Past exam of the mathematics course of the University of Oxford / Undated

Source: released examination PDF, bundle pages 126–133.

The cover gives a Long Vacation sitting on Thursday 9 September, with no printed year. It is retained as undated.

Solutions cover all nine questions and every subpart. The digit example in Question 4(c) is inconsistent with its definitions; Question 5(a)(ii) is solved with the printed $u_2=0$. Questions 7–9 are included despite the absent Section C heading in the source.

Divide by $x>1$. The product rule turns the linear ordinary differential equation intoEquivalently, $\ln x$ is an integrating factor after division by $x\ln x$. An antiderivative gives $\ln xy=\frac{1}{2}(\ln x{)}^2+C$. The initial condition at $e$ gives $C=-1/2$, henceIts derivative is $y^{\prime }=[1+(\ln x{)}^{-2}]/(2x)$, so substitution gives $x\ln x{y}^{\prime }+y=\ln x$, and $y(e)=0$. Every solution of a differential equation has the integrated form above, and the initial condition fixes its only constant. This agrees with the uniqueness theorem for ordinary differential equations on the stated interval.

The product rule gives $y^{\prime }=x^n{z}^{\prime }+n{x}^{n-1}z$ and $y^{\prime \prime }=x^n{z}^{\prime \prime }+2n{x}^{n-1}{z}^{\prime }+n(n-1)x^{n-2}z$. On an interval avoiding $x=0$, substitution and division by $x^{n+2}$ giveFor a constant-coefficient differential equation, the coefficient of $z^{\prime }$ forces $2n+4=0$. At $n=-2$, the remaining coefficient $n^2+3n+2$ also vanishes. ThusThe characteristic equation of a constant-coefficient differential equation is $r^2+2r+2=0$, with complex conjugate roots of a polynomial $-1\pm i$. Its real solutions $e^{-x}\cos x$ and $e^{-x}\sin x$ are linearly independent. Therefore the general solution isThe constants can be chosen independently on the two sides of the singular point. The power substitution reducing a differential equation to constant coefficients here is invertible wherever $x\ne 0$.

Use the change of variables $u=x+y$. Then $u^{\prime }=1+y^{\prime }=1+\cos u=2{\cos }^2(u/2)$, a separable differential equation. Where $u$ is not an odd multiple of $\pi$, separation of variables givesIndeed the derivative of $\tan (u/2)$ with respect to $u$ is $1/(1+\cos u)$. The nonconstant branches can also be written $y=2\arctan (x+C)+2k\pi -x$, $k\in \mathbb{Z}$. Dividing by $1+\cos u$ excluded its zeros: the missing equilibrium solutions for $u$ give the additional solutions of a differential equationThey satisfy $y^{\prime }=-1=\cos (x+y)$ directly. The transformed right side is continuously differentiable, so the uniqueness theorem for ordinary differential equations prevents a nonconstant branch from crossing an equilibrium solution at finite $x$.

Put $h(s)=\sin x(s)-\cos y(s)$. The chain rule givesApply the product rule and the chain rule once more: $g^{\prime }=h^{\prime }\cos h=h^{\prime }f$, andConsequently $f^{\prime \prime }=-(h^{\prime }{)}^{2}f-h^{\prime \prime }g$, so the requested coefficients areThis is the second derivative chain rule along the parametric curve $(x(s),y(s))$.

The Jacobian determinant isThe change of variables is defined only for $x\ne 0$. On the interior of the integration region, $x>0$ and $y>0$, so this Jacobian determinant is negative and nonzero.

The inverse change of variables is $x=v/(1+u)$, $y=uv/(1+u)$. Thus the triangular region becomesThe upper limit depends on $u$: the new region is not a rectangle. The absolute inverse Jacobian determinant is $x^2/(x+y)$, so the change of variables formula cancels the rational factor in the integrand:Although the original integrand is undefined at the vertex $x=y=0$, its nonnegative improper integral exists. One can apply the change of variables formula on $x\ge \epsilon$ and let $\epsilon \downarrow 0$ by the monotone convergence theorem; the transformed integrand has the finite integral just computed.

