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Use the independent-mark interpretation: the Brownian motions attached to the atoms are independent of one another and of the initial Poisson random measure. Fix $t$ and defineConditionally on the initial atoms, each atom is retained as a dangerous atom if its own path meets the target by time $t$. These tests are independent, with position-dependent retention probability $p_t(x)$. The permitted Poisson thinning property makes the dangerous atoms a Poisson random measure of intensity $p_t(x)dx$.

Its total intensity is computed without invoking a marking or displacement theorem. The event in the definition of $p_t(x)$ says that $x$ belongs toBy the Tonelli theorem,Symmetry of Brownian motion gives $-\xi \stackrel{d}{=}\xi$ as processes. It therefore identifies the last expectation with $\mathbb{E}\mathrm{vol}(W(t))$, where the sign of the deterministic path $f$ is unchanged. This swept set is a Wiener sausage with an added deterministic path.

The total intensity is finite on each bounded time interval. To see this explicitly, write $K_t=1+{\sup }_{s\le t}|f(s)|$ and $R_t={\sup }_{s\le t}|\xi (s)|$. The sausage is contained in a ball of radius $K_t+R_t$. On a finite time grid, splitting according to the first crossing of a positive level $u$ by a one-dimensional Brownian motion shows that the chance of ending above $u$, conditional on the crossing, is at least $1/2$: the remaining independent increment is symmetric. Thus the grid maximum exceeds $u$ with probability at most twice the final Gaussian upper tail. Dense grids and continuity give the same bound for the path maximum. Applying this to each coordinate and its negative givesThe last inequality follows from $\mathbb{E}{e}^{\theta N(0,t)}=e^{{\theta }^{2}t/2}$ and minimizing the exponential Markov bound. Integration of this tail gives finite moments of $R_t$ of every positive order, so $\mathbb{E}\mathrm{vol}(W(t))<\infty$.

There are consequently only finitely many dangerous atoms on each compact time interval, almost surely. For closed balls, continuity of the paths makes each finite hitting-time infimum attained. The event $T>t$ is therefore the absence of dangerous atoms up to time $t$.

The open-ball convention has the same survival probability at each fixed $t$. Indeed for any compact path range $K$, the difference between its closed and open radius-one neighborhoods is the shell $\{x:\mathrm{dist}(x,K)=1\}$, which has zero Lebesgue measure. To see this, choose a nearest $y\in K$ for a shell point $x$. For every sufficiently small $\epsilon >0$, the ball of radius $\epsilon /2$ centered at $x-\epsilon (x-y)$ lies in the open neighborhood and inside the ball of radius $3\epsilon /2$ about $x$. Thus the shell has a fixed proportional hole at every scale; the Lebesgue density theorem forces its measure to be zero. By the Tonelli theorem, closed-touch and open-entry retention probabilities have equal integrals. Independent thinning therefore gives no atom that touches only the boundary before this fixed horizon, almost surely. This also excludes a difference at an open-entry infimum equal to $t$.

The zero-count probability of a Poisson distribution is $e^{-\lambda }$, and henceThis survival among independently moving Poisson traps formula has been derived using only independent thinning and the defining Poisson zero-count probability, with the expectation/volume step supplied by Tonelli's theorem.

For a nonnegative submartingale $(M_k{)}_{0\le k\le n}$ and $M_n^{\ast }={\max }_{0\le k\le n}{M}_k$, the weak form of the Doob maximal inequality isTo prove it, let $A_k$ be the event that $k$ is the first index with $M_k\ge u$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $A=\{M_n^{\ast }\ge u\}$. The submartingale property gives $\mathbb{E}[M_n\mid {\mathcal{F}}_k]\ge M_k$. Multiplying by $1_{A_k}$, taking expectations and summing yields $\mathbb{E}[M_n{1}_A]\ge {\sum }_k\mathbb{E}[M_k{1}_{A_k}]\ge u\mathbb{P}(A)$. Nonnegativity gives the second inequality.

