Worked Block-Maxima and Threshold-Excess Calculations

From a Parent Model to Maxima and Excesses

Extreme-value calculations often begin with one parent distribution but ask two different questions. A block-maxima analysis studies the largest observation in a block; a threshold analysis studies how far an observation exceeds a specified level, given that it exceeds that level at all. Both calculations require careful attention to normalization, conditioning, and support.

The standard GEV and GPD parameterizations used here make the scale and shape checks explicit. In particular, the GEV and GPD shape parameter is denoted by $\xi$, and the scale parameters are positive. A matching shape parameter does not, by itself, mean that a finite-sample maximum and a threshold excess have the same distribution.

Definition: A GEV distribution with location $\mu$, scale $\sigma>0$, and shape $\xi$ has CDF $$ G(x)=\exp\left\{-\left[1+\xi\frac{x-\mu}{\sigma}\right]^{-1/\xi}\right\} $$ where the bracket is positive. When $\xi=0$, the continuous limiting form is $$ G(x)=\exp\left\{-\exp\left(-\frac{x-\mu}{\sigma}\right)\right\}. $$ A GPD with scale $\beta>0$ and shape $\xi$ has excess CDF $$ H(y)=1-\left(1+\xi\frac{y}{\beta}\right)^{-1/\xi},\qquad y\geq 0, $$ when $1+\xi y/\beta>0$. For $\xi=0$, $H(y)=1-\exp(-y/\beta)$. If $\xi<0$, the GPD has finite upper endpoint $-\beta/\xi$; the CDF there is 1.

The Block-Maximum Calculation

Let $X_1,\ldots,X_n$ be independent observations with common CDF $F$, and let $M_n=\max(X_1,\ldots,X_n)$. Independence gives an exact finite-sample formula. For constants $a_n>0$ and $b_n$, the CDF of the normalized maximum $(M_n-b_n)/a_n$ follows by evaluating the parent CDF at the corresponding point.

Result: For every real $z$, $$ P\left(\frac{M_n-b_n}{a_n}\leq z\right)=F(b_n+a_n z)^n. $$ This identity is exact; a limiting GEV distribution, when one applies, is obtained by taking a limit of its right-hand side.

Proof. Since $a_n>0$, the event $(M_n-b_n)/a_n\leq z$ is the event $M_n\leq b_n+a_nz$. A maximum is at most a value precisely when every observation is at most that value. Thus $$ P(M_n\leq b_n+a_nz) =P(X_1\leq b_n+a_nz,\ldots,X_n\leq b_n+a_nz) =F(b_n+a_nz)^n, $$ where the last equality uses independence and the common distribution $F$. $\square$

Worked Example: Exponential Block Maxima

Suppose $X_1,\ldots,X_n$ are independent exponential random variables with rate 1, so $F(x)=1-e^{-x}$ for $x\geq0$. Use $b_n=\log n$ and $a_n=1$, and set $Z_n=M_n-\log n$. Find the limiting law and check a finite-$n$ value.

For a fixed real $z$, once $n$ is large enough that $\log n+z\geq0$, the exact CDF is $$ P(Z_n\leq z) =F(\log n+z)^n =\left(1-\frac{e^{-z}}{n}\right)^n. $$ Since $(1-c/n)^n\to e^{-c}$ for fixed $c>0$, this converges to $$ G(z)=\exp(-e^{-z}). $$ This is the Gumbel GEV law with $\mu=0$, $\sigma=1$, and $\xi=0$. The limiting law has support on the whole real line; the finite-sample normalized maximum has the additional lower bound $Z_n\geq-\log n$.

For $n=100$ and $z=0$, the exact probability is $P(Z_{100}\leq0)=0.99^{100}\approx0.3660$. The limiting probability is $G(0)=e^{-1}\approx0.3679$. The close values illustrate the limit, not an exact equality at finite $n$.

