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Let $V_{[1]}\ge \cdots \ge V_{[n]}$ be descending order statistics, with $w_n=0$ and nonnegative decreasing prizes. Decompose the prize vector into awards of $w_k-w_{k+1}$ to each of the best $k$ players. The corresponding truthful multi-unit auction charges each winner the next value $V_{[k+1]}$. Its total payment is $k(w_k-w_{k+1})V_{[k+1]}$. Revenue equivalence transfers the expected payment to the all-pay contest, because the interim allocations and lowest-type utilities coincide. Summing the layers proves the formula.

For $1\le k<n$ and a positive threshold $t=X_{(n-k)}$, the Hill estimator averages the logarithmic ratios of the top $k$ order statistics to $t$. For a Fréchet distribution domain with shape $\alpha$, it estimates the reciprocal tail index $1/\alpha$. At an untied threshold it equals the empirical distribution function plug-in tail integral ${\int }_t^{\infty }(1-F_n(x))/(1-F_n(t))dx/x$. At a tied threshold the plug-in denominator counts strictly larger observations and the identity needs adjustment. Fixed $k$ does not by itself yield a consistent tail-index estimate.

For observations $X_1,\dots ,X_n$, the nearest neighbour distance of order $k$ at $x$ is the $k$th order statistic of the distances $|X_i-x|$. An atomless distance distribution function ensures ties occur with probability zero.

There is a parity qualification in the printed claim. With the usual strict-majority definition, an even electorate need not have a Condorcet winner. On the axis $a<b<c$, the two orders $a\succ b\succ c$ and $c\succ b\succ a$ are both single-peaked preferences, but every pairwise contest is tied. Thus no alternative strictly defeats every other one. We first prove the intended strict-winner result for odd $n$.

Let $n=2k+1$ and order the peaks along the common axis. Their median $w$ has at least $k+1$ peaks at or to each side. For any $y<w$, the $k+1$ voters whose peaks are at or to the right of $w$ prefer $w$ to $y$ by single-peakedness. For any $y>w$, the corresponding $k+1$ voters on the left prefer $w$. Thus the median peak is the unique Condorcet winner.

The corresponding median voter rule is strategyproof. Fix the other $2k$ peaks $q_1\le \cdots \le q_{2k}$. As one voter reports a peak $p$, the selected median iswhere minimum and maximum refer to the common axis. All attainable outcomes lie between $q_k$ and $q_{k+1}$. If the true peak lies inside this interval, truthful reporting obtains the voter's top. If it lies left of the interval, truthful reporting obtains its left endpoint, which the voter's single-peaked order prefers to every larger attainable outcome. The case right of the interval is symmetric. No report improves the outcome. For one voter the rule simply selects its peak.

There is also a proof that does not depend on knowing the axis. If a voter truly prefers a proposed new winner $z$ to the current strict Condorcet winner $w$, that voter already opposes $w$ in the contest $w$ versus $z$. The strict majority supporting $w$ in that contest therefore consists of other voters and is unaffected by its report. Thus $z$ cannot become a strict Condorcet winner after a profitable misreport. This proves strategyproofness on the domain of profiles for which the selected strict winner exists, including all admissible odd-electorate single-peaked profiles.

For even $n$, a weak Condorcet winner is guaranteed: every alternative between the two middle peaks weakly defeats each other alternative, allowing ties. With a fixed common axis, consistently choosing the lower median or consistently choosing the upper median gives a single-valued strategyproof rule, by the same interval-clamping argument for an order statistic. Arbitrary tie selection is not asserted to have this property. The corrected conclusion is therefore a unique strict winner and its strategyproof selection for odd electorates; a weak winner with a specified median rule for even electorates.

Let $w_m=w(m/n)$, so a type $v$ values a prize at $v{w}_m$. In a symmetric increasing Bayes-Nash equilibrium of this rank-order contest, a player reporting type $t$ wins when at most $m-1$ opponents have larger types. Its winning probability isHere the count of opponents above $t$ has a binomial distribution. Differentiation, or the associated order statistic density, givesDefine the effort by the all-pay effort identityThis also verifies equilibrium globally. The derivative of a type $v$'s payoff from reporting $t$ is $w_m(v-t)q_m^{\prime }(t)$, positive before $v$ and negative after $v$. Thus truthful reporting is a best response. Bidding above the maximal equilibrium effort gains no additional winning probability. Type zero chooses zero effort.

By exchanging the two integrations, the expected value of total effort isThe last integral is the moment $\mathbb{E}[X(1-X)]$ for a Beta distribution with parameters $n-m$ and $m$, namely $m(n-m)/(n(n+1))$. This is the uniform-value multi-prize all-pay effort formula.

Put $x=m/n$ and $h(x)=w(x)x(1-x)$. The positive constant $n^2/(n+1)$ multiplying $h(x)$ does not affect the maximizing $m$. Sincethe assumed inequality makes $h$ nonincreasing on the feasible interval. Hence one prize maximizes expected total effort.

For the power family, the PDF gives $w(x)=x^{-\alpha }$. ThenIf $\alpha \ge 1$, this derivative is strictly negative for $0<x<1$: the bracket is affine and its values at the endpoints are $1-\alpha \le 0$ and $-1$. Thus $m=1$ is optimal.

