Yes, it is a uniformly most powerful test for the one-sided alternative. For every fixed ${\sigma }_1^2>{\sigma }_0^2$, the likelihood ratio is increasing in the same statistic $S$. The size-$\alpha$ threshold found in part (c) depends only on the null variance and $\alpha$, not on the selected alternative variance. Thus one and the same test is most powerful against each simple member of the composite alternative, which is exactly uniform most powerfulness.

For $S={\sum }_i{x}_i^2$, the likelihood ratio in favor of the larger variance isBecause ${\sigma }_1>{\sigma }_0$, this is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper-tail rejection region. Under the null, $S/{\sigma }_0^2\sim {\chi }_n^2$, so size $\alpha$ is obtained withContinuity of the chi-squared distribution makes the boundary have probability zero. This is a fixed-mean normal variance test.

The Neyman-Pearson lemma compares two simple hypotheses with densities $f_0,f_1$. Among all tests of size at most $\alpha$, a most powerful test rejects when $f_1(x)/f_0(x)>c$, accepts when it is below $c$, and may randomize on equality to attain the prescribed size. The threshold and any boundary randomization are chosen so the rejection probability under $H_0$ is $\alpha$.

The key inequality is $({\varphi }^{\ast }-\varphi )(f_1-c{f}_0)\ge 0$ pointwise for the likelihood-ratio test ${\varphi }^{\ast }$. Integration shows its power is at least that of any competing test with no larger size. With continuous likelihood-ratio statistics, boundary randomization is unnecessary.

The standardized observations $X_i/\sigma$ are independent standard normal random variables. By the definition of the chi-squared distribution,Its distribution is continuous, so $F_n(c_2)-F_n(c_1)=1-\alpha$ gives the required probability. Put $S={\sum }_i{X}_i^2$. Solving $c_1<S/{\sigma }^2<c_2$ for the positive variance reverses the reciprocal order. The exact confidence interval isThe mean is known to be zero, so the degrees of freedom are $n$, not $n-1$. Coverage is exactly $1-\alpha$ under the stated normal model.

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ on the set where this expectation is finite. For independent, identically distributed summands, independence factors the expectation:The identity holds wherever the factors are finite; in the remaining parts the bounded summands make them finite for every real $t$.

Write a first return time as $2J$, where $\mathbb{P}(J=j)=2^{-j}$ for $j\ge 1$. It has mean $2\cdot 2=4$ and period two, so the stationary probability at one is $1/4$. The periodic convergence theorem would give limits $1/2$ along even times and zero along odd times.

Here one can obtain them directly. Put $u_n={\mathbb{P}}_1(X_{2n}=1)$, with $u_0=1$. Splitting according to the first return gives the renewal equationInduction gives $u_n=1/2$ for every $n\ge 1$: the right side is $2^{-n}+\frac{1}{2}{\sum }_{j=1}^{n-1}{2}^{-j}=1/2$. Odd-time returns are impossible because every return interval is even. Therefore the exact set of limit points isThe sequence does not converge to the stationary mass $1/4$, precisely because it is periodic.

The stationary balance equations on the closed class $\{1,\dots ,M\}$ are ${\pi }_j={\pi }_{j+1}+{\pi }_1/M$, with ${\pi }_{M+1}=0$. Working backward gives ${\pi }_j={\pi }_1(M-j+1)/M$.

Normalization yields $1={\pi }_1(M+1)/2$, soThese probabilities sum to one and satisfy each balance equation. In particular ${\pi }_1=2/(M+1)$ is the reciprocal of the mean return time $(M+1)/2$, consistent with part (d).

The first return time ${\tau }_1$ has distribution ${\mathbb{P}}_1({\tau }_1=k)={\alpha }_k$. It is finite almost surely, so state one is recurrent. It is positive recurrent precisely when its mean return time is finite:By irreducibility the same recurrence class contains every state. Thus the condition is finiteness of the expected value of the positive-integer distribution $({\alpha }_k)$. If that mean is infinite, the chain is null recurrent, not transient.

After a jump from one to $k$, the first return to one occurs exactly $k$ steps later, with probability ${\alpha }_k$. Later return times are sums of such positive-support values. Thus their greatest common divisor isIn an irreducible chain this is the common period of every state. Hence the chain is aperiodic exactly when $d=1$. A self-loop at one is sufficient but not necessary; support at two and three also has greatest common divisor one.