The associated Doob Lp maximal inequality, for $p>1$ and $M_n\in L^p$, isIndeed $M_k\le \mathbb{E}[M_n\mid {\mathcal{F}}_k]$, so conditional Jensen inequality first ensures finite $p$th moments for all $M_k$ and hence for the finite maximum. Integrating the weak inequality against $p{u}^{p-2}du$ and using the Tonelli theorem givesThe last step is Holder inequality. Dividing, unless the maximum is zero, proves the stated norm bound. For an arbitrary martingale, its absolute value is a nonnegative submartingale by conditional Jensen inequality, so both corresponding absolute-maximum bounds follow.

For $0<b<\lambda =2$, integrate the identity $e^{bz}-1={\int }_0^{z}b{e}^{bt}dt$ for $z\ge 0$. The Tonelli theorem and the tail bound in (c) giveFor $b=0$ the expectation is $1$, and for $b<0$ the fact $Z\ge 0$ gives $0<e^{bZ}\le 1$. ThereforeThis is an exponential moment bound for a Gaussian random-walk maximum. The upper-tail bound proves the requested sufficient range; it alone does not establish a converse at the endpoint.

Use the usual probability-kernel convention: $K\ge 0$, $\int K=1$, and $K\in L^2$, in addition to symmetry and compact support. These conditions are needed for the stated finite bound; symmetry and compact support by themselves do not impose normalization or nonnegativity. Interpret the differentiability assumption in the usual Sobolev space sense $f\in H^2(\mathbb{R})$, or assume classical regularity sufficient for the integral Taylor remainder. The kernel density estimator isThe bias-variance decomposition of mean squared error, followed by Tonelli theorem, yieldsIndependence and a change of variables give the exact integrated variance of a kernel density estimatorAll changes of integration order are justified either by nonnegativity or by the displayed finite integrals.

For the bias of a kernel density estimator, symmetry gives $\int uK(u)du=0$. Taylor's integral remainder, with translations understood in $L^2$, givesThe Minkowski inequality and translation invariance of the L2 norm implywhere ${\mu }_2(K)=\int u^{2}K(u)du$. Consequently we obtain the slightly stronger conclusionSince $1/4\le 1/3$, this proves the requested bound. For signed kernels, the same argument has $\int u^2|K(u)|du$ in place of ${\mu }_2(K)$; the printed bound is not generally justified merely by signed-kernel symmetry. For example, let $U_a=(2a{)}^{-1}{1}_{[-a,a]}$ and $K=(4/3)U_1-(1/3)U_2$. This symmetric compactly supported signed kernel has integral one and ${\mu }_2(K)=0$, but its fourth moment is nonzero. For a normal density its convolution has nonzero bias at any fixed positive bandwidth. Sending $n\to \infty$ would therefore contradict a bound containing only the integrated variance term.

Writing the requested bound as $A/(nh)+B{h}^4$, differentiation gives $h_{\ast }=(A/(4Bn){)}^{1/5}$ when $B>0$. Thus $h\asymp n^{-1/5}$ gives mean integrated squared error of order $n^{-4/5}$. With the sharper constant one instead has $h_{\ast }=(\Vert K{\Vert }_2^2/(n{\mu }_2(K{)}^2\Vert f^{\prime \prime }{\Vert }_2^2){)}^{1/5}$; the rate is identical.

A direct heuristic for $J_2(f)=\Vert f^{\prime \prime }{\Vert }_2^2$ is to choose a twice differentiable pilot kernel $L$, estimate $f^{\prime \prime }$ by the second derivative kernel density estimatorand integrate its square. Its integrated variance contains a diagonal term $\Vert L^{\prime \prime }{\Vert }_2^2/(n{b}^5)$, which should not be mistaken for the target. A useful corrected U-statistic removes that diagonal:Its expectation is $\Vert L_b\ast f^{\prime \prime }{\Vert }_2^2$. Thus smoothing bias of an estimator, rather than the diagonal contribution, remains. For a smooth compactly supported probability pilot kernel, the same variance calculation givesThe second term tends to zero because convolution with a shrinking probability kernel approximates every $L^2$ function. Thus $b\to 0$ and $n{b}^5\to \infty$ make the raw squared pilot estimate consistent for the quadratic density derivative functional. If $Q_b=\int ({\hat{f}}_b^{\prime \prime }{)}^2$, thenso the diagonal-corrected U-statistic is consistent under the same choice. Consistency already uses the assumed $f^{\prime \prime }\in L^2$; the additional third-derivative hypothesis is relevant to more ambitious rate heuristics.

