Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III (A10140W1)

Source: released examination PDF, pages 191–199.

Solutions to all nine questions of the examination sat on 22 June 2023, including all optional questions.

With the change of variables $z=y^2$, multiplication of the differential equation by $2y$ gives $z^{\prime }+2xz=2x$. The integrating factor is $e^{x^2}$, soThe initial condition gives $C=3$. Since $y$ starts negative and $z\ge 1$, continuity selects the negative square root throughout:This never reaches the singular value $y=0$.

Separation of variables givesHere $x>1$ makes the argument of the second logarithm positive. Substituting the initial condition yields $3y^2+1=4(x^2-1{)}^3$. The positive branch is thereforeThis is the maximal connected interval containing $\sqrt{2}$ on which this branch solves the original differential equation. At its left endpoint $y=0$, the original right-hand side has a zero denominator, so one cannot extend the solution through that point merely by continuing the squared equation.

For $y=x$, the first and second derivatives are one and zero, giving $-2x+2x=0$. To find an independent solution, use reduction of order and write $y=xv(x)$. Cancellation of the terms proportional to $v$ leavesWriting $w=v^{\prime }$ and using separation of variables givesSince $1/[x^2(1-x^2)]=x^{-2}+(1-x^2{)}^{-1}$, integration yields the general solutionThe second factor divided by $x$ has nonzero derivative $1/[x^2(1-x^2)]$, so the two solutions are linearly independent.

The multivariable chain rule gives $u_s=2s{u}_x+2t{u}_y$ and $u_t=-2t{u}_x+2s{u}_y$. Thereforeso $\boxed{m=2}$. The $u_y$ terms cancel identically, independently of the particular function $u$.

The partial derivatives areThus the Jacobian determinant isThis vanishing reflects the functional dependence found next.

The tangent addition formula givesHence $\boxed{f(v)=\tan v}$. The restriction $xy\ne 1$ excludes the poles of tangent that can arise in the possible range $-\pi <v<\pi$. One should not replace this identity by $v=\arctan u$ globally: the principal inverse tangent can differ from $v$ by an integer multiple of $\pi$.

A scalar potential is $U(x,y)=x^{3}y+y^4$, since $\nabla U=(3x^{2}y,x^3+4y^3)$. The fundamental theorem for line integrals therefore reduces the line integral to its endpoints. The curve starts at $(0,1)$ and ends at $(2,1)$, so

The region is the part of the disk $x^2+y^2\le a^2$ in the first quadrant above $y=x$. In polar coordinates, it is $0\le r\le a$ and $\pi /4\le \theta \le \pi /2$. The integrand becomes $r$, and the area Jacobian determinant contributes another $r$. Thus

The Taylor expansions $\sin t=t+O(t^3)$ and $e^t=1+t+t^2/2+O(t^3)$ giveHence the second-order Taylor polynomial isThe sine contributes no quadratic term at zero.

For a twice continuously differentiable function, an interior local extremum requires $f_x=f_y=0$. At such a critical point, put $D=f_{xx}{f}_{yy}-f_{xy}^2$. The second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. When $D=0$, the test is inconclusive. More generally the Hessian matrix is necessarily positive semidefinite at a local minimum and negative semidefinite at a local maximum, but these non-strict signs alone are not sufficient.

For this polynomial,Subtracting the two stationarity equations gives $y(2x-y)=0$. If $y=0$, then $x=0$; if $y=2x$, the remaining equation gives $x=0$ or $x=3/2$. Thus the only critical points are $(0,0)$ and $(3/2,3)$.

At $(3/2,3)$, $\det H=-18<0$, so it is a saddle point. At $(0,0)$ the determinant is zero, but along $y=-x$ the value is $-x^3$, which takes both signs arbitrarily near zero. ThereforeIn particular a vanishing Hessian determinant does not exclude a saddle point.

Assume $f,F$ are continuously differentiable near the point and $F_z\ne 0$. The implicit function theorem supplies a local function $z(x,y)$ withAt an interior extremum of $g(x,y)=f(x,y,z(x,y))$, the multivariable chain rule givesMultiply these identities by $F_y$ and $F_x$, respectively, and subtract. The terms containing $f_z$ cancel, givingNo assumption $f_z\ne 0$ is needed for this elimination; dividing by $f_z$ would unnecessarily exclude valid cases. If $F_z=0$, this particular local graph argument is unavailable and a different chart or the full Lagrange multiplier condition is needed.

On the constraint $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtainTherefore the two global minima areEvery pair $(y,z)$ is allowed on either smooth branch $x=\pm \sqrt{1+z^2}$, so these are also the only local extrema. There is no local maximum or global maximum: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part i.

The joint probability mass function isIt is nonnegative and its sum over $S\times S$ is one. Summing over one coordinate gives the other marginal probability mass function.

For $\mathbb{P}(X=x)>0$, the conditional probability mass function isThe denominator is the sum of the numerator over all possible $y$, so these conditional probabilities sum to one.

A partition of a sample space is a finite or countable family $(A_j)$ of pairwise disjoint events whose union is the whole sample space. If $X$ is integrable, the law of total expectation statesThis follows by summing $X{1}_{A_j}$ over the partition of a sample space; absolute integrability justifies the countable sum. For nonnegative $X$, the identity also holds with infinite expectations. Zero-probability parts contribute zero without requiring a conditional value on them.

The first swap necessarily removes one red ball from urn 1 and replaces it with a blue ball. Thus $\boxed{R_1=n-1}$ with probability one.

Conditional on $R_k=r$, urn 1 has $r$ red and $n-r$ blue balls, while urn 2 has $n-r$ red and $r$ blue balls. The independent selections therefore giveEvery other conditional probability is zero. The out-of-range terms at $r=0$ or $r=n$ already have zero probability. These are the transition probabilities of the Bernoulli-Laplace diffusion model.

Only upward and downward moves change the value, so the conditional expectation isThus $\boxed{\alpha =1,\beta =1-2/n}$.

Apply the law of total expectation to the partition of a sample space given by the possible values of $R_k$. Since the conditional mean is affine,Here $r_0=n$. This uses linearity of expectation and does not assert independence of successive urn counts.

The equilibrium of the first-order linear recurrence is $n/2$. Subtracting it givesConsequentlyFor $n=2$, interpret the formula at $k=0$ through the initial condition: $r_0=2$, while $r_k=1$ for $k\ge 1$. For every $n\ge 2$, $r_k\to n/2$.

With independent Bernoulli trials of success probability $p$, $N$ counts the failures before the first success. There are $n$ failures followed by one success with probability $(1-p{)}^{n}p$, giving the stated geometric distribution on $\{0,1,2,\dots \}$. It counts failures, rather than the total number of trials including the success.

Writing $q=1-p$, the probability generating function is a geometric series:The displayed restriction ensures absolute convergence.

Differentiate the probability generating function inside its radius of convergence:Since these are $\mathbb{E}[N]$ and $\mathbb{E}[N(N-1)]$, respectively,

Take independent Bernoulli random variables $X_i$ of success probability $r$, independently of $N$. On the event $N=n$, the random sum ${\sum }_{i=1}^N{X}_i$ has the binomial distribution with parameters $(n,r)$, including the empty sum zero at $n=0$. It therefore has exactly the specified joint distribution with $N$.

Thus one may realize $\boxed{Z=\sum _{i=1}^N{X}_i}$ on a suitable probability space. This is a distributional construction; it does not require a pre-existing $Z$ to be independent of $N$.

For an independent nonnegative integer-valued count $N$ and identically distributed independent summands, the random sum probability generating function is $G_Z(s)=G_N(G_{X_1}(s))$. Conditioning on $N=n$ proves this, since the conditional probability generating function is $G_{X_1}(s{)}^n$.

Here $G_{X_1}(s)=1-r+rs$, soThis identity holds throughout $|s|\le 1$, and its power-series extension has radius $[p+(1-p)r]/[(1-p)r]$.

Put $p_{\ast }=p/[p+(1-p)r]$. ThenUniqueness of the probability generating function therefore givesThis is Bernoulli thinning of a geometric count, using the convention of counting failures before success.

