Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 (AS2, XMST 4306)

Source: Mathematics release, supplied bundle Part A 2007 11-1.pdf, pages 241–258. The eighteen-page Mathematics and Statistics paper is dated 24 June 2010. Its editable source transcription is retained privately.

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Regard $K$ as a vector space over $F$. Since $K$ is finite, a maximal linearly independent list has finite length $d$ and is a basis. Every element is uniquely a linear combination of its $d$ basis vectors, with $q$ choices for each coefficient. Hence $\boxed{|K|=q^d}$, with $d=[K:F]$. This includes $d=1$ when the extension is trivial.

Write $q=p^r$, where $p$ is the characteristic. In a splitting field of $X^{q^m}-X$ over $F$, let $K$ be its set of roots. The derivative of this polynomial is $-1$, so all $q^m$ roots are distinct. The Frobenius power is additive and multiplicative: $(x+y{)}^{q^m}=x^{q^m}+y^{q^m}$ and $(xy{)}^{q^m}=x^{q^m}{y}^{q^m}$. Thus roots are closed under addition, negatives and multiplication; for a nonzero root, $(x^{-1}{)}^{q^m}=x^{-1}$. They include $0,1$ and form a field.

Every $a\in F$ satisfies $a^q=a$, by Lagrange's theorem on $F^{\times }$ and separately at zero. Iterating gives $a^{q^m}=a$, so $F\subseteq K$. The constructed field therefore contains $F$ and has exactly $\boxed{q^m\text{ elements}}$.

By the allowed cyclicity theorem, $K^{\times }$ has a generator $z$. Every nonzero element of $K$ is a power of $z$, and zero also belongs to $F(z)$. Thus $K\subseteq F(z)\subseteq K$, giving $\boxed{K=F(z)}$. This is a primitive element argument for finite fields using the multiplicative group, not a claim that the additive group is cyclic.

For $\alpha \ne 0$, Lagrange's theorem in the multiplicative group of order $q^m-1$ gives ${\alpha }^{q^m-1}=1$; at zero the desired identity is immediate. Thus all elements satisfy ${\alpha }^{q^m}-\alpha =0$.

For an irreducible degree-$m$ polynomial $f\in F[X]$, the quotient $L=F[X]/(f)$ is a field of dimension $m$, hence has $q^m$ elements by part 1. Its element $\alpha =[X]$ therefore satisfies ${\alpha }^{q^m}-\alpha =0$. The evaluation homomorphism $F[X]\to L$ has kernel $(f)$, so

Given $N⊴G$ and $H\le G$, the conditions are $\boxed{G=NH,N\cap H=\{1\}}$. Every element then has a unique expression $nh$. The extra datum is the action $\alpha :H\to \mathrm{Aut}(N)$ given by conjugation, ${\alpha }_h(n)=hn{h}^{-1}$. On pairs $(n,h)$ the reconstructed multiplication isFor example $S_3=⟨(123)⟩⋊⟨(12)⟩$. The order-three subgroup is normal, the intersection is trivial, and their sizes multiply to six. Conjugation by $(12)$ inverts $(123)$, so the action is nontrivial. Consequently this is not the direct product of these subgroups: $C_3\times C_2$ is Abelian, whereas $S_3$ is not.

Compose the second map first. Direct substitution givesThe result is affine because the product matrix is invertible. The identity is $(I,0)$ and the inverse is $(A^{-1},-A^{-1}b)$. Associativity follows from composition of maps, or by expanding this product three times. Distinct pairs give distinct maps, since evaluating at zero gives $b$ and evaluating the difference on basis vectors gives all columns of $A$. Thus these pairs faithfully describe a group.

The map $A\mapsto (A,0)$ is an injective group homomorphism, so its image is the required copy of ${\mathrm{GL}}_n(F)$. Translations are the subgroup $N=\{(I,b):b\in F^n\}$, isomorphic to the additive group of $F^n$. For any affine map,Thus conjugation preserves translations, proving $N⊴G$.