For the graph of a differentiable function, the arc length is $L=\int \sqrt{1+(y^{\prime }{)}^2}dx$. Since $y^{\prime }=2x$,Use the hyperbolic substitution $2x=\sinh t$: then $dx=\frac{1}{2}\cosh tdt$ and $\sqrt{1+4x^2}=\cosh t$. The hyperbolic functions ${\cosh }^{2}t=(1+\cosh 2t)/2$ yieldsHere $\mathrm{arsinh}t=\ln (t+\sqrt{1+t^2})$ is the inverse hyperbolic sine.

The middle component suggests a scalar potential with $-{\varphi }_y=2y{\cosh }^2(xz)$, so $\varphi =-y^2{\cosh }^2(xz)+h(x,z)$. The chain rule and $2\sinh t\cosh t=\sinh (2t)$ giveMatching the other two components requires $h_x=h_z=0$. ThereforeIts negative gradient is the given vector field in all three components; the additive constant has no effect on that gradient.

A parametrization of the straight line segment is $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then $\mathbf{F}(\mathbf{r}(t))=(t,t^2,t)$ and ${\mathbf{r}}^{\prime }(t)=(1,1,1)$. By the line integral formula,

Evaluate the line integral separately on the three oriented line segments. Their parametrizations and contributions areThusThe different value in i demonstrates path dependence. The vector field in this part is different from the conservative vector field in part (a).

Write $A=a^2>0$ and $E=e^{-(x^2+y^2)/A}>0$. The partial derivatives factor asA critical point has both partial derivatives zero. If $x,y\ne 0$, this would require $x^2-2y^2=A$ and $x^2-2y^2=-2A$, which is impossible. On the axes the remaining conditions giveAt either critical point with $x\ne 0$, the Hessian matrix isIt is negative definite, so the second-derivative test shows that both $(\pm |a|,0)$ are strict local maxima, of value $a^2/e$. They also attain the global maximum: $f\le x^2{e}^{-x^2/A}\le A/e$. For completeness, the Hessian matrix at $(0,\pm |a|)$ is $\mathrm{diag}(6/e,8/e)$, giving strict local minima, while at the origin it is $\mathrm{diag}(2,-4)$, giving a saddle point.

A discrete random variable on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ is a real-valued measurable function $X:\Omega \to \mathbb{R}$ whose probability distribution is concentrated on some finite or countable set $S$:Measurability means $\{\omega :X(\omega )\le t\}\in \mathcal{F}$ for every real $t$. Its probability mass function is $p_X(x)=\mathbb{P}(X=x)$, and ${\sum }_{x\in S}{p}_X(x)=1$. Requiring a countable image everywhere, rather than outside a null set, is an equivalent convention after harmless modification on a null set.

Use the ordered outcomes of the two rolls as the sample space:Thus each elementary outcome has probability $1/36$. This is the usual probability space for two independent fair dice.

For an outcome $\omega =(i,j)$, the two random variables are the coordinate mapsTheir marginal probability mass functions are $p_X(i)=p_Y(j)=1/6$ on $\{1,\dots ,6\}$, and their joint probability mass function satisfiesOff that support, the same factorization holds with zeros. Hence $X$ and $Y$ are independent random variables, by the joint probability mass function criterion for independence.

There are three even faces, so $\mathbb{P}(A)=1/2$. The sum is even precisely when both faces have the same parity, giving $9+9=18$ outcomes and $\mathbb{P}(B)=1/2$. Both $A$ and $B$ occur precisely when both faces are even, giving nine outcomes. ThereforeThus $A$ and $B$ are independent events.

No: pairwise independence does not imply that the three are independent events. Take $C=\{Y\text{ is even}\}$. Then $\mathbb{P}(C)=1/2$, and $\mathbb{P}(A\cap C)=9/36=1/4$ by independence of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb{P}(B\cap C)=1/4$. Thus all three pairs of events are independent.

But $A\cap B\cap C$ also consists of the nine even-even outcomes, soThis violates the triple-intersection requirement for independent events.

The printed numerical example reverses the digits and contradicts both definitions. With the stated last-digit definition, $[Z]=17$ gives $X=7$, $Y=1$. Use those definitions throughout.

