Past exam of the probability, statistics and options I course of the University of Oxford / 2008 (AS1, XMST 4305)

The hypothesis is a two-transitive group action. Given $\alpha ,\beta \ne {\omega }_1$, apply it to $({\omega }_1,\alpha )$ and $({\omega }_1,\beta )$. The resulting element fixes ${\omega }_1$ and sends $\alpha$ to $\beta$, so the stabilizer subgroup acts transitively on the remaining $n-1$ points.

The action on all of $\Omega$ is also transitive. By the orbit-stabilizer theorem,Thus $\boxed{n(n-1)\mid |G|}$. The diagonal pairs form one orbit, while two-transitivity makes all ordered unequal pairs a second orbit. Equality is preserved by a group action, so these cannot mix. There are exactly two orbits on $\Omega \times \Omega$.

Eisenstein's criterion says that an integer polynomial with leading coefficient not divisible by a prime $\ell$, all other coefficients divisible by $\ell$, and constant coefficient not divisible by ${\ell }^2$ is irreducible over the rationals, provided it is primitive; the usual Gauss's lemma supplies the passage from integer to rational factors.

Apply it with $\ell =2$ toAll nonleading coefficients are even and the constant is not divisible by four. A translation of the variable preserves reducibility, so $x^8+1$ is irreducible over $\mathbb{Q}$.

Let $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. The minimal polynomial of $\zeta$ is $x^8+1$, so $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Since $2c=\zeta +{\zeta }^{-1}$, the number $\zeta$ satisfies $t^2-2ct+1=0$ over $\mathbb{Q}(c)$. This real field cannot contain the nonreal $\zeta$, so that extension has degree exactly two. The tower law for field extensions gives

The quadratic reciprocity states that for distinct odd primes $a,b$,For fixed odd prime $r$, first suppose that $p,q\ne r$. The congruence $p\equiv q(\mathrm{mod}4r)$ gives the same residues modulo $r$ and modulo four. Reciprocity yieldsIf $p=r$, the congruence forces $r\mid q$, so $q=r$ and both symbols are zero; the same handles $q=r$.

For $n={\prod }_j{r}_j^{a_j}$, where all $r_j$ are odd primes, $4r_j$ divides $4n$. Thus $p\equiv q(\mathrm{mod}4n)$ implies equality of each prime-numerator Legendre symbol. Multiplicativity givesThis also covers symbols that vanish; for $n=1$ both sides are one.

A Lebesgue null set $E\subset \mathbb{R}$ can, for every $ϵ>0$, be covered by countably many open intervals whose total length is less than $ϵ$. Equivalently, its Lebesgue outer measure is zero.

If $E_j$ is null, choose for it such a cover of total length below $ϵ2^{-j}$, $j\ge 1$. The union of these countably many countable covers covers ${\bigcup }_j{E}_j$ and has total length below $ϵ$. Since $ϵ$ is arbitrary, a countable union of null sets is null.

For positive integers $n,m$, let $E_{n,m}=\{x:f(x)\ge 1/n\}\cap [-m,m]$. This measurable set has finite measure, and $f\ge n^{-1}{1}_{E_{n,m}}$. Monotonicity of the Lebesgue integral givesso every $E_{n,m}$ is a Lebesgue null set. The set where $f>0$ is the countable union of these sets over $n,m$. Part (a) makes it null, henceThe bounded truncation ensures that only integrable indicator functions are used with the stated comparison rule.

There is no singularity at the lower endpoint one. Put $t=\log (1+x)$, so $dx=e^{t}dt$. The nonnegative integral becomesFor $\alpha =1$, it converges as a reciprocal-square integral. For $\alpha >1$, exponential decay makes it converge. For $\alpha <1$, its integrand eventually grows without bound, so its integral diverges. ThusThis is the logarithmic borderline in a power-tail integrability test.

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. Since $(0,1)$ has finite Lebesgue measure, comparison gives ${\int }_0^{1}g\le 1$. Therefore $g$ is integrable on $(0,1)$, despite oscillating infinitely often near zero.

For $x>1$, $1/x\in (0,1)$ and ${\cos }^2(1/x)\ge {\cos }^2(1)>0$. Its integral over the infinite interval is bounded below by the integral of a positive constant. Therefore $g$ is not integrable on $(1,\infty )$. In fact it tends to one at infinity.

