Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 (AS2, XMST 4306)

Source: Oxford mathematics examination release, private bundle Part A 2007 11-1.pdf, pages 45–59. The printed cover is for mathematics and statistics. The complete statements and editable transcription are retained in the private examination archive.

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

An element $\alpha$ is an algebraic element over $F$ if some nonzero polynomial in $F[X]$ vanishes at $\alpha$. A field extension is an algebraic extension if each of its elements is an algebraic element over $F$. If $[K:F]=d<\infty$, the $d+1$ vectors $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent, producing such a polynomial. Thus every finite field extension is an algebraic extension.

The converse fails for the field of algebraic numbers $\overline{\mathbb{Q}}$. It is an algebraic extension of $\mathbb{Q}$ by definition, but it contains primitive $p$th roots of unity of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the cyclotomic polynomial. The tower law would make every such degree at most $[\overline{\mathbb{Q}}:\mathbb{Q}]$ if the latter were finite, a contradiction.

The sixth roots of unity all belong to $\mathbb{Q}({\zeta }_6)$, where ${\zeta }_6=(1+\sqrt{-3})/2$. Its minimal polynomial of an algebraic element is $X^2-X+1$, which has no rational root. Thus the splitting field is $\mathbb{Q}(\sqrt{-3})$ and

A primitive seventh root of unity ${\zeta }_7$ generates all roots. Its minimal polynomial of an algebraic element is the permitted irreducible cyclotomic polynomial $1+X+\cdots +X^6$. Hence the splitting field has

A primitive eighth root of unity is ${\zeta }_8=(1+i)/\sqrt{2}$, and $\mathbb{Q}({\zeta }_8)=\mathbb{Q}(i,\sqrt{2})$. Indeed $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$, while the displayed expression gives the reverse containment. Since $\mathbb{Q}(\sqrt{2})\subset \mathbb{R}$, adjoining $i$ has degree two; adjoining $\sqrt{2}$ to $\mathbb{Q}$ also has degree two. By the tower law,

The polynomial $X^3+X+1$ has no root in ${\mathbb{F}}_2$, so it is an irreducible polynomial. The quotient ring of ${\mathbb{F}}_2[X]$ by the ideal it generates is therefore a field with eight elements. The seven nonzero elements form a group of order seven, so each satisfies $x^7=1$ by Lagrange's theorem. Thus this field splits $X^7-1$. A root different from one has order seven; it cannot lie in ${\mathbb{F}}_2$ and generates this degree-three field extension, since three is prime and the tower law applies. Therefore

In characteristic two, the binomial theorem gives $X^8-1=(X-1{)}^8$. The only root is already in ${\mathbb{F}}_2$, so the splitting field is the original field, despite the repeated root:

Wilson's theorem states that an integer $n>1$ is prime if and only if $(n-1)!\equiv -1(\mathrm{mod}n)$. For a prime $p$, every nonzero residue has a multiplicative inverse. Pair each residue with its inverse; the only self-inverse ones satisfy $(a-1)(a+1)=0$ in ${\mathbb{F}}_p$ and hence are $1,-1$ (the same residue when $p=2$). The resulting product is $-1$. Conversely, if $(n-1)!\equiv -1(\mathrm{mod}n)$, no divisor $d$ of $n$ can satisfy $1<d<n$: it would divide both $(n-1)!$ and $n$, so could not divide $(n-1)!+1$. Hence $n$ is prime.

For a composite $n\ge 6$ which is not the square of a prime, choose a proper factorization $n=ab$ with $1<a<b<n$ if possible. Both factors occur in $(n-1)!$, proving divisibility by $n$. If the initially chosen factors coincide but their common value is composite, choose a different proper divisor to obtain unequal factors; equivalently, the only composite integer all of whose proper nontrivial divisors coincide is a prime square. For $n=p^2\ge 9$, the distinct factors $p$ and $2p$ lie below $p^2$ and their product is divisible by $p^2$. Thus $(n-1)!\equiv 0(\mathrm{mod}n)$ for every composite $n\ge 6$.

