For a locally integrable periodic function, its Fourier series coefficients are $c_m={\int }_0^{1}f(x)e^{-2\pi imx}dx$. Window the function to one period by $w(x)=H(x)-H(x-1)$. With the stated Fourier transform convention,There is no extra factor of $2\pi$: the Fourier coefficient already integrates over a period of length one.

For a completely arbitrary periodic distribution, multiplication by a discontinuous window need not be defined, for example at a derivative of a delta supported on a period boundary. The sharp-window assertion therefore presupposes a windowing convention or enough regularity at the endpoints. The smooth-window Fourier coefficients of a periodic distribution use a smooth compactly supported partition-of-unity window $\chi$ with ${\sum }_{j\in \mathbb{Z}}\chi (x+j)=1$; then $c_m=\widehat{\chi f}(2\pi m)$ in the distribution sense. The logarithmic example below is locally integrable, so the printed sharp window is legitimate and no ambiguity affects it.

For $f(x)=\log |x-1/2|$, put $u=x-1/2$. Symmetry gives, for integer $m\ne 0$,An integration by parts has no endpoint term: $u\log u\to 0$ at zero and $\sin (\pi m)=0$ at the upper endpoint. It follows thatwhere the sine integral is $\mathrm{Si}(z)={\int }_0^z\sin t/tdt$. Its limiting value $\pi /2$ follows from the Abel-regularized calculation in part (ii); integration by parts of its tail gives $\mathrm{Si}(\pi m)=\pi /2-(-1{)}^m/(\pi m)+O(m^{-3})$ for positive integer $m$. Hence the Fourier coefficients of a periodic logarithmic singularity haveThe first term is the requested leading behaviour. Its alternating sign records the singularity at half a period, and its $m^{-1}$ decay records the logarithmic cusp. The $m^{-2}$ correction comes from the derivative mismatch across the periodically joined endpoints. Also $c_0=-1-\log 2$ and $c_{-m}=c_m$, consistent with the real symmetric profile.

On the tensor product state space, the total angular momentum is $\mathbf{J}={\mathbf{J}}^{(1)}\otimes I+I\otimes {\mathbf{J}}^{(2)}$, conventionally abbreviated ${\mathbf{J}}^{(1)}+{\mathbf{J}}^{(2)}$. Operators on different factors commute. Define $J_{\pm }^{(i)}=J_1^{(i)}\pm i{J}_2^{(i)}$. Expanding the square givesThe first three terms commute with $J_3=J_3^{(1)}+J_3^{(2)}$. In a mixed term one subsystem's magnetic quantum number increases and the other's decreases, so its commutator with total $J_3$ is zero. Hence $\boxed{[{\mathbf{J}}^2,J_3]=0}$.

Use the standard identities $[J_3,J_{\pm }]=\pm J_{\pm }$ and $J_{\pm }|j,m⟩=\sqrt{(j\mp m)(j\pm m+1)}|j,m\pm 1⟩$. The PDF has the commutator in the reversed order with unchanged sign and prints $m+1$ also for lowering; these are sign/label slips. The standard identities are consistent with the meaning of raising and lowering and are used here.

Put $s=j_1+j_2$. The unique product state of magnetic quantum number $-s$ isFor $j_1,j_2>0$, define the orthonormal product states $|a⟩=|j_1,-j_1+1⟩|j_2,-j_2⟩$ and $|b⟩=|j_1,-j_1⟩|j_2,-j_2+1⟩$. Raising the preceding state givesThe normalized orthogonal combination is annihilated by $J_{-}$, because its two lowering amplitudes cancel. It is consequently the lowest-weight state with $J=s-1$:up to an overall phase. For $j_1=3,j_2=1$, the outcome $m_1=-3,m_2=0$ is the $|b⟩$ component. The Born rule therefore gives

For the order-18 group, the index-two subgroup $P$ is normal and $G=⟨P,t⟩$. Restrict a complex irreducible representation $V$ to the finite abelian group $P$. Simultaneous diagonalization decomposes it into character spaces $V_{\chi }$. The inversion relation makes $t$ exchange $V_{\chi }$ and $V_{{\chi }^{-1}}$. The sum of spaces belonging to any such character orbit is $G$-invariant, so irreducibility leaves only one orbit.