A continuous map $f:X\to Y$ is one for which $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$. This defines continuity using the two topologies, without a metric or a coordinate description.

For metric spaces, continuity at $x_0$ means that for every $ϵ>0$ there is $\delta >0$ such thatThe choice of $\delta$ may depend on both $ϵ$ and $x_0$.

A basis of a topology is a collection $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every open $U$ and $y\in U$ there is $B\in \mathcal{B}$ with $y\in B\subseteq U$.

A map $f:X\to Y$ is continuous exactly when $f^{-1}(B)$ is open in $X$ for every basic set $B$ of $Y$. Necessity follows from the definition of a continuous map. For sufficiency, express an arbitrary open set as a union of basic sets and use preservation of unions by inverse images.

Every open interval is open in the usual metric topology. Conversely, if $U\subset \mathbb{R}$ is open and $x\in U$, there exists ${ϵ}_x>0$ with $(x-{ϵ}_x,x+{ϵ}_x)\subseteq U$. Taking the union of these intervals over all $x\in U$ recovers $U$. Thus open intervals form a basis of the usual topology.

If $f$ is continuous, the two sets are inverse images of the open rays $(-\infty ,\lambda )$ and $(\lambda ,\infty )$, so they are open. Conversely, suppose both families of inverse images are open. For any interval $(a,b)$,which is open. By part (c)(i) and the basis of a topology criterion from part (b), all inverse images of open sets are open. Hence $f$ is continuous.

A connected set is one whose relative topology admits no decomposition into two disjoint nonempty open sets. A path-connected set permits, between any two points, a continuous map $\gamma :[0,1]\to A$ with the prescribed endpoints.

Path connectedness implies connectedness, because a continuous image of an interval is connected. The converse can fail, for example for the topologist's sine curve. On open subsets of Euclidean spaces ${\mathbb{R}}^n$, the two properties are equivalent; more generally this holds in locally path-connected spaces.

If $f(A)$ had a separation into two nonempty relatively open disjoint pieces, their inverse images under the restriction $f{|}_A$ would be two nonempty relatively open disjoint subsets covering $A$. This would contradict connectedness of $A$. Thus a continuous image of a connected set is connected.

The unit circle is path-connected, since it is the continuous image of the interval $[0,2\pi ]$ under $\theta \mapsto (\cos \theta ,\sin \theta )$. It is therefore connected, and part (ii) makes $f(C)$ connected. A connected subset of $\mathbb{R}$ is an interval: omission of a point lying between two of its points would separate it by the two open rays about that point.

Moreover, the circle is compact, so the continuous image of a compact space is compact. Hence here the image is a closed bounded interval, possibly a singleton.

For a differentiable map $f:\Omega \subset {\mathbb{R}}^n\to {\mathbb{R}}^k$, the Jacobian matrix represents its derivative and has entries $(Df{)}_{ij}=\partial f_i/\partial x_j$. Differentiating the two components gives

The inverse function theorem states that if $f$ is continuously differentiable near $a\in {\mathbb{R}}^n$ and $Df(a)$ is invertible, then neighborhoods of $a$ and $f(a)$ exist on which $f$ is a bijection with a continuously differentiable inverse, satisfying $D(f^{-1})(f(a))=Df(a{)}^{-1}$.

Here $a=(\pi /2,\pi )$, $f(a)=(0,0)$, andThe hypotheses hold because both components are smooth. Inverting the matrix yields

With fixed endpoint values, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=\text{constant}$.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so $y^{\prime \prime }=y$. The extremals arewhere the two fixed endpoint values determine $A,B$. The Beltrami constant is $y^2-y^{\prime 2}$, consistent with this solution.

For a nonzero smooth perturbation $\eta$ vanishing at the endpoints, expand the quadratic functional. The first variation vanishes at an extremal, whileThe linear term $\int y^{\prime }$ depends only on the fixed endpoints. Consequently the extremal is a strict minimum and cannot be a maximum.

If $L$ is independent of $q_1$, the Euler-Lagrange equation makes the generalized momentum $p_1=\partial L/\partial {\dot{q}}_1$ constant. If $L$ is independent of time, the energy function $E={\sum }_a{\dot{q}}_a(\partial L/\partial {\dot{q}}_a)-L$ is constant; differentiating it gives $\dot{E}=-L_t$ along solutions.

Set $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$. Their derivatives are $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The radial and angular equations areEquivalently,Two conserved quantities areThe second equals $L$ here because the Lagrangian is homogeneous quadratic in velocities. For a constant-radius orbit with nonzero angular motion, the radial equation requires $r-3m=0$. ThusIt lies outside $2m$ and is the circular photon orbit. If both velocities were zero, the differential equations would also admit static constant configurations at any radius; these are degenerate zero-energy paths rather than the intended moving photon trajectory.

The integral Gauss's law isThe divergence theorem makes the left side ${\int }_V\nabla \cdot \mathbf{E}dV$. If the differential relation $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ holds, this proves the integral law. Conversely, if the integral identity holds for every suitable volume and the fields are sufficiently regular, a nonzero value of $\nabla \cdot \mathbf{E}-\rho /{ϵ}_0$ would persist in sign on a small neighborhood and contradict its zero integral. The differential law follows.

For a sphere of radius $r$, spherical symmetry makes the flux $4\pi r^{2}E(r)$ and the enclosed charge $4\pi {\int }_0^r{s}^2\rho (s)ds$. HenceThere is no extra point charge at the origin in the stipulated density distribution. Outside the charge support,

Use the two-dimensional streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. HereThe divergence is $u_x+v_y=-\cos x\sin y+\cos x\sin y=0$, so the flow is incompressible.

The vorticity is perpendicular to the plane:It is generally nonzero, so this is a rotational flow even though it has a streamfunction.

For steady inviscid flow of constant density without a body-force potential, Bernoulli's theorem givesThe constant can differ between streamlines, because this flow has nonzero vorticity. Dotting the steady Euler equation with $\mathbf{u}$ gives $\mathbf{u}\cdot \nabla (p/\rho +|\mathbf{u}{|}^2/2)=0$, which proves precisely the along-streamline form.

Direct differentiation gives the convective accelerationThe steady Euler equations therefore require $p_x=-\rho \sin x\cos x$ and $p_y=\rho \sin y\cos y$. Integrating both derivatives consistently givesThe additive constant is arbitrary. With a corresponding redefinition of this constant, the Bernoulli function is $p/\rho +|\mathbf{u}{|}^2/2=C_0-{\psi }^2$, making its constancy along streamlines explicit.

Condition (1b) is the kinematic boundary condition: surface particles remain on the moving free-surface graph. Condition (1c) is the dynamic boundary condition for an inviscid interface from unsteady Bernoulli's equation, with the surface pressure fixed at the atmospheric value and absorbed into the potential's time gauge. Condition (1d) is impermeability of the fixed horizontal bed, so its normal velocity vanishes.

Take the undisturbed surface to be $y=0$ and treat $\eta$, $\varphi$, and their induced velocities as first order in wave amplitude. Products of first-order quantities are second order and are discarded. Evaluating surface conditions at $y=\eta$ differs from evaluation at zero only by second-order terms. Thus the linearized surface gravity wave problem isHere $u={\varphi }_x$, $v={\varphi }_y$. The linear free-surface conditions can also be combined into ${\varphi }_{tt}+g{\varphi }_y=0$ at zero.

For $k>0$, phase velocity is the speed of a fixed phase, $c_p=\omega /k$. The group velocity $c_g=d\omega /dk$ transports a narrow-band wave envelope. With $q=kH$,As $H\to 0$ with $k$ fixed, $\tanh (kH)/(kH)\to 1$. ThereforeBoth velocities tend to the shallow-water speed $\sqrt{gH}$ at leading order.

The inverse change of variables is $x=2u+v$, $y=v$, with absolute Jacobian determinant two. Positivity of $x,y$ givesThe change of variables for probability densities givesFor fixed $u$, integrate over $v>\max (0,-2u)$:Thus $\boxed{f_U(u)=\frac{1}{2}{e}^{-|u|}}$, the standard Laplace distribution.

The Markov property says that, conditional on the present state, the next state is independent of the earlier history:Two states communicate if each is reachable from the other with positive probability in finitely many steps. The equivalence classes are communicating classes; a class is closed if no positive-probability transition leaves it.

The displayed probabilities require $0\le p\le 1/2$; negative values allowed by the printed one-sided inequality are not transition probabilities. For $p=0$, every state is an absorbing singleton, so all singleton classes are closed. For $0<p\le 1/2$, neighboring states communicate in both directions, so there is one closed irreducible class $S$ and no open classes.

