplicity rather than merely invoking it. The nonidentity even cycle types in $S_5$ are three-cycles, products of two disjoint transpositions and five-cycles. In $A_5$, a three-cycle has centralizer of order three, so its conjugacy class has size $20$. A double transposition has centralizer of order four in $A_5$, giving class size $15$. A five-cycle has centralizer its cyclic subgroup of order five, giving class size $12$; the $24$ such elements therefore split into two classes of size $12$. These account for all $60$ elements with the identity.

A normal subgroup is a union of whole conjugacy classes, containing the identity, and its order must divide $60$. No nonempty proper sum of the class sizes $15,20,12,12$, plus one, is a divisor of $60$ greater than one: the possible proper nontrivial sums are $13,16,21,25,28,33,36,40,45,48$. Only the full union has order $60$. Hence $\boxed{A_5\text{ is simple}}$, and so is $G$.

Choose one regular tetrahedron inscribed in the dodecahedron and let $H$ be its stabilizer among rotations of the dodecahedron. The twelve rotations of that tetrahedron preserve the dodecahedron; geometrically these are its identity, eight rotations by $\pm 2\pi /3$ about tetrahedral vertex axes and three half-turns about opposite tetrahedral edge axes. They act faithfully on its four vertices and give the twelve even permutations. Thus $\boxed{H\cong A_4,|H|=12}$ without using part (ii).

To check the core condition geometrically, the fifteen half-turns of the dodecahedron, about opposite-edge-midpoint axes, are conjugate under its rotations. A normal subgroup contained in $H$ cannot contain even one of these half-turns, because normality would then force all fifteen into a group of order twelve. On the other hand every nontrivial normal subgroup of $A_4$ contains a half-turn: if it contains a three-cycle, its conjugates and their inverses generate $A_4$; otherwise its nonidentity elements are among the three double transpositions, which are conjugate in $A_4$. Apply this observation to any normal subgroup of the full rotation group lying inside $H$. It must be trivial, proving exactly the required condition.

For an odd prime $p$, the Legendre symbol is $\boxed{(a/p)=0}$ if $p\mid a$, $\boxed{(a/p)=1}$ if $a$ is a nonzero square modulo $p$, and $\boxed{(a/p)=-1}$ otherwise. It depends only on the residue class of $a$.

Euler's criterion states that for $p\nmid a$,For a multiple of $p$, both the Legendre symbol and that positive power vanish modulo $p$, so the congruence extends to that case as well.

Gauss's lemma for quadratic residues states: if $p\nmid a$ and $m$ is the number of least positive residue classes among $a,2a,\dots ,((p-1)/2)a$ that exceed $p/2$, then $\boxed{(a/p)=(-1{)}^m}$.

For $a=2$ the displayed multiples are $2,4,\dots ,p-1$, already least positive residue classes. Those greater than $p/2$ correspond to $j>p/4$. Thus their number is $m=(p-1)/2-\lfloor p/4\rfloor$. For $p\equiv 1,3,5,7(\mathrm{mod}8)$, its parity is respectively even, odd, odd, even. Gauss's lemma therefore gives

Euler's criterion gives $(-1/p)=(-1{)}^{(p-1)/2}$. The Legendre symbol is multiplicative, so multiplying this sign by the result of (iv) yields

Suppose there are only finitely many primes $p_1,\dots ,p_r\equiv 3(\mathrm{mod}8)$, and let $N$ be their product. This is odd, so $N^2+2\equiv 3(\mathrm{mod}8)$. No listed prime divides this number, since reduction modulo any $p_i$ gives two. Every prime divisor $q$ is odd and does not divide $N$, and satisfies $N^2\equiv -2(\mathrm{mod}q)$. Thus $(-2/q)=1$, forcing $q\equiv 1$ or $3(\mathrm{mod}8)$ by (v).

A product of prime factors all congruent to one modulo eight could not be three modulo eight. More precisely, the product of factors congruent to three is three modulo eight exactly when their total exponent is odd. Thus at least one prime divisor $q\equiv 3(\mathrm{mod}8)$ occurs. It is not in the alleged complete list, a contradiction. Hence there are infinitely many primes congruent to three modulo eight.

The function is nonnegative and measurable. For $\alpha \ne -1$, integrate from $\epsilon$ to one to get $(1-{\epsilon }^{\alpha +1})/(\alpha +1)$. The limit is finite exactly when $\alpha +1>0$. At $\alpha =-1$ the integral is $-\log \epsilon \to \infty$. Thus $\boxed{x^{\alpha }\in L^1(0,1)⟺\alpha >-1}$.