If the character is trivial, all of $P$ is in the representation's kernel. Otherwise $\chi$ has order three and $\chi \ne {\chi }^{-1}$. The only spaces are $V_{\chi }$ and $V_{{\chi }^{-1}}$, and both have the same kernel in $P$. A nontrivial character of $P\cong C_3\times C_3$ maps onto $C_3$, so that kernel has order three. It acts trivially on all of $V$. Therefore the group has no faithful irreducible complex representation. In fact choosing nonzero $v\in V_{\chi }$ makes $\mathrm{span}(v,tv)$ invariant, so these nontrivial-character irreducibles have degree two.

The inductive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the order sum obtained by putting a copy of $\beta$ after a disjoint copy of $\alpha$.

These agree by transfinite induction on the second argument. Appending no points gives $\alpha$; appending one new greatest point gives a successor. At a limit stage, the order sum is exhausted by the increasing initial segments obtained by appending each $\beta <\lambda$, so its order type is their supremum. Thus the synthetic construction satisfies all three recursion clauses, and uniqueness of transfinite recursion proves equivalence.

The adjoint operators, orthonormal singular vectors and squared-norm normal equation for a linear inverse problem conditions require Hilbert spaces. For arbitrary normed spaces the Banach-space adjoint takes values in the dual spaces and need not produce this formula. Interpret the given singular value system in Hilbert spaces, with $u_j$ in the unknown space and $v_j$ in the data space, as in the printed identities. Positive singular values tending to zero provide the compact ill-posed case. A general bounded linear operator need not have this property; the identity operator is a well-conditioned counterexample. Components in the kernel must be fixed by a minimum-norm convention, and the unregularized inverse is defined only for data satisfying the corresponding range condition.

Off northeast Greenland, the Northeast Water Polynya is organized by the interaction of local currents and winds with coastal geometry and relatively fixed ice barriers. A regional northward coastal current encounters the Norske Øer shelf-ice barrier; divergence and shelter on its lee side help create the southern opening. This local northward current should not be replaced in the explanation by assuming uniform southward flow along the entire Greenland coast. In summer, weak winds, the current and the barrier can maintain the opening, while solar heating and the seasonal heat budget allow it to expand northward without continuous new-ice formation.

Winter conditions differ. Strong northerly winds can remove newly formed sea ice from the northern region around Ob Bank, with the Ob Bank ice barrier and Kronprins Christian Land controlling shelter and divergence. Large air–sea temperature differences then sustain rapid freezing and export. Thus the recurring geographical opening has a changing balance of dynamical export and seasonal thermodynamics; long-lived does not mean identical formation conditions throughout the year. Ice barriers themselves can also change between years.

The North Water Polynya, or Pikialasorsuaq, occupies the Smith Sound and northern Baffin Bay region between Greenland and the Canadian Arctic. An ice arch across the narrow upstream channel restricts the arrival of thick drifting Arctic ice. Northerly winds and predominantly southward transport through the channel remove newly formed ice into Baffin Bay. The combination of blocked upstream ice supply and continuing downstream export maintains low ice concentration despite strong winter cooling. The location and stability of the arch therefore matter as well as the local wind. Warm Atlantic-origin water carried along the Greenland side can provide a sensible-heat contribution, with tidal and other mixing helping deliver heat upward; the entire opening should not be attributed either exclusively to warm-water convection or exclusively to wind.

