Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1

Expand $e_{r+1}$ in the given spanning set:Some $b_j$ must be nonzero, since otherwise $e_{r+1}$ would lie in the span of the earlier $e_i$, contradicting linear independence. Reorder the remaining $f_j$ so $b_{r+1}\ne 0$. Solving this equation for $f_{r+1}$ expresses it in terms of $e_1,\dots ,e_{r+1},f_{r+2},\dots ,f_m$. Every member of the old spanning set is therefore in the new span, so the replacement set still spans $V$. This proves the needed step of the Steinitz exchange lemma directly.

Starting with $f_1,\dots ,f_m$, apply that replacement successively to $e_1,e_2,\dots$. After $r\le m$ steps the spanning list is $e_1,\dots ,e_r$ together with $m-r$ unreplaced $f$'s. If $n>m$, after $m$ replacements the list $e_1,\dots ,e_m$ spans $V$, forcing $e_{m+1}$ to be a linear combination of it. That contradicts linear independence. Hence $\boxed{n\le m}$.

The normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. It is a subgroup: the identity normalizes $H$, products of normalizing elements normalize it, and conjugating the equality by $g^{-1}$ proves inverse closure. The mapis well-defined and onto. Two images are equal exactly when $g_2^{-1}{g}_1\in N_G(H)$, exactly the condition that their cosets coincide. Thus the number of conjugates is $\boxed{[G:N_G(H)]}$.

By the permitted conjugacy assertion for Sylow subgroups, the number $n_p$ of Sylow $p$-subgroups equals $[G:N_G(P)]$ for one such subgroup $P$. Lagrange's theorem makes this divide $|G|$, and $P\subseteq N_G(P)$ also makes it divide $|G|/|P|$.

For order $72=8\cdot 9$, a Sylow $3$-subgroup has order nine, so $n_3$ divides eight. To obtain the required congruence without assuming an additional Sylow counting theorem, let $P$ act by conjugation on the set of Sylow $3$-subgroups. Its orbit sizes are powers of three. A fixed subgroup $Q$ is normalized by $P$, so $PQ$ is a subgroup and its order $|P||Q|/|P\cap Q|$ is a power of three. Since nine is the largest power of three dividing $72$, this forces $PQ=P=Q$. There is exactly one fixed point, namely $P$ itself. All other orbits have sizes divisible by three, giving $n_3\equiv 1(\mathrm{mod}3)$. Among $1,2,4,8$, only one and four satisfy this. Therefore $\boxed{n_3\in \{1,4\}}$.

For a piecewise continuously differentiable curve $\gamma (t)=(x(t),y(t))$ in the upper half-plane model, the given Riemannian metric gives hyperbolic lengthThe metric matrix has determinant $y^{-4}$, so its area element is $y^{-2}dxdy$. Thus a measurable region $D$ has hyperbolic area ${\int }_D{y}^{-2}dxdy$. For the specified strip the integral isThe infinite Euclidean height is compatible with finite hyperbolic area because the area density decreases quadratically with height.

Uniform convergence means that for every $\epsilon >0$ there is an integer $N$, independent of $x$, such that $|F_n(x)-F(x)|<\epsilon$ for all $n\ge N$ and every $x\in (0,1)$. Equivalently ${\sup }_{0<x<1}|F_n(x)-F(x)|\to 0$. In pointwise convergence, the required $N$ may depend on the chosen $x$.

At every fixed $x$, $F_n(x)=e^x/n\to 0$. Moreover ${\sup }_{0<x<1}{e}^x/n=e/n\to 0$. Thus the pointwise limit is zero and convergence is uniform.

For every fixed $x>0$, $n{x}^2\to \infty$, so the pointwise limit is zero. But ${\sup }_{0<x<1}{e}^{-n{x}^2}=1$ for every $n$, since $x$ can approach zero. Hence convergence is not uniform. For an explicit moving-point test take $x_n=n^{-1/2}$ for $n>1$, which gives $F_n(x_n)=e^{-1}$.

