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Let $ϵ$ be the group parameter. The infinitesimal generator of a Lie point symmetryhas flow equationsTherefore it generates the scaling group

Here ${\xi }^t=t$, ${\xi }^x=x$, and $\eta =0$. The total-derivative formulas for the second prolongation of a Lie point symmetry giveandThusFor the wave equation expression $F=u_{tt}-u_{xx}$,It therefore vanishes whenever $F=0$, proving that $V$ generates the simultaneous spacetime scaling symmetry of the wave equation.

The independent invariant of the scaling orbits is the similarity variablewhile $u$ itself is invariant. Seek a group-invariant solution $u(t,x)=f(z)$. Direct partial differentiation givesThe wave equation reduces toHence, on any interval avoiding $z=\pm 1$,

For $E\to \infty$, the outer part of the action integral for a linear-to-square-root potential dominates:The WKB quantization condition therefore givesTaking the power $2/5$ yields the asymptotic equivalenceHence

For fixed $n$, the right-hand side of the quantization equation tends to zero with $\epsilon$. Ifthe unique energy satisfies $E_n<1$, so the first branch of the action integral for a linear-to-square-root potential applies exactly. SolvinggivesThus this formula is valid whenever

By the evenness of $V$, the semiclassical action integral is twice its integral over the positive half-axis. If $0\le E\le 1$, then

If $E\ge 1$, split the integral at $x=4$:The first integral isIn the second, put $r=\sqrt{x}$ and then $u=E+1-r$. It becomesAfter simplification,This is the action integral for a linear-to-square-root potential.

The graph begins at the origin, is continuous and strictly increasing, passes through $(1,16/3)$, and tends to infinity. Therefore for every $n\ge 0$ and $\epsilon >0$, the equationhas exactly one solution $E=E_n$ by the intermediate value theorem and strict monotonicity.

The endpoints are the two classical turning points, determined by $V(x)=E$. Since the potential is even, they are symmetric. For $0\le E\le 1$, they lie in the linear part of the potential and satisfy $|x|/4=E$; for $E\ge 1$, they lie in the outer part and satisfy $\sqrt{|x|}-1=E$. HenceThe two-turning-point WKB quantization condition uses

Statement (iv) is true. Apply the Weierstrass approximation theorem separately to the real and imaginary parts of the continuous function $f$ on the compact interval $[0,1]$. Combining the two real polynomial sequences gives complex polynomials converging uniformly to $f$ on $I$.

Statement (iii) is false. Let $\Omega$ be a thin open annulus about the unit circle, let $f(z)=1/z$, and split the circle into two proper closed arcs $K_1,K_2$ whose union is the whole circle. Each arc has connected complement, so polynomial approximation of the reciprocal on a proper circular arc gives uniform polynomial approximations to $f$ on each $K_j$ separately.

If polynomials $P_n$ converged uniformly to $1/z$ on $K_1\cup K_2$, then uniform convergence on the unit circle and the Cauchy integral theorem would givea contradiction.

Statement (ii) is true. LetEach $K_n$ is compact and has connected complement. By the polynomial Runge theorem, choose a polynomial $P_n$ withEvery fixed $z\in E$ belongs to $K_n$ for all sufficiently large $n$, so $P_n(z)\to f(z)$. Thus the convergence is pointwise on $E$.

The polynomial form of Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ such that

Statement (i) is false. Take the entire function $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged uniformly to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty )$, since its values at zero and near infinity differ by one.

By continuity of Brownian motion, the event that the last strict crossing of $b$ occurs by time $t$ isApply part (b) with $a$ and $b$ interchanged and its time parameter replaced by $1/t$. This turns the right-hand side intoPart (c), now with drift $b$, barrier $a$, and horizon $1/t$, givesThis is the distribution function of the last passage time above a level for Brownian motion with negative drift. Taking $t\downarrow 0$ givesthe probability that the negatively drifted Brownian path never exceeds $b$, consistent with the infinite-horizon crossing probability for Brownian motion with negative drift.

Let $X_s=W_s-as$ and $M_t={\sup }_{0\le s\le t}{X}_s$. Split the crossing event according to its endpoint:Under the Cameron-Martin theorem for a linear drift, the law of $X$ relative to standard Wiener measure has endpoint densityOn paths that cross $b$ and end at $x\le b$, the Brownian reflection principle reflects the path after its first hit of $b$ and changes the endpoint to $2b-x\ge b$. Under this reflection the density becomesThe remaining exponential is the Cameron--Martin density for drift $+a$. Thereforewhere the last equality uses the symmetry of the normal distribution. Subtracting the crossing probability from one givesEquivalently, in terms of the standard normal distribution function,This is the finite-horizon maximum of Brownian motion with negative drift formula.

Use the time-inverted Brownian motion $B_u=u{W}_{1/u}$. For $s>0$, set $u=1/s$. ThenAs $0<s\le t$ corresponds exactly to $u\ge 1/t$, and the inequality at $s=0$ is automatic because $b>0$, this is the pathwise identitySince $B$ and $W$ have the same law by time inversion of Brownian motion,

For every finite collection of positive times, $(B_{t_1},\dots ,B_{t_m})$ is a linear transformation of the corresponding values of $W$, so $B$ is a centered Gaussian process. If $0<s\le t$, thenThus $B$ has the covariance of Brownian motion.