Put $q=e^{-\lambda }\in (0,1)$ and $N=\lfloor Z\rfloor$. The exponential distribution givesThe Euclidean division $N=10Y+X$ has unique remainder $X\in \{0,\dots ,9\}$ and quotient $Y\in \{0,1,\dots \}$. Thus the joint probability mass function isand zero otherwise. This is decimal digit decomposition of a geometric random variable. Following the erroneous example instead would require different definitions and would not answer the stated question.

Summing the joint probability mass function over $y$ with the geometric series gives the marginal probability mass functionDivide the joint probability mass function by this positive marginal probability mass function to obtain the conditional distribution:It is the geometric distribution counting failures before the first success, with success probability $1-q^{10}$. For the other nonnegative integers $x\ge 10$, $\mathbb{P}(X=x)=0$; the elementary conditional probability given $X=x$ is undefined and supplies no additional case.

Summing over the last digit yields the marginal probability mass functionTogether with the other marginal probability mass function, this givesThe factorization holds on the whole support and with zeros elsewhere, proving that $X$ and $Y$ are independent. The conditional distribution in ii likewise does not depend on the permissible value $x$.

For the homogeneous linear recurrence relation, try $u_n=r^n$. Its characteristic equation of a linear recurrence isThe distinct roots of a polynomial $1,-4$ give the complementary terms $A+B(-4{)}^n$. A constant cannot absorb the forcing because $1$ is already a root of a polynomial of the characteristic equation; try a particular solution $cn$. Substitution gives $c=-4c+1$, hence $c=1/5$. The general solution is $u_n=A+B(-4{)}^n+n/5$.

The initial conditions give $A+B=0$ and $A-4B=4/5$. Hence $A=4/25$, $B=-4/25$, and

The homogeneous characteristic equation of a linear recurrence isThe repeated characteristic root of a linear recurrence $-1$ supplies the independent terms $(-1{)}^n$ and $n(-1{)}^n$. A constant particular solution is $1/2$, since $1/2=-2(1/2)-1/2+2$. Thus $u_n=1/2+(A+Bn)(-1{)}^n$.

The scan prints $u_2=0$, not $u_1=0$. These initial conditions are consistent. They give $A=1/2$ from $u_0=1$, then $0=1/2+A+2B$, so $B=-1/2$. ThereforeIn particular $u_1=1/2$ and $u_2=0$. Direct substitution verifies the recurrence relation. This is an example of nonconsecutive initial values for a linear recurrence; replacing the printed subscript by 1 would change the answer.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable. Expand and use linearity of expectation:The same identity holds whenever the quantities involved exist with sufficient integrability.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

Absolute integrability now requiresFor $\alpha \le 0$, the integrand is bounded by $z{e}^{-z^2/2}$, whose integral is finite. For $\alpha >0$, once $z^2\ge 1/\alpha$, the exponent is at least $\alpha z^4/2$, so the integral diverges. HenceAgain the zero follows from the odd function and symmetry; outside the integrable range both signed parts diverge, so the expected value is undefined.

Let $A=\{|Z|>2\}$ and $B=\{Z>0\}$. The standard normal distribution is symmetric and assigns no probability mass to zero. Thus $\mathbb{P}(B)=1/2$, and its equal two-sided tails giveThereforeso the two events are independent. The permitted approximation gives $\mathbb{P}(A)\approx 0.05$ and $\mathbb{P}(A\cap B)\approx 0.025$, but independence is exact.

A joint probability density function integrates to one. HeresoThis value also makes the probability density function nonnegative everywhere.

Obtain each marginal density by integrating the joint probability density function over the other variable:Each marginal density is zero off its displayed interval and has total integral one.

On the product support $[0,1{]}^2$, the joint probability density function factors asThe equality also holds outside the support with both sides zero. The probability density function factorization criterion proves that $U$ and $V$ are independent random variables.

For each nonnegative integer $z$, the events $\{X=z-y,Y=y\}$, $0\le y\le z$, are pairwise disjoint and their union is $\{X+Y=z\}$. By countable additivity and then independence,This is the discrete convolution of the two probability mass functions. If the convention for natural numbers excludes zero, set both probability mass functions equal to zero there; the formula remains valid. Also $p_Z(z)=0$ for negative $z$.

