Reverse-Weibull Limits for Finite Endpoints

Why a Finite Endpoint Changes the Limit

For a heavy-tailed distribution, a sample maximum can grow without bound as the sample size increases. If observations instead have a finite upper endpoint, the maximum approaches that endpoint. The key question is then not how quickly the maximum grows, but how quickly its gap from the endpoint shrinks. The probability of observations lying close to the endpoint determines the scale of that gap and the limiting law.

This tutorial derives the reverse-Weibull limit under a power-law approximation to the upper tail near the endpoint. The derivation makes the sign convention explicit: a centered and scaled maximum has a limiting distribution on the nonpositive half-line, while the endpoint deficit has an ordinary Weibull distribution on the nonnegative half-line.

Finite Endpoints and the Tail Condition

Definition: For a random variable \(X\) with distribution function \(F\), its finite upper endpoint is \[ x_F=\sup\{x:F(x)<1\}, \] when this supremum is finite. We say the distribution has a power-law upper tail at the endpoint with constants \(c>0\) and \(\alpha>0\) if \[ 1-F(x_F-t)\sim c t^\alpha \qquad (t\downarrow 0). \] Here \(g(t)\sim h(t)\) means \(g(t)/h(t)\to 1\).

The condition says that the probability of an observation exceeding a point \(t\) below the endpoint is approximately \(c t^\alpha\) when \(t\) is small. The exponent \(\alpha\) describes how quickly probability mass thins near the endpoint; \(c\) sets the scale. This condition also rules out an atom at \(x_F\), since such an atom would prevent \(1-F(x_F-t)\) from tending to zero.

Let \(X_1,\ldots,X_n\) be independent and identically distributed with distribution function \(F\), and write \(M_n=\max(X_1,\ldots,X_n)\). For any real \(z\),

P(M_n\le z)=F(z)^n.

The maximum is at a distance of order \(n^{-1/\alpha}\) from \(x_F\): a gap \(t\) has exceedance probability about \(ct^\alpha\), and among \(n\) observations the expected number above \(x_F-t\) is about \(nct^\alpha\). Setting this quantity to one gives the natural scale.

The Reverse-Weibull Limit

Theorem: Suppose \(X_1,X_2,\ldots\) are independent and identically distributed with finite upper endpoint \(x_F\), and suppose \[ 1-F(x_F-t)\sim c t^\alpha \] for \(c>0\), \(\alpha>0\), as \(t\downarrow0\). Set \[ a_n=(cn)^{-1/\alpha},\qquad Z_n=\frac{M_n-x_F}{a_n}. \] Then \(Z_n\xrightarrow{d}Z\), where the limiting distribution function is \[ H(x)= \begin{cases} \exp\!\left[-(-x)^\alpha\right],&x<0,\\ 1,&x\ge 0. \end{cases} \]

Proof. Fix \(x<0\), and put \(t_n=-a_nx=a_n(-x)>0\). Since \(a_n\to0\), we have \(t_n\downarrow0\), so the endpoint-tail condition gives

q_n:=1-F(x_F+a_nx) \sim c\{a_n(-x)\}^{\alpha} =\frac{(-x)^\alpha}{n}.

Thus \(q_n\to0\) and \(nq_n\to(-x)^\alpha\). Independence now yields

P(Z_n\le x) =P(M_n\le x_F+a_nx) =F(x_F+a_nx)^n =(1-q_n)^n.

For \(0\le q<1\), \(\log(1-q)/q\to-1\) as \(q\to0\). Therefore \(n\log(1-q_n)\to-(-x)^\alpha\), and exponentiating gives \(P(Z_n\le x)\to\exp[-(-x)^\alpha]\). If \(x>0\), then \(x_F+a_nx>x_F\), so \(F(x_F+a_nx)=1\) and \(P(Z_n\le x)=1\). At \(x=0\), \(M_n\le x_F\) almost surely, hence \(P(Z_n\le0)=1\). These limits give \(H(x)\) at every \(x\); \(H\) is continuous, including at zero. This proves convergence in distribution. \(\square\)

Worked Examples

Worked Example: Uniform Observations on a Bounded Interval

Let \(X_i\sim\operatorname{Uniform}(0,1)\). Find the limiting distribution of the normalized maximum \(M_n\).

For \(0\le x\le1\), \(F(x)=x\), so the upper endpoint is \(x_F=1\). For \(0<t<1\),

$$ 1-F(1-t)=1-(1-t)=t. $$

Thus \(c=1\), \(\alpha=1\), and \(a_n=1/n\). For a fixed \(x<0\), once \(n\) is large enough that \(x\ge -n\), the exact distribution calculation is

P\!\left(\frac{M_n-1}{1/n}\le x\right) =P(M_n\le 1+x/n) =(1+x/n)^n.

As \(n\to\infty\), this tends to \(e^x\). Hence the limit CDF is \(e^x\) for \(x<0\) and \(1\) for \(x\ge0\), exactly the reverse-Weibull law with exponent \(1\). The endpoint gap \(n(1-M_n)\) consequently has limiting survival function \(e^{-y}\) for \(y\ge0\).

Worked Example: A Beta Distribution with a Quadratic Endpoint Tail

Let \(X_i\) have a \(\operatorname{Beta}(1,2)\) distribution, whose density on \([0,1]\) is \(f(x)=2(1-x)\). Determine the normalization and limit.