Its paths are continuous for $t>0$. As $t\downarrow 0$, put $u=1/t$. The given almost-sure limit givesThe Gaussian-process characterization of Brownian motion now proves the time inversion of Brownian motion:

For $0<\pi <1$, let ${\delta }_{\pi }$ be the Bayes classifier with class-zero prior $\pi$, and define its two class-conditional error probabilities byWriting the likelihood ratio as $L=f_1/f_0$, the rule chooses class one whenThe Gaussian likelihood-ratio level sets have probability zero, so dominated convergence shows that $e_0$ and $e_1$ depend continuously on $\pi$. As $\pi \downarrow 0$, the threshold tends to zero and the classifier chooses class one almost surely, givingAs $\pi \uparrow 1$, it chooses class zero almost surely, giving the opposite limit $(0,1)$. The intermediate value theorem therefore provides ${\pi }^{\ast }\in (0,1)$ such that

The rule ${\delta }_{{\pi }^{\ast }}$ is an equalizer rule with worst-case risk $r$. For any classifier $\delta$,because ${\delta }_{{\pi }^{\ast }}$ minimizes the integrated risk for its prior. Hence ${\delta }_{{\pi }^{\ast }}$ is a minimax decision rule, as summarized by the minimax Gaussian Bayes classifier from equal class errors.

The prior is indeed least favorable. Its Bayes risk is $r$, while for any other prior $q$ the Bayes risk is at most the integrated risk of the same equalizer rule ${\delta }_{{\pi }^{\ast }}$, namelyThus no prior has larger Bayes risk.

Condition on the observed value $X=x$. If a classifier reports class zero, its conditional probability of error is $\mathbb{P}(Y=1\mid X=x)$; if it reports class one, the conditional error is $\mathbb{P}(Y=0\mid X=x)$. Therefore the smaller conditional error is attained by class one exactly whenand by class zero exactly when the reverse strict inequality holds. Integrating these pointwise conditional errors proves that ${\delta }_{{\pi }_0}$ is a Bayes classifier.

The only possible nonuniqueness lies on the tie setBecause the two Gaussian class distributions are distinct, $D_{{\pi }_0}$ is not identically zero. It is a nonzero polynomial of degree at most two, so its zero set has Lebesgue measure zero. Both Gaussian laws have densities with respect to Lebesgue measure, hence assign probability zero to $B$. Every Bayes rule must therefore agree with ${\delta }_{{\pi }_0}$ almost surely. This proves the uniqueness of a Bayes classifier up to the standard null-set equivalence of decision rules.

Let $f_i$ be the multivariate normal density of $N_p({\mu }_i,{\Sigma }_i)$. Under zero-one loss, the Bayes classifier chooses the class with the larger posterior probability. ThusEquivalently, it chooses class one when the log posterior oddsis nonnegative. The decision boundary is $D_{{\pi }_0}(x)=0$.

If ${\Sigma }_0={\Sigma }_1=\Sigma$, the $x^T{\Sigma }^{-1}x$ terms cancel, leavingan affine function. This is the linear boundary of linear discriminant analysis. If ${\Sigma }_0\ne {\Sigma }_1$, the quadratic part iswhich is nonzero, so the boundary is a possibly degenerate quadratic hypersurface, as in quadratic discriminant analysis. This is the Gaussian Bayes classifier.

For an M/M/1 queue, the first service lasts on average $1/\mu$. During it, the mean number of arrivals is $\lambda /\mu$, and each arrival initiates another statistically identical busy-period family. Thus, when $\lambda <\mu$,so the mean busy period of an M-M-1 queue isFor $\lambda =\mu$ the busy period is almost surely finite but has infinite mean, while for $\lambda >\mu$ it is infinite with positive probability; in either case its expectation is infinite.

For an M-M-infinity queue, stationary occupancy is Poisson with mean $\rho =\lambda /\mu$, so the long-run probability of an empty system isThe process alternates between idle periods and busy periods. An idle period waits for the next Poisson arrival, so its mean is $1/\lambda$. The renewal-reward theorem, with reward equal to time spent empty, givesSolving yields the mean busy period of an M-M-infinity queue:

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M/M/1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

Write the interval as $I=(c,d]$, where $0<c<d<1$, to avoid confusing its left endpoint with the rotation parameter $\alpha$. For every sufficiently large integer $k$, define an inner and outer interval on the circle byApply part (b) simultaneously to this countable collection of intervals. The intersection $G$ of the corresponding full-measure sets still has full Lebesgue measure, and is therefore dense in the circle.

Fix an arbitrary $x$. For each sufficiently large $k$, choose $y_k\in G$ with circle distance less than $1/k$ from $x$. An irrational circle rotation preserves this distance, so for every $j\ge 0$,Averaging and using part (b) for $y_k$ givesLetting $k\to \infty$ proves the everywhere interval frequency under an irrational rotation: