Generalized Pareto Models for Threshold Excesses

Why Model Threshold Excesses?

When unusually large observations matter, it can be useful to focus on the amount by which an observation exceeds a high threshold. For example, if $X$ is a measured river level and $u$ is a flood threshold, the excess is $X-u$, conditional on $X$ being greater than $u$. A generalized Pareto distribution (GPD) is a flexible model for this nonnegative excess. Its shape parameter allows the modeled tail to be unbounded or to end at a finite upper limit.

The GPD is a model for the conditional distribution of excesses, not automatically for the distribution of every observation $X$. In particular, probabilities under the conditional model describe what happens given that the threshold was crossed; they do not by themselves give the probability that a crossing occurs.

Definition, Parameters, and Support

Definition: Let $u$ be a fixed threshold and let $Y=X-u$ conditional on $X>u$. A generalized Pareto model for $Y$ has scale parameter $\sigma>0$, shape parameter $\xi\in\mathbb{R}$, and survival function $$ P(Y>y\mid X>u)= \begin{cases} (1+\xi y/\sigma)^{-1/\xi}, & \xi\ne 0,\\ \exp(-y/\sigma), & \xi=0, \end{cases} $$ for $y\ge 0$ such that $1+\xi y/\sigma>0$ when $\xi\ne 0$. For $\xi<0$, the excess is bounded above by $-\sigma/\xi$; the survival probability is zero at and beyond that endpoint. The corresponding cumulative distribution function is $F_Y(y)=1-P(Y>y\mid X>u)$ on the support, and is zero for $y<0$.

Here $\sigma$ is a scale: it has the same units as the excess $Y$. The shape $\xi$ determines the form and extent of the tail. When $\xi>0$, the support is unbounded and the tail decreases as a power of $y$. When $\xi=0$, the excess has an exponential distribution with mean $\sigma$. When $\xi<0$, the support ends at the finite upper endpoint $-\sigma/\xi$. These parameter interpretations concern the conditional excess model.

For $\xi\ne 0$, the density in the interior of the support is

f_Y(y)=\frac{1}{\sigma}\left(1+\frac{\xi y}{\sigma}\right)^{-1/\xi-1}.

For $\xi=0$, the density is $f_Y(y)=\sigma^{-1}\exp(-y/\sigma)$ for $y\ge 0$. When $\xi<0$, the density formula applies for $0\le y<-\sigma/\xi$; the distribution assigns no probability beyond the endpoint. Values of a density at individual boundary points do not change the probabilities it determines.

Why These Formulas Define a Distribution

The survival formula decreases from one at $y=0$ to zero at the far end of the support. The density is its negative derivative on the interior. The following calculation verifies that it integrates to one, including for a finite-endpoint model.

Result: For $\sigma>0$ and $\xi\in\mathbb{R}$, the GPD density integrates to one over its support. Its survival function is the one given in the definition.

Proof. For $\xi>0$, use $t=1+\xi y/\sigma$, so $dy=(\sigma/\xi)\,dt$. The density integral is

\int_0^\infty f_Y(y)\,dy =\frac{1}{\xi}\int_1^\infty t^{-1/\xi-1}\,dt =\frac{1}{\xi}\left[-\xi t^{-1/\xi}\right]_1^\infty =1.

For $\xi<0$, write $a=-1/\xi>0$. The support ends at $-\sigma/\xi$, and the same substitution takes $y=0$ to $t=1$ and the upper endpoint to $t=0$. Accounting for the reversed limits gives

\int_0^{-\sigma/\xi} f_Y(y)\,dy =a\int_0^1 t^{a-1}\,dt =a\left[\frac{t^a}{a}\right]_0^1 =1.

For $\xi=0$, $\int_0^\infty \sigma^{-1}\exp(-y/\sigma)\,dy=1$. In each case, integrating the density from $0$ to $y$ gives $F_Y(y)=1-(1+\xi y/\sigma)^{-1/\xi}$ for $\xi\ne0$, or $F_Y(y)=1-\exp(-y/\sigma)$ for $\xi=0$. Thus the density and survival formulas are consistent, and the total probability is one. $\square$

Probabilities and Quantiles

A survival probability is often the most direct way to answer questions about large excesses. If $y$ is in the support, $P(Y>y\mid X>u)$ follows from the defining formula. For an interval $a<Y\le b$, where $0\le a<b$ are in the support, subtract the two survival probabilities:

P(a<Y\le b\mid X>u) =P(Y>a\mid X>u)-P(Y>b\mid X>u).