The finite geometric series gives $F_n(x)=(1-x^{n+1})/(1-x)$, so the pointwise limit is $F(x)=1/(1-x)$. The error is $x^{n+1}/(1-x)$, unbounded as $x\uparrow 1$ for every fixed $n$. Thus convergence is not uniform on $(0,1)$. It is uniform on any smaller interval bounded away from one, since the error is at most $b^{n+1}/(1-b)$ for $0<x\le b<1$.

Factor the denominator as $z^2(1+z^2)$. Near zero,so zero is a double pole with residue zero. At $z=\pm i$ the poles are simple and the denominator derivative is $2z+4z^3=-2z$. ThereforeThese are all the poles.

The only pole is the simple one at $z=1$, with $\boxed{{\mathrm{Res}}_1=e}$. The point $z=0$ is an essential singularity, not another pole: multiplying $e^{1/z^2}$ by a nonvanishing analytic factor cannot remove its infinite principal part. For completeness its residue can also be read fromThe coefficient of $z^{-1}$ is $-{\sum }_{k\ge 1}1/k!=1-e$. This extra residue belongs to the essential singularity; it does not change the list of poles.

The zeros of the denominator satisfy $e^z=\ell \pi$ with nonzero integer $\ell$, since an exponential never vanishes. Equivalently the poles areAt such a point $e^{z_{m,k}}=(-1{)}^{k}m\pi$, and the derivative of the denominator is $e^z\cos (e^z)=(-1{)}^{m+k}m\pi \ne 0$. All these poles are simple, with

In three dimensions, the general rank-two isotropic tensor is $a{\delta }_{ij}$ and the general rank-three tensor invariant under proper rotations is $b{ϵ}_{ijk}$. Here $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The latter is orientation-sensitive: an ordinary third-rank tensor invariant under all orthogonal transformations is zero instead. This distinction does not change the stress conclusion.

For an isotropic material, the constitutive tensor must consequently have $A_{ijk}=b{ϵ}_{ijk}$ in the proper-rotation convention. Hence ${\sigma }_{ij}=b{ϵ}_{ijk}{B}_k=-{\sigma }_{ji}$. Mechanical stress in the model is also symmetric, so ${\sigma }_{ij}=-{\sigma }_{ij}$ and every component vanishes. Thus nonzero stress linear in the field requires anisotropic material coefficients. This is the isotropic linear magnetostriction vanishes restriction, not a claim that all higher-order magnetostriction is impossible in isotropic materials.

In vacuum SI notation, Maxwell's equations areTaking the divergence of the last equation, using that the divergence of a curl vanishes and then the first equation, givesThis is local conservation of electric charge. For spatially uniform electrical conductivity, Ohm's law $J=\sigma E$ gives $\nabla \cdot J=\sigma \rho /{ϵ}_0$, soFor a homogeneous dielectric conductor with permittivity $ϵ$, replace ${ϵ}_0$ by $ϵ$. This charge relaxation is a statement about bulk charge; charge transported to a boundary is still accounted for by the conservation law.

For a real potential, the Time-dependent Schrödinger equation and its complex conjugate giveThe potential terms cancel. The expression on the right is a total derivative, givingFor the plane wave, ${\psi }_t=-i\omega \psi$ and ${\psi }_{xx}=-k^2\psi$. With zero potential, substitution givesIt is a free-particle momentum eigenstate with momentum $\hslash k$, energy $\hslash \omega$ and uniform probability flux. On the whole real line it is a generalized state, not a normalized square-integrable wavefunction; wave packets or box normalization give the corresponding physical interpretation.

The general continuity equation is ${\rho }_t+\nabla \cdot (\rho u)=0$, or $D\rho /Dt+\rho \nabla \cdot u=0$. An incompressible flow has $D\rho /Dt=0$, so for positive density $\boxed{\nabla \cdot u=0}$.