Integrating from one to $R$ gives $(R^{\alpha +1}-1)/(\alpha +1)$ for $\alpha \ne -1$. It has a finite limit as $R\to \infty$ exactly when $\alpha +1<0$, while at $\alpha =-1$ the logarithm diverges. Nonnegativity equates finiteness of this integral with Lebesgue integrability. Hence $\boxed{x^{\alpha }\in L^1(1,\infty )⟺\alpha <-1}$.

On $(0,1)$ its absolute value is at most $x|\log x|$, which has integral $1/4$, by integration by parts or the substitution $x=e^{-t}$. The function is measurable and this majorant is integrable, so $\boxed{x\log x/(1+x{)}^2\text{ is integrable on }(0,1)}$.

For $x\ge 1$, $(1+x{)}^2\le 4x^2$, so $x\log x/(1+x{)}^2\ge (\log x)/(4x)\ge 0$. The latter integral diverges because its primitive is $(\log x{)}^2/8$. Thus the absolute integral diverges and the answer is $\boxed{\text{not integrable on }(1,\infty )}$.

The integrand is nonnegative and measurable. It is bounded by $x$ near zero and by $x{e}^{-x}$ for $x>0$, so its integral is finite. For every $x>0$ the geometric expansion gives $x/(e^x+1)={\sum }_{n\ge 1}(-1{)}^{n-1}x{e}^{-nx}$. To justify termwise integration, note thatTonelli gives an integrable sum of absolute values; dominated convergence then applies to the signed partial sums. ConsequentlyIf evaluated further, this is $(1-2^{-1})\zeta (2)={\pi }^2/12$.

For product measure spaces, Tonelli's theorem states that for a nonnegative measurable function the product-space integral equals either iterated integral, allowing the value $+\infty$. Fubini's theorem states that if a measurable function has finite integral of its absolute value, then its sections are integrable almost everywhere, both iterated integrals exist and equal the product-space integral. The absolute-integrability hypothesis is what permits this conclusion for a signed function. These statements apply to the Lebesgue product spaces used here, which are sigma-finite.

The function $g$ is measurable, and for fixed positive $y$ both terms are integrable over $0<x<1$ by linear substitution and integrability of $f$. Those substitutions giveThe strict sign follows from positivity of $f$ on a nonempty interval, since $a>1$. Similarly each section in $y$ is absolutely integrable for fixed $x>0$, andIf $g$ were absolutely integrable on the stated rectangle, Fubini would make the two finite iterated integrals equal. The first would be strictly negative and the second strictly positive, which is impossible. Therefore $\boxed{g\notin L^1((0,1)\times (1,\infty ))}$. The one-dimensional section integrability does not rescue the missing product-space hypothesis.

The function $h$ is measurable and nonnegative. In first-quadrant polar coordinates, including the Jacobian, its integral isThe radial factor is $1/2$ by the supplied fact and the angular integrand is at most one, giving finiteness. The nonnegative change-of-variables formula is valid before finiteness is known, and this bound proves $h\in L^1$.

Fubini may now factor the Cartesian integral. Substituting $u=x^2$, $v=y^2$ in the respective factors givesBoth are finite: their powers at zero exceed $-1$, and the exponential controls infinity. Comparing with the polar expression proveswhich is exactly the product of integrals requested.

A subset $U$ of a metric space is open if for every $x\in U$ there exists $r>0$ with $B_d(x,r)\subseteq U$, where $B_d(x,r)=\{y:d(x,y)<r\}$. The radius may depend on the point. The empty set is open vacuously.

Continuity at $x$ means that for every $\epsilon >0$ there is $\delta >0$ such that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$ for every $y\in X$. Equivalently, the inverse image of each neighborhood of $f(x)$ is a neighborhood of $x$.

The map is continuous if it is continuous at every point of its domain. Equivalently the inverse image of every open subset of $Y$ is open in $X$. These equivalent formulations use the given metrics' induced topologies.

Write $r_x=d(x,a)$. The prescribed function is symmetric and nonnegative. If $x\ne y$, both $r_x,r_y$ could be zero only if $x=y=a$, so $d_a(x,y)>0$. For the triangle inequality, if any two of $x,y,z$ coincide it is immediate. If all three are distinct,Thus $\boxed{d_a\text{ is a metric}}$. Also $d_a(a,x)=d(a,x)$, a fact useful in the following parts.