Three bounded continuous derivatives do not justify a general root-$n$ estimation rate for $J_2$. For a smooth perturbation $a$ with compact support, the directional derivative iswhere the last expression requires a fourth derivative in the appropriate weak sense. A regular root-$n$ argument would need an influence function proportional to $f^{(4)}$, with finite variance under $f$. Three derivatives do not supply that condition.

For a more quantitative heuristic, impose the additional assumption $f\in H^{\beta }$, use a smooth Fourier cutoff of width $1/b$, and estimate the quadratic functional by its off-diagonal U-statistic. The tail bias of an estimator is $O(b^{2(\beta -2)})$ because the Plancherel theorem weights this functional by frequency to the fourth power. The degenerate part of its standard deviation is $O(n^{-1}{b}^{-9/2})$: the squared $L^2$ norm of the quadratic kernel scales as $b^{-9}$. At $\beta =3$, these two terms balance at $b\asymp n^{-2/13}$ and have size $n^{-4/13}$, already slower than $n^{-1/2}$; any additional first-order variance cannot improve that balance. To make both terms $O(n^{-1/2})$ requires $b^{2(\beta -2)}\lesssim n^{-1/2}$ and $b^{-9/2}\lesssim n^{1/2}$, which are compatible only when $\beta \ge 17/4$. This is an illustrative smoothness calculation under extra Sobolev and moment assumptions, not a rate theorem asserted solely from bounded third derivatives. Particular much smoother probability density functions may of course permit root-$n$ estimation.

Let $T_j=\inf \{n\ge 1:X_n=j\}$. By the Strong Markov property, the probability of hitting $j$ and subsequently returning to $i$ is $f_{ij}{f}_{ji}$. This event implies a return to $i$, so $f_{ii}\ge f_{ij}{f}_{ji}$. For $i=j$, interpret the event as two successive returns.

Set $T_i^{(0)}=0$ and let $T_i^{(r)}$ be the $r$th successive return. Repeated use of the Strong Markov property gives ${\mathbb{P}}_i(T_i^{(r)}<\infty )=f_{ii}^r$. The total number $N_i$ of visits, including time zero, is both ${\sum }_{n\ge 0}{1}_{\{X_n=i\}}$ and ${\sum }_{r\ge 0}{1}_{\{T_i^{(r)}<\infty \}}$. By Tonelli theorem,The identity is valid with value $+\infty$. In particular, divergence of the return-probability sum is equivalent to Markov-chain recurrence.

In the classical risk model write $U(t)=u+ct-{\sum }_{j=1}^{N(t)}{X}_j$, where the Poisson process $N(t)$ has rate $\lambda$ and is independent of the claim sizes. Define the ruin time and ultimate ultimate ruin probability byThe relative safety loading is $\rho =c/(\lambda \mu )-1>0$, so $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the nonzero positive solutionExistence and uniqueness follow from the secant-slope existence criterion for an adjustment coefficient. Explicitly, $\lambda (M(r)-1)-cr$ has derivative $\lambda \mu -c<0$ at zero and is strictly convex for positive claims. The assumed divergence of $M$ makes it cross zero once at a positive $R<r_{\infty }$. When $r_{\infty }=\infty$, an exponential lower bound from any positive tail event shows that $M(r)/r\to \infty$. In particular $R$ is inside the finite-transform domain, not at its endpoint.

Put $\overline{F}=1-F$. Replacing $\phi =1-\psi$ in the survival renewal equation for a classical risk model yieldsBy the tail integral formula for moments, $\mu ={\int }_0^{\infty }\overline{F}(x)dx$. Therefore this is the defective renewal equationThe original kernel has mass $\lambda \mu /c=1/(1+\rho )<1$. For the exponential tilt of the ruin renewal kernel, setMultiplying by $e^{Ru}$ gives the required proper renewal equationTo verify that this is a probability renewal kernel, Tonelli theorem gives, for $r>0$ in the finite-transform domain,The adjustment coefficient equation consequently implies ${\int }_0^{\infty }{k}_R(x)dx=1$. Its expected value isIt is finite because $R$ is interior to the finite-transform domain and is positive because the strict convex crossing has derivative $\lambda M^{\prime }(R)-c>0$.