Every pair factors as $(A,b)=(I,b)(A,0)$, and a linear map which is also a translation fixes zero and has linear part $I$, hence is the identity. Thus $G=NH$ and $N\cap H=\{1\}$. The conjugation formula in (c) says that the action needed for reconstruction is simply the usual matrix action $A:t\mapsto At$ on the additive group $F^n$. Therefore

There are $p^n$ choices for the translation. Choose the columns of an invertible matrix successively: the first is any nonzero column, the second avoids its span, and at stage $j+1$ one avoids a $j$-dimensional span of size $p^j$. ConsequentlyThe successive linear independence conditions are exactly the criterion that the columns form a basis.

A primitive root modulo $p$ is a generator of the cyclic multiplicative group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, so it has order $p-1$. If $g$ is one generator, the generators are exactly $g^k$ with $\gcd (k,p-1)=1$. Hence there are $\boxed{\varphi (p-1)}$. For $p=2$ the group is trivial and its unique element is its generator, consistently with $\varphi (1)=1$.

For $n\ge 1$, the zero residue contributes zero. Multiplication by any nonzero $\alpha$ permutes the nonzero residues, so the sum $S$ satisfies $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $p-1\nmid n$, choose a primitive root $\alpha$; then ${\alpha }^n\ne 1$, forcing $S=0$. If $p-1\mid n$, every nonzero term equals one, so $S=p-1=-1$ in the field.

The printed display also mentions $n=0$, although its introductory assumption is $n\ge 1$. For that extra case the sum of the $p$ integer values $x^0$ for $1\le x\le p$ is $p\equiv 0$, so that branch is consistent too. The result is

The Legendre symbol being one means $a$ is nonzero and a square modulo $p$. Take one root $r\ne 0$. Factoring $x^2-r^2=(x-r)(x+r)$ in the field shows that any root is $r$ or $-r$. Since $p$ is odd these are distinct. Thus there are exactly $\boxed{2}$ solutions.

Each nonzero quadratic residue is attained twice as $x^2$ for $1\le x\le p-1$. Therefore twice the sum of those residues is ${\sum }_{x=1}^{p-1}{x}^2$, which is zero modulo $p$ by (ii), since $p-1>2$. As two is invertible modulo $p$, the sum of the residues is zero. The sum of all nonzero residues is $p(p-1)/2\equiv 0$, so subtracting gives the same result for the nonresidues. Hence $\boxed{\text{both indicated sums are }0(\mathrm{mod}p)}$.

By (ii), failure of divisibility is equivalent to $q-1\mid 30$. The positive divisors of $30$ are $1,2,3,5,6,10,15,30$; adding one gives $2,3,4,6,7,11,16,31$. Retaining only primes yields

The monotone convergence theorem states that if nonnegative measurable functions $h_n$ increase almost everywhere to $h$, then $\int h={\lim }_n\int h_n$, allowing infinity. The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ for an integrable $g$, then $h$ is integrable and $\int h_n\to \int h$.

Apply monotone convergence to $A_n={\sum }_{k=1}^n|a_k|$. Its limit $A={\sum }_{k\ge 1}|a_k|$ has integral ${\sum }_k\int |a_k|<\infty$, so $A<\infty$ almost everywhere. Thus the original series converges absolutely almost everywhere. Its partial sums are bounded in absolute value by the integrable function $A$, and their limit $s$ satisfies $|s|\le A$. Dominated convergence givesAssigning any value on the null exceptional set defines an integrable representative of $s$.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. All terms after multiplication by $(\log x{)}^2$ are nonnegative, so monotone convergence permits integration term by term. Substituting $x=e^{-t}$ and integrating by parts twice givesConsequently the integral is ${\sum }_{k\ge 1}2/k^2$, and the supplied sum gives $\boxed{{\pi }^2/3}$.

A sufficient differentiation theorem is the following local version. Suppose $u(t,\cdot )$ is integrable, $u(\cdot ,x)$ is continuously differentiable for almost every $x$, and for each compact parameter interval there is an integrable $g(x)$ bounding $|{\partial }_{t}u(t,x)|$ uniformly there. Then $I^{\prime }(t)={\int }_E{\partial }_{t}u(t,x)dx$. The mean value theorem bounds the difference quotients by $g$, and dominated convergence proves the formula. Integrability at one parameter value together with the derivative bound also suffices locally.

The $k$th derivative of the integrand is $(\cos x{)}^k\cos (t\cos x+k\pi /2)$, whose absolute value is at most one. The integration interval has finite measure. Repeated application of the differentiation theorem therefore givesDominated convergence also gives continuity of every derivative, proving infinite differentiability.