If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$. The strict open-ball inequality therefore excludes every such $x$, while including $b$. Hence $\boxed{B_{d_a}(b,d(b,a))=\{b\}}$. Every point other than $a$ is isolated in this metric.

If continuity holds everywhere, it holds at $a$. Conversely, if it holds at $a$, it holds at every other point automatically: the singleton neighborhood from (ii) maps entirely to the single value at that point. Thus continuity everywhere in $d_a$ is equivalent to continuity at $a$ in $d_a$.

For every $r>0$ one has $B_{d_a}(a,r)=B_d(a,r)$ because $d_a(a,x)=d(a,x)$. The epsilon–delta definition at $a$ is therefore identical for the two domain metrics. In particular continuity at $a$ in $d_a$ implies continuity there in $d$.

The same equality of balls gives the reverse implication: continuity at $a$ in $d$ implies continuity at $a$ in $d_a$. The latter implies continuity everywhere by the first part. These implications complete the equivalence of all three printed conditions.

If $a\notin A$, every point of $A$ is isolated by (ii). The set $A$ is the union of these open singletons, so $\boxed{A\text{ is open in }(X,d_a)}$, including the empty case.

If $A$ is open and contains $a$, there is an epsilon ball at $a$ contained in it. This ball is exactly $B_d(a,\epsilon )$ by the equality of balls proved above. Conversely such a ball supplies a neighborhood at $a$, and all other points of $A$ have their singleton neighborhoods inside $A$. Thus the stated ball condition is necessary and sufficient.

An isolated point outside $B$ cannot belong to its closure. Thus the only possible extra closure point is $a$. The balls at $a$ meet $B$ exactly when arbitrarily small original-metric balls at $a$ do so. ConsequentlyIn the first case $a$ is already included. In the second case a ball smaller than the positive distance separates it from $B$; in the last case every ball meets $B$. If $B$ is empty its closure is empty, with distance convention $d(a,\varnothing )=+\infty$.

A subset is connected if it has no decomposition into two disjoint nonempty sets both relatively open in it. Equivalently its only relatively clopen subsets are itself and the empty set. Connectedness here refers to the subset's induced topology.

Suppose $f(X)=U\cup V$ were a separation into two disjoint nonempty relatively open sets. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$, and they are disjoint, nonempty and cover $X$. This contradicts connectedness of $X$. Hence $\boxed{f(X)\text{ is connected}}$.

The set $X/R$ consists of the equivalence classes, with canonical surjection $q:x\mapsto [x]$. The quotient topology declares $V\subseteq X/R$ open exactly when $q^{-1}(V)$ is open in $X$. This makes $q$ continuous and is the finest topology on the quotient set doing so.

Let $A$ be clopen and $C$ an equivalence class. The intersections $C\cap A$ and $C\setminus A$ are complementary relatively open subsets of $C$. Connectedness of $C$ means one must be empty. Thus every equivalence class is entirely in $A$ or entirely outside it, proving $\boxed{q^{-1}(q(A))=A}$. This is precisely saturation.

If $X$ is connected, its continuous surjective image under $q$ is connected by (a)(ii). Conversely suppose $X/R$ is connected and $A\subseteq X$ is clopen. By (i) both $A$ and its complement are saturated. The quotient definition then makes $q(A)$ open, and its complement $q(X\setminus A)$ open too. Thus $q(A)$ is clopen in the connected quotient, so is empty or the whole quotient. Pulling back gives $A=\varnothing$ or $X$. By the clopen criterion $X$ is connected. Hence $\boxed{X\text{ connected}⟺X/R\text{ connected}}$.

The difference map $(z,w)\mapsto z-w$ is continuous and $X$ is its inverse image of the open set $\mathbb{C}\setminus \{0\}$, so $X$ is open in ${\mathbb{C}}^2$. For fixed $z$, the equivalence class is $\{z\}\times (\mathbb{C}\setminus \{z\})$, homeomorphic to a punctured plane. It is path-connected: two points can be joined by a polygonal path avoiding the missing point. Thus all equivalence classes are connected.

The projection $\pi$ is continuous and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in ${\mathbb{C}}^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi (U)$. Hence $\pi (U)$ is open.

Its fibers are precisely the equivalence classes, so it induces a bijection $\bar{\pi }:X/R\to \mathbb{C}$. Continuity follows from the quotient topology and $\pi =\bar{\pi }\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar{\pi }(V)=\pi (q^{-1}(V))$ is open. Thus the inverse is continuous too, proving $\boxed{X/R\cong \mathbb{C}}$ as topological spaces. The plane is connected, so the connected-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.