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Let $h(i)$ be the probability of reaching state $3$ before state $0$. The boundary values are $h(0)=0$ and $h(3)=1$. First-step analysis at states $1$ and $2$ givesTherefore $h(1)=h(1)/4+1/4$, so the requested hitting probability is

Let $g(0)=0$. First-step analysis givesand, because state $3$ moves to $2$ or remains at $3$ with equal probabilities,The last equation gives $g(3)=2+g(2)$. Substitution into the second gives $g(2)=4+g(1)$, and the first then gives $g(1)=3+g(1)/2$. HenceThese are the expected hitting times of the absorbing state.

Part (a) shows that $f-p_n^{\ast }$ is orthogonal to $p_n^{\ast }\in {\mathcal{P}}_n$. Sincethe Pythagorean theorem in an inner-product space immediately givesThis is the norm decomposition associated with orthogonal projection onto a finite-dimensional subspace.

The nonzero orthogonal polynomials $Q_0,\dots ,Q_n$ have distinct degrees and form an orthogonal basis of ${\mathcal{P}}_n$. DefineFor each $j\le n$,Thus the residual $f-p_n^{\ast }$ is orthogonal to all of ${\mathcal{P}}_n$.

For any $p\in {\mathcal{P}}_n$, writeThe two terms are orthogonal, so the Pythagorean theorem in an inner-product space givesEquality holds only for $p=p_n^{\ast }$. This proves the formula and uniqueness of the least-squares polynomial in an orthogonal-polynomial basis.

Here $N=2$ and $M=1$. At $(x,0,0)$, the $z$-components of the fields of the two outer charges cancel by reflection symmetry. By Coulomb's law, the remaining field isMultiplying by the test charge $-Q$ givesThe quantity in parentheses is positive. Thus for $x>0$ the force points in the negative $x$ direction, while for $x<0$ it points in the positive $x$ direction. It is always attractive toward the origin, as in the transverse force from a collinear electric quadrupole.

The monopole vanishes exactly whenand the dipole vanishes exactly whenSince $N$ and $M$ are positive integers, both vanish precisely for

Assume now that the monopole has been cancelled. If $Qd$ is held fixed while $d\to 0$, the quadrupole coefficient $Q{d}^2=(Qd)d$ tends to zero, whereas the dipole coefficient $(M-1)Qd$ has a finite limit. For $M\ne 1$ this is a point-dipole limit; for $M=1$ the limiting external potential vanishes.

If instead $Q{d}^2$ is held fixed, the dipole coefficient isIt diverges unless $M=1$. When $M=1$ and $N=2$, both lower multipoles vanish and the quadrupole coefficient has a finite nonzero limit. These are the dipole and quadrupole scaling limits of three collinear charges.

Let $\theta$ be the angle between $r$ and the positive $z$-axis. By superposition of the electric potential of point charges,For $r>d$, the Legendre polynomial expansion givesThereforeThe three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is $Q(N-M-1)$ and the electric dipole moment is $Qd(M-1)\hat{z}$. This is the multipole expansion of three collinear charges.

For a particle in the scalar potential energy $U(x)$,The canonical commutation relation $[x,p]=i\hslash$ givesThe Ehrenfest theorem with $O=x$ therefore givesSince $[p^2,p]=0$ and direct action on a wavefunction gives $[U(x),p]=i\hslash U^{\prime }(x)$,Finally $H$ has no explicit time dependence and $[H,H]=0$, so

The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that $⟨U^{\prime }(x)⟩\simeq U^{\prime }(⟨x⟩)$. These conclusions are summarized by the classical equations from Ehrenfest theorem.

Write the expectation value asDifferentiating and retaining the possible explicit time dependence of the observable givesThe Schrödinger equation and its adjoint arewhere the Hamiltonian operator is Hermitian. Substitution yieldsThis is the proof of Ehrenfest theorem from the Schrodinger equation.

The domains are conformally equivalent. The Möbius transformationmaps the upper half of the unit disc onto the first quadrant: its diameter $(-1,1)$ maps to the positive real axis and its upper semicircle maps to the positive imaginary axis. Squaring maps the first quadrant bijectively and conformally onto the upper half-plane. Thereforeis the required equivalence, as recorded by the conformal equivalence between the upper half-disc and upper half-plane.

The domains are not conformally equivalent. If $F:\mathbb{C}\to \mathbb{H}$ were a conformal equivalence, composing it with a Cayley transform from the upper half-plane $\mathbb{H}$ to the unit disc would produce a bounded nonconstant entire function. This contradicts the Liouville theorem, proving that the complex plane is not conformally equivalent to the upper half-plane.

The domains are not conformally equivalent. Suppose that $F:\mathbb{C}\setminus \{0\}\to \{w:0<|w|<1\}$ were a conformal equivalence. Since $F$ is bounded near zero, the Riemann removable singularity theorem extends it holomorphically across zero. The extension is a bounded entire function, so the Liouville theorem makes it constant, contradicting bijectivity. This is the punctured plane is not conformally equivalent to the punctured unit disc obstruction.

Both alternatives have larger $\theta$, so use upper-tail rejection regions. Let $z_{0.95}={\Phi }^{-1}(0.95)\simeq 1.645$ be the 95th quantile of the standard normal distribution.

For the likelihood estimator,When $S_1$ is large, the normal approximation to the Poisson distribution gives, under $H_0$,An approximate size-$0.05$ test therefore rejects $H_0$ when

When every $x_i$ is large, $Y_i\approx N(\theta x_i,\theta x_i)$ independently. A linear combination of independent normal variables is normal, so under $H_0$,The corresponding approximate size-$0.05$ test rejects whenIn both cases the null rejection probability is approximately $0.05$, while values near the alternative mean $2$ increasingly fall in the rejection region as the total exposure grows. These are the normal-approximation tests for a Poisson exposure model.

PutIndependence and $\mathrm{var}(Y_i)=\theta x_i$ givewhereas ${\sum }_i{Y}_i\sim \mathrm{Poisson}(\theta S_1)$ givesThe Cauchy-Schwarz inequality, applied to $(x_i^{1/2})$ and $(x_i^{3/2})$, yieldsConsequentlyEquality holds exactly when all positive exposures $x_i$ are equal. This is the variance comparison for Poisson exposure estimators.

Ignoring terms independent of $\theta$, the log-likelihood isIts score function isSince ${\ell }^{\prime \prime }(\theta )=-({\sum }_i{Y}_i)/{\theta }^2\le 0$, the interior critical point is the maximum, givingThe sum of independent Poisson distributions is Poisson with mean $\theta {\sum }_i{x}_i$, soThus the maximum-likelihood estimator is also unbiased.

Write the linear mean model asThe errors are independent but heteroscedastic. Minimizing the unweighted residual sum of squaresgives the normal equationThus, provided ${\sum }_i{x}_i^2>0$,Since $\mathbb{E}{Y}_i=\theta x_i$,Hence it is an unbiased estimator. This is the least-squares part of the estimators for a Poisson exposure model.