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The cumulative distribution function is $F(x)=P(X\le x)$. For the differentiable distribution in the question, its probability density function is $f(x)=F^{\prime }(x)$, so probabilities over intervals are obtained by integrating $f$. The survival probability is $P(X>x)=1-F(x)$. Differentiating immediately givesThe strict inequality is consistent with the definition using $X\le x$. This relation connects the probability density function with the decline of the survival probability.

Start with a finite subcover of the compact set $K$. Among its remaining open balls, retain one of greatest radius and discard every ball intersecting it. Repeat until no balls remain. The retained family $\mathcal{F}$ is finite and pairwise disjoint. If a discarded ball $B(c,r)$ met the retained ball $B(c_0,R)$, then $r\le R$ and $|c-c_0|<r+R$. Therefore, for any $y\in B(c,r)$,Every original ball lies in the concentric triple of one retained ball. This proves the Wiener covering lemma for the finite subcover. By Lebesgue measure scaling and subadditivity,Disjointness givesThe empty compact set is covered by the empty selection.

The spherical derivative of a measure at $x$ is the limitwhen it exists, where $v_d$ is the unit-ball volume. We prove that this limit is zero outside a Lebesgue-null set for the singular measure.

First derive the needed maximal estimate from the selection lemma. For a finite positive Borel measure $\eta$, putThis is the uncentered maximal function of a finite measure. For $\alpha >0$, its strict superlevel set is the union of all balls with $\eta (B)>\alpha {\lambda }_d(B)$, so it is open. Cover any compact subset $H$ of this set by such balls and apply the finite Wiener covering lemma. The selected disjoint balls giveBy inner regularity of Lebesgue measure, this proves the uncentered maximal weak-type inequality

Since $\mu$ and ${\lambda }_d$ are mutually singular measures, there is a Borel set $N$ with ${\lambda }_d(N)=0$ and $\mu ({\mathbb{R}}^d\setminus N)=0$. By regularity of finite Borel measures on Euclidean space, for each $\epsilon >0$ choose compact $L\subset N$ with $\mu ({\mathbb{R}}^d\setminus L)<\epsilon$. Define the remainder measure ${\eta }_{\epsilon }(A)=\mu (A\setminus L)$.

For $x\notin L$, distance from $x$ to this compact set is positive. Thus sufficiently small centered balls avoid $L$, andWrite ${\overline{D}}_s\mu (x)$ for the upper limit of the ratios. If it exceeds $\alpha$, then either $x\in L$ or $M_u{\eta }_{\epsilon }(x)>\alpha$. HenceThe right side has Lebesgue measure at most $3^d\epsilon /\alpha$, since ${\lambda }_d(L)=0$. This even bounds the outer measure of the left side, without a separate measurability argument for the upper density. Let $\epsilon \downarrow 0$, and then take the countable union over positive rational $\alpha$. We obtain ${\overline{D}}_s\mu =0$ Lebesgue almost everywhere. The ratios are nonnegative, so their lower limit is also zero andThis proves that the spherical derivative of a singular measure vanishes. It is important to approximate the null carrier by compact subsets: a null carrier can be dense, so points outside it need not have any neighbourhood avoiding it.

In one dimension, consider the cumulative distribution function $F$. For either sign of $h\ne 0$, monotonicity gives a nonnegative difference quotient, and the relevant half-open interval lies inside $B(x,2|h|)$. In detail, the numerator is $\mu ((x,x+h])$ for $h>0$, while after reversing both signs it is $\mu ((x+h,x])$ for $h<0$. ThusAt each point where the spherical derivative of a measure is zero, this bound tends to zero from both sides. The enlarged open interval avoids any endpoint ambiguity from atoms. Therefore the ordinary two-sided derivative exists andThis is why singular distribution functions have zero derivative almost everywhere without having to be constant: almost-everywhere differentiation recovers increments only under extra hypotheses such as absolute continuity of a function.

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.

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.

Integrate each supplied predictive density to get its cumulative distribution function $F_t$, then form the sequential probability integral transformFor a correct continuous one-step conditional predictive model, $\mathbb{P}(U_t\le u\mid y_1,\dots ,y_{t-1})=u$. Iterating this identity shows that the $U_t$ are independent uniform variables. A histogram or quantile plot checks uniformity; serial plots and autocorrelations check for temporal structure left unexplained by the forecasts.

Also inspect empirical coverage of central prediction intervals, tail exceedances and interval widths, assessing calibration together with sharpness. These checks need only the supplied forecasts and observations. Compare chosen predictive discrepancy statistics with simulated uniform reference sequences. A total log score alone is a relative reward and does not provide a universal absolute goodness-of-fit threshold.