If $0<\alpha <1$, the unique continuous maximizer isThe objective strictly increases before $x_{\ast }$ and decreases after it. Therefore its discrete maximizer lies amongCompare the surviving candidates using $m^{1-\alpha }(n-m)$, since $R_m=n^{\alpha }{m}^{1-\alpha }(n-m)/(n+1)$. If $n{x}_{\ast }$ is an integer, that integer is the unique maximizer; if its floor is zero, the only feasible candidate is $1$. This proves the discrete prize-count optimization for a power-valued contest including the endpoint cases.

The sufficient condition need only hold on $[1/n,(n-1)/n]$. The power family is undefined at zero, so the printed endpoint $x=0$ cannot apply to it. More generally, a positive finite value $w(0)$ would make the displayed inequality fail at zero. The design argument uses only positive feasible prize fractions and needs no value there.

Let $U_{(1)}=\min (U_1,U_2)$ and $U_{(2)}=\max (U_1,U_2)$ be the order statistics. For the symmetric uniform distribution, each observation is above or below $\theta$ with probability $1/2$, independently. ThusSo $\boxed{[U_{(1)},U_{(2)}]}$ is an exact 50% confidence interval.

The observed interval $[10,11.5]$ contains values impossible under the sampling model. Both observations must lie within one unit of the parameter, so the interval of all compatible values is $[U_{(2)}-1,U_{(1)}+1]=[10.5,11]$. It contains the true parameter almost surely. A support-restricted confidence interval for a uniform location parameter can therefore reportMore precisely, the procedure $[U_{(1)},U_{(2)}]\cap [U_{(2)}-1,U_{(1)}+1]$ retains exact 50% unconditional coverage, since the second interval contains $\theta$ almost surely. The intersection is a nonempty interval for every sample in the model and equals $[10.5,11]$ for these data. Reporting the full compatible interval alone is also valid and has coverage one. On this sample's large-spread event it lies wholly between the observations; this does not turn the unconditional 50% procedure into a generally 100% procedure.

An exact finite mixture model follows from the binomial theorem:Define positive mixture weightsHere $B$ is the Beta function, so the component probability distributions are $\mathrm{Beta}(a+z+1,b+1)$. All coefficients are computable from the observed integer counts.

Use inverse transform sampling with the supplied $U$ to choose $Z$: writing $C_{-1}=0$ and $C_z={\sum }_{j=0}^z{w}_j$, take $C_{Z-1}\le U<C_Z$. Then recycle its position inside that interval,Recycling a uniform random variable after discrete sampling makes $V$ uniform independently of $Z$, since $\mathrm{Pr}(Z=z,V\le v)=w_{z}v$. It is also independent of all supplied random variables with gamma distributions. Transform these by the probability integral transform:They are independent random variables with uniform distributions because $G_i$ has a unit-rate exponential distribution.

Conditional on $Z=z$, put $k=a+z+b+1$ and $r=a+z+1$. Form the $k$ values $V,W_1,\dots ,W_{k-1}$ and return their $r$th order statistic. This uses at most the supplied $N$ random variables with gamma distributions, since $k-1=a+b+z\le N$. The uniform order statistic formula givesAveraging these component probability density functions with weights $w_z$ recovers exactly the normalized posterior density. Unused inputs can be discarded. When $N=0$, $Z=0$, $k=r=1$, and the output is $U$, as required. Interval endpoints and ties have probability zero; any consistent convention there is harmless. Exactness here concerns the mathematical algorithm; finite-precision arithmetic can be stabilized with logarithmic mixture weights.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

LetIgnoring the immaterial choice of density values at the sample points, the likelihood function isBetween consecutive order statistics, $N(\theta )$ is constant while $(1+1/\theta {)}^{N(\theta )}$ decreases with $\theta$. A maximum can therefore be moved to the left endpoint of one of these intervals. With an equivalent version using $1_{\{x\le \theta \}}$, the maximum is attained at a sample order statistic. Thus the maximum-likelihood estimator coincides with one of $X_1,\dots ,X_n$.

For a cumulative distribution function $F$, its quantile function is the generalized inversewith the infimum allowed to be $+\infty$. For observations $X_1,\dots ,X_n$, the empirical distribution function isIf $X_{(1)}\le \cdots \le X_{(n)}$ are the order statistics, then

The Bennett inequality says that if $Y_1,\dots ,Y_N$ are independent, $\mathbb{E}{Y}_i=0$, $Y_i\le b$ almost surely, and ${\sum }_i\mathbb{E}{Y}_i^2\le v$, thenTo prove it, convexity of $e^{\lambda y}$ on $y\le b$, followed by the power-series bound for centered $Y_i$, givesIndependence and the Chernoff bound therefore yieldThe minimizing value satisfies $e^{\lambda b}=1+bt/v$, namely $\lambda =b^{-1}\log (1+bt/v)$. Substitution gives the stated exponent.

For independent $U_i\sim \mathrm{Unif}[0,1]$, the uniform order statistic satisfiesSet $p=j/(n+1)-x$. The event $U_{(j)}\le p$ means that at least $j$ sample points lie in $[0,p]$. If this occurs, some $j$-element subset consists entirely of such points. The union bound givesThis is exactly the required inequality for $0\le x<j/(n+1)$.