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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 Schrödinger 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.

The literal PDF, however, only places the packet centres ${\mathbf{y}}_i$ in $Y$; it does not say that the packets themselves fit inside $Y$. Since it permits $r>a$, the two numerical approximations just displayed do not follow for every allowed packet and region. For example, take all ${\mathbf{y}}_i=0$, let $Y$ be a very small ball, and let $|\varphi {|}^2$ be uniform on a ball of radius $r\gg 10a$. Translate the other branch to a similarly small region $Z$ at distance much larger than $100r$. ThenIf these equal packet probabilities are denoted by $\eta$ and their repeated-label integral by $R$, the next probability iswhich approaches $\eta$, not one, for large $N$. Thus packet containment is needed for the stated near-region event to have the usual branch probabilities. The literal general answer is given by the integrals above.

The physical GRW amplification mechanism remains intact: with properly chosen macroscopic branch regions, spontaneous localization selects the alternatives with approximately the squared branch amplitudes and stabilizes the selected record, on a time scale $\tau /N$. Gaussian tails are not exactly eliminated, so the persistence is approximate. The containment qualification concerns how the region is defined, rather than whether the dynamics selects a macroscopically separated branch.

The proposed probability density function is nonnegative. To check its integral, put $I={\int }_{\mathbb{R}}{e}^{-x^2/2}dx$. The Gaussian integral can be evaluated by squaring and using polar coordinates:Since $I>0$, $I=\sqrt{2\pi }$ and $\int \varphi =1$. Thus $X$ has the standard normal distribution. Completing the square in its moment-generating function gives, for every real $t$,Differentiation under the integral is justified near every finite $t$ by Gaussian decay. Expanding $e^{t^2/2}={\sum }_{m\ge 0}{t}^{2m}/(2^{m}m!)$ gives all moments:

To obtain the two-term bounds for Mills ratio, use ${\varphi }^{\prime }(t)=-t\varphi (t)$. Integration by parts gives, for $x>0$,The positive remainder proves $r(x)<1/x$. A second integration by parts givesSubstitution and division by the strictly positive $\varphi (x)$ showThe last integral is strictly positive for every finite $x>0$, proving $\boxed{1/x-1/x^3<r(x)<1/x}$. Both remainders are finite for every fixed positive $x$; the lower bound remains valid even when it is negative.

For $x\ge 0$, the monotonicity of the logarithm and the uniform distribution giveFor $x<0$ the cumulative distribution function is zero. Differentiation gives the probability density function $f(x)=\lambda e^{-\lambda x}$ for $x>0$, and zero elsewhere. The distribution is $\boxed{\mathrm{Exp}(\lambda )}$, with rate $\lambda$. This is inverse transform sampling of an exponential distribution.

Condition on the fair coin, taken independent of $U$. For the squared uniform distribution, the cumulative distribution function of $U^2$ is $\sqrt{x}$ on $0<x<1$, giving probability density function $1/(2\sqrt{x})$. Reflection about $1/2$ gives the probability density function of $1-U^2$ as $1/(2\sqrt{1-x})$. Averaging the two conditional probability density functions produces this mixture model:There are no atoms at the endpoints; each integrable singularity contributes half the total mass. Equivalently the cumulative distribution function is $[\sqrt{x}+1-\sqrt{1-x}]/2$ on the interval.

For fixed rational $0\le t_1<t_2<t_3<t_4$, writeBy independent increments, $Z$ is independent of $(V,W)$ and has normal distribution $N(0,t_3-t_2)$. Conditional on $(V,W)$, the displayed difference has a continuous probability density function; in particular it equals zero with probability zero. Taking the countable intersection over all such rational quadruples proves that, on one event of probability one, the maxima over any two separated rational closed intervals are distinct.

Suppose a local maximum at $t$ were not a strict local maximum. There is a neighbourhood in which every value is at most $B_t$, and, arbitrarily close to $t$, some other time $u$ strictly inside that neighbourhood has $B_u=B_t$. Choose separated rational closed intervals containing $t$ and $u$, both lying inside that neighbourhood. When one time is zero, use a first interval with left endpoint zero. Both interval maxima equal $B_t$, contrary to the preceding event.

Therefore every local maximum of Brownian motion is a strict local maximum almost surely, simultaneously over all times. This countable-interval argument avoids an invalid intersection of probability-one events over uncountably many candidate times.