We derive the probabilities for a sequential elimination all-pay contest from a discounted subgame perfect equilibrium, using backward induction, and only then take $\delta \uparrow 1$. This preserves the selection supplied by discounting.

First consider a two-player all-pay auction with effective prizes $A_1\ge A_2>0$, so the incremental payoff is $A_i$ times winning probability minus effort. For $0\le b\le A_2$, the independent mixed strategies have effort cumulative distribution functionsThe second player has an atom $1-A_2/A_1$ at zero. For positive bids on this support, the first player's payoff is $A_1{G}_2(b)-b=A_1-A_2$ and the second's is $A_2{G}_1(b)-b=0$. Bids above $A_2$ cannot improve either payoff. The first has no zero atom, so the second's zero bid also earns zero. If the first deviates to zero, it can win only when the second bids zero; even with every such tie resolved in its favor, its payoff is at most $A_1(1-A_2/A_1)=A_1-A_2$. The two-player complete-information all-pay equilibrium therefore has winning probabilitiesThese follow by integrating $G_2$ against the uniform $G_1$. A third player with effective prize at most $A_2$ cannot profit by entering: for $0<b\le A_2$, its winning probability is $G_1(b)G_2(b)\le b/A_2$, giving payoff at most zero; above $A_2$ its effort exceeds its prize.

For the dynamic induction, relabel any remaining subgame's valuations as $v_1>\cdots >v_k$, with $r<k$ prizes left. Define the backward-induction threshold in an elimination all-pay contestFor $r=1$ the sum is empty and $C_1=v_2$. The discounted continuation value in an elimination contest is obtained from the following net utilities:The base case is the one-prize all-pay auction: only the two highest valuations need positive effort, with prizes $v_1,v_2$ and payoffs $v_1-v_2,0$. Every lower player has a nonprofitable deviation by the preceding calculation.

For $r\ge 2$, if either of the top two wins, the other becomes the highest player in a subgame with $r-1$ prizes. The continuation threshold in either such subgame is the same numberThe remaining top player's continuation payoff is $v_i-D$. Its effective prize in a sequential contest, net of the discounted payoff from losing, is thereforeThus $A_1>A_2$ for $\delta <1$, and the two-player distributions above apply. Adding the losing-state baselines giveswhich agree with the proposed formulas.

For a player of rank $i\ge 3$, the inductive continuation utility after either top player wins is identical: its new rank is $i-1$, and the utility expression depends only on its own value and the lower-valued tail. Thus its current zero-effort payoff is that common continuation value multiplied by $\delta$, exactly the stated $J_i$. Its effective prize for deviating to win now is $A_i=v_i-J_i$. For $3\le i\le r$, direct subtraction givesFor $i\ge r+1$, $A_i=v_i\le v_{r+1}\le C_r$, since the coefficients in $C_r$ are nonnegative and sum to one. Such a player cannot gain by entering against the two active players. There is one additional zero-bid deviation to check for the highest player. If both active bids become zero, the tie could award a prize to any remaining player. Losing to any player below rank $2$ gives no greater continuation utility than losing to rank $2$: deleting rank $2$ makes the ordered remaining rival list componentwise smallest, and its continuation threshold is a nonnegative weighted sum of that list. Thus even resolving every all-zero tie in the highest player's favor gives payoff at most its usual losing baseline plus $A_1(1-A_2/A_1)=A_1-A_2$. No tie rule can improve this deviation. This verifies all best responses and the continuation-utility formulas. Applying the construction to every remaining-player set proves a subgame perfect equilibrium of the discounted contest, not merely an on-path prescription.

Now take the vanishing-discount limit of an elimination all-pay contest. At every nonfinal subgame,Thus the two highest remaining players win with equal probabilities in a nonfinal stage. In the final stage the effective prizes are their actual valuations, so its lower-valued participant beats the higher one with probability equal to half their valuation ratio.

Return to the original ranks. Only the original top $m+1$ can ever win; at most $m-1$ higher players can have left before the final stage, so players of rank $m+2$ or worse never enter its top pair. Player $1$ must lose $m-1$ fair nonfinal contests to remain unawarded until the final stage, where its opponent has value $v_{m+1}$. HenceA player $2\le i\le m$ first enters the top pair at stage $i-1$, after $i-2$ higher-ranked winners have left. To remain unawarded, it must lose the $m-i+1$ nonfinal contests from then on, followed by the final contest against rank $m+1$. ThereforeThere are exactly $m$ distinct winners, so the expected value of their indicator sum is $m$. Combining these calculations givesThis proves the ranked winning probabilities in an undiscounted elimination contest. When $m=1$, it reduces to the ordinary two-player all-pay winning probabilities; when $m>1$, the repeated fair stages explain each power of $1/2$.

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.