Worked Example: Pareto Block Maxima

Suppose the observations have a Pareto distribution with $P(X>x)=x^{-2}$ for $x\geq1$. Normalize the block maximum as $Z_n=M_n/n^{1/2}$, and identify its limiting GEV parameters.

For fixed $z>0$, if $n$ is large enough that $n^{1/2}z\geq1$, the exact CDF is $$ P(Z_n\leq z) =\left(1-\frac{1}{n z^2}\right)^n \longrightarrow \exp(-z^{-2}). $$ To express the limit in GEV form, take $\mu=1$, $\sigma=1/2$, and $\xi=1/2$. Then $$ 1+\xi\frac{z-\mu}{\sigma} =1+\frac{1}{2}\frac{z-1}{1/2} =z, \qquad -\frac{1}{\xi}=-2. $$ The resulting GEV CDF is $\exp(-z^{-2})$, on its support $z>0$, as required. The finite-$n$ normalized maximum instead has support $z\geq n^{-1/2}$. For example, at $n=100$ and $z=2$, its CDF is $0.9975^{100}\approx0.7783$, compared with the limiting value $\exp(-1/4)\approx0.7788$.

Threshold Excesses and Their Parameters

For a threshold $u$, the excess is $Y=X-u$ conditional on $X>u$. Its conditional survival probability is $$ P(Y>y)=P(X>u+y\mid X>u)=\frac{P(X>u+y)}{P(X>u)},\qquad y\geq0, $$ provided $P(X>u)>0$. This ratio is the starting point for calculating a GPD model from a specified parent distribution.

Worked Example: Exponential Threshold Excess

Let $X$ have an exponential distribution with rate 1, and choose $u=2$. Calculate the excess distribution and the probability that an excess exceeds 1.

For $y\geq0$, conditioning gives $$ P(Y>y)=\frac{e^{-(2+y)}}{e^{-2}}=e^{-y}. $$ Therefore $Y$ is exactly GPD with $\xi=0$ and $\beta=1$, not merely approximately GPD. In particular, $$ P(Y>1)=e^{-1}\approx0.3679, \qquad P(Y\leq1)=1-e^{-1}\approx0.6321. $$ The support check is immediate: exponential excesses can be any $y\geq0$, with no finite upper endpoint. The threshold $u=2$ changes which observations are retained, but not the conditional excess distribution.

Worked Example: Pareto Threshold Excess

Let $X$ have Pareto survival $P(X>x)=x^{-2}$ for $x\geq1$, and choose $u=3$. Find the GPD parameters for $Y=X-3\mid X>3$, and calculate $P(Y\leq3)$.

For $y\geq0$, $$ P(Y>y)=\frac{P(X>3+y)}{P(X>3)} =\frac{(3+y)^{-2}}{3^{-2}} =\left(1+\frac{y}{3}\right)^{-2}. $$ The GPD survival function is $(1+\xi y/\beta)^{-1/\xi}$. Matching the exponent gives $1/\xi=2$, so $\xi=1/2$; matching the expression inside the power gives $\xi/\beta=1/3$, so $\beta=3/2$. Thus $$ P(Y\leq3)=1-\left(1+\frac{(1/2)3}{3/2}\right)^{-2} =1-(1+1)^{-2} =\frac{3}{4}. $$ The support condition holds for every $y\geq0$, because $1+y/3>0$. The excess is unbounded, consistent with the positive shape parameter.

Why the GEV and GPD Shapes Agree

For many models, block maxima and high-threshold excesses have the same limiting shape parameter. The Pareto examples show this directly: the maximum limit has $\xi=1/2$, and the exact excess law also has $\xi=1/2$. The GPD excess law need not be exact for a general parent distribution. For a distribution in a GEV maximum domain of attraction, it describes the limiting behavior of excesses over high thresholds.