In each trial the presence indicators of $A$ and $B$ are independent Bernoulli random variables, with probabilities $p$ and $r$. Independence across trials makes their two entire indicator sequences independent. Let $N$ be the number of trials before the first $A$; then $N\sim {\mathrm{Geom}}_0(p)$. Among these $N$ preceding trials, the $B$ indicators are independent with probability $r$.

Thus the count is the random sum from part i, andAn $AB$ outcome on the stopping trial contributes nothing, because the question counts only outcomes strictly before the first $A$.

The cumulative distribution function isThe total integral of the nonnegative probability density function is one.

For a random variable with a probability density function, a singleton has zero probability:Equivalently, its cumulative distribution function is continuous, so it has no jump at $x$.

The integers form a countable set, and the events $\{X=k\}$ are disjoint. By countable additivity and part ii,Countability is essential: one cannot use the same sum argument over an arbitrary uncountable collection of zero-probability events.

Normalization of the probability density function requiresHence $\boxed{c=t-1}$. The distribution is a Pareto distribution with minimum one and shape parameter $t-1$.

The condition $t>2$ guarantees a finite first moment. Direct integration gives

For an integer $k\ge 1$, $\lfloor X\rfloor \ge k$ is the same event as $X\ge k$. Integration for the Pareto distribution givesThe second identity subtracts two successive tail probabilities, equivalently integrating over $[k,k+1)$. If $\mathbb{N}$ includes zero, the extra case is $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely.

For a nonnegative integer-valued random variable $W$, the tail-sum formula states $\mathbb{E}[W]={\sum }_{k=1}^{\infty }\mathbb{P}(W\ge k)$. It follows by writing $W={\sum }_{k\ge 1}{1}_{\{W\ge k\}}$ and using nonnegativity to interchange sum and expectation.

With $W=\lfloor X\rfloor$, this givesThe Riemann zeta function is finite here because $t-1>1$.

By part iii, $X$ is an integer with probability zero. Thus $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely, andThe pointwise inequalities $\lfloor X\rfloor \le X\le \lceil X\rceil$ implyPutting $s=t-1>1$ and rearranging yieldsThis extracts bounds for an infinite series from simple rounding inequalities.

For the exponential distribution with rate $\theta$, the change of variables $u=\theta x$ givesThe integrals are one and two by integration by parts. Hence $\boxed{\mathbb{E}[X]=1/\theta ,\mathrm{Var}(X)=1/{\theta }^2}$.

For an estimator $T$ of $\theta$, its bias and mean squared error areWhen the second moment exists, expanding around ${\mathbb{E}}_{\theta }[T]$ gives the bias-variance decomposition ${\mathrm{MSE}}_{\theta }(T)={\mathrm{Var}}_{\theta }(T)+b_{\theta }(T{)}^2$.

For the independent random sample, the likelihood function on its support is $L(\theta )={\theta }^n{e}^{-\theta Y}$, where $Y=\sum X_i$. Its log-likelihood is $\ell (\theta )=n\log \theta -\theta Y$, withAlmost surely $Y>0$, and the unique stationary point is consequently the unique global maximum:The exceptional all-zero sample has probability zero; if $Y=0$ were nevertheless supplied, there would be no finite maximizing rate.

Use the gamma density and the normalization integral for shape $n-1$:The same argument, valid for $n\ge 3$, gives $\mathbb{E}[Y^{-2}]={\theta }^2/[(n-1)(n-2)]$.

The inverse moments in part i giveThus the bias is positive, and direct expansion of the mean squared error yieldsFor comparison, the variance is $n^2{\theta }^2/[(n-1{)}^2(n-2)]$; adding the squared bias gives the same answer by the bias-variance decomposition.

The central limit theorem givesWrite $z=z_{1-\alpha /2}$ for the indicated standard normal quantile. Inverting the approximate pivot inequality and substituting $Y/n=1/\hat{\theta }$ gives the approximate confidence intervalIts coverage is approximately $1-\alpha$ for large $n$. The intersection respects the positive rate parameter; asymptotically the lower endpoint is positive. This is an asymptotic interval, not an exact finite-sample confidence interval.