Let $u$ be the infection time and $s=t-u\ge 0$ its incubation delay, with independent delays having probability density function $f(s)$. Each infection contributes to the symptom-onset intensity according to its delay density. Back-calculation of infection incidence uses the convolutionObserved onset intensity is infection intensity convolved with the incubation distribution. If infections begin at a known $t_0$, take $h(u)=0$ for $u<t_0$, reducing the first integral's lower limit to $t_0$. Otherwise past infections must be included; the observation window's start need not be the infection process's start.

Choose a proposal probability density function $q$ positive wherever $|h|f$ is nonzero, up to null sets, and draw independent and identically distributed random variables $Y_1,\dots ,Y_n$ from $q$. The ordinary importance sampling estimator isThe support condition for importance sampling and absolute integrability give ${\mathbb{E}}_q[h(Y)f(Y)/q(Y)]=\int hf=\mu$, so this is an unbiased estimator. Its variance, possibly infinite, isLet $A=\int |h|f$. If $A>0$, the Cauchy-Schwarz inequality yieldsEquality holds for $q$ proportional to $|h|f$. Thus the minimum-variance importance distribution and the minimum variance areFor a constant-sign integrand this is zero variance. If $A=0$, the integrand vanishes almost everywhere and the zero estimator already has zero variance. Requiring the proposal to cover all of the target's support, including where $h=0$, can exclude $q_{\ast }$; in that stricter class, $(1-ϵ)q_{\ast }+ϵf$ approaches the same infimum as $ϵ\downarrow 0$. This distinguishes coverage of the integral from coverage of every target event.

The moment-generating function of a random variable $Z$ is $m_Z(\theta )=\mathbb{E}[e^{\theta Z}]$, at the real values of $\theta$ for which that expectation is finite. For a standard normal distribution variable $X$, combining the exponentials in its probability density function yieldsThe last equality follows by scaling the Gaussian integral. The variables $X_i^2$ are independent random variables, so the expectation of their product factorizes. ThereforeThis is the moment-generating function of a chi-squared distribution with $n$ degrees of freedom. For positive $n$ the defining integral diverges at and above $\theta =1/2$.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

For a nonnegative continuous event time $T$, the survivor function is $F(t)=\mathbb{P}(T>t)$, and its probability density function isThe hazard function is the instantaneous event rate conditional on being event-free at that time:If the event is certain to occur in finite time, the density is proper: ${\int }_0^{\infty }f(t)dt=1$, so $F(t)\to 0$. Since $F(0)=1$ and $F^{\prime }=-hF$, the cumulative hazard function is $H(t)=-\log F(t)$. Therefore a certain eventual event requires unbounded integrated hazard:If the lifetime has a finite upper endpoint, the same divergence occurs as that endpoint is approached, and $H$ is interpreted as infinite beyond it.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

For the maximum-entropy regularization functional, differentiation of the scalar integrand gives $(s\log s-s{)}^{\prime }=\log s$. Thus the Fréchet derivative at a positive reference function $y$ is $E^{\prime }(y)=\log y$, and the stated differentiability result gives $\partial E(y)=\{\log y\}$. Consequently,This is the generalized Kullback–Leibler divergence. When both functions integrate to the same mass, the last two terms cancel after integration; for probability density functions it becomes the usual Kullback-Leibler divergence. For two positive functions, adding the reverse distance givesA sufficient setting for the differentiability argument is $L^{\infty }(\Omega )$ with $y$ bounded away from zero. For nonnegative $x$, use $0\log (0/y)=0$ wherever appropriate. Positivity alone in $L^2$ does not automatically provide an open domain on which the Fréchet derivative exists.

Let $s=Kx$ and assume $s>0$. The scalar Poisson data fidelity integrand has derivative $1-y/s$. Applying the permitted interchange of differentiation and integration in a direction $z$ givesThus the ambient-space Fréchet derivative, or its Hilbert space gradient when the pairing is an inner product, isThe quotient is pointwise, and the adjoint operator transports it back from the data domain to the source domain. The chosen spaces must ensure the integrals and derivative pairing are finite; an output bounded away from zero is a useful sufficient condition.

The printed set of probability density functions and positive cone are not vector spaces, so the notation for a bounded linear map between them is understood as a positive bounded linear operator on ambient function spaces, restricted to admissible inputs. If unit mass is a constraint, admissible directions satisfy ${\int }_{\Omega }z=0$. The displayed gradient is then a representative modulo an additive constant; optimization with that constraint additionally includes its normal cone.

Use the rate convention for the exponential distribution. By convolution of independent random variables, the probability density function of $S=X+Y$ is, for $s\ge 0$,It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage hypoexponential distribution. As a check, its limit as $\mu \to \lambda$ is ${\lambda }^{2}s{e}^{-\lambda s}$, the shape-two gamma distribution.