Probability, statistics and options I course of the University of Oxford (AS1)

1

a

Solution

In field characteristic

p

, the binomial theorem gives

(X - b)^{p} = X^{p} - b^{p}

, since the intermediate binomial coefficients are divisible by the prime number

p

. If

b \in F

is a root of

X^{p} - a

, then

b^{p} = a

, and hence

X^{p} - a = (X - b)^{p} in F [X] .

b

Solution

Let

α

be a root in a splitting field. The preceding characteristic-

p

identity gives

X^{p} - a = (X - α)^{p}

, so there is only one distinct root. If the polynomial were reducible over

F

, choose a monic proper factor

g \in F [X]

of degree

d

, with

1 \leq d < p

. In the splitting field it must be

g = (X - α)^{d}

. Its coefficient of

X^{d - 1}

- d α

, which belongs to

F

. Since

d

is nonzero and thus invertible in the field

F

, this forces

α \in F

, contradicting the absence of a root there. Therefore $X^{p} - a$ is irreducible over $F$ .

c

Solution

Suppose a rational function

P (t) / Q (t) \in F_{p} (t)

, with nonzero polynomials

P, Q

, had

p

th power

t

. Then

P (t)^{p} = tQ (t)^{p}

, so comparison of degrees would give

p de g P = 1 + p de g Q

, impossible modulo

p

. Thus

X^{p} - t

has no root in

K

, and part (b) proves it is irreducible.

In a splitting field containing

α

with

α^{p} = t

X^{p} - t = (X - α)^{p} .

It has one distinct root with multiplicity

p

, so it is inseparable. This is the purely inseparable polynomial over a rational function field.

B (Group Theory)

1

a

Solution

For a group action, the orbit and stabilizer are

G x = {gx : g \in G}, G_{x} = {g \in G : gx = x} .

The orbit-stabilizer theorem for a finite group states

∣ G ∣ = ∣ G x ∣ ∣ G_{x} ∣.

b

Solution

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are

0, 2 π /3, 4 π /3

, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving

∣ G_{1} ∣ = 4 \cdot 3 = 12

by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

∣ G_{1} ∣ = 12, ∣ G_{2} ∣ = 24.

c

Solution

Realize the cube's vertices as

(\pm 1, \pm 1, \pm 1)

and take the four with coordinate product

+ 1

. They form a regular tetrahedron, and their negatives form the other one. Every proper rotation of the first tetrahedron also permutes the negatives, hence is a cube rotation. This embeds

G_{1}

as a subgroup of

G_{2}

The embedded group preserves each of these two four-vertex sets and acts transitively on each. Thus the resulting action on all eight cube vertices has two orbits, each of size four.

C (Number Theory)

1

a

i

Solution

A prime number is an integer

p > 1

whose only positive divisors are

1

and

p

. The restriction

p > 1

excludes the unit

1

ii

Solution

The fundamental theorem of arithmetic says that every integer

n > 1

is a product of prime numbers, uniquely up to reordering. Equivalently,

n = \prod_{p} p^{a_{p}}

for a unique family of nonnegative integer exponents, only finitely many of which are nonzero.

iii

Solution

If the complete list of prime numbers were

p_{1}, \dots, p_{k}

, then

N = p_{1} \dots p_{k} + 1 > 1

would have a prime divisor by the fundamental theorem of arithmetic. No

p_{i}

divides it, since

N \equiv 1 (mod p_{i})

. This contradicts completeness of the list. Hence there are infinitely many prime numbers.

b

i

Solution

Every integer has one of the six residues

0, 1, 2, 3, 4, 5

modulo

6

. A prime number other than

2

3

is divisible by neither

2

nor

3

, eliminating residues

0, 2, 3, 4

. Thus

p \equiv 1 or - 1 (mod 6) .

ii

Solution

Suppose there were only finitely many prime numbers congruent to

5

modulo

6

, and list them as

p_{1}, \dots, p_{k}

. The list is nonempty since it contains

5

. Let

N = 6 p_{1} \dots p_{k} - 1

. It is greater than one and divisible by neither

2

nor

3

. Every prime factor therefore has residue

1

- 1

modulo

6

. Since

N \equiv - 1 (mod 6)

, at least one prime factor

q

must have residue

- 1

. But

N \equiv - 1 (mod p_{i})

for every listed prime, so

q

is not on the list. This contradiction proves there are infinitely many primes congruent to $5$ modulo $6$ .