For a prime $n$, write $(n-1)!=kn-1$. Then ${\sin }^2((n-1)!\pi /n)={\sin }^2(\pi /n)$, so the summand is one. For a composite $n\ge 6$ it is zero. The only remaining composite index is four, whose summand equals ${\sin }^2(3\pi /2)/{\sin }^2(\pi /4)=2$. Therefore the sum is the prime-counting function plus two whenever $x\ge 4$; subtraction of two gives the claimed exact formula.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open subset $T(U)\subset {\mathbb{R}}^2$, with continuously differentiable inverse. Equivalently, it is a $C^1$ diffeomorphism; its Jacobian determinant never vanishes. The inverse function theorem supplies the local inverse, while global injectivity is an additional requirement.

The change of variables formula for a Lebesgue integrable function isThe absolute Jacobian determinant accounts for area regardless of orientation. The formula also holds for nonnegative measurable functions with possibly infinite integral.

Put $u=x-y$ and $v=x+y$. The square becomes $(-1,1{)}^2$, with $x=(u+v)/2$, $y=(v-u)/2$ and absolute Jacobian determinant $1/2$. On its support the denominator $v+2$ exceeds one and $|u|<1$, so the nonnegative integrand is bounded by one on a set of finite area; it is therefore Lebesgue integrable. The change of variables formula and Fubini's theorem give

The dominated convergence theorem states: if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le g$ almost everywhere for an integrable $g\ge 0$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$.

To derive it from the monotone convergence theorem, first prove Fatou's lemma. For nonnegative measurable $h_n$, the functions $k_N={\inf }_{n\ge N}{h}_n$ increase to $\mathrm{liminf}{h}_n$ and satisfy $\int k_N\le {\inf }_{n\ge N}\int h_n$. Monotone convergence yields $\int \mathrm{liminf}{h}_n\le \mathrm{liminf}\int h_n$.

Now $|f|\le g$ almost everywhere, so $f$ is integrable. Apply Fatou's lemma to the nonnegative functions $2g-|f_n-f|$, which converge almost everywhere to $2g$:Hence $\int |f_n-f|\to 0$, and $|\int f_n-\int f|\le \int |f_n-f|$ proves the assertion. Common exceptional null sets may be removed at the beginning since the sequence is countable.

For fixed $x>0$, $f_n(x)\to e^{-x}\sin x$; for $x\le 0$, the limit is zero. On $0<x<n$, the supplied inequality gives $(1-x/n{)}^n\le e^{-x}$, soThis is the given integrable dominating function. By the dominated convergence theorem,

A topological space is compact if every open cover has a finite subcover. It is connected if it cannot be written as the union of two disjoint nonempty open subsets. It is Hausdorff if every two distinct points have disjoint open neighborhoods. These definitions apply to the space itself and, with the subspace topology, to its subsets.

True. If open sets cover $Y$, their inverse images under the continuous surjection $f$ cover $X$. Compactness of $X$ gives a finite subcover; surjectivity then makes the corresponding finite family cover $Y$. This proves the compactness of a continuous image of a compact space.

True. A separation of $Y$ into two disjoint nonempty open subsets would pull back under the continuous surjection $f$ to a separation of $X$. The inverse images are nonempty by surjectivity, contradicting connectedness of $X$. Thus a continuous image of a connected space is connected.

True. The Hausdorff assumption makes $\{y\}$ closed in $Y$. Its inverse image under the continuous map $f$ is closed in the compact space $X$, and a closed subset of a compact space is compact. This is the compactness of the fiber with its subspace topology.

False. Take $X=[0,1]$, $Y=S^1$ and the surjection $f(t)=e^{2\pi it}$. The interval is compact and connected, and the circle is Hausdorff. But $f^{-1}(1)=\{0,1\}$ is a disconnected two-point subspace. Thus a continuous surjection need not have connected fibers.

False. Let $X$ be a two-point set with the indiscrete topology, let $Y$ be a one-point space, and let $f$ be the unique map. The space $X$ is finite and hence compact; its only nonempty open subset is all of $X$, so it is connected. It is not Hausdorff, whereas $Y$ is Hausdorff and $f$ is a continuous surjection.