For the displayed velocity field, on $x^2+y^2>0$,The sum is zero. A stream function with the required sign convention isIts derivatives give exactly $u=({\psi }_y,-{\psi }_x)$. The origin is excluded because the prescribed velocity is singular there.

Assume independent samples $X_1,\dots ,X_n$ and $Y_1,\dots ,Y_m$, each consisting of independent normal random variables, with respective means ${\mu }_1,{\mu }_2$ and a common unknown variance ${\sigma }^2>0$. Take $n,m\ge 2$. Normality, independence and equal variances are the assumptions giving the exact pooled Student's t-test; unequal variances would require a different test. We test $H_0:{\mu }_1={\mu }_2$ against an unrestricted difference.

Put $N=n+m$ andThe normal likelihood is proportional to $({\sigma }^2{)}^{-N/2}\exp [-S/(2{\sigma }^2)]$, where $S$ is the residual sum of squares. Under the unrestricted model, the maximizing means are $\bar{X},\bar{Y}$ and $S=W$. Under $H_0$, the maximizing common mean is $(n\bar{X}+m\bar{Y})/N$ and decomposition about the sample means gives $S=W+B$. For either model, maximizing over variance gives ${\hat{\sigma }}^2=S/N$. Thus the generalized likelihood-ratio test statistic isSmall $\Lambda$ is therefore equivalent to large $|T|$.

Under the null, $Z=(\bar{X}-\bar{Y})/[\sigma \sqrt{1/n+1/m}]$ is standard normal. In each sample an orthogonal change of Gaussian coordinates separates the sample-mean coordinate from the centered residual coordinates; those coordinates are independent. Consequently $V=W/{\sigma }^2$ has a chi-squared distribution with $N-2$ degrees of freedom and is independent of $Z$. Therefore $T=Z/\sqrt{V/(N-2)}$ has the Student's t-distribution with $N-2$ degrees of freedom. At significance level $\alpha$, reject exactly when $\boxed{|T|>t_{N-2,1-\alpha /2}}$, the two-sample pooled t-test.

An irreducible Markov chain has the property that for every pair of states $i,j$ there is an integer $n\ge 0$ with $(P^n{)}_{ij}>0$. Equivalently every state can reach every other state along a finite path of positive-probability transitions.

A recurrent state $i$ satisfies ${\mathrm{Pr}}_i(T_i^{+}<\infty )=1$, where $T_i^{+}=\inf \{n\ge 1:X_n=i\}$. Saying that the transition matrix is recurrent means that every state is recurrent. In an irreducible Markov chain it suffices that one state is recurrent, since recurrence is a communicating-class property.

Irreducibility and the presence of another state imply $P_{ii}<1$: otherwise $i$ would be absorbing and could not reach that other state. Hence the proposed matrix has nonnegative entries and row sums ${\sum }_{j\ne i}{P}_{ij}/(1-P_{ii})=1$.

Observe the original Markov chain only at times when it changes state. Starting from $i$, the next different state is $j\ne i$ with probabilityEach holding run has a finite geometric length almost surely. The Strong Markov property at the successive change times therefore identifies the observed process as the discrete-time jump chain with transition matrix $\tilde{P}$.

Any positive-probability path of the original chain gives a path of this new chain after repeated consecutive states are deleted. Every remaining edge has positive $\tilde{P}$ probability, so $\tilde{P}$ is irreducible. For recurrence, starting at any state $i$, repeated application of the Markov property at return times shows that the original recurrent chain visits $i$ infinitely often almost surely. A single sojourn cannot account for infinitely many visits, because every sojourn is finite almost surely. There must therefore be infinitely many separate visits to $i$ in the observed chain too. Thus $\tilde{P}$ is recurrent, proving that recurrence survives deletion of self-transitions.

