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Use the shape-rate convention for the gamma distribution. A shape $r$ and rate $\beta$ give expected value $r/\beta$ and variance $r/{\beta }^2$. The two moments here force $r=2$ and $\beta =2/\mu$. Thus $M_X(t)={\beta }^2/(\beta -t{)}^2$, and the geometric-sum moment-generating function becomesPut $r_0=\sqrt{1-p}$, $a=\beta (1-r_0)$ and $b=\beta (1+r_0)$. Then $0<a<b$, $ab=p{\beta }^2$, andConsequently the geometric sum of shape-two gamma variables has the same probability distribution as the sum of two independent exponential distributions with rates $a,b$. Their convolution of independent random variables yieldsThe probability density function is zero for $s<0$. It is nonnegative because $a<b$, and its integral isThis verifies normalization directly. An alternative check uses the conditional gamma distribution with shape $2n$:which is exactly the same density. The expected value $1/a+1/b=\mu /p$ supplies a further check.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

A credibility estimate blends the risk's own sample mean with a population or prior expected value. In the basic equal-exposure form it is $Z\overline{X}+(1-Z)\mu$, where the credibility factor $Z\in [0,1]$ is the weight attached to the individual experience. More data ordinarily increase that weight; greater within-risk variation relative to between-risk variation reduces it.

Write $m(\theta )=\mu (\theta )$ for the conditional claim expected value, keeping $\mu$ for the prior-level parameter. Normalization of the probability density function givesAt interior points of its finite domain, differentiation under the integral sign yieldsThis is an instance of the exponential-family derivative identities for natural parameter $-\theta$. The minus sign is present in the PDF and is lost in the converted TeX. The conditional variance is likewise $(\log q{)}^{\prime \prime }(\theta )$.

Differentiate the log of the natural conjugate prior:Integrate on a compact subinterval and then let its endpoints approach those of the parameter interval. The prescribed vanishing of the prior density eliminates the boundary term. Since $m(\theta )\ge 0$, monotone convergence theorem justifies the limit of its expectation, and the identity givesIn particular the assumed proper prior and boundary conditions themselves establish this first-moment integrability.

For $T={\sum }_{i=1}^n{x}_i$ and $\overline{x}=T/n$, conditional independence makes the likelihood function proportional to $q(\theta {)}^{-n}{e}^{-\theta T}$. Multiplication by the prior distribution gives the Bayesian posteriorThus this is a conjugate prior family, with updated parameters

The integration-by-parts calculation for the posterior mean also needs vanishing posterior boundary values. Here that property can be checked, rather than assumed. Let $a,b$ be the essential lower and upper endpoints of the claim support. Every nondegenerate tilted claim law has $a<m(\theta )<b$, so the established prior mean satisfies $a<\mu <b$. Supported observations have $\overline{x}\in [a,b]$; hence $a<{\mu }_n<b$. At a finite parameter endpoint the prior's vanishing forces $q(\theta )\to \infty$, because its exponential factor has a finite positive limit; the increased power $q^{-k_n}$ makes the posterior vanish there too. At $\theta \to +\infty$, choose $a<d<{\mu }_n$ with positive base-measure mass below $d$. Then $q(\theta )\ge C{e}^{-d\theta }$ and the unnormalized posterior is at most $C^{-k_n}{e}^{-k_n({\mu }_n-d)\theta }\to 0$. At $\theta \to -\infty$, choose ${\mu }_n<d<b$ with positive mass above $d$ to obtain the corresponding bound $C^{-k_n}{e}^{k_n(d-{\mu }_n)\theta }\to 0$. These bounds also ensure posterior propriety at infinite endpoints. This is the endpoint control for a Laplace-family conjugate posterior.

Apply the previous score integration to the Bayesian posterior. The natural conjugate credibility identity isThus $Z=n/(n+k)$ is the credibility factor. This is an exact posterior mean, rather than an affine approximation to one.

For the specific shape-two gamma distribution, takeThen $m(\theta )=2/\theta$ and $\mathrm{Var}(X\mid \theta )=2/{\theta }^2$. The gamma rate gamma conjugacy prior isthat is, $\Theta \sim \mathrm{Gamma}(2k+1,\text{rate }k\mu )$. Its normalizing constant in the notation of the question is $c(\mu ,k)=\Gamma (2k+1)/(k\mu {)}^{2k+1}$. The gamma distribution density vanishes at both zero and infinity for $k>0$. The posterior has shape $2k+2n+1$ and rate $k\mu +T$, so its conditional-mean posterior mean is again $(k\mu +T)/(k+n)$.

For the last calculation, gamma distribution integration givesThe second identity requires the printed $k>1/2$. Therefore the expected process variance and variance of hypothetical means, denoted by $a,b$ in this question, areConsequently $a/b=k$, and the Bühlmann credibility factor agrees with the exact Bayesian weight:This is the exact Bühlmann credibility for gamma claims, expressed using the reciprocal rate as the random scale. The symbols $a,b$ here are the process and between-risk variances, respectively; their roles should not be interchanged when comparing with other notation for the Bühlmann model.

Expand around the expectation $\mu$:Thus variance as the minimum mean squared error of a constant givesThe finite second moment also guarantees the finite first absolute moment needed below.

For a probability density function $f$, splitting at $a$ givesLet $F$ be the cumulative distribution function. Since a density gives no mass at a singleton, differentiating under this integral yields $H^{\prime }(a)=F(a)-(1-F(a))=2F(a)-1$. This derivative is nondecreasing, so $H$ is convex andThis is the median minimizes expected absolute loss property. The minimizer need not be unique: a gap in the density can give an interval of medians, all with the same loss.

Under $\mathbb{Q}$, $X_T=x+W_T^{\mathbb{Q}}$ has density ${\phi }_T(z-x)=(2\pi T{)}^{-1/2}\exp (-(z-x{)}^2/(2T))$. The reciprocal Radon-Nikodym derivative depends only on $X_T$:Consequently the probability density function under the original measure isIt integrates to one because ${\mathbb{E}}_{\mathbb{Q}}\cosh X_T=e^{T/2}\cosh x$. Completing the square also gives the useful mixture distribution formThus the terminal law is a mixture of $N(x+T,T)$ and $N(x-T,T)$ with the displayed positive weights. The diffusion with hyperbolic tangent drift density also exhibits the Doob h-transform with $h(z)=\cosh z$.

For quota share reinsurance the insurer retains the same fraction of each claim, soThe annual retained loss is consequently $T_I=\alpha T$. Each transformed risk severity has probability density function $f_i(y/\alpha )/\alpha$ for $y>0$. The mixed transformed severity, with the same weights ${\lambda }_i/({\lambda }_1+{\lambda }_2)$, still gives a compound Poisson distribution. Scaling the expected value and variance givesThis agrees with the general retained-claim formulas because $\mathbb{E}g(X_i)=\alpha {\mu }_i$ and $\mathbb{E}[g(X_i{)}^2]={\alpha }^2({\sigma }_i^2+{\mu }_i^2)$.

For excess of loss reinsurance the insurer pays each claim up to its retention level:The cap applies separately to every claim. In particular the retained annual loss is ${\sum }_j\min (X_j,M)$, rather than a single cap on the annual total.

Let $F_i$ be the cumulative distribution function for the claim size on risk $i$, and put ${\overline{F}}_i=1-F_i$. The retained severity on that risk has the original probability density function on $0<y<M$ and an atom of a measure at $M$ of mass ${\overline{F}}_i(M)$. Thus $T_I$ has a compound Poisson distribution with rate ${\lambda }_1+{\lambda }_2$ and the mixture of these capped severity laws. The mixture's mass at $M$ is ${\sum }_i{\lambda }_i{\overline{F}}_i(M)/({\lambda }_1+{\lambda }_2)$.

For the capped claim moments, use the tail integral formula for moments. Since $\mathbb{P}(\min (X_i,M)>x)={\overline{F}}_i(x)$ for $0\le x<M$ and is zero for $x\ge M$,Substitution into the compound Poisson distribution moment formulas givesEquivalently, the integrals are ${\int }_0^{M}x{f}_i(x)dx+M{\overline{F}}_i(M)$ and ${\int }_0^M{x}^2{f}_i(x)dx+M^2{\overline{F}}_i(M)$. The annual variance uses the retained raw second moments; subtracting their squared means would omit the variation in the Poisson distribution count.

