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For $V_1={\partial }_x$, one has $\xi =1$, $\eta =0$, and hence $D_x\xi =0$. The recursion from part a gives ${\eta }_j=0$ for every $j$.

For $V_2=x{\partial }_x$, one has $D_x\xi =1$. Starting from ${\eta }_0=0$, induction gives

For $V_3=x^2{\partial }_x$, one has $D_x\xi =2x$. The first coefficients are ${\eta }_1=-2x{u}_1$ and ${\eta }_2=-2u_1-4x{u}_2$, and induction givesThus the three prolongations of the projective vector fields on the line are

Write $u_j=u^{(j)}$. The transformation ${\psi }_s:(x,u)\mapsto (\tilde{x},\tilde{u})$ induces a transformation of the $n$th jet space of a scalar ordinary differential equation by differentiating $\tilde{u}$ with respect to $\tilde{x}$ through order $n$. The infinitesimal generator of this induced action is the prolongation of a vector fieldTo determine its coefficients, put ${\eta }_0=\eta$ and use the total derivative operatorDifferentiating transformed derivatives byand then differentiating with respect to the group parameter at $s=0$ gives the recursionThis also proves the displayed form of the prolongation by mathematical induction on the derivative order.

Part d leaves only the region above $a=-1$ and to the right of $b=-1$ lying below the hyperbola $ab=1$. Part c removes the two opposite-sign quadrants, and its displayed divergence also removes either coordinate axis except at the origin. Hence periodic orbits might still occur only inIn the $a$-$b$ plane these are the portion of the first quadrant below $ab=1$, the open square $(-1,0)\times (-1,0)$, and the origin. The origin is not merely possible: part a gives its family of periodic orbits.

An interior equilibrium point of a dynamical system must satisfyIf $ab\ne 1$, the unique candidate isIt belongs to $\Lambda$ precisely in either of the two casesAt this equilibrium the Jacobian matrix has determinantIn the first region the equilibrium has Poincaré index $+1$, so index considerations alone do not rule out a periodic orbit. In the second region it is a saddle equilibrium of index $-1$. Everywhere else there is no interior equilibrium, except at $a=b=-1$, where the segment $x+y=1$ consists of equilibria.

Every simple closed curve in the convex set $\Lambda$ bounds a region contained in $\Lambda$. A periodic orbit has index $+1$, equal to the sum of the indices of the equilibria that it encloses. Thus a region with no equilibrium has total index zero, while the region with the single saddle has total index $-1$; both contradict the index of a planar periodic orbit. At $a=b=-1$, a periodic orbit cannot cross the equilibrium segment and must lie in one of its two complementary convex regions, where it encloses no equilibrium, giving the same contradiction. Therefore index considerations exclude periodic orbits except possibly when

Use the Dulac theorem on the simply connected positive quadrant with $\varphi (x,y)=1/(xy)$. If $F=(\dot{x},\dot{y})$, thenand henceFor $ab<0$, both terms have the same strict sign throughout $\Lambda$. The Bendixson-Dulac criterion therefore gives no periodic orbit in $\Lambda$. The same computation also excludes the case $ab=0$ unless $a=b=0$.

Keep the same function $V$. For arbitrary $a,b$ its orbital derivative isLet $\gamma$ be a nonconstant orbit of the unperturbed system with period $T$. Sinceintegration over one period givesConsequently,If a perturbed periodic orbit persisted and converged to $\gamma$ as $(a,b)\to (0,0)$, its net change in $V$ over one period would vanish. The averaged first-integral obstruction to persistence of a periodic orbit would therefore requireWhen $ab<0$, the two leading terms have the same nonzero sign and their magnitude is bounded below by $r\min (A,B)(|a|+|b|)$. Thus no unperturbed periodic orbit persists for sufficiently small $a,b$ with $ab<0$.

For $a=b=0$, take the Logarithmic first integral of the Lotka-Volterra equationsAlong a trajectory,Thus $V$ is a first integral. The scalar function $h(s)=s-\log s-1$ is nonnegative for $s>0$, vanishes only at $s=1$, and tends to infinity as $s\downarrow 0$ or $s\to \infty$. Moreover,is positive definite, so $V$ is strictly convex. It follows that $(1,1)$ is its unique critical point and every level set $V=c>0$ is a compact smooth simple closed curve in the positive quadrant. The vector field is nonzero on such a level curve, so uniqueness of solutions makes each connected level curve one periodic orbit. The zero level consists of the fixed point $(1,1)$. Hence every trajectory in $\Lambda$ is either periodic or the fixed point $(1,1)$.

PutThe two endpoints $t=1$ and $t=-1$ both contribute. Near $t=1$, set $u=1-t$. ThenExpanding the last amplitude in powers of $u$ and applying the Complex Watson lemma after rotating onto a decaying endpoint ray gives one exponential series. Repeating at $t=-1$ gives the other. After multiplication by the prefactor in the integral representation, the two endpoint contributions combine aswhereSeparating the even and odd powers gives the large-argument asymptotic expansion of the Bessel function of the first kind:Its first terms areFor fixed $\nu$, the expansion is uniform in every closed sectorwith $\delta >0$. The exclusion of the negative real axis fixes the branch of $z^{1/2}$ and stays away from its Stokes boundary.