For $m=3$, the detailed balance relations give $2{\pi }_0={\pi }_1={\pi }_2=2{\pi }_3$. Normalization yieldsFor $0<p<1/2$, positive holding probabilities make the finite irreducible chain aperiodic, so its transition probabilities converge to ${\pi }_j$ from every initial state. At $p=1/2$, every step changes parity and the period is two; convergence from a fixed initial state fails. A stationary initial mixture still has distribution $\pi$ at every time, but this does not give convergence of the transition probabilities asked for.

A Poisson process of rate $\lambda >0$ starts from zero, has independent stationary increments, has unit jumps, and satisfies $N_t-N_s\sim \mathrm{Poisson}(\lambda (t-s))$ for $t>s$. Its inter-arrival times are independent with the exponential distribution of rate $\lambda$: any finite joint density is ${\prod }_i\lambda e^{-\lambda x_i}$ on positive coordinates.

The second arrival is a sum of two such variables. Convolution givesThis is a gamma distribution of shape two and rate $\lambda$. If $S$ is the first arrival of the independent rate-$\mu$ process, exponential independence gives

For observed counts $O_{ij}$ in an $r$ by $c$ contingency table, Pearson's chi-squared test statistic for independence isUnder the null model and the usual large-expected-count conditions, its reference distribution is ${\chi }^2$ with $(r-1)(c-1)$ degrees of freedom.

The pooled CC count is $83+50=133$, the case total is 1939, and the grand total is 3874. Under the null, the common estimated CC proportion is $133/3874$. Multiplication by the fixed case total givesThis is the fitted expected count, rather than a separate observation.

Under the null, the two groups share one vector of three category probabilities. The sum-to-one constraint leaves two free parameters. Under the unrestricted alternative, each group has its own such probability vector, leaving four free parameters. The fixed case and control totals introduce no further parameters; the difference is two, giving the chi-squared reference degrees of freedom.

The null reference is ${\chi }_2^2$, because the alternative has two additional parameters. The supplied statistic satisfies $8.42>5.99$, so reject no association at the 5% level. For two degrees of freedom the upper-tail probability is $e^{-8.42/2}\approx 0.0148$.

The result is evidence that genotype proportions differ between cases and controls. It is a statistical association test; it does not by itself identify a causal mechanism.

Under the recessive alternative, CT and TT have the same status as non-CC individuals, so their separate labels do not distinguish the proposed risk groups. Pooling them makes the intended comparison CC versus non-CC. The table becomesThe independence test on this two-by-two table has one degree of freedom. Since $8.41>3.84$, reject no association at the 5% level; its chi-squared upper-tail probability is approximately $0.0037$. CC is more frequent among cases, consistent with the stated directional alternative. The supplied Pearson statistic is conventionally compared with its one-degree-of-freedom chi-squared reference.

Differentiating the least-squares residual norm gives the normal equationsIf $X$ has full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$, and thereforeHere $\bar{x}=0$ and $\bar{z}=2$, so the design matrix and its Gram matrix areThe centered predictor columns are perpendicular, givingBoth estimates are linear combinations of the jointly normal errors, hence jointly normal. Zero covariance therefore also gives independence. The normality assumption is what permits this last step; orthogonality alone would only prove uncorrelatedness.

The Neyman-Pearson lemma says that for testing a simple null density $f_0$ against a simple alternative $f_1$, a size-$\alpha$ test that rejects for the largest values of $f_1/f_0$ is most powerful. Randomization on a likelihood-ratio boundary can be used to achieve the exact size.

The source writes the same ${\lambda }_0$ in both hypotheses but immediately introduces ${\lambda }_1>{\lambda }_0$. Use the intended alternative $\lambda ={\lambda }_1$. If read literally as two identical hypotheses, there is no such distinct alternative to test.

For Weibull shape two, the sample likelihood is proportional toThuswhich is increasing in $S={\sum }_i{X}_i^2$. The most powerful critical region is thereforeIndeed, $2S/{\lambda }_0^2\sim {\chi }_{2n}^2$, so $C=({\lambda }_0^2/2)q_{2n,1-\alpha }$. The critical value depends only on the null and size, not on the chosen larger alternative. Hence the same test is uniformly most powerful for $\lambda >{\lambda }_0$, and in particular for the printed restricted range $\lambda >{\lambda }_1$.