For the unit-width box kernel, the convolution and kernel density estimator areandIn particular $\mathbb{E}{\hat{f}}_{n,h}^K=g_h$. The scaling gives $\Vert K_h{\Vert }_2^2=h^{-1}\Vert K{\Vert }_2^2=h^{-1}$. Even if $f$ is not square-integrable, Young's convolution inequality gives $g_h\in L^2$ because $f\in L^1$ and $K_h\in L^2$.

Using independence to eliminate the cross terms in the centred estimator, and Tonelli theorem to integrate the nonnegative variance, gives the exact integrated variance of a kernel density estimator:The Cauchy-Schwarz inequality now yieldsThis constant comes from the unit $L^2$ norm of the unscaled unit-width box kernel.

For the piecewise constant probability density function, integrability forces the constants on both unbounded outer intervals to be zero. Thus $f$ is bounded, has compact support, and has finitely many jumps. Let ${\Delta }_r=f(x_r+)-f(x_r-)$ denote the jump at $x_r$. For $h$ smaller than the least gap between consecutive breakpoints, the smoothing regions $[x_r-h/2,x_r+h/2]$ do not overlap. Outside these regions the convolution equals $f$.

Within a region, put $v=x-x_r$. For $-h/2<v<0$, the bias is ${\Delta }_r(1/2+v/h)$; for $0<v<h/2$, it is $-{\Delta }_r(1/2-v/h)$. Endpoint values do not affect an $L^2$ norm. Integrating these two triangular errors givesThis is the box-kernel bias of a piecewise constant density. By the triangle inequality, with $C_f=({\sum }_r{\Delta }_r^2/12{)}^{1/2}$,Balancing this bias-variance tradeoff with $h_n=n^{-1/2}$ gives, for all sufficiently large $n$,

Let $H$ denote a ridge's peak sea-ice draft, reserving $h$ for the draft at a randomly sampled position. In the exponential ridge-draft model, normalization by the line density $\mu$ givesThe mean peak sea-ice draft isConsequentlywith $h_m>h_0$, $b$ having dimensions inverse length and $B$ inverse length squared. The normalized peak probability density function is a shifted exponential distribution.

For the triangular argument, interpret the common ridge shape as geometrically similar triangles with common along-track slope $\tan \delta$ and variable peak height. Literal congruence would require identical sizes and could not coexist with an exponential peak-draft distribution. Each side of a triangle has $dx=|dh|/\tan \delta$. A ridge reaching draft $H\ge h$ therefore contributes $2\cot \delta dh$ of horizontal track in the interval $[h,h+dh]$. Summing this occupation length over all qualifying peaks proves the triangular ridge occupation identity:ThusThis is a tail relation for sampled draft occupation, not an instruction to normalize $n$ and $g$ identically. Below $h_0$, the ideal triangles contribute $2\mu \cot \delta$ rather than the same exponential; level ice and gaps contribute their own draft distributions. If triangular keels are referenced to a level-ice base, the vertical coordinate must be shifted consistently. We also require nonoverlapping occupation: arbitrary choices of $\mu$, mean draft and slope can otherwise demand more than the available track length.

Observed mean keel slopes are typically of order $20^{\circ }$–$30^{\circ }$, with broad individual variation rather than a single universal angle. Orientation matters: if a track crosses a straight ridge at angle $\psi$ to the crest, $\tan {\delta }_{\mathrm{track}}=\tan {\delta }_{\perp }|\sin \psi |$. The track slope can therefore approach zero at a grazing crossing. A sonar morphology study found location-dependent mean slopes about $22^{\circ }$–$27^{\circ }$ after correcting for ridge orientation.

Young sea-ice pressure ridges often have recognizably triangular sections with angular, porous rubble and comparatively continuous crests. Melting, refreezing and repeated cracking modify older ridges: their blocks can become rounded and consolidated, and their keel or crest can fragment into separated hummocks rather than retain one triangular shape. A pre-exam multibeam study found first-year sea ice ridge slopes averaging roughly $27^{\circ }$, while multi-year sea ice ridges often consisted of irregular separated smooth blocks. Multi-year sections can be broader or locally shallower, but age alone does not determine one slope angle. The constant-slope triangle is a useful statistical idealization, not a faithful shape for every old ridge.