Both regions permit much greater winter ocean-to-atmosphere heat transfer than nearby thick ice, and both can concentrate ice formation and brine rejection. The production of saline shelf water affects local density stratification and exchanges with surrounding waters; it is not automatically equivalent to deep or abyssal water formation. In spring and summer the early availability of light and open water promotes biological production and supports rich marine food webs. These persistent or recurrent openings also provide habitat and access of particular importance to Arctic animals and coastal communities. The key Arctic mechanisms are ice-barrier control of supply, current- and wind-driven divergence, and a seasonally varying contribution from ocean heat.

One convenient version of Watson's lemma is the following. Suppose $G(t)$ is integrable away from zero and has, as $t\downarrow 0$, an asymptotic expansion $G(t)\sim {\sum }_{j\ge 0}{c}_j{t}^{\beta +j-1}$, where $\beta >0$. On a finite interval, or on an infinite interval with suitable exponential growth control,To prove a finite truncation, split the integral at a fixed small $\delta >0$. The part away from zero is exponentially small. On $(0,\delta )$ subtract the first $N$ local terms and bound the remainder by $C{t}^{\beta +N-1}$; its integral is $O({\lambda }^{-\beta -N})$. Extending each polynomial integral to infinity changes it only exponentially, and the substitution $u=\lambda t$ produces the Gamma integral. A little-o local remainder gives the corresponding little-o asymptotic remainder. This is a proof by localization and remainder bounds, rather than merely formal integration of a series.

For the monotone-phase integral, assume the derivatives needed below exist and are bounded near $a$, and the rest of the integral is controlled. Put $t=\varphi (a)-\varphi (x)$. Since ${\varphi }^{\prime }(a)<0$, this is a regular endpoint coordinate andDifferentiating with $dx/dt=-1/{\varphi }^{\prime }$ gives $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Watson's lemma therefore gives the two-term monotone-endpoint Laplace expansionAn $O(e^{\lambda \varphi (a)}{\lambda }^{-3})$ remainder follows, for example, from a sufficiently smooth transformed amplitude with bounded second derivative near zero. The displayed second coefficient can vanish for special amplitudes without changing the expansion.

The unheaded transition request concerns a cubic endpoint-to-saddle transition. Near $x=0$ its exponent has the expansionAt the cubic endpoint $x$ has size ${\lambda }^{-1/3}$. Requiring the linear term to contribute on the same scale gives $1-\alpha =O({\lambda }^{-2/3})$. Thus the distinguished limit and its uniform leading integral areFor fixed $\nu$, the substitution $x={\lambda }^{-1/3}t$ makes the omitted exponent terms $O({\lambda }^{-2/3})$ on bounded $t$ intervals. Cubic decay controls the remaining tail, while the original interval away from zero is exponentially negligible. The estimate is uniform for $\nu$ in compact sets. In the printed range $\alpha \le 1$, $\nu \ge 0$; the same transition integral also makes sense for negative fixed $\nu$.

At $\nu =0$, the Gamma integral gives $h(0)=6^{1/3}\Gamma (1/3)/3$, recovering the cubic endpoint regime. For $\nu \to +\infty$, scale $t=u/\nu$ and expand $e^{-u^3/(6{\nu }^3)}$ to obtain $h(\nu )={\nu }^{-1}-{\nu }^{-4}+O({\nu }^{-7})$. Its leading term gives ${\lambda }^{-1/3}/\nu =1/[\lambda (1-\alpha )]$, recovering the ordinary endpoint regime in their overlap. Both matching limits are therefore checked explicitly.

Interpret the differential expression as $Lu=(p{u}^{\prime \prime }{)}^{\prime \prime }+qu$, with a suitable differential-operator domain and the stated clamped boundary conditions. For sufficiently smooth $u,v$ in that domain, twice integrating by parts givesThe boundary term vanishes because $v=v^{\prime }=0$ at both ends. For real coefficients this form is symmetric/Hermitian. In particularIndeed nonnegativity follows from $p>0$ and $q\ge 0$. If the integral were zero, $u^{\prime \prime }=0$ almost everywhere in the interval, so $u$ would be affine. Its endpoint values then force $u=0$. This proves strict positivity on the intended clamped domain and gives the weak formulation of a clamped fourth-order equation.

