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For a sample of size $n$ from a continuous probability distribution, this interval covers the median $F^{-1}(1/2)$ with probability $1-2{\sum }_{r=0}^{j-1}\left(\genfrac{}{}{0ex}{}{n}{r}\right)2^{-n}$, for $1\le j<n/2$. Count observations below the median: their number has the binomial distribution with parameters $n,1/2$. No strict increase of the distribution function is needed.

Independent and identically distributed random variables with a continuous probability distribution realize each of their possible strict orderings with equal probability.

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.

A non-degenerate probability distribution is not concentrated at a single point. In extreme value theory, a non-degenerate distribution function $F$ is max-stable if, for each integer $n\ge 1$, constants $a_n>0,b_n\in \mathbb{R}$ satisfy $F(a_{n}x+b_n{)}^n=F(x)$ for every $x$. This says that a normalized sample maximum has the same law as one observation. A distribution function $F$ belongs to the maximum domain of attraction of a non-degenerate $G$ if there are $a_n>0,b_n$ such thatat every continuity point of $G$. Equivalently, $(M_n-b_n)/a_n$ has convergence in distribution to $G$.

The extremal types theorem says that any such non-degenerate limit is max-stable and, up to a positive affine change of variable, is exactly one of the following:These are respectively the Gumbel distribution, Fréchet distribution, and negative Weibull distribution. The theorem is also known as the Fisher–Tippett–Gnedenko theorem.

Useful sufficient conditions can be expressed using the survival function $\overline{F}=1-F$ and the right endpoint of a distribution $x_F$. An infinite endpoint with $\overline{F}(tx)/\overline{F}(t)\to x^{-\alpha }$ for every $x>0$ gives the Fréchet distribution domain. A finite endpoint with $\overline{F}(x_F-sx)/\overline{F}(x_F-s)\to x^{\alpha }$ as $s\downarrow 0$ gives the negative Weibull distribution domain. For the Gumbel distribution, it suffices that a positive Gumbel auxiliary function $a(t)$ gives $\overline{F}(t+a(t)x)/\overline{F}(t)\to e^{-x}$ for every real $x$ as $t\uparrow x_F$. These are regular variation and exponential tail-ratio conditions; they need no proof here.

In case (i), ${\overline{F}}_1(x)=(1-x{)}^2$ on $0<x<1$, so the finite-endpoint ratio is exactly $x^2$. Therefore the domain is negative Weibull with shape $2$. With $a_n=n^{-1/2}$ and $b_n=1$,and the limit is one for $x\ge 0$.

In case (ii), ${\overline{F}}_2(t)=e^{-\lambda t}$ has exponential tail ratio with the constant Gumbel auxiliary function $1/\lambda$. Therefore the domain is Gumbel. Choosing $a_n=1/\lambda$ and $b_n=\log n/\lambda$ givesIn case (iii), ${\overline{F}}_3(t)=(1+t{)}^{-\lambda }$ has regular variation of index $-\lambda$, so the domain is Fréchet with shape $\lambda$. The convenient choices $a_n=n^{1/\lambda }$ and $b_n=-1$ givewith limit zero for $x\le 0$.

Finally, define the empirical distribution function $F_n(x)=n^{-1}{\sum }_{i=1}^n{1}_{\{X_i\le x\}}$. On a sample with positive threshold $t=X_{(n-k)}$ and exactly $k$ observations strictly above $t$, its survival function has $1-F_n(t)=k/n$. Using the Tonelli theorem to integrate the finite nonnegative sum givesThis is the Hill estimator as an empirical plug-in version of the tail-integral limit. It does not prove consistency for fixed $k$; that is a separate issue.

There is a finite-sample qualification because the question permits ties and does not assume positive observations. The formula requires $1\le k<n$ and $t>0$. If $r=\#\{i:X_i>t\}<k$ because the threshold is tied, direct substitution giveswhere terms equal to the threshold contribute zero. This is generally $k/r$ times the displayed Hill estimator, rather than the same estimator. If $r=0$, the empirical plug-in denominator is zero. For a continuous probability distribution the no-tie condition holds almost surely; for a Fréchet distribution domain and fixed $k$, the threshold is positive with probability tending to one. Thus the stated plug-in identity holds at a positive untied threshold, and the literal claim for all samples from an arbitrary distribution function needs this qualification.

The independent and identically distributed random variables have a continuous probability distribution, so any tie has probability zero. Their joint probability density is unchanged by a permutation of their indices: this is order symmetry of independent random variables. The $n!$ strict orderings are disjoint and exhaust the sample space apart from the zero-probability ties. Each ordering therefore has the same probability, and their probabilities sum to one.

Under the null hypothesis $\theta =0$, $X$ has the uniform distribution on $[0,1]$. Against $\theta =1$, the likelihood ratio is $2x$, increasing on that interval. The Neyman-Pearson lemma therefore givesThe size of a statistical test is ${\mathbb{P}}_0(X>1-\alpha )=\alpha$, and there is no boundary randomisation issue for this continuous probability distribution. Its statistical power against $\theta =1$ is ${\int }_{1-\alpha }^{1}2xdx=2\alpha -{\alpha }^2$.

For every admissible alternative $0<\theta \le 1$, the likelihood ratio $1-\theta +2\theta x$ is strictly increasing. The same upper-tail rejection set has size $\alpha$ and is a most powerful test against each fixed such alternative, again by the Neyman-Pearson lemma. Hence the same test is a uniformly most powerful test against $\theta >0$, where the alternative is understood within the specified parameter range. Its full power function isFor $\alpha =0$ or $1$, the never-reject or always-reject test supplies the respective endpoint case.