We quote the key renewal theorem in the following form: for a nonarithmetic distribution $K$ of positive increments with finite positive mean $m$, and a directly Riemann integrable nonnegative function $h$, the locally bounded solution of $z=h+z\ast K$ satisfies $z(u)\to m^{-1}{\int }_0^{\infty }h$. It has renewal representation $z=h\ast {\sum }_{j\ge 0}{K}^{\ast j}$; this follows by iterating the equation, since the probability that arbitrarily many positive increments have sum at most a fixed $u$ tends to zero.

Here $K_R(dx)=k_R(x)dx$ is absolutely continuous, and hence nonarithmetic distribution, even if the original claim law has atoms. The function $h_R$ is continuous. Choose $\delta >0$ with $M(R+\delta )<\infty$. The Markov inequality gives $\overline{F}(x)\le M(R+\delta )e^{-(R+\delta )x}$, and hence $h_R(u)\le C_{\delta }{e}^{-\delta u}$. Continuity on compact intervals and this exponential bound make the upper Riemann sums finite with uniformly vanishing tails, proving direct Riemann integrability.

A further application of Tonelli theorem evaluates the forcing integral:All hypotheses of the key renewal theorem are now checked, so the interior adjustment coefficient ruin prefactor isThis proves the requested Cramér–Lundberg ruin asymptotic, including its constant.

For the two-component hyperexponential distribution, conditioning on the chosen exponential component givesIts moment-generating function and derivative areThe moment-generating function diverges at $1/2$, while the derivative of the adjustment equation at zero is negative. Strict convexity therefore places its unique positive adjustment coefficient inThe adjustment equation, divided by $R$, becomesUsing this identity to subtract $(1+\rho )\mu$ from $M^{\prime }(R)$ givesThus the asymptotic constant in terms of $\rho$ and $R$ isThe denominator is positive on the identified domain. If desired, $R$ is the smaller root of $6(1+\rho )R^2-(5+9\rho )R+3\rho =0$; the other algebraic root lies outside the positive finite-transform interval and is not an adjustment coefficient.

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 $n$,

For a deterministic $s>0$, the positive part $(s-K{)}_{+}$ vanishes for $K\ge s$, so direct integration givesAt $s=0$ both sides vanish. Applying this pointwise to the nonnegative random variable proves the power payoff static call representationBy Tonelli theorem, the corresponding moment identity isincluding the possibility that both sides are infinite. No higher-moment assumption is needed to interchange these nonnegative integrals.

Write $R_j$ for the rectangles, $x_j$ for their centers and $e_j$ for their long-axis directions. For the finite exponent $p$ in the displayed estimate, take smooth rotated cap functions ${\psi }_j$, with $0\le {\psi }_j\le 1$, equal to one on angular distance at most $\delta /C$ from $e_j$ and supported within $2\delta /C$. Choose $C$ sufficiently large once and for all. The direction separation makes these cap supports disjoint, and $\int |{\psi }_j{|}^{p}d\sigma \lesssim \delta$.

DefineThe Fourier modulation and translation identity gives the second equality. Rotating the circle cap Fourier lower bound then gives $|h_j(x)|\gtrsim \delta$ on $R_j$. The half-side lengths of $R_j$ are no larger than the two frequency bounds used in part (b).

Let ${\epsilon }_j$ be independent Rademacher random variables. Because the input cap supports are disjoint, for every choice of signswhere $M=\#\mathcal{R}$. Apply the assumed Fourier extension estimate to the sum. Average over signs and use the Khintchine inequality pointwise, followed by the Tonelli theorem:There is no requirement that the spatial rectangles be disjoint; disjointness is used only for the input caps on the unit circle. Their spatial overlaps are precisely what the square function measures. The cap lower bounds now implySince each rectangle has area ${\delta }^{-3}$, the restriction-to-rectangle overlap principle givesThe constants are independent of $\delta$, the centers and the collection. The finite-$p$ interpretation is the one for which the printed power integral is defined. A single cap also shows that the assumed diagonal Fourier extension estimate can hold only for $p\ge 4$: its output contributes at least ${\delta }^{p-3}$ to the $p$th-power norm, whereas its input contributes at most a constant times $\delta$.