There is a genuine domain issue in the literal wording. The closure of clamped smooth functions in the stated L2 norm is all of $L^2(-1,1)$, since it contains the dense set $C_c^{\infty }(-1,1)$. Boundary traces are not retained by that closure, and a fourth-order expression does not define an operator on every element of $L^2$. For example $p=(1-x^2{)}^2$, $q=0$ satisfy the given sign conditions, but $u=1_{(-1/2,1/2)}$ belongs to this closure and makes $(p{u}^{\prime \prime }{)}^{\prime \prime }$ a distribution rather than an $L^2$ function. Even a general twice-smooth function need not have the four derivatives required by the classical expression.

A rigorous classical repair is to take smooth enough coefficients, for example $p\in C^2([-1,1])$, $q\in C([-1,1])$, and use a clamped domain such as $H^4\cap H_0^2$ in $L^2$, with the appropriate operator realization. A weak repair uses the clamped second-order Sobolev space $H_0^2$ and the form just displayed, with hypotheses ensuring its boundedness and closure; uniform positivity $p\ge p_0>0$ makes it coercive there. The integration-by-parts and strict-positivity proof does not need that stronger uniform bound, but existence and completion statements must not be inferred from the original $L^2$ closure alone. The printed $L[f]$ with an expression in $u$ is also a harmless variable-name mismatch; the calculation consistently uses $Lu$.

Yes: this is an A-stable third-order method. Its second derivative is essential; the Second Dahlquist barrier applies to ordinary first-derivative linear multistep methods, not this multiderivative multistep method.

Apply the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification polynomial isWe verify the root condition for a multistep method throughout $\mathrm{Re}z\le 0$. For a quadratic $a{\zeta }^2+b\zeta +c$ with $|a|>|c|$, the complex quadratic Schur criterion isfor both roots to be strictly inside the unit disk. Here is a brief justification of the criterion rather than just a root plot. Set $P^{\#}(\zeta )={\zeta }^2\overline{P(1/\overline{\zeta })}$. The polynomial $\overline{a}P-c{P}^{\#}$ is $\zeta (D\zeta +E)$, where $D=|a{|}^2-|c{|}^2$ and $E=\overline{a}b-c\overline{b}$. Its two roots are inside precisely when $|E|<D$. On the unit circle $|P^{\#}|=|P|$, and $|c|<|a|$, so the Rouche theorem equates its interior root count with that of $P$. A boundary zero of $P$ would also be a boundary zero of the transformed polynomial, which the strict inequality excludes. This proves the strict criterion; the non-strict form follows by continuity, with any unit root simple because $|c/a|<1$.

For the present polynomial it is enough to show $8|Q-1|<|Q{|}^2-1$, except at $z=0$. Write $z=-r+iy$, $r\ge 0$, and put $k=r^2+3r$, $s=y^2$. ThenDirect expansion gives the nonnegative factorizationEvery term is nonnegative, and the sum is strictly positive unless $r=y=0$. Also $|Q{|}^2=X+1>1$, so the criterion applies. For every nonzero $z$ in the closed left half-plane both amplification roots are strictly inside the unit disk. At $z=0$ they are $1$ and $1/7$, and the unit root is simple. ThereforeThis proof establishes stability of both the principal and parasitic modes, including on the imaginary axis.

First use the usual real-valued, measurable interpretation of the transition data. The complementary squares then imply $|f|\le 1$, and the compact support gives $f\in L_2$. Consider the periodization on $[-\pi ,\pi ]$. For $|t|<2\pi /3$, only $f(t)=1$ contributes. For $2\pi /3\le t\le \pi$, only $f(t)$ and $f(t-2\pi )$ can contribute, and their squares sum to one. For $-\pi \le t\le -2\pi /3$, apply the complementary identity at $t+2\pi$ to obtain $f(t{)}^2+f(t+2\pi {)}^2=1$. Therefore the orthonormal translates and Fourier periodization identity holds almost everywhere.