From (a), combine the three derivative integrals:The integrand is $-\frac{d}{dx}[\sin x\sin (t\cos x)]$. This bracket vanishes at both endpoints, so the integral is zero. Hence $\boxed{t{J}_0^{\prime \prime }+J_0^{\prime }+t{J}_0=0}$, also at $t=0$ by the same formula.

A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.

A subset $A$ is connected if, with its subspace topology, it cannot be written as $A=U\cup V$ with $U,V$ disjoint nonempty relatively open sets. Equivalently its only subsets both open and closed relative to $A$ are $\varnothing ,A$. This definition includes the empty set and singletons as connected sets.

The connected subsets of the real line are exactly the intervals, including singletons, unbounded intervals and the empty set. Equivalently, whenever $x<y$ belong to the set, every $z$ with $x<z<y$ also belongs to it. Endpoints may or may not be included.

Use the separation definition in (i). If $A={\bigcup }_i{A}_i$ had a separation $A=U\cup V$, the intersections with each connected $A_i$ show that $A_i$ lies wholly in one side. Choose one nonempty $A_{i_0}$; every other $A_i$ intersects it and must lie on that same side. Thus the other side is empty, a contradiction. If the union is empty it is connected by definition. Therefore $\boxed{\bigcup _i{A}_i\text{ is connected}}$.

Let $K_a$ be the union of all connected subsets containing $a$. This collection is nonempty because it includes $\{a\}$. Any two of its members meet at $a$, so (a)(iii) makes their full union connected. It contains $a$ and by its definition contains every connected subset through $a$. Thushas exactly the required largest-set property.

If $K_a\cap K_b\ne \varnothing$, their union is connected by (a)(iii). It contains $a$, so maximality gives $K_a\cup K_b\subseteq K_a$ and hence $K_b\subseteq K_a$. Interchanging the roles gives the reverse inclusion. Thus the two components coincide; otherwise they are disjoint.

The inclusion of $X$ into $\mathbb{R}$ is continuous, so a connected component $K$ of $X$ is an interval in $\mathbb{R}$. For $x\in K$, openness of $X$ gives $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq X$. This interval is connected and meets $K$, so their union is connected and maximality forces the interval into $K$. Thus $K$ is open in $\mathbb{R}$. A nonempty open connected subset of the line is an open interval, allowing an infinite endpoint or the whole line. Hence every component is an open interval.

The components cover $X$ and are pairwise disjoint by (b)(ii); by (iii) they are open intervals. Each nonempty interval contains a rational number. Enumerate the rationals and assign to each interval the first rational it contains. Disjointness makes these assigned rationals distinct, giving an injection from the set of components into $\mathbb{N}$. Thus $\boxed{X\text{ is a countable union of pairwise disjoint open intervals}}$, with the empty union for $X=\varnothing$.

A subset $K$ is compact if every cover of $K$ by sets open in the ambient space has a finite subcover. Equivalently, every relatively open cover of the subspace $K$ has a finite subcover. These agree because relatively open sets are intersections with ambient open sets.

A subset $B$ of a metric space is bounded if it lies in a ball of finite radius: there exist a point $x_0$ and $R<\infty$ such that $d(x,x_0)\le R$ for every $x\in B$. Equivalently its diameter ${\sup }_{x,y\in B}d(x,y)$ is finite, with the empty set bounded by convention.

In any topological space a closed subset of a compact set is compact: add its open complement to an open cover and take a finite subcover of the compact set. However, compact subsets need not be closed in a general space. In the indiscrete two-point space a singleton is compact but not closed. Conversely a closed subset need not be compact: $\mathbb{R}$ itself is closed in $\mathbb{R}$ but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover.

In a Hausdorff space every compact subset $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; compactness selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply compactness, as the same real-line example shows. It does imply compactness if the ambient space is compact.

In a metric space compactness implies boundedness: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is bounded, but its singleton cover has no finite subcover. It is even closed and complete, so closedness plus boundedness alone does not suffice in general metric spaces.

By the Heine–Borel theorem a subset of $\mathbb{R}$ is compact exactly when it is closed and bounded. Consequently the nonempty compact intervals are exactly $\boxed{[a,b](-\infty <a\le b<\infty )}$. This includes the degenerate interval $[a,a]$. The empty interval is compact as well; nonclosed bounded intervals and all unbounded intervals fail the criterion.