D (Integration)

1

a

Solution

The Lebesgue outer measure is

m^{*} (E) = in f {\sum_{j} ∣ I_{j} ∣ : E \subseteq ⋃_{j} I_{j}, I_{j} open intervals} .

The Lebesgue inner measure is

m_{*} (E) = sup {m (K) : K \subseteq E, K compact}

, with

m_{*} (\emptyset) = 0

. For a bounded set inside a finite interval

I

, this is equivalently

∣ I ∣ - m^{*} (I ∖ E)

A set is Lebesgue measurable if it satisfies Carathéodory's criterion:

m^{*} (A) = m^{*} (A \cap E) + m^{*} (A ∖ E) for every A \subseteq R .

For finite outer measure this is equivalent to equality of inner and outer measure. For a general unbounded set, an equivalent local criterion is

m_{*} (E \cap [- n, n]) = m^{*} (E \cap [- n, n])

for every positive integer

n

. Merely writing

m_{*} (E) = m^{*} (E) = \infty

would not establish measurability.

b

Solution

A null set has Lebesgue outer measure zero. Equivalently, for every

ε > 0

it can be covered by countably many open intervals whose total length is less than

ε

. Such a set is automatically Lebesgue measurable and has Lebesgue measure zero.

c

Solution

A \subseteq E

and

E

is a null set, every interval cover of

E

also covers

A

. Thus monotonicity of Lebesgue outer measure gives

0 \leq m^{*} (A) \leq m^{*} (E) = 0

. Hence every subset of a null set is null, and in particular is Lebesgue measurable.

d

Solution

A real measurable function is one whose inverse image of every open subset of

R

is Lebesgue measurable. Equivalently,

{x : f (x) > a} is Lebesgue measurable for every real a .

These upper rays generate the Borel sigma-algebra, making this equivalent to requiring measurable inverse images of all Borel sets.

e

Solution

a \geq 0

, the set

{f > a}

is contained in

E

, since

f = 0

outside

E

. It is therefore a null set and is Lebesgue measurable. If

a < 0

, its complement

{f \leq a}

is contained in

E

and is measurable for the same reason. Complements of measurable sets are measurable. Thus every upper level set is measurable, proving $f$ is a measurable function.

f

Solution

Let

d = h - g

. It vanishes outside the null set

E

, so part (e) proves

d

is a measurable function, regardless of its values on

E

. Since the sum of two measurable functions is measurable,

h = g + d

is measurable. This is the completeness of Lebesgue measure under changes on null sets.

2

Solution

For sigma-finite measures, Tonelli's theorem states that the double integral of a nonnegative measurable function equals either iterated integral, allowing

+ \infty

. Fubini's theorem states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.

Assign any value at

(0, 0)

, which does not affect Lebesgue integrability. Near that corner set

s = x + y

and

t = (x - y) / (x + y)

. Then

x = s (1 + t) /2

y = s (1 - t) /2

, and the absolute Jacobian determinant is

s /2

. On

0 < s < 1

, the full range is

- 1 < t < 1

, so

\iint_{x + y < 1} ∣ F (x, y) ∣ d x d y = \frac{1}{2} \int_{0}^{1} s^{2 - α} d s \int_{- 1}^{1} ∣ t ∣ d t = \frac{1}{2} \int_{0}^{1} s^{2 - α} d s .

The remaining part of the unit square has

x + y \geq 1

and is harmless. The improper power integral is finite precisely when

α < 3

. Under the given restriction,

F is Lebesgue integrable exactly for 2 < α < 3.

For those values, the change

(x, y) \mapsto (y, x)

preserves the square and changes the sign of

F

. Absolute integrability justifies this change of variables, giving

\iint_{R^{2}} F (x, y) d x d y = 0.

For

α \geq 3

the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.

E (Topology)

1

Solution

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on

A \subseteq Y

consists of sets

A \cap O

with

O

open in

Y

Let

{U_{j}}

cover

f (X)

by relatively open sets, and write

U_{j} = f (X) \cap O_{j}

with

O_{j}

open in

Y

. By continuity,

{f^{- 1} (O_{j})}

is an open cover of

X

. A finite subcollection covers

X

by compactness, and its corresponding

U_{j}

cover

f (X)

. Thus a continuous image of a compact space is compact.