Put $Q(u)=\inf \{x\in \mathbb{R}:F(x)\ge u\}$ for $0<u<1$. This is the quantile function, rather than an ordinary inverse requiring strict increase. The limits of the distribution function at infinity make $Q(u)$ finite, and the fact that $F$ is a right-continuous function ensures $F(Q(u))\ge u$. ConsequentlyIncreasing $u$ shrinks the set in the infimum, so the quantile function is a monotone function. To prove the left continuity of the quantile function, fix $u$ and let $L={\sup }_{v<u}Q(v)\le Q(u)$. If $L<Q(u)$, choose $L<y<Q(u)$. Then $F(y)<u$, so some $v\in (F(y),u)$ satisfies $Q(v)>y$, a contradiction. Thus $Q(v)\uparrow Q(u)$ as $v\uparrow u$. At a flat stretch of the distribution function, the quantile function may jump immediately to the right, consistent with this left-continuous convention.

For the uniform order statistic, count the observations at most $u$. That count has the binomial distribution with parameters $n,u$, givingDifferentiating makes adjacent terms telescope: use $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ and $(n-r)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. The probability density function is thereforeIn particular, $U_{(j)}$ has the Beta distribution with parameters $j,n-j+1$.

Write $m=Q(1/2)$ for the specified median. Continuity of the distribution function gives $F(m)=1/2$ and no mass at $m$, even if the distribution function has a flat stretch there. Thus $B={\sum }_i{1}_{\{X_i<m\}}$ has the binomial distribution with parameters $n,1/2$. Except on a null event, the order-statistic confidence interval for a median covers $m$ exactly when $j\le B\le n-j$. Symmetry of the binomial distribution makes its two failure probabilities equal. Alternatively, the probability integral transform givesHence the coverage is exactly $1-\alpha$, withThe asymmetric open/closed endpoint convention does not change the coverage because the continuous probability distribution assigns no mass to the median.

For independent and identically distributed random variables sampled from a probability density function $f$, a kernel density estimator with kernel for density estimation $K$ and smoothing bandwidth $h>0$ isThe usual kernel for density estimation is nonnegative; finite $\int K^2$ makes the integrated variance finite. The mean integrated squared error and its bias-variance decomposition of mean squared error areThe interchange is justified by the Tonelli theorem.

For the exponential distribution and the specified unit-width uniform kernel, put $g_h(x)=\mathbb{E}{\hat{f}}_h(x)$. The expected value is $h^{-1}\mathbb{P}\{x-h/2\le X\le x+h/2\}$. Intersecting this interval with the positive half-line givesOn $[-h/4,0)$ the true probability density function vanishes, whereasHere $1-e^{-z}\ge z/2$ for $0\le z\le 1/4$. This leakage across the support boundary proves the integrated squared bias from a density jump:In particular, $h_0=1$ and $c=1/256$ suffice. The phenomenon differs from a smooth interior bias of a kernel density estimator: the order-one boundary error persists over a region of width proportional to $h$.

By independence, $\mathrm{Var}({\hat{f}}_h(x))=n^{-1}\{\mathbb{E}{K}_h(x-X{)}^2-g_h(x{)}^2\}$. Integrating the first term and using $\int K^2=1$ yieldsAlso $I(h)\le \int f^2=1/2$: the kernel density estimator mean is an average of translates of $f$, so the Jensen inequality followed by the Tonelli theorem bounds its squared integral.

To optimize over every $h>0$, we must establish that useful smoothing bandwidths approach zero, rather than optimize only a formal small-$h$ expression. An exact calculation supplies that justification. The autocorrelation integral of the exponential distribution density isAveraging over the two uniform windows, or integrating the piecewise expression for $g_h$, givesThese formulas show that $B$ is continuous on $(0,\infty )$, is strictly positive there because $g_h>0$ on $(-h/2,0)$, and tends to $1/2$ as $h\to \infty$. Consequently, $B$ is bounded away from zero on $[\delta ,\infty )$ for every $\delta >0$. Its expansion agrees with the supplied $B(h)=h/12+o(h)$.