For the canonical surjection $p:X\to X/\sim$, define a subset $W$ of $X/\sim$ to be open exactly when $p^{-1}(W)$ is open in $X$. This is the quotient topology. The empty set and the whole quotient have open inverse images. Moreover inverse images commute with arbitrary unions and finite intersections, so the corresponding axioms of a topology follow directly from those on $X$.

The map $p$ need not be an open map. For example, on $X=\mathbb{R}$ identify every point of $[0,1]$ to one point and leave all other classes singletons. The set $U=(-1/2,1/2)$ is open, butis not open. The definition of the quotient topology therefore shows that $p(U)$ is not open. The obstruction is that saturating an open set by equivalence classes need not preserve openness.

Treat the two squares as a single edge-pairing construction, including the identifications between different squares. Number the corners of the left square counterclockwise from its lower left corner by $A,B,C,D$, and similarly use $E,F,G,H$ on the right. The single arrows identify $D$ with $H$ and $C$ with $E$. The double arrows identify $G$ with $F$ and $H$ with $G$. The triple arrows identify $A$ with $E$ and $B$ with $F$. The quadruple arrows identify $A$ with $C$ and $D$ with $B$.

Thus there are precisely two vertex classes, $\{A,C,E\}$ and $\{B,D,F,G,H\}$. There are four edge classes and two two-dimensional cells, so the resulting closed connected surface has Euler characteristicIt is a nonorientable surface: on the left square the quadruple-arrow sides have the same orientation relative to a traversal of its boundary, giving a twisted band after gluing. Equivalently, no choice of orientations for the two faces makes the induced directions opposite on every paired edge. The vertex links are circles, so these edge identifications produce a closed surface, rather than a singular cell space. The classification of compact surfaces now givesA torus also has Euler characteristic zero, but is excluded by the failed orientation test.

Differentiability at $a$ means that there is a linear map $L:V\to V_1$ withThis map is the total derivative $D{f}_a$. In the Frobenius norm, $|AB|\le |A||B|$, by the Cauchy-Schwarz inequality applied to matrix entries. Expanding with the order of all factors preserved givesThe remainder after the terms linear in $H$ has norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. HenceFor $|C|<1$, the Neumann series $I-C+C^2-\cdots$ converges absolutely in the finite-dimensional space of matrices. Multiplying its partial sums by $I+C$ and passing to the limit shows that it is $(I+C{)}^{-1}$. Its remainder after $I-C$ has norm at most $|C{|}^2/(1-|C|)=o(|C|)$, proving the required expansion.

If $A$ is an invertible matrix and $|H|<1/|A^{-1}|$, then $A+H=A(I+A^{-1}H)$ is an invertible matrix. Thus the general linear group is an open subset of the matrix space. The preceding expansion also givesApply the chain rule to $g(X)=(X^{-1}{)}^3$, using the cubic total derivative at $A^{-1}$ and this total derivative of inversion:The three terms generally cannot be combined by commuting $A$ and $H$.

Parametrize the surface of revolution byThe magnitude of the cross product ${\mathbf{r}}_x\times {\mathbf{r}}_{\theta }$ is $y\sqrt{1+y^{\prime 2}}$. Integrating the surface area element over $\theta$ givesPositivity of $y$ removes an otherwise necessary absolute value in the area element.

For $F(y,y^{\prime })=y\sqrt{1+y^{\prime 2}}$, the Euler-Lagrange equation isSince $F$ has no explicit $x$ dependence, the Beltrami identity givesSolving this first integral, or substituting into the second-order equation, yieldsThe constants satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. There need not be a solution for arbitrary endpoint data, and when there is one these equations can have more than one pair of constants. These are the possible stationary catenoids, so the axisymmetric second variation of a catenoid still matters when deciding whether they minimize surface area.