For an image that is not closed, take

X

to be a singleton and

Y = {0, 1}

with the indiscrete topology. Send the sole point to

0

. This map is continuous and

X

is compact, while

f (X) = {0}

is not closed in

Y

, whose only closed sets are

\emptyset

and

Y

. A Hausdorff hypothesis on the target would exclude this example.

2

a

Solution

A sequence

x_{n}

converges to

x

if every open neighbourhood of

x

contains all sufficiently late terms. A Hausdorff space separates any two distinct points by disjoint open neighbourhoods.

If a sequence had distinct limits

x, y

in a Hausdorff space, choose disjoint open neighbourhoods

U

x

and

V

y

. Convergence to each limit implies that all sufficiently late terms lie in both

U

and

V

, contradicting

U \cap V = \emptyset

. Hence a convergent sequence in a Hausdorff space has a unique limit.

b

Solution

The product topology on

X \times Y

has as a basis of a topology the rectangles

U \times V

with

U

open in

X

and

V

open in

Y

; arbitrary unions of such rectangles are its open sets.

For distinct points

(x_{1}, y_{1}), (x_{2}, y_{2})

, at least one coordinate differs. If

x_{1} \neq = x_{2}

, disjoint open neighbourhoods

U_{1}, U_{2}

in the Hausdorff space

X

give disjoint open neighbourhoods

U_{1} \times Y, U_{2} \times Y

of the two points. If the second coordinates differ instead, use

X \times V_{1}, X \times V_{2}

. Thus a product of two Hausdorff spaces is Hausdorff.

c

Solution

For the quotient map

q : X \to X / \sim

, the quotient topology is defined by

U \subseteq X / \sim open ⟺ q^{- 1} (U) open in X .

Take the Hausdorff space

X = R

and identify two numbers precisely when they are both rational or both irrational. The quotient has two points. The inverse images of its singleton subsets are

Q

and

R ∖ Q

, neither of which is open in

R

, since each is dense and has dense complement. Hence the quotient has the indiscrete topology and is not Hausdorff.

F (Multivariable Calculus)

1

a

i

Solution

k

-dimensional embedded submanifold of

R^{n}

is a subset

M

such that near each

p \in M

there is a smooth diffeomorphism of ambient open neighbourhoods that sends

M

to the coordinate plane

R^{k} \times {0}^{n - k}

. More precisely, for some open

U ∋ p

and diffeomorphism

Φ : U \to V \subseteq R^{n}

Φ (M \cap U) = V \cap (R^{k} \times {0}^{n - k}) .

Thus locally the subset has

k

freely varying smooth coordinates.

ii

Solution

A tangent vector to

M

p

is a velocity

γ^{'} (0)

of a smooth curve

γ : (- ε, ε) \to M

with

γ (0) = p

. The tangent space is the set of all such velocities:

T_{p} M = {γ^{'} (0) : γ (0) = p, γ a smooth curve in M} .

It is a

k

-dimensional linear subspace of

R^{n}

; for a local parametrization

ϕ

, it is the image of the injective derivative

D ϕ

at the corresponding parameter point.

b

i

Solution

Let

g (x, y, z) = x^{2} + y^{2} - z^{2}

. The upper cone away from the apex is the part of

g^{- 1} (0)

with

z > 0

. There

\nabla g = (2 x, 2 y, - 2 z) \neq = 0.

The regular level set theorem says that a level set of a smooth scalar function with nonzero gradient is an embedded submanifold of codimension one. Intersecting it with the open half-space

z > 0

preserves this local property. Thus $M$ is a two-dimensional submanifold of $R^{3}$ .

ii

Solution

The tangent space to a regular level set is the kernel of its derivative. Writing

z_{0} = x_{0}^{2} + y_{0}^{2}

gives

T_{v_{0}} M = {(ξ, η, ζ) : x_{0} ξ + y_{0} η - z_{0} ζ = 0} .