The image $f(K)$ is compact: pull any open cover back by the continuous map, use a finite subcover of $K$, then map forward. For a general topological target, closedness is not guaranteed; a continuous image into the indiscrete two-point space can be a nonclosed singleton. No connectedness assertion follows merely from compactness of $K$.

If $Y$ is metric, $f(K)$ is compact, therefore closed in $Y$, bounded and totally bounded. It is also complete in its induced metric and sequentially compact. For the latter properties, compact metric spaces have a convergent subsequence of every sequence; a Cauchy sequence with one convergent subsequence converges to that same limit. Total boundedness follows by selecting finitely many balls of each prescribed radius. These are properties of the image itself, irrespective of whether $Y$ is complete.

A sequence is Cauchy ifThe comparison is between sufficiently late terms, without presupposing a limiting point in the space.

A metric space is complete if every Cauchy sequence of its points converges to a point of that same space. The requirement on the location of the limit distinguishes, for example, $\mathbb{R}$ from its incomplete subspace $\mathbb{Q}$.

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

Continuity and nonvanishing on the compact interval implyHence $m\Vert f-g{\Vert }_{\infty }\le d_{\alpha }(f,g)\le M\Vert f-g{\Vert }_{\infty }$. A $d_{\alpha }$-Cauchy sequence is therefore uniformly Cauchy, so (i) gives a continuous uniform limit; the upper bound then gives convergence in $d_{\alpha }$. Thus $\boxed{(C([0,1]),d_{\alpha })\text{ is complete}}$. The positive lower bound is what makes the argument work.

Differentiability at $x_0$ means that a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$. Since finite-dimensional linear maps are bounded, the right side tends to $f(x_0)$ as $h\to 0$, proving continuity.

For the specified scalar function, $0\le f(x,y)\le x^2\le x^2+y^2$. Dividing by $\sqrt{x^2+y^2}$ gives a bound tending to zero at the origin. Consequently $\boxed{df(0,0)=0}$ and the function is differentiable there.

For the final product assertion, write $f(x_0+h)=Lh+r(h)$, where $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,Cauchy–Schwarz and boundedness of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=⟨df(x_0)h,g(x_0)⟩}$; differentiability of $g$ is unnecessary because $f(x_0)=0$.

A $C^1$ embedded hypersurface $M\subseteq {\mathbb{R}}^n$ is locally carried by a $C^1$ coordinate diffeomorphism to an open subset of ${\mathbb{R}}^{n-1}\times \{0\}$. Equivalently, near each of its points it has the form $F^{-1}(0)$ for a $C^1$ function with nonzero gradient there.

The Lagrange multiplier theorem states that if a differentiable function $g$ has a local extremum on such a hypersurface at $x_0$, then $dg(x_0)$ annihilates the tangent space. In a regular local level-set description this isfor some real $\lambda$. The regularity condition $\nabla F(x_0)\ne 0$ is essential; it is what identifies the normal line and permits the local constraint chart.

Work on the positive orthant, where the real powers defining $g$ are smooth. Its closure under the constraint is the compact weighted simplex $\overline{M}=\{x_i\ge 0,\sum p_i{x}_i=1\}$, since $0\le x_i\le 1/p_i$. The product extends continuously there, is zero on its boundary when $n>1$, and is one at $(1,\dots ,1)$. Thus a maximum is attained in the positive interior.

At an interior maximizer the constraint gradient $(p_1,\dots ,p_n)$ is nonzero. The multiplier equations giveAll coordinates are equal, and the weighted constraint forces each to equal one. Therefore the maximum is one and $\boxed{g(x)\le 1}$ on $M$. For $n=1$ the constraint itself gives $x_1=1$, so the same conclusion holds directly. Equality occurs exactly at the all-ones point.

Set $A={\sum }_i{p}_i{a}_i>0$ and $x_i=a_i/A$. Then ${\sum }_i{p}_i{x}_i=1$, so part 1 gives ${\prod }_i(a_i/A{)}^{p_i}\le 1$. Since ${\sum }_i{p}_i=1$, this becomesEquality holds exactly when $x_i=1$ for all $i$, equivalently all the $a_i$ are equal. This is the weighted arithmetic–geometric mean inequality.