Every state reaches state one by deterministic downward steps. From one, the chain can reach a given state $j$ exactly when it can jump to some $k\ge j$ with ${\alpha }_k>0$ and then count down to $j$. ThereforeIf the support is bounded by $M$, states above $M$ cannot be reached from one, so the chain on the stated infinite state space is not irreducible. Its smaller closed class can still have a stationary distribution, as in part (e).

A Markov chain is irreducible if every state communicates with every other: for any $i,j$ there is an integer $n\ge 0$ with $p_{ij}^{(n)}>0$. Thus any state can be reached from any other in finitely many steps with positive probability.

The Gaussian integral gives ${\int }_{{\mathbb{R}}^3}{e}^{-\gamma |y{|}^2}{d}^{3}y=(\pi /\gamma {)}^{3/2}$. Thus the normalized ground state is, up to an overall phase,Its probability density is even in each shifted coordinate, so all three expectations $⟨y_i⟩$ vanish by oddness. ThereforeThe mean position is the minimum of the displaced potential, rather than the original coordinate origin.

Put $\gamma =m\omega /\hslash$. Substitution of $\psi =F(y)e^{-\gamma y^2/2}$ in the one-dimensional Schrödinger equation with energy $(n+1/2)\hslash \omega$ givesFor degrees zero, one and two, coefficient comparison yields, up to independent nonzero constants,For the shifted coordinates $y_i=x_i-(r_0{)}_i$, let $G(y)=e^{-\gamma |y{|}^2/2}$. A basis of wave functions for the first three energy levels is:

These are products of the one-dimensional polynomial solutions. Arbitrary linear combinations within each listed eigenspace are also stationary states. Their counts $1,3,6$ agree with part (a), so the degeneracies have all been included.

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

For the surface of revolution $r(\eta ,\zeta )=(\eta ,f(\eta )\cos \zeta ,f(\eta )\sin \zeta )$, the coordinate tangent vectors have squared lengths $1+f^{\prime 2}$ and $f^2$, and their dot product is zero. Hence its first fundamental form isUse the energy Lagrangian $L=A{\dot{\eta }}^2+f^2{\dot{\zeta }}^2$. Its Euler-Lagrange equations areFor a meridian $\zeta ={\zeta }_0$, choose arc length as parameter, so $\dot{\eta }=1/\sqrt{A}$. Differentiating gives $2A\ddot{\eta }+A^{\prime }{\dot{\eta }}^2=0$, while the second equation is automatic. Part (b) makes this path an extremal of length. Thus every meridian is a geodesic.

For a nonconstant parallel $\eta ={\eta }_0$, choose constant nonzero $\dot{\zeta }$. The second equation holds, and the first reduces to $f({\eta }_0)f^{\prime }({\eta }_0){\dot{\zeta }}^2=0$. Since $f>0$, this is equivalent toConversely that condition makes both equations hold at constant speed, hence supplies a length extremal. Geometrically, a parallel is geodesic exactly at a stationary radius of the surface.

The autonomous Lagrangian has the Beltrami identityHere $L$ is homogeneous of degree two in the velocities, so $\dot{u}{L}_{\dot{u}}+\dot{v}{L}_{\dot{v}}=2L$. Thus $L$ itself is constant, say $C\ge 0$.

For $C>0$, put $\ell =\sqrt{L}$. Its Euler-Lagrange expression along the curve isThe first term vanishes by stationarity for $L$, and the second by constancy of $L$. Thus the same path is stationary for its length functional $\int \sqrt{L}dt$.

On a regular segment described as a graph $v=v(u)$, reparametrize by $u$. Since $\dot{v}=v_u\dot{u}$,Choose the orientation with increasing $u$; reversing it does not change stationarity. Invariance of length under regular reparametrization therefore givesThis is the principle of length extremals from a constant quadratic energy. The stated nonnegativity also permits the degenerate case $C=0$. Then the length is zero, its absolute minimum since the integrand is nonnegative, so it is an extremal in the minimizing sense. The differentiated square-root Euler-Lagrange equation above is only used where $L>0$; it would not be legitimate to divide by $\sqrt{L}$ on a null curve. For a genuine surface metric and a regular nonconstant path, $C>0$ automatically.

For fixed endpoint variations, the Euler-Lagrange equations areIndeed the first variation is $\int {\sum }_i(L_{q_i}\delta q_i+L_{{\dot{q}}_i}\delta {\dot{q}}_i)dt$. Integration by parts removes the derivatives from the variations; their zero endpoint values remove the boundary term. The fundamental lemma then makes every remaining coefficient vanish.