If an affine tangent plane through

v_{0}

is intended, its equation is

x_{0} (x - x_{0}) + y_{0} (y - y_{0}) - z_{0} (z - z_{0}) = 0

. Since

x_{0}^{2} + y_{0}^{2} = z_{0}^{2}

, this simplifies to the same plane

x_{0} x + y_{0} y - z_{0} z = 0

, which passes through the apex.

iii

Solution

Every point of the indicated ray has the form

w_{0} = (r, 2 r, 5 r)

with

r > 0

; the apex is excluded. Substitution into the preceding tangent space formula gives

T_{w_{0}} M = {(ξ, η, ζ) : ξ + 2 η - 5 ζ = 0},

independent of

r

. The affine tangent plane has the same equation, since the ray lies in this plane. Thus both interpretations of the requested tangent plane are constant along this generator of the cone.

G (Calculus of Variations)

1

Solution

Let

P = F_{y^{'}}

and

Q = F_{y^{''}}

. For a smooth variation

y + ε η

, the first variation is

δ J = \int_{a}^{b} (F_{y} η + P η^{'} + Q η^{''}) d x = \int_{a}^{b} (F_{y} - P^{'} + Q^{''}) η d x + [(P - Q^{'}) η + Q η^{'}]_{a}^{b} .

Here the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:

F_{y} - \frac{d}{d x} F_{y^{'}} + \frac{d ^{2}}{d x ^{2}} F_{y^{''}} = 0.

If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionals

Q = 0, P - Q^{'} = 0 at a, b .

If an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For

H = y^{''} Q - y^{'} (Q^{'} - P) - F

, differentiate using the chain rule:

H^{'} = y^{'} (P^{'} - Q^{''} - F_{y}) - F_{x} .

(4)

The Euler-Lagrange equation cancels the first term. Consequently

H^{'} = - F_{x}

, and if there is no explicit

x

dependence,

H = y^{''} F_{y^{''}} + y^{'} (F_{y^{'}} - (F_{y^{''}})^{'}) - F is constant .

(5)

This is a Beltrami identity for a second-derivative functional.

H (Classical Mechanics)

1

Solution

The inertia tensor about the centroid has entries

J_{ij} = \int_{B} ρ (r^{2} δ_{ij} - x_{i} x_{j}) d V

. Reflection symmetry makes every off-diagonal entry zero. Under

x = a ξ

y = a η

z = c ζ

, the ellipsoid becomes the unit ball. Its three coordinate second moments are equal, and

\int_{ball} ξ^{2} d V = \frac{1}{3} \int_{ball} r^{2} d V = \frac{4 π}{15} .

Using the stated mass gives

\int_{B} ρ x^{2} d V = M a^{2} /5

, the same for

y

, and

\int_{B} ρ z^{2} d V = M c^{2} /5

. Therefore

J^{(G)} = \frac{M}{5} diag (a^{2} + c^{2}, a^{2} + c^{2}, 2 a^{2}) .

For the parallel coordinate axes based at

P = (0, 0, c)

, the parallel axis theorem in tensor form is

J^{(P)} = J^{(G)} + M (∣ d ∣^{2} I - d d^{T})

, where

d = P - G

. Thus

J^{(P)} = \frac{M}{5} diag (a^{2} + 6 c^{2}, a^{2} + 6 c^{2}, 2 a^{2}) .

J (Quantum Theory)

1

Solution

The one-dimensional time-independent Schrödinger equation is

- \frac{ℏ ^{2}}{2 m} ψ^{''} (x) + V (x) ψ (x) = E ψ (x) .

For the infinite square well on

[0, a]

V = 0

inside and the wave function is zero outside. The boundary conditions are

ψ (0) = ψ (a) = 0

. For positive integer

n

ψ_{n} = A sin (nπ x / a)

satisfies them, and

ψ_{n}^{''} = - (nπ / a)^{2} ψ_{n}

. Substitution gives the energy levels

E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m a ^{2}} .

Normalization requires

1 = \int_{0}^{a} ∣ ψ_{n} ∣^{2} d x = A^{2} a /2

. The requested real positive constant is therefore

A = \frac{2}{a} .

K (Fluid Dynamics and Waves)

1

Solution

A streamline is a curve whose tangent is parallel to the instantaneous fluid velocity field. In steady flow one may parametrize it by

d r / d s = u (r)

. Dotting the steady Euler equations with

u

eliminates the cross-product term, since

u \cdot ((\nabla \times u) \times u) = 0

. Hence

u \cdot \nabla (p + \frac{1}{2} ρ ∣ u ∣^{2}) = 0,

which is Bernoulli's principle along each streamline.