Both candidate sets contain zero. The sum of subspaces is closed under linear combinations because $a(u_1+w_1)+b(u_2+w_2)=(a{u}_1+b{u}_2)+(a{w}_1+b{w}_2)$, with the two terms in $U$ and $W$. The intersection is closed because any linear combination of vectors lying in both subspaces still lies in both. Thus each is a vector subspace.

Take a basis $s_1,\dots ,s_r$ of the intersection, extend it by $u_1,\dots ,u_a$ to a basis of $U$, and by $w_1,\dots ,w_b$ to a basis of $W$. The combined list $s,u,w$ spans $U+W$. To prove independence, a zero combination makes the $u$ combination equal a vector in both $U$ and $W$, so it is a combination of the $s$'s. Independence of the basis of $U$ makes all $u$ coefficients zero; independence of the basis of $W$ then makes all remaining coefficients zero. Hence $\dim (U+W)=r+a+b$, giving the dimension formula for a sum of subspaces

For the concrete kernels, the equations $Ax=0$ giveApplying the two rows of $B$ to this expression gives $4(s-t)$ and $5(s-t)/3$. Thus the intersection has basis $v_0=(0,2,1,1)$. Choose $v_1=(5,-1,3,0)$ in $U$ and $v_2=(-4,-2,1,0)$ in $W$. ThenIndeed $U$ has dimension two because $A$ has rank two, and $v_0,v_1$ are independent. Likewise $W$ has dimension two and $v_0,v_2$ are independent, so the last list spans the sum. Its first three coordinates have determinant $-48$, confirming independence and dimension three.

The Fundamental theorem of finitely generated abelian groups states that every such group is isomorphic towith uniquely determined rank and invariant factors, allowing an empty finite part. Modulo $pA$, the free summand contributes $(\mathbb{Z}/p\mathbb{Z}{)}^r$. If $A/pA=0$ for some prime, this forces $r=0$, so $A$ is finite. Conversely, if $A$ is finite, choose a prime $p$ coprime to $|A|$. Bezout's identity gives an integer $b$ with $bp\equiv 1(\mathrm{mod}|A|)$. Since $|A|$ kills every element, multiplication by $p$ is onto, with inverse multiplication by $b$. Hence $pA=A$. This proves $\boxed{A\text{ finite}⟺A/pA=0\text{ for some prime }p}$ under finite generation.

For the requested nonzero example use the additive divisible group $A=\mathbb{Q}$. Each rational $q$ equals $p(q/p)$, so $p\mathbb{Q}=\mathbb{Q}$ for every prime. It is not finitely generated: a finite list of rational generators has a common denominator $D$, and every integral combination then lies in $D^{-1}\mathbb{Z}$. The rational $1/(2D)$ does not lie there. This proves non-finite-generation directly and shows why the hypothesis in finite generation detected by a prime quotient is essential.

In the upper half-plane model, a hyperbolic line is either a vertical line or a Euclidean semicircle perpendicular to the real axis. For the vertical line $x=a$, set $R_L(z)=2a-\bar{z}$. For the circle $|z-a|=r$ with real $a$, setThe vertical formula preserves $y$ and Euclidean speed. For the circular formula, $\mathrm{Im}{R}_L(z)=r^2\mathrm{Im}z/|z-a{|}^2$, while its local Euclidean length multiplier is $r^2/|z-a{|}^2$. Thus both preserve $|dz|/\mathrm{Im}z$, the hyperbolic metric, and consequently all curve lengths and distances. Each is an involution preserving the half-plane; each fixes its specified line pointwise and moves points off that line. This establishes the required hyperbolic reflection explicitly.

To factor an arbitrary isometry $g$, first recall why a bisector reflection exchanges any distinct points $P,Q$. The hyperbolic distance formula isEquating the distances to $P$ and $Q$ yields a vertical line or a circle with centre on the real axis, hence a hyperbolic line. Substitution in the reflection formulas above shows that its reflection exchanges $P$ and $Q$. Choose a base point $P$. If $gP\ne P$, compose $g$ on the left with their bisector reflection, obtaining $h$ that fixes $P$; otherwise take $h=g$.