The functional is unbounded above. Let $\ell$ be the positive affine interpolant of the endpoint values and choose a nonnegative, nonzero, continuously differentiable $h$ vanishing at both endpoints. The admissible functions $y_M=\ell +Mh$ satisfyThis excludes a global maximum, but does not prove that a stationary curve is a minimum. The concluding assertion is false without additional conditions.

An explicit counterexample uses $x_1=-2$, $x_2=2$, $y_1=y_2=\cosh 2$ and the stationary function $y=\cosh x$. Its second variation, for a perturbation $h$ vanishing at the endpoints, isChoose $h=1-x^2/4$. Since ${\mathrm{sech}}^{2}x\le 4e^{-2|x|}$, the positive contribution is at most ${\int }_{\mathbb{R}}{x}^2{e}^{-2|x|}dx=1/2$. On $[-1,1]$, $h^2\ge 9/16$, so the magnitude of the negative contribution is at least $(9/8)\tanh 1>9/16>1/2$. Thus ${\delta }^{2}J[h]<0$. Small perturbations in this direction preserve positivity and the endpoint data while decreasing surface area. Consequently this catenoid is not a local minimum. The valid conclusion from part (ii) is a necessary stationary form; minimization requires a separate stability and existence analysis.

Write $q=(x,X{)}^{\mathsf{T}}$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The Euler-Lagrange equations give $M\ddot{q}+Kq=0$. A normal mode is a motion $q=v\cos (\omega t+\phi )$ with fixed nonzero vector $v$; its normal frequency is $\omega >0$. Substitution gives $(K-{\omega }^{2}M)v=0$. Thus the squared normal frequencies are the assumed positive distinct roots ofFor either root, solve the associated two linear equations for the mode ratio.

For the rods, the small angles are $\theta =x/a$ and $\varphi =(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second rod's center has horizontal velocity $(\dot{x}+\dot{X})/2$ and its angular velocity is $(\dot{X}-\dot{x})/(2a)$. Its center-of-mass moment of inertia is $2m{a}^2/3$, givingAdding these kinetic energies gives $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The first center rises by $a{\theta }^2/4$, and the second by $a{\theta }^2/2+a{\varphi }^2/2$. Multiplying by the respective masses and $g$ givesTherefore the stated Lagrangian is $T-V$.

Set $\mu =2a{\omega }^2/(3g)$. The mode equation isHenceThe mode ratio is $r=X/x=(6-3\mu )/(1+\mu )$, so the ratio of the rods' angles is $\varphi /\theta =(r-1)/2=(5-4\mu )/[2(1+\mu )]$. It is positive for ${\mu }_{-}=(17-3\sqrt{21})/10$ and negative for ${\mu }_{+}=(17+3\sqrt{21})/10$. Thus the lower-frequency normal mode has the rods in phase, and the higher-frequency normal mode has them out of phase.

In vacuum, the source-free Maxwell equations arePut $\psi =\omega (t-z/c)$ and denote the constant electric amplitude by $E_0$. A corresponding traveling-wave magnetic field isBoth divergences vanish because both the electric field and the magnetic field depend only on $z,t$ and have zero $z$ component. Direct differentiation gives $\nabla \times \mathbf{E}=(\omega E_0/c)(\cos \psi ,\sin \psi ,0)=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{B}=(\omega E_0/c^2)(-\sin \psi ,\cos \psi ,0)=c^{-2}{\partial }_t\mathbf{E}$. A time-independent background magnetic field satisfying the homogeneous equations could be added; the formula is the wave component.

Introduce the magnetic vector potential and electric potential by $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. These identities automatically give the divergence equation for $\mathbf{B}$ and Faraday's law. For this wave choose $\varphi =0$ and $\mathbf{A}=(c/\omega )\mathbf{B}$; the derivatives just calculated verify both potential equations.

The electromagnetic force is the Lorentz force, $Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. On the smooth constraint plane $z=z_0$, the constraint force is normal to the plane. Both $\mathbf{v}$ and $\mathbf{B}$ are tangential, so $\mathbf{v}\times \mathbf{B}$ is normal and is balanced by that constraint force; it does not affect the tangential motion. ThusA particular motion is ${\mathbf{r}}_{\parallel }={\mathbf{r}}_0-[Q{E}_0/(m{\omega }^2)](\cos \psi ,\sin \psi )$, a circle of radiusThe general planar motion also includes a constant drift velocity; the particle moves on a circle when that drift is zero. The constraint force can do no work because it is perpendicular to the velocity.