For the supplied field,

xu + y v = 0

, so along a streamline

\frac{d}{d s} (x^{2} + y^{2}) = 2 xu + 2 y v = 0.

Writing

r^{2} = x^{2} + y^{2}

, the tangential speed is

v_{θ} = - b r + c / r

and the radial speed is zero. Both

r

and

∣ u ∣^{2} = (- b r + c / r)^{2}

are constant around each nonstagnant circular streamline. The Bernoulli quantity is constant there, so the pressure is constant on every such circle.

If a circle is wholly stagnant, the same conclusion follows directly from the steady Euler equations: on it

u = 0

and

\nabla∣ u ∣^{2} = 0

, hence

\nabla p = 0

. Equivalently the radial momentum equation gives

\nabla p = ρ (- b r + c / r)^{2} e_{r} / r

and no angular pressure variation, including at zero tangential speed. Thus $p$ is constant on $x^{2} + y^{2} = a^{2}$ for every nonzero $a$ , with no exception for a stagnation circle.

2

Solution

Use the convention that positive vortex strength is counterclockwise and complex velocity is

u - i v

. A complex potential of a two-dimensional potential flow for the superposed uniform flow and two point vortices is

W (z) = U z + \frac{Γ}{2 πi} lo g (z - a) - \frac{Γ}{2 πi} lo g (z + a) .

Local choices of logarithm branch change no velocity. Differentiation gives

W^{'} (z) = U - \frac{i Γ a}{π ( z ^{2} - a ^{2} )} .

For nonzero vortex strength and

U > 0

, its two zeros are

z_{\pm} = \pm a^{2} + \frac{i Γ a}{π U} .

The number under the square root is nonzero, so these are distinct stagnation points, neither at a vortex.

A free point vortex moves with the velocity induced by the other vortex and the background, omitting its own singular field. At either vortex, the other induces vertical velocity

- Γ/ (4 πa)

, while the background contributes

(U, 0)

. Their velocities are therefore equal. Their separation stays

(2 a, 0)

, making this induced velocity constant for all time. Both travel along parallel straight lines with

V = (U, - \frac{Γ}{4 πa}), ∣ V ∣ = U^{2} + \frac{Γ ^{2}}{16 π ^{2} a ^{2}} .

(4)

λ \geq 0

has

P (X = k) = e^{- λ} \frac{λ ^{k}}{k !}, k = 0, 1, 2, \dots; E [X] = Var X = λ .

For

λ = 0

it is concentrated at zero. For

λ > 0

, shifting the index in the defining series gives

E [X] = λ

and

E [X (X - 1)] = λ^{2}

, so

Var X = λ^{2} + λ - λ^{2} = λ

b

Solution

Independence and convolution give

P (X + Y = k) = e^{- (μ + λ)} \sum_{j = 0}^{k} \frac{μ ^{j} λ ^{k - j}}{j ! ( k - j )!} = e^{- (μ + λ)} \frac{( μ + λ ) ^{k}}{k !},

by the binomial theorem. Thus

X + Y \sim Poisson (μ + λ) .

c

Solution

By the addition rule for independent Poisson random variables,

X_{n}

has the same distribution as a sum of

n

independent variables

Z_{j} \sim Poisson (λ)

. Each has expectation and variance

λ

. The central limit theorem therefore gives

\frac{X _{n} - nλ}{n} d N (0, λ) .

For

λ = 0

, every variable is zero, and the conclusion is the degenerate normal law with zero variance.

N_{0} = 0

, has independent increments and stationary increments, and has

N_{t + h} - N_{t} \sim Poisson (2 h)

for

h \geq 0

when time is measured in minutes. Equivalently it is a counting process with independent stationary increments satisfying, uniformly in the start time,

P (Δ N = 1) = 2 h + o (h), P (Δ N \geq 2) = o (h)

h ↓ 0

. Thus an average rate alone is insufficient; the independence and increment assumptions are part of the model.

ii

Solution

The one-minute increment has Poisson distribution with parameter two, independently of where that interval occurs. Hence

P (N_{11 : 01} - N_{11 : 00} = 0) = e^{- 2} \approx 0.1353.