Let $R$ denote the red die. Conditional on $R=r$, each white roll succeeds independently with probability $p_r=(7-r)/6$. Consequently $T\mid R=r$ has the geometric distribution on $\{1,2,\dots \}$ and expected value $1/p_r=6/(7-r)$. The law of total expectation givesThis is $2.45$ rolls; the averaging must be over conditional expected values, not over success probabilities before taking a reciprocal.

By the law of total probability,Bayes' theorem therefore gives the conditional probability mass functionThe six probabilities sum to one. A small red outcome is more likely after an immediate success because more white faces can meet its threshold.

For a geometric distribution on positive integers, the probability generating function is $ps/[1-(1-p)s]$. Conditional averaging over the red die yieldsAs a power series, its exact domain of absolute convergence isIndeed the slowest decaying component has failure probability $5/6$, so $\mathbb{P}(T=t)\ge (1/36)(5/6{)}^{t-1}$. At $|s|=6/5$, the terms of the power series fail to tend to zero; outside that disk they grow. All six component series converge inside it. In particular, the probability generating function is defined on the usual real interval $0\le s\le 1$, with $G_T(1)=1$. The rational expression beyond the convergence disk is only an algebraic continuation, not the defining expected value.

The least-squares estimators choose the intercept and slope to minimize the sum of squared regression residuals:After observation, replace $Y_i$ by $y_i$ to obtain the numerical estimates. Their definition does not require a normal distribution for the errors. A unique slope requires some variation in the fixed predictors; under ${\sum }_i{x}_i=0$, this is $S_{xx}={\sum }_i{x}_i^2>0$, assumed below. If all $x_i=0$, the slope is unidentifiable.

Differentiate the ordinary least squares objective in a. Its normal equations for linear least squares areUsing ${\sum }_i{x}_i=0$ and $S_{xx}={\sum }_i{x}_i^2>0$ givesThe Hessian matrix of the objective in $(a,b)$ is $2\mathrm{diag}(n,S_{xx})$, which is positive-definite, so this is its unique global minimum. For the observed estimates, replace $Y_i$ by $y_i$.

The centered design gives $\hat{\beta }=\beta +S_{xx}^{-1}{\sum }_i{x}_i{ϵ}_i$. By independence of the errors, their off-diagonal covariances vanish. Thus variance additivity for independent random variables givesA larger spread of predictor values decreases the slope variance for fixed error variance.

Here $\hat{\alpha }=\alpha +n^{-1}{\sum }_i{ϵ}_i$. By linearity of covariance and independence,Thus the two estimators are uncorrelated random variables.

They need not be independent. For a counterexample take $n=2$, $x_1=-1$, $x_2=1$, and independent errors taking $-1,1$ with equal probability. ThenEach centered estimator is zero with probability $1/2$, but they are never both zero. Their zero events therefore violate independence.

A sufficient additional assumption is that the errors have a common normal distribution. The two estimators are then Jointly Gaussian, being linear combinations of the independent Gaussian random variables. Their zero covariance, established in iii, implies independence of uncorrelated jointly normal variables. Normal errors are a sufficient condition here, not a claim that no other error probability distribution could ever work.

For the random vector $\mathbf{ϵ}=({ϵ}_1,\dots ,{ϵ}_n{)}^{\top }$, its covariance matrix isIts diagonal entries are variances, and its off-diagonal entries measure pairwise covariance. Finite second moments make every entry well defined.

Independence of the centered standard normal random variables gives $\mathbb{E}{ϵ}_i=0$ for every $i$. AlsoFor nonadjacent indices, the underlying sets of $\eta$ variables are disjoint, so the errors are independent and have zero covariance. For adjacent indices with $i\ge 2$,The first adjacent pair separately gives $\mathbb{E}[{ϵ}_1{ϵ}_2]=\mathbb{E}[{\eta }_1^2{\eta }_2]=0$. HenceThese adjacent products of independent Gaussian variables are uncorrelated random variables even though adjacent errors share a factor and are dependent.