Orient $C$ counterclockwise around the solid body and let $\mathbf{n}$ point from the solid into the fluid. The pressure force is $-{\int }_{C}p\mathbf{n}ds$. Since $\mathbf{n}ds=(dy,-dx)$,For steady irrotational flow, Bernoulli's equation gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant term integrates to zero. The boundary is a streamline, so $dw=w^{\prime }dz$ is real there, giving $w^{\prime }dz=\overline{w^{\prime }}d\overline{z}$. Thereforewhich is Blasius force theorem with the specified orientation.

The prescribed stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. The zero-circulation solution from the Milne-Thomson circle theorem isOn $|z|=a$ the two terms are conjugates, so the boundary stream function is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.

Differentiating givesThe coefficient of $z^{-1}$ in $(w^{\prime }{)}^2$ is $36{\alpha }^{2}c{a}^2(a^2+c^2)$. The residue theorem and Blasius force theorem giveThe zero-circulation convention is relevant: if an independently imposed circulation $\Gamma$ is allowed, adding $\Gamma \log z/(2\pi i)$ changes $F_y$ to $-3\rho \alpha \Gamma c^2$. Thus the claimed zero transverse force belongs to the standard circle-theorem construction, or to a single-valued exterior complex potential of a two-dimensional potential flow, which excludes that extra circulation.

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

The independent identically distributed central limit theorem states that for independent identically distributed random variables $U_k$ with finite mean $\mu$ and positive finite variance ${\sigma }^2$,For $0<u<1$, continuity, strict monotonicity and the endpoint limits of the cumulative distribution function give a unique $x=F^{-1}(u)$. Then $P(F(X)\le u)=P(X\le x)=F(x)=u$. Thus $Y=F(X)$ has the uniform distribution on $(0,1)$, by the probability integral transform.

For $W=-\log (1-Y)$, $P(W\le w)=1-e^{-w}$ when $w\ge 0$, and it is zero otherwise. Therefore its probability density function is $e^{-w}{\chi }_{(0,\infty )}(w)$: the exponential distribution of rate one. By integration by parts, $E[W]=1$ and $E[W^2]=2$, so $\mathrm{Var}(W)=1$.

The variables $W_k=-\log (1-F_k(X_k))$ are independent because they are separate measurable functions of independent random variables. The probability integral transform makes all of them identically distributed as $\mathrm{Exp}(1)$, even though the $X_k$ need not be identically distributed. The displayed $Z_n$ is exactly $n^{-1/2}{\sum }_{k=1}^n(W_k-1)$. The central limit theorem therefore gives

A Poisson process of rate $\lambda$ is a counting process starting at zero, with independent stationary increments, such that an increment over an interval of length $t$ has the Poisson distribution of parameter $\lambda t$. It has right-continuous paths with unit jumps almost surely.

Independence of the two processes makes increments of $A+B$ independent on disjoint intervals. The sum of independent Poisson random variables with parameters $\alpha t$ and $\beta t$ is Poisson with parameter $(\alpha +\beta )t$, as follows by convolution:Thus $A+B$ is a Poisson process of rate $\alpha +\beta$. Its first red and green waiting times are independent exponential random variables of rates $\alpha ,\beta$, so

The state-zero stationary distribution equation is ${\pi }_0=q{\pi }_0+q{\pi }_1$, so ${\pi }_1=(p/q){\pi }_0$. The state-one equation is ${\pi }_1=p{\pi }_0+q{\pi }_2$, giving ${\pi }_2=(p/q){\pi }_1$. More generally, the stationary probability flow across the cut $\{0,\dots ,i\}$ is balanced, so $p{\pi }_i=q{\pi }_{i+1}$. This also follows inductively from ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$.