Result: For a GPD excess $Y$ with scale $\beta>0$ and shape $\xi$, raising the threshold by $v\geq0$ gives a conditional excess $Y-v\mid Y>v$ that is GPD with the same shape $\xi$ and scale $\beta_v=\beta+\xi v$, whenever $P(Y>v)>0$. For $\xi=0$, this reads $\beta_v=\beta$.

Proof. For $y\geq0$ in the conditional support, divide the GPD survival at $v+y$ by the survival at $v$: $$ P(Y-v>y\mid Y>v) =\frac{(1+\xi(v+y)/\beta)^{-1/\xi}} {(1+\xi v/\beta)^{-1/\xi}} =\left(1+\frac{\xi y}{\beta+\xi v}\right)^{-1/\xi}. $$ This is the GPD survival with scale $\beta_v=\beta+\xi v$ and unchanged shape $\xi$. The scale is positive for thresholds with positive exceedance probability. When $\xi=0$, the ratio is $e^{-y/\beta}$, which gives the same result with $\beta_v=\beta$. The conditional support is inherited from the original GPD support. $\square$

A corresponding GPD form also explains the usual GEV-to-GPD parameter check. For a GEV model, write $t(x)=1+\xi(x-\mu)/\sigma$, with $t(x)>0$. In the relevant upper tail, the GEV survival $1-\exp\{-t(x)^{-1/\xi}\}$ is asymptotic to $t(x)^{-1/\xi}$, because $1-e^{-w}\sim w$ as $w\to0$. For $\xi=0$, the analogous upper-tail approximation is $\exp\{-(x-\mu)/\sigma\}$. Taking the ratio of these tail approximations at $u+y$ and $u$ gives the GPD form with shape $\xi$ and scale $$ \beta_u=\sigma+\xi(u-\mu). $$ This is an asymptotic tail relation, not a claim that a general GEV distribution has exactly GPD excesses at every finite threshold. The threshold and the excess values must remain in the GEV support; in particular, $\beta_u>0$.

Support Checks and a Common Mistake

Support checks are part of the calculation, not a final technicality. For a GEV model, verify $1+\xi(x-\mu)/\sigma>0$. For a GPD model, verify $y\geq0$ and $1+\xi y/\beta>0$. A negative shape gives a finite endpoint: the GEV upper endpoint is $\mu-\sigma/\xi$, and the GPD excess endpoint is $-\beta/\xi$. A proposed value beyond the endpoint is not a value at which the displayed density or interior CDF formula can be used.

Common mistake: treating $F(b_n+a_nz)^n$ as the limiting GEV CDF for every finite $n$. Correction: it is the exact finite-sample CDF; the GEV expression is its limit when the chosen normalization and parent distribution justify that limit.
Common mistake: matching only the exponents in a GPD survival function. Correction: match both the exponent and the scale inside the parentheses. In the Pareto example, $\xi=1/2$ follows from the exponent, while $\beta=3/2$ follows from the scale $y/3$.
Common mistake: assuming that a matching GEV and GPD shape makes the two finite-sample models identical. Correction: the Pareto excess is exactly GPD, but the GEV-to-GPD relationship for a general parent model describes high-threshold tail behavior and is generally approximate.

Key takeaway: Derive maxima from $F(x)^n$, derive excesses from a conditional survival ratio, and then check support and parameter values. These steps distinguish exact finite-sample calculations from limiting extreme-value approximations.

Check Your Understanding

Use the formulas and support checks above to answer these questions.

For independent exponential observations with rate 1, what normalization of $M_n$ gives a standard Gumbel limit?
For Pareto survival $P(X>x)=x^{-3}$, what are the GPD shape and scale for excesses over threshold $u=2$?
A GPD has $\xi=-1/2$ and $\beta=2$. What is its upper endpoint, and which excess values are in its support?
Why is the CDF of a normalized maximum at finite $n$ not automatically a GEV CDF?
For a GEV model with $\mu=0$, $\sigma=2$, and $\xi=1/2$, calculate $\beta_u$ at $u=2$ using the high-threshold GPD parameter relation.