Define a periodic MRA low-pass filter by prescribing it on $[-\pi ,\pi ]$:In the central band, $f(t)=1$, so $f(2t)=m(t)f(t)$. In the rest of this fundamental interval, both $m(t)$ and $f(2t)$ vanish. If $\pi <|t|<4\pi /3$, reducing $t$ modulo $2\pi$ puts it in an outer band where $m=0$, and again $f(2t)=0$. For $|t|\ge 4\pi /3$, both $f(t)$ and $f(2t)$ vanish. Endpoint choices affect only a null set. Thus the complementary transition bands for a scaling function give both Fourier conditions, with this explicitly constructed periodic mask.

A concrete real-valued choice, illustrating that the transition constraints are consistent, isOn the positive transition band put $u=3t/(2\pi )-1\in [0,1]$. The other translate has parameter $1-u$, so the sum is ${\cos }^2(\pi u/2)+{\sin }^2(\pi u/2)=1$.

There is a genuine qualification in the printed formulation: it uses ordinary squares without explicitly declaring $f$ real-valued. If arbitrary complex values are allowed, the claimed verification is false. Take $f=1$ for $|t|\le 2\pi /3$, $f=0$ for $|t|\ge 4\pi /3$, and setThe printed complementary-square rule holds, including its endpoints: in the interior $i^2+(\sqrt{2}{)}^2=1$, while the endpoints pair one with zero. Nevertheless, for $2\pi /3<t<\pi$,This compactly supported $L_2$ function is the Fourier transform of an $L_2$ function, so it is a counterexample within the natural complex Hilbert space. The verification is valid for real-valued $f$, or with absolute squares in the transition rule; it is not valid for unrestricted complex $f$ with ordinary squares.

For unit-spaced knots, the quadratic cardinal B-spline with support $[i,i+3]$ is $N_i(t)=B(t-i)$, where the Cox-de Boor recurrence givesIn particular $B(1)=B(2)=1/2$ and $B(0)=B(3)=0$. At $x_i=i+2$, the only nonzero entries in row $i$ are $A_{ii}=1/2$ and, if $i<n$, $A_{i,i+1}=1/2$. Write $U$ for the upper shift matrix, with ones immediately above its diagonal. Thensince $U^n=0$. Thus $(A_x^{-1}{)}_{ij}=2(-1{)}^{j-i}$ for $j\ge i$, and zero otherwise. The largest absolute row sum is the first, giving $\Vert A_x^{-1}{\Vert }_{{\ell }_{\infty }}=2n$.

Using (1) and the permitted $d_3=3$ proves the stronger linear growth of shifted quadratic spline interpolation estimateTherefore $\Vert P_x\Vert =\Theta (n)$ and is not uniformly bounded. An $O(n)$ upper bound alone would not imply unboundedness; the lower bound is essential.

For the usual spline interpolation problem the sites are distinct and ordered; the Schoenberg–Whitney theorem then makes $A_x$ invertible under the stated support positivity. More generally, the following argument applies whenever the interpolation map in the question is well defined and $A_x$ is invertible.

Factor the spline interpolation operator as $P_x=B{A}_x^{-1}R$, whereSampling has operator norm at most one from the supremum norm to ${\ell }_{\infty }$. Nonnegativity and the subpartition of unity for B-splines giveConsequently $\Vert P_x\Vert \le \Vert A_x^{-1}{\Vert }_{{\ell }_{\infty }}$.