Integrating the density gives \(F(x)=2x-x^2\) for \(0\le x\le1\). Thus, for \(0<t<1\),

1-F(1-t)=1-\{2(1-t)-(1-t)^2\}=t^2.

Here \(x_F=1\), \(c=1\), and \(\alpha=2\), so \(a_n=n^{-1/2}\). For fixed \(x<0\), when \(n\) is sufficiently large that \(1+x/\sqrt n\in[0,1]\),

P\!\left(\sqrt n(M_n-1)\le x\right) =F(1+x/\sqrt n)^n =\left(1-\frac{x^2}{n}\right)^n \longrightarrow e^{-x^2}.

The limiting CDF is therefore \(\exp(-x^2)\) for \(x<0\) and \(1\) for \(x\ge0\). The square-root normalization differs from the uniform case because the probability of lying within \(t\) of the endpoint is of order \(t^2\), not \(t\).

Worked Example: A Bounded Density with an Asymptotic Power Tail

Suppose \(X_i\) has density \(f(x)=\tfrac32(1-x^2)\) on \([0,1]\). Find the scaling and limiting law, showing how the tail approximation leads to the limit even though the endpoint tail is not exactly a power.

The density is nonnegative on \([0,1]\), and it integrates to

\int_0^1 \frac32(1-x^2)\,dx =\frac32\left(1-\frac13\right)=1.

For \(0<t<1\), substituting \(y=1-x\) in the endpoint-tail integral gives

1-F(1-t) =\int_{1-t}^1 \frac32(1-x^2)\,dx =\frac32t^2-\frac12t^3 \sim \frac32t^2.

Consequently \(x_F=1\), \(c=3/2\), \(\alpha=2\), and \(a_n=(2/(3n))^{1/2}\). Fix \(x<0\) and let \(y=-x>0\). The tail probability at the normalized threshold satisfies

n\{1-F(1+a_nx)\} =n\left(\frac32a_n^2y^2-\frac12a_n^3y^3\right) =y^2-\frac12n a_n^3y^3 \longrightarrow y^2,

because \(n a_n^3=(2/3)^{3/2}n^{-1/2}\to0\). Therefore the same maximum calculation as in the theorem gives

P\!\left(\frac{M_n-1}{a_n}\le x\right) =F(1+a_nx)^n \longrightarrow \exp(-y^2)=\exp(-x^2).

The cubic term affects the finite-sample tail but vanishes at the limiting scale. The endpoint exponent, determined by the leading quadratic term, controls the limit.

Reading the Limit and Avoiding a Sign Error

The normalized maximum \(Z_n=(M_n-x_F)/a_n\) is nonpositive, which is why its limit is called a reverse-Weibull distribution. The nonnegative endpoint deficit \(D_n=(x_F-M_n)/a_n\) uses the opposite sign. For \(y\ge0\),

P(D_n>y)=P(Z_n<-y)\ \longrightarrow\ \exp(-y^\alpha),

at continuity points of the limiting distribution. Equivalently, \(D_n\) converges to a Weibull distribution with CDF \(1-\exp(-y^\alpha)\) on \(y\ge0\). This is the same endpoint behavior expressed with a different variable; it is not a different extreme-value regime.

A useful consequence of the scaling is that, for fixed \(y>0\), the probability the maximum falls at least \(a_n y\) below the endpoint is asymptotically \(\exp(-y^\alpha)\). Increasing \(n\) shrinks the physical gap through \(a_n\), while the scaled gap retains a nondegenerate distribution. The exponent \(\alpha\) controls that rate: larger \(\alpha\) gives the scale \(n^{-1/\alpha}\), with the constant \(c\) included in \(a_n\).

Common mistake: reversing the sign but keeping the same CDF. Since \(M_n-x_F\le0\), the reverse-Weibull CDF is supported on nonpositive values. If instead using \(x_F-M_n\), the variable is nonnegative and its CDF is \(1-\exp(-y^\alpha)\), not \(\exp(-y^\alpha)\).
Common mistake: using \(n^{-1}\) for every finite endpoint. The correct order is \(n^{-1/\alpha}\), with the constant \(c\) setting the scale. The uniform case has \(\alpha=1\), while the \(\operatorname{Beta}(1,2)\) example has \(\alpha=2\).
Common mistake: treating the tail approximation as an exact formula. An asymptotic equivalence suffices because the proof only requires \(n\{1-F(x_F+a_nx)\}\to(-x)^\alpha\). It need not hold exactly for finite \(n\).

Key takeaway: If \(1-F(x_F-t)\sim ct^\alpha\), the maximum approaches the finite endpoint on the scale \(a_n=(cn)^{-1/\alpha}\), and \((M_n-x_F)/a_n\) converges to the reverse-Weibull law with CDF \(\exp[-(-x)^\alpha]\) for \(x<0\).

Check Your Understanding

Use the endpoint-tail condition and the sign convention to answer these questions.

If \(1-F(x_F-t)\sim 4t^3\), what normalization \(a_n\) does the theorem prescribe?
For a uniform distribution on \([2,5]\), identify \(x_F\), \(c\), and \(\alpha\) in the endpoint-tail condition.
Why does the reverse-Weibull CDF equal \(1\) for positive arguments?
For the density \(f(x)=\tfrac32(1-x^2)\) on \([0,1]\), what is the leading term of \(1-F(1-t)\) as \(t\downarrow0\)?
How does the limiting distribution change in form if the normalized variable is the endpoint deficit rather than the centered maximum?