Choose $Q\ne P$. The points $hQ,Q$ have the same distance from $P$, so their bisector contains $P$. If they differ, its reflection corrects $hQ$ while retaining $P$. We obtain an isometry $k$ fixing both $P,Q$ after at most two reflections.

An isometry fixing two points is either the identity or reflection in their joining line. To see this directly, a point with prescribed distances from $P,Q$ lies at the intersection of two hyperbolic circles; these are Euclidean circles in this model and have at most two intersections. When there are two, reflection in the line $PQ$ interchanges them. Pick a third point off that line. Either $k$ fixes it, or composing $k$ with that line reflection fixes it. An isometry fixing three noncollinear points is the identity: the first two distances leave only the reflected pair, and the distance to the third point distinguishes that pair. Thus $k$ is the identity or one additional reflection. Reversing the compositions proves every isometry is a product of at most three hyperbolic reflections.

A norm $p$ is nonnegative with $p(x)=0$ exactly when $x=0$, satisfies $p(ax)=|a|p(x)$, and obeys $p(x+y)\le p(x)+p(y)$. The Euclidean norm has positivity and absolute homogeneity immediately. For the triangle inequality, positivity of $\Vert x-ty{\Vert }_2^2$ for all real $t$ gives the Cauchy-Schwarz inequality $|x\cdot y|\le \Vert x{\Vert }_2\Vert y{\Vert }_2$ (with $y=0$ immediate). Hencewhich proves the triangle inequality after taking square roots.

For the prescribed set, use the Minkowski functionalSince $U$ is open and contains zero, some Euclidean ball of radius $r>0$ lies inside it. Since it is bounded, it lies inside a ball of radius $R$. These inclusions imply $\Vert x{\Vert }_2/R\le p_U(x)\le \Vert x{\Vert }_2/r$, so the functional is finite and positive for nonzero $x$; it is zero at zero. Symmetry gives $p_U(-x)=p_U(x)$, and substitution in the infimum gives $p_U(ax)=a{p}_U(x)$ for $a>0$, hence absolute homogeneity for all real scalars.

Convexity and $0\in U$ imply $sU\subseteq U$ for $0\le s\le 1$. Consequently, for $a>p_U(x)$ and $b>p_U(y)$, one has $x/a,y/b\in U$. Convexity then givesso $p_U(x+y)\le a+b$. Letting $a\downarrow p_U(x)$ and $b\downarrow p_U(y)$ proves the triangle inequality.

Finally, if $p_U(x)<1$, there is a $t<1$ with $x\in tU\subseteq U$. Conversely, if $x\in U$ and $x\ne 0$, openness permits $(1+\delta )x\in U$ for some $\delta >0$, so $p_U(x)\le 1/(1+\delta )<1$; zero is immediate. Thus $\boxed{U=\{x:p_U(x)<1\}}$, completing the norm from a bounded symmetric convex neighbourhood construction.

Let $F(z)=p(z)/q(z)$. Its only possible finite poles are the distinct nonreal roots ${\alpha }_j$, and the residue of $F(z)e^{iz}$ there is $p({\alpha }_j)e^{i{\alpha }_j}/q^{\prime }({\alpha }_j)$; this is zero if the numerator cancels that pole. Close the real segment by a positively oriented semicircle in the upper half-plane, where $|e^{iz}|=e^{-\mathrm{Im}z}$ decays.

For sufficiently large radius $R$, all roots lie strictly inside or outside the contour as appropriate, and the degree bound gives $|F(z)|\le C/R$ uniformly on the arc. Parametrizing by $z=R{e}^{i\theta }$ bounds the arc integral byHere $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$, and symmetry handles the other half. The residue theorem therefore givesTo justify that this is the ordinary improper integral rather than just its symmetric principal value, note $F(x)=O(|x{|}^{-1})$ and $F^{\prime }(x)=O(|x{|}^{-2})$. On either tail, integration by parts writes the integral as a vanishing endpoint term $F(x)e^{ix}/i$ minus an absolutely convergent integral involving $F^{\prime }$. Thus both one-sided limits exist. When the degree bound is stricter the original integral is already absolutely convergent. This proves the Fourier integral of a rational function with simple nonreal poles formula with all tail and arc limits accounted for.

The fixed-end wave equation has spatial modes $\sin (n\pi x)$ and temporal factors $\cos (n\pi t),\sin (n\pi t)$. The zero initial velocity removes every temporal sine term. Expand the initial displacement in a Fourier sine series, whose coefficients areTwo integrations by parts give the last equality; the endpoint terms from $x(1-x)$ vanish and the remaining derivative endpoints distinguish even from odd $n$. ConsequentlyThis solves the initial-boundary problem in the finite-energy sense, with convergent first-derivative series. At $t=0$ the energy in the question is ${\int }_0^1(1-2x{)}^{2}dx=1/3$. At any time, sine/cosine orthogonality gives the contribution of mode $n$ as $(n\pi {)}^2{b}_n^2/2$: its kinetic and elastic pieces have factors ${\sin }^2(n\pi t)$ and ${\cos }^2(n\pi t)$ summing to one. ThereforeThe square summability of the derivative coefficients justifies the orthogonality calculation and its limit.

Using $E=-\nabla \varphi$ and the electrostatic Poisson equation, $\rho =-{ϵ}_0{\nabla }^2\varphi$, integration by parts givesThe boundary potential is zero, so the surface term vanishes. Hence $\boxed{W=({ϵ}_0/2){\int }_D{E}^{2}dV}$.

For the capacitor use the usual parallel-plate approximation, neglecting edge fields. The two gaps have widths $H+a$ and $H-a$, and their field magnitudes are $V/(H+a)$ and $V/(H-a)$. The field-energy expression givesThe two faces of the middle plate carry charges $Q_{-}={ϵ}_{0}AV/(H+a)$ and $Q_{+}={ϵ}_{0}AV/(H-a)$. Its total charge is their sum; the grounded plates have zero potential and contribute zero to $\frac{1}{2}\sum Q_j{\varphi }_j$. The charge-potential expression therefore gives $W=\frac{1}{2}V(Q_{-}+Q_{+})$, exactly the same result. Since $H^2-a^2$ is largest at $a=0$, the energy at fixed $V$ is minimized when the middle plate is centered, with value ${ϵ}_{0}A{V}^2/H$.

This is the electrostatic energy of a three-plate capacitor in the intended uniform-field model. Finite circular plates have fringing corrections; no small-gap-to-radius ratio is explicitly stated in the source, so the simple formula should not be described as an exact finite-disc solution.

The angular momentum commutation relations are $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$. The commutator product rule givessince the bracketed expression is symmetric in $j,k$ and ${ϵ}_{ijk}$ is antisymmetric. Alsoproving $\boxed{L^2=L_{-}{L}_{+}+L_3^2+\hslash L_3}$.

For orbital angular momentum, $L_i=-i\hslash {ϵ}_{ijk}{x}_j{\partial }_k$. A differentiable radial function has ${\partial }_{k}f(r)=f^{\prime }(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+i{x}_2$. The coordinate differential operators give $L_{3}s=\hslash s$, whileannihilates every power of $s$, because $({\partial }_1+i{\partial }_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yieldsSubstitution into the expression for $L^2$ gives the highest-weight complex-coordinate orbital wavefunction resultAs $[L^2,L_{-}]=0$, lowering preserves this eigenvalue whenever the lowered function is nonzero. Explicitly,For $n=0$ the lowered function is zero, so it is not an eigenfunction. For $n\ge 0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne 0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_{+}$.

In the inviscid incompressible irrotational model, write the total velocity as $(U+{\varphi }_x,{\varphi }_y)$. The perturbation potential obeys Laplace's equation in $-h<y<\eta (x,t)$. The complete boundary conditions, neglecting surface tension and taking the air pressure constant, areThe first surface condition expresses that the free surface is material; the second is the Bernoulli equation with atmospheric pressure and the constant background kinetic energy absorbed into the potential's time gauge.

At first order, evaluation is on $y=0$ and products of perturbations vanish. HenceFor a mode $e^{i(\omega t-kx)}$, the bottom condition selects $\varphi =C\cosh [k(y+h)]e^{i(\omega t-kx)}$. Let $\Omega =\omega -kU$. The linear surface conditions areEliminating $C$ gives the dispersion relationIt is the intrinsic gravity-wave frequency squared, with the uniform current supplying the Doppler shift. For either sign of real $k$, $k\tanh (kh)$ is nonnegative.

The Rao-Blackwell theorem states that if an estimator $D$ has finite second moment and $T$ is a sufficient statistic, then $D^{\ast }(T)=\mathbb{E}[D\mid T]$ can be chosen without knowing the parameter. It has the same expectation as $D$ and no larger variance. In particular an unbiased estimator remains unbiased and its squared-error risk cannot increase. Sufficiency makes the conditional distribution, hence the function of $T$ used here, parameter-independent.

The tower property gives $\mathbb{E}{D}^{\ast }=\mathbb{E}D$. The law of total variance givesso the variance decreases, with equality exactly when $D$ is already a function of $T$ almost surely. Since the bias is unchanged, the same identity proves the squared-error risk claim. More generally the corresponding convex-loss inequality follows from conditional Jensen's inequality.

For the given geometric distribution, the joint mass is $(1-p{)}^n{p}^{\sum x_j-n}$ on positive-integer sample vectors. Thus $T={\sum }_j{X}_j$ is a one-dimensional sufficient statistic. Directly, conditional on $T=t$, every positive-integer composition of $t$ into $n$ parts has the same probability; there are $\left(\genfrac{}{}{0ex}{}{t-1}{n-1}\right)$ such compositions, independently of $p$.

The simple estimator $D=1_{\{X_1>1\}}$ is unbiased for $p$, since $\mathrm{Pr}(X_1>1)=p$. For $n\ge 2$, compositions with $X_1>1$ are counted by subtracting one from their first part, giving $\left(\genfrac{}{}{0ex}{}{t-2}{n-1}\right)$ when $t>n$. Consequently the Rao-Blackwell estimator of a geometric failure probability isAt $T=n$ its value is zero, as it must be when all observations are one. For the possible one-observation case, use $\hat{p}=1_{\{T>1\}}$; the displayed fraction would otherwise be undefined at $T=1$. The conditional-expectation argument proves unbiasedness without needing to sum the negative-binomial mass explicitly.

The directed positive-transition graph gives the communicating classes $\{1,3,6\}$, $\{2,4\}$ and $\{5\}$. The first and last classes are closed; $\{2,4\}$ is open. In particular the PDF's fourth row has all six entries equal to $1/6$. The converted TeX incorrectly drops the last entry; that incomplete row is not a transition distribution.

The finite irreducible closed class $C=\{1,3,6\}$ reaches state six almost surely. For example, from every state in $C$ there is a uniformly positive probability of hitting six within a bounded number of steps, so the probability of avoiding it for successive blocks tends to zero. Put $h_i={\mathrm{Pr}}_i(T_6<\infty )$, with $h_1=h_3=h_6=1$ and $h_5=0$. First-step conditioning at states two and four givesSolving $4h_2-h_4=2$ and $5h_4-h_2=3$ yields $\boxed{h_2=13/19}$, with $h_4=14/19$.