For the reverse bound, select a row $r$ of $A_x^{-1}$ with maximal absolute row sum, and choose $y_j=\mathrm{sgn}((A_x^{-1}{)}_{rj})$, assigning any value of modulus at most one when the entry is zero. Then $\Vert y{\Vert }_{{\ell }_{\infty }}=1$ andAt distinct sites there is a continuous piecewise linear function $f$ with $f(x_j)=y_j$ and $\Vert f{\Vert }_{\infty }=1$: interpolate linearly between the ordered sites and extend constantly to the endpoints. The given uniform-norm stability of a B-spline basis therefore impliesThus the B-spline interpolation operator norm satisfiesPositivity $N_i(x_i)>0$ alone, without the usual distinct-site convention, is insufficient to define an interpolation operator. For example, take order-two B-splines with knots $0,1,2,3$ and $x_1=x_2=3/2$. Both diagonal values are $1/2$, but the B-spline collocation matrix has identical rows $(1/2,1/2)$ and is singular. The displayed proof uses the well-defined interpolation map posited in the question; ordered distinct sites supply its standard existence guarantee.

At temperatures above a few MeV, weak interactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ maintain neutron-proton chemical equilibrium. Negligible lepton chemical potentials give the neutron-to-proton ratio $r_{\mathrm{eq}}=n_n/n_p=\exp [-\Delta /(k_{B}T)]$, where $\Delta =(m_n-m_p)c^2\simeq 1.293\mathrm{MeV}$. At high temperature this ratio is nearly one.

During radiation domination, the interaction time grows as $T^{-5}$, faster than the expansion time $T^{-2}$. Their crossing therefore produces cosmological weak freeze-out. If $\Theta =T/(10^{10}\mathrm{K})$, equating the displayed order-of-magnitude times gives ${\Theta }_f^{-5}\sim g_{\ast }^{-1/2}{\Theta }_f^{-2}$, soUsing $H^{-1}=2t$ instead of $t$ changes only the order-one coefficient. Before electron-positron annihilation, the usual effective relativistic count is $g_{\ast }=2+(7/8)(4+6)=10.75$. The factor $7/8$ weights fermionic relative to bosonic energy density.

After freeze-out, free neutrons decay while protons increase correspondingly. Define the neutron baryon fraction $X_n=n_n/(n_n+n_p)$. In the instantaneous-freeze-out estimate,where ${\tau }_n=t_{1/2}/\ln 2$ is the neutron mean lifetime. Hence the ratio at the onset of nuclear synthesis isThis expression accounts for the protons produced by neutron decay.

Nuclear burning initially waits at the deuterium bottleneck: although deuterium is bound by about $2.2\mathrm{MeV}$, the many photons per baryon can still photodissociate it. Roughly, ${\eta }_b^{-1}{e}^{-B_D/(k_B{T}_N)}\sim 1$ sets the bottleneck temperature, with refinements from reaction rates and phase space. Around $k_B{T}_N\sim 0.1\mathrm{MeV}$, deuterium survives, reactions through deuterium, tritium and helium-3 rapidly build tightly bound helium-4, and nearly all surviving neutrons are captured.

Thus $n_{\mathrm{He}}\simeq n_n/2$ and the primordial helium mass fraction isFor the standard estimate $r_N\simeq 1/7$, neutron and proton constituent fractions are $1/8$ and $7/8$, respectively, and $Y_{\mathrm{He}}\simeq 1/4$. Free neutrons are negligible after capture. The remaining hydrogen mass fraction is about $3/4$, and the helium-to-hydrogen number ratio is about $1/12$.

The printed numerical time coefficients are schematic, not a precision prediction of this quarter fraction: their literal crossing with $g_{\ast }=10.75$ gives $k_B{T}_f\simeq 1.28\mathrm{MeV}$. Keeping physical weak-rate coefficients and the gradual freeze-out instead gives an effective freeze-out around $0.7$–$0.8\mathrm{MeV}$, followed by decay over a few minutes. The abundance formulas above explain the standard quarter fraction and its parameter dependence without treating the rounded time formulas as exact nuclear-rate data.