Watson lemma states that ifand the Laplace integral has suitable growth control away from zero, then

The Taylor series of the amplitude isTermwise application of Watson lemma therefore gives the asymptotic expansionIn particular, the leading asymptotic approximation is $I(x)\sim 2x^{-3}$.

The digital call option and digital put option indicators partition the possible terminal stock prices:Thus the digital put-call parity isAt time zero, part b gives $D_{\mathrm{call}}(0,S_0)=e^{-rT}\Phi (d_{-})$, so

For $0\le t<T$, set $\tau =T-t$ andUnder the equivalent martingale measure, conditional log-normality givesThe risk-neutral pricing value of the digital call option is consequentlyAt $t=T$ this converges to the stated payoff away from $S_T=K$, with the payoff convention specifying the boundary value.

The delta hedge holds the derivative of the claim value with respect to the current stock price. Sincethe number of risky-asset units for $t<T$ isThis is the Black-Scholes digital option formula. The hedge becomes singular close to maturity near the strike, reflecting the discontinuity of the payoff.

PutThe call payoff is $(S_T-K{)}^{+}$, and it is positive whenWriting $\varphi$ for the standard normal density, the given risk-neutral pricing formula becomesCompleting the square givesso the two tail integrals are $\Phi (d_{+})$ and $\Phi (d_{-})$, where $\Phi$ is the standard normal distribution function. Therefore the Black-Scholes formula is

The payoff identityreplicates the call-minus-put position by one stock and borrowing the present value of $K$. Hence put-call parity givesand therefore

The strong law of large numbers gives $S_n/n=\overline{X}\to 1/\theta$ almost surely, and hence in probability. Thereforeby the continuous mapping theorem. Thus the Bayes estimator has consistency.

It differs from the maximum-likelihood estimator bySince $S_n/n\to 1/\theta$, the numerator is $O_p(n)$ and the denominator is $O_p(n^2)$, so this difference is $O_p(n^{-1})=o_p(n^{-1/2})$. The limit in part a and Slutsky theorem now give

Multiplying the likelihood by the gamma distribution prior density givesThus Gamma-exponential conjugacy giveswhere the second parameter is the rate. Under squared-error loss, the Bayes estimator under squared error loss is the posterior mean, hence

Let $S_n={\sum }_{i=1}^n{X}_i$. The log-likelihood iswith score function $n/\theta -S_n$ and second derivative $-n/{\theta }^2<0$. The unique maximum-likelihood estimator is thereforeFor an exponential distribution of rate $\theta$,The central limit theorem givesApplying the delta method to $h(x)=1/x$, for which $h^{\prime }(1/\theta )=-{\theta }^2$, yields the exponential-rate maximum-likelihood estimator limit

Put $K=N+s$, the total capacity. The customer count is an M-M-s-K queue, hence a birth-death process on $\{0,1,\dots ,K\}$. Its nonzero transition rates areThe first rate is zero at $K$ because an arrival finding the shop full is rejected. The second rate is the number of busy servers times the individual service rate.

Write $\rho =\lambda /\mu$. The detailed balance for a birth-death process recursion isThereforewhereThis is the stationary distribution of an M-M-s-K queue.

The state space is finite, and every state communicates with every other through successive births and deaths because $\lambda ,\mu >0$. The chain is therefore an irreducible Markov chain, so its invariant distribution is unique. Finally the adjacent-state identitiesare detailed balance, while all nonadjacent rates vanish in both directions. Part a now shows that the queue is reversible in equilibrium.

A continuous-time Markov chain is a reversible Markov chain in equilibrium if, when started in a stationary distribution $\pi$, the process $(X_t:0\le t\le T)$ has the same finite-dimensional distributions as $(X_{T-t}:0\le t\le T)$ for every $T>0$.

For transition rates $q_{ij}$, the detailed balance for a continuous-time Markov chain equations areThey imply invariance directly. For each state $j$,because $q_{jj}=-{\sum }_{k\ne j}{q}_{jk}$. Thus $\pi Q=0$, so $\pi$ is an invariant distribution of a continuous-time Markov chain.

The displayed density is that of a gamma distribution with shape $k+1$ and rate one. PutBy the additivity of independent gamma distributions with a common rate,Equivalently, the independence of random variables makes the Laplace transform of a nonnegative random variable of the sum equalTherefore the required probability density function is

SetThese are bounded random variables with values in $(0,1]$, and for every positive integer $n$,Part c shows that $U$ and $V$ have the same law. Since $x\mapsto e^{-x}$ is injective on $[0,\infty )$ with inverse $u\mapsto -\log u$, taking the pushforward measure by this inverse shows that $X$ and $Y$ have the same law. This proves the uniqueness theorem for Laplace transforms of nonnegative random variables.

Choose $M<\infty$ such that $|X|,|Y|\le M$ almost surely. Equality of the positive moments, together with equality of the zeroth moments $\mathbb{E}[1]=1$, givesfor every polynomial $p$. By the Weierstrass approximation theorem, every continuous function on $[-M,M]$ is a uniform limit of polynomials. Expectations preserve this limit, so $X$ and $Y$ have equal expectations for every continuous function on their common compact support. The moment determinacy on a compact interval therefore gives equality of their laws.