Past exam of the probability, statistics and options I course of the University of Oxford / 2010 (AS1, XMST 4305)

Source: released examination bundle, Part A 2007 11-1.pdf, pages 228–240. The printed title is Probability, Statistics and Options I, for mathematics and statistics. Every question in every printed section is covered. No mark allocations are printed.

In field characteristic $p$, the intermediate binomial coefficients in the $p$th power vanish. HenceRepeated translation gives all $p$ roots $\alpha +j$, $j\in {\mathbb{F}}_p$. They are distinct because the prime field embeds in $F$. They exhaust the roots by degree; also the formal derivative of $X^p-X-a$ is $-1$, so this is a separable polynomial. If one root belongs to $F$, they all do, andThis is the root-translation property of an Artin–Schreier polynomial.

The intended hypothesis must be a proper monic factor, with $1\le r\le p-1$. Without that restriction the printed deduction is false: every polynomial has the displayed trivial factorization with $r=p$, even when it has no root in $F$.

By part 1, the roots in a splitting field are distinct and have the form $\alpha +j$. The degree-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using Vieta's formulas givesBoth $a_1$ and the sum of prime-field elements lie in $F$. Since $1\le r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha =(-a_1-{\sum }_{j\in J}j)/r\in F$, and part 1 proves the polynomial splits into distinct linear factors over $F$. In particular an Artin–Schreier polynomial of prime degree is either irreducible or splits completely.

Reduce the monic integer polynomial modulo $p$. Every $u\in {\mathbb{F}}_p$ satisfies $u^p=u$, so $u^p-u-1=-1\ne 0$. Thus the reduction has no root in ${\mathbb{F}}_p$. If it had a proper factor, part 2 would force such a root, a contradiction. Hence it is irreducible over ${\mathbb{F}}_p$.

The criterion of irreducibility by degree-preserving reduction modulo a prime, which follows from Gauss's lemma, now applies because the polynomial is monic and its degree is preserved on reduction. Consequently $X^p-X-1$ is irreducible in $\mathbb{Q}[X]$.

The Burnside lemma states that the number of orbits of a finite group action isFor $m=2n-1\ge 3$, the dihedral group consists of $m$ rotations $r^j$ by $2\pi j/m$, and $m$ reflections $r^{j}s$, each across the line through one vertex and the midpoint of the opposite edge. Thus $|G|=2m=4n-2$.

For the pentagon, the identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations has a single cycle on vertices and a single cycle on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two vertex pairs, and one fixed edge and two edge pairs, fixing $v^3{e}^3$ colourings. Applying the Burnside lemma to simultaneous vertex and edge colourings givesVertex colours and edge colours may be chosen independently; each symmetry must preserve both assignments.

The Legendre symbol isIt depends only on the residue class of $a$ modulo the odd prime $p$.

Euler's criterion statesFor $p\nmid a$, the sign is $+1$ precisely for a quadratic residue and $-1$ for a quadratic nonresidue. For $p\mid a$, both sides are zero.

Write $s=(a/p)$. If $p\mid a$, both sides of the proposed identity are zero. Otherwise $s=\pm 1$. The multiplicativity of the Legendre symbol and Euler's criterion giveFor the middle equality, Euler's congruence determines the sign exactly, since both the symbol and $s^{(p-1)/2}$ are in $\{1,-1\}$ and $p$ is odd.

Again $s=(a/p)\in \{1,-1\}$. By Euler's criterion, $(s/p)=s^{(p-1)/2}$. If $p\equiv 1(\mathrm{mod}4)$, the exponent is even and the answer is one. If $p\equiv 3(\mathrm{mod}4)$, it is odd and the answer is $s$. Hence

Use the countable identityIf $f(x)>b$, some $n$ has $1/n\le f(x)-b$, proving the nontrivial inclusion. Each preimage on the right is measurable by statistical hypothesis, so closure of a sigma-algebra under countable unions proves the strict-superlevel preimage is measurable.

For each real $a$,The final strict-superlevel set is measurable by part 1 and the given criterion for measurable functions. A complement relative to the measurable domain is measurable, so $-g$ is measurable.

If $f$ is measurable, then for $a\le 0$, $\{f^{+}\ge a\}=E$, and for $a>0$, $\{f^{+}\ge a\}=\{f\ge a\}$. Thus the given superlevel criterion proves $f^{+}$ measurable. Part 2 makes $-f$ measurable, and the same argument applied to $(-f{)}^{+}$ proves $f^{-}$ measurable.

Conversely $f=f^{+}-f^{-}$. If both parts are measurable, part 2 and the permitted closure under addition prove $f$ measurable. Therefore $f$ is measurable exactly when its positive and negative parts are measurable.

Since $|f|=f^{+}+f^{-}$, the previous part and closure under addition prove measurability of the absolute value.

The converse fails. Choose a nonmeasurable subset $N$ of $[0,1]$ and define $f=1$ on $N$, $f=-1$ on its complement. Then $|f|\equiv 1$ is measurable, whereas $f^{-1}([0,\infty ))=N$ is not. Thus measurability of $|f|$ does not imply measurability of $f$.

On $(0,1)$, $|g(x)|\le 1$, so ${\int }_0^1|g|\le 1$. For the second function,Both functions are measurable, and therefore both restrictions are Lebesgue integrable. The oscillation of $g$ near zero does not affect absolute integrability on this finite interval.

For $x\ge 1$, the bound $|\sin u|\le |u|$ gives $|g(x)|\le x^{-3/2}$, whose tail integral is finite. Thus $g{1}_{[1,\infty )}$ is integrable.

Near one, $h(x)=[(x-1)(x+1){]}^{-1/2}$ has an integrable inverse-square-root singularity. At infinity, however, $h(x)\ge 1/x$ for $x>1$, and the harmonic tail diverges. Thus $h{1}_{[1,\infty )}$ is not integrable. Assigning $h(1)=0$ changes neither conclusion, since one point has Lebesgue measure zero.

A topology $\mathcal{T}$ on $X$ contains $\varnothing$ and $X$, is closed under arbitrary unions, and is closed under finite intersections. Its members are the open sets. A subset $F\subseteq X$ is closed precisely when $X\setminus F$ is open.

The subspace topology on $Y$ isIf $Y$ is open in $X$, each such intersection is open in $X$. Conversely, an open subset of $X$ contained in $Y$ is its own intersection with $Y$. Thus the open sets of $Y$ are exactly the open subsets of $X$ contained in $Y$ in this case.

The closure $\overline{A}$ is the intersection of all closed subsets of $X$ containing $A$. It is the smallest such closed subset. Equivalently, $x\in \overline{A}$ exactly when every open neighborhood of $x$ meets $A$.

We prove (i)$\Rightarrow$(ii). If $B=U\cap F$ with $U$ open and $F$ closed, then $\overline{B}\subseteq F$. Henceso $B=U\cap \overline{B}$. By the subspace topology definition this makes $B$ open in its closure.

We prove (ii)$\Rightarrow$(iii). If $B$ is open in $\overline{B}$, its complement $\overline{B}\setminus B$ is closed in the subspace $\overline{B}$. A relatively closed subset of a closed subset of $X$ is closed in $X$: it is the intersection of that closed subset with another closed subset of $X$. Since a closure is closed, $\overline{B}\setminus B$ is closed in $X$.

We prove (iii)$\Rightarrow$(i). If $C=\overline{B}\setminus B$ is closed in $X$, set $U=X\setminus C$ and $F=\overline{B}$. Then $U$ is open, $F$ is closed, andTogether the three implications prove the equivalence. These are the standard characterizations of a locally closed subset.

A Hausdorff space is one in which any distinct points $x,y$ have disjoint open neighborhoods: there exist open sets $U,V$ with $x\in U$, $y\in V$ and $U\cap V=\varnothing$.

A map $f:X\to Y$ is continuous if the inverse image of every open subset of $Y$ is open in $X$:

For the natural projection $p:X\to X/R$, the quotient topology declares $U\subseteq X/R$ open exactly when $p^{-1}(U)$ is open in $X$. It is the finest topology on the quotient making $p$ continuous.

Every map has $f(x)=f(x)$, so the relation is reflexive. Equality is symmetric, so $f(x)=f(y)$ for all the specified maps is equivalent to $f(y)=f(x)$; the relation is symmetric. If $x\mathcal{R}y$ and $y\mathcal{R}z$, every continuous map to every Hausdorff space has $f(x)=f(y)=f(z)$, proving $x\mathcal{R}z$. Thus $\mathcal{R}$ is an equivalence relation.

If $x\in A=g^{-1}(B)$ and $y\mathcal{R}x$, the definition of the relation applies in particular to the fixed continuous map $g$ into the fixed Hausdorff space $Y$. It gives $g(y)=g(x)\in B$, hence $y\in A$. Thus $A$ is a union of whole equivalence classes, also called a saturation of a subset.

Always $A\subseteq p^{-1}(p(A))$. Conversely, a point in $p^{-1}(p(A))$ shares its class with some $x\in A$, and the preceding argument places it in $A$. Hence

Take distinct classes $p(x)\ne p(y)$. The failure of $x\mathcal{R}y$ supplies a Hausdorff space $Y$ and a continuous map $g:X\to Y$ with $g(x)\ne g(y)$. Choose disjoint open neighborhoods $V,W$ of these two images. Their inverse images $A=g^{-1}(V)$ and $B=g^{-1}(W)$ are disjoint open subsets of $X$, and part (ii) says both are saturated.

The quotient topology therefore makes $p(A)$ and $p(B)$ open, since their inverse images are precisely $A,B$. They are disjoint as well: an equivalence class meeting both would, by saturation, be contained in both. They contain $p(x)$ and $p(y)$ respectively. Thus $X/\mathcal{R}$ is Hausdorff. This construction is the Hausdorff reflection by continuous maps.

The inverse function theorem says that a continuously differentiable map on an open set, with invertible derivative at a point, restricts to a $C^1$ diffeomorphism between neighborhoods of that point and its image. The inverse derivative is the inverse matrix of the original derivative at the corresponding preimage.

Put $s=1+x+y\ne 0$. Differentiating each coordinate givesThe determinant never vanishes on $\Omega$, so the theorem applies at every point. The image point $(1/3,1/3)$ has preimage $(1,1)$: the equations $x/s=y/s=1/3$ imply $x=y=1$. Therefore

For fixed $x,y$, the map $z\mapsto z^3+z+xy-1$ is strictly increasing because its derivative is $3z^2+1>0$, and it tends to opposite infinities at the two ends of the real line. The intermediate value theorem gives exactly one root, defining $g(x,y)$.

For $F(x,y,z)=z^3+z+xy-1$, $F_z=3z^2+1$ never vanishes. The implicit function theorem supplies a $C^1$ solution locally around every $(x,y)$. Uniqueness makes all these local functions agree on overlaps, proving $g\in C^1({\mathbb{R}}^2)$. Implicit differentiation givesAt $(1,1)$ the root is zero, so $Dg(1,1)=(-1,-1)$.

Independent endpoint-fixed variations of the two coordinates give the Euler-Lagrange equationsDifferentiating the sphere parametrization gives the Riemannian metric $d{s}^2=a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its energy functional is the stated $I$. Its stationary paths are geodesics with affine, constant-speed parametrization; conversely a geodesic can be so parametrized. This explains the use of energy rather than arc length.

For $F=a^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$, the equations areThe second gives the conserved generalized momentum ${\sin }^2\theta \dot{\varphi }=h$. Since $F$ has no explicit time dependence, its Beltrami identity gives constant $\dot{\theta }{F}_{\dot{\theta }}+\dot{\varphi }{F}_{\dot{\varphi }}-F=F$. Dividing by $a^2/2$ gives the second first integral,Thus $E$ is the squared speed divided by $a^2$, while $h$ records the rotational symmetry about the polar axis. The coordinate equations apply away from the polar coordinate singularities.