For an expanding universe, cosmic inflation requires $\ddot{a}>0$. Including any vacuum contribution in the total density and pressure, the acceleration equation gives $\boxed{\rho +3P<0}$, or $w=P/\rho <-1/3$ for positive density. Equivalently the comoving Hubble radius $(aH{)}^{-1}={\dot{a}}^{-1}$ decreases. Since $|{\Omega }_{\mathrm{total}}-1|=|k|/(aH{)}^2$, prolonged acceleration suppresses curvature and addresses the flatness problem. A sufficiently long inflationary phase also stretches a region that was initially causally connected to encompass the observed region, addressing the horizon problem. The condition on acceleration must last long enough; an arbitrarily short interval is insufficient.

For a homogeneous canonical scalar field $\Phi$ with potential $V(\Phi )$,Thus $V>{\dot{\Phi }}^2$ is sufficient for inflation. A suitably flat positive potential gives slow-roll inflation: Hubble damping in $\ddot{\Phi }+3H\dot{\Phi }+V_{,\Phi }=0$ keeps the kinetic energy small compared with $V$, so $P_{\Phi }\simeq -{\rho }_{\Phi }$.

The Euler-Lagrange equation gives $\ddot{\theta }=0$. Lift the path to the real line beginning at ${\theta }_i$. Its endpoint can be any ${\theta }_f+2\pi m$, where $m\in \mathbb{Z}$ labels the relative winding number. With $\Delta ={\theta }_f-{\theta }_i$, every classical path and its action areThese exhaust the constant-velocity paths in the different winding sectors.

Set $h=\mathbb{E}\theta >0$ and $w=\mathrm{Var}(\theta )>0$. Keeping $h$ fixed in the gamma distribution family gives $\alpha =h^2/w$ and $\lambda =h/w$. The ratio of expected process variance to variance of hypothetical means becomesAs the prior variance $w$ decreases, $K$ increases, so the credibility factor decreases. In the limit $w\downarrow 0$, $Z\downarrow 0$: the risk parameter is already essentially known from the prior, leaving little need to use the noisy sample mean. The population claim expected value $h/(k-1)$ remains fixed throughout this comparison.

The two curves have different destinations, so the equality of their initial segments is not an equality of their complete laws. At the first contact with $[1,\infty )$, the initial hull separates the target $1$ from infinity. Except for the null event of hitting the fixed point $1$ itself, this happens by a contact at a real point strictly larger than $1$. The fact that a fixed nonzero real point is almost surely not visited is the SLE does not hit a fixed nonzero boundary point property; it is consistent with the strictly positive probability of swallowing an entire interval at once in Question 3.

The original curve, whose target is infinity, continues in the closure of the unbounded surviving component. The curve targeted at $1$ continues in the surviving component whose boundary contains its target, which is now separated from infinity. Its domain normalization must follow that component. Ultimately one curve tends to infinity and the other to $1$. This is target-dependent continuation of SLE6.

The precise failure of the argument is the loss of the conformal conjugacy between the same surviving components. In the transformed construction, at time $S$ the pole $-1$ is swallowed; its image is infinity. The composition $g_t^{\ast }\circ \Phi \circ g_t^{-1}$ was a conformal automorphism of the upper half-plane only while both maps described corresponding surviving domains. After the boundary contact, the image of the original target-containing domain is the component containing $1$, whereas $g_t^{\ast }$ in the ordinary half-plane-capacity convention maps out the hull relative to the component containing infinity. These are no longer the same component. Equivalently, the swallowed pole collides with the growing hull, and the analytic-neighbourhood assumption at the Loewner driving function cannot be continued by the preceding argument.

Hence neither the given analytic identities nor the Brownian motion description obtained from their capacity clock licenses a continuation of the comparison beyond the stopping time. The zero drift before contact does not remove this geometric obstruction. The restriction to pre-contact segments is essential, even though the curves can continue in their respective target components.