The choice $h_n=\sqrt{12/n}$ givesChoose smoothing bandwidths within $1/n$ of the infimum. Their mean integrated squared error tends to zero, and nonnegativity of the integrated variance gives $B(h_n)\to 0$, hence $h_n\to 0$. For any $\epsilon \in (0,1/12)$, eventuallyThe last step is the arithmetic-geometric mean inequality. Taking the lower limit and then $\epsilon \downarrow 0$ matches the upper bound. ThusThe original PDF has this constant; the TeX aid's $1/\sqrt{2}$ is a transcription error.

The GRW model takes a normalized wave function in $L^2({\mathbb{R}}^{3N})$. Between jumps it obeys the Schrodinger equation, $i\hslash {\partial }_t\Psi =H\Psi$, with the usual many-particle Hamiltonian operator, for example $H={\sum }_i[-{\hslash }^2{\nabla }_i^2/(2m_i)]+V({\mathbf{x}}_1,\dots ,{\mathbf{x}}_N)$. Each particle has an independent Poisson process of collapse times with rate $1/\tau$. Thus the total jump rate is $N/\tau$, the mean wait is $\tau /N$, and the jumping label is uniform on $\{1,\dots ,N\}$.

At a jump of particle $i$, use the GRW localization operatorIts centre has probability density function $p_i(\mathbf{c})=\Vert L_i(\mathbf{c})\Psi {\Vert }^2$. The post-jump quantum state is $L_i(\mathbf{c})\Psi /\Vert L_i(\mathbf{c})\Psi \Vert$. The normalization is consistent because $\int d^3\mathbf{c}{L}_i(\mathbf{c}{)}^2=I$. The collapse times and labels have these state-independent rates; the centres depend on the wave function through the Born rule-like squared norm.

For precision about the stated spatial event, let $B=Y_{10a}=\{\mathbf{c}:\mathrm{dist}(\mathbf{c},Y)\le 10a\}$. With $H=0$, the GRW collapse-centre event probabilities are exactlyThe last expression is defined only for $p_1(B)>0$. It includes the possibility that the next jump hits the same particle.

Write $f_i^Y(\mathbf{x})=|\varphi (\mathbf{x}-{\mathbf{y}}_i){|}^2$, and similarly $f_i^Z$. Putwith the same definitions for $Z$. The macroscopic separation and negligible packet tails make the two branches essentially orthogonal, so their interference contribution can be neglected. The first-event probability is thereforeThe $Z$ contribution is extremely small: its actual positions are within $r$ of $Z$, whereas centres in $B$ are within $10a$ of $Y$, leaving a distance much larger than $a$ in the Gaussian. Packet-tail errors add to this Gaussian-tail error.

For the next-event calculation, two distinct labels have independent position densities within either product branch, giving factors ${\eta }_i{\eta }_j$. A repeated label instead gives $R_i$, since both localization Gaussians multiply the same position variable. ConsequentlyThese formulas, with $p_{2\mid 1}=p_{12}/p_1$, answer the spatial questions without an extra containment assumption. If the first event is dominated by the $Y$ branch, they simplify toNeglecting the $Z$ term in this conditional expression requires it to be small relative to the retained first-event probability, not merely small in absolute value. The exact operator formulas above cover rare-event exceptions as well.

The intended GRW branch persistence versus geometric containment approximation is obtained if $Y$ and $Z$ contain essentially all the position mass of their corresponding branches. Then positions in $Y$ have a Gaussian-centre margin of $10a$ inside $B$, so ${\eta }_i^Y\simeq R_i^Y\simeq 1$, while ${\eta }_i^Z,R_i^Z\simeq 0$. This givesFor the Gaussian convention used here, the three-dimensional probability of a displacement exceeding $10a$ is $\mathrm{erfc}(10)+20e^{-100}/\sqrt{\pi }$, an extremely small number. A first centre near $Y$ also suppresses the $Z$ branch relative to the $Y$ branch by the very small ratio $|\beta {|}^2{q}_i^Z(\mathbf{c})/(|\alpha {|}^2{q}_i^Y(\mathbf{c}))$ when the latter branch has appreciable weight. All particles are correlated with the same pointer alternative, so one localization selects the macroscopic branch even when the next particle label differs.