For small oscillations about equilibrium, put $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$. The potential has no linear term at equilibrium; to quadratic order the Lagrangian is $L_2={\dot{x}}^{\mathsf{T}}M\dot{x}/2-x^{\mathsf{T}}Kx/2$. Its equations are $M\ddot{x}+Kx=0$. A normal mode $x=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem $(K-{\omega }^{2}M)v=0$. For a positive-definite mass matrix, diagonalization $M^{-1/2}K{M}^{-1/2}$ gives the squared frequencies and independent normal coordinates. A strictly positive Hessian gives positive frequencies; zero eigenvalues require separate higher-order stability analysis.

Here the spring extensions are $x_1$, $x_2-x_1$, and $-x_2$. ThusThe stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and $M=mI$. Its eigenvectors $(1,1)$ and $(1,-1)$ have stiffness eigenvalues $k$ and $3k$, givingIn the first mode the masses move together with equal displacements and the middle spring keeps its length. In the second they move oppositely with equal amplitudes, stretching and compressing the middle spring twice as much as either end spring.

For a real potential, the Time-dependent Schrödinger equation isMultiply this equation by ${\psi }^{\ast }$, multiply its conjugate by $\psi$, and subtract. The potential terms cancel and the product rule converts the remaining terms into a divergence. ConsequentlyThis is the probability current and its continuity equation.

Put $R(r)=\sin (kr)/r$. The supplied radial Laplacian formula gives ${\nabla }^{2}R=-k^{2}R$ for $r>0$. Also $i\hslash {\partial }_t\psi =E\psi$. Therefore the free-particle equation is satisfied whenThe apparent singularity at zero is removable: $R(r)=k-k^3{r}^2/6+\cdots$ extends smoothly there and satisfies the same equation. Since $R$ is real and the time dependence is only a common phase, ${\psi }^{\ast }\nabla \psi =R\nabla R$ is real, so $\mathbf{j}=0$. The density $R^2$ is time-independent, proving the continuity equation explicitly. This is a standing spherical wave with zero net radial current.

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

Seek nonzero time-harmonic waves of the form $\varphi =\mathrm{Re}[F(y)e^{i(kx-\omega t)}]$ and $\eta =\mathrm{Re}[A{e}^{i(kx-\omega t)}]$, taking $k>0$. Laplace's equation reduces to $F^{\prime \prime }-k^{2}F=0$. The impermeable-bed condition $F^{\prime }(-h)=0$ gives $F(y)=C\cosh (k(y+h))$.

The linearized kinematic and dynamic linearized free-surface boundary conditions at zero becomeEliminating $A$ and the nonzero amplitude $C$ givesConversely, if this dispersion relation holds, either amplitude can be chosen nonzero and the other set by the second boundary equation, producing a wave satisfying every condition. The expression also holds for negative $k$ because $k\tanh (kh)$ is even.

For an exponential distribution with rate $\lambda >0$, direct integration gives the moment-generating functionIt is finite in a neighborhood of zero, so its derivatives there give moments. Differentiating three times gives $M_X^{(3)}(t)=6\lambda /(\lambda -t{)}^4$, hence

Take $Y=a{X}_1+b$. From the definition of a moment-generating function, $\mathbb{E}{e}^{tY}=e^{bt}\mathbb{E}{e}^{at{X}_1}=e^{bt}{M}_1(at)$, for $t$ where the expectation is finite.

Take $Y=X_1+X_2$. Independence makes the joint exponential expectation factor:This is the product rule for the moment-generating function of an independent sum.

Take $Y=X_1+X_1^{\prime }$, where $X_1^{\prime }$ is an independent copy of $X_1$. The preceding factorization gives $\mathbb{E}{e}^{tY}=M_1(t{)}^2$. Using $2X_1$ would instead give $M_1(2t)$, so an independent copy is required.

Take $Y=U{X}_1$, where $U$ is uniform on $(0,1)$ and independent of $X_1$. Conditioning on $U$ givesFor $t$ near zero this is finite because the moment-generating function of $X_1$ is finite on a neighborhood of zero; Tonelli's theorem justifies the conditioning integral for the nonnegative exponential.