For a probability level $p$ with $0<p<1$, the $p$-quantile $q_p$ solves $F_Y(q_p)=p$. Inverting the CDF gives

Quantile formula: For a GPD excess, $$ q_p= \begin{cases} \dfrac{\sigma}{\xi}\bigl((1-p)^{-\xi}-1\bigr), & \xi\ne0,\\[6pt] -\sigma\log(1-p), & \xi=0. \end{cases} $$ For $\xi<0$, this quantile lies below the finite endpoint $-\sigma/\xi$.

To derive the formula when $\xi\ne0$, set $1-(1+\xi q_p/\sigma)^{-1/\xi}=p$. Then $(1+\xi q_p/\sigma)^{-1/\xi}=1-p$, so $1+\xi q_p/\sigma=(1-p)^{-\xi}$, which yields the stated expression. For $\xi=0$, solve $1-\exp(-q_p/\sigma)=p$, giving $q_p=-\sigma\log(1-p)$. The CDF is continuous and strictly increasing across the support, so each such $p$ has a unique quantile there. $\square$

Worked Example: Positive Shape and an Unbounded Excess

Suppose the excess above a threshold has a GPD model with $\sigma=10$ units and $\xi=0.2$. Find the conditional probability that an excess is greater than 15 units, and find the 95th percentile.

The value $15$ is in the nonnegative support. Its survival probability is

P(Y>15\mid X>u) =\left(1+\frac{0.2(15)}{10}\right)^{-1/0.2} =(1.3)^{-5} =\frac{1}{3.71293} \approx 0.2693.

Thus about 26.9% of threshold exceedances are more than 15 units. For the 95th percentile, use $p=0.95$:

q_{0.95} =\frac{10}{0.2}\left((0.05)^{-0.2}-1\right) =50\left(20^{0.2}-1\right) \approx 50(1.82056-1) \approx 41.03.

The model therefore places 5% of the conditional excess distribution above approximately 41.03 units. The support is unbounded because $\xi>0$.

Worked Example: Negative Shape and a Finite Endpoint

Let $\sigma=8$ units and $\xi=-0.25$. Find the largest possible excess under the model, the probability that $Y>12$, and the 90th percentile.

The finite endpoint is

-\frac{\sigma}{\xi}=-\frac{8}{-0.25}=32.

Since $12<32$, the survival formula applies:

P(Y>12\mid X>u) =\left(1+\frac{(-0.25)(12)}{8}\right)^{-1/(-0.25)} =(1-0.375)^4 =(0.625)^4 =0.152587890625.

For the 90th percentile,

q_{0.90} =\frac{8}{-0.25}\left((0.10)^{0.25}-1\right) =-32\left(\sqrt[4]{0.10}-1\right) \approx -32(0.56234-1) \approx 14.01.

This is inside the support: $0<14.01<32$. The model assigns zero probability to excesses of 32 units or more, even though excesses near 32 are possible.

Worked Example: Zero Shape and the Exponential Case

Suppose $\xi=0$ and $\sigma=6$ units. Calculate the probability that an excess is above 9 units and the 80th percentile.

At zero shape, the survival function is exponential:

P(Y>9\mid X>u) =\exp(-9/6) =\exp(-1.5) \approx 0.2231.

The 80th percentile is

q_{0.80} =-6\log(1-0.80) =-6\log(0.20) \approx 9.6566.

About 22.3% of exceedances are greater than 9 units, while 80% are at or below approximately 9.66 units.

Interpreting Shape and Scale Carefully

The three shape cases can be compared directly. In each row, the support refers to the excess $Y$, not to the original observation $X$.

Shape	Support for $Y$	Tail behavior
$\xi>0$	$y\ge0$, unbounded above	Power-decreasing survival tail
$\xi=0$	$y\ge0$, unbounded above	Exponential survival tail
$\xi<0$	$0\le y\le-\sigma/\xi$	Finite upper endpoint

A simple way to see why the positive-shape tail is described as power-decreasing is to rewrite its survival function, for $\xi>0$, as

P(Y>y\mid X>u) =\left(1+\frac{\xi y}{\sigma}\right)^{-1/\xi}.

For large $y$, the quantity in parentheses grows proportionally to $y$, so the survival probability decreases at a power rate. By contrast, at $\xi=0$, it decreases exponentially. For $\xi<0$, the bracket reaches zero at $y=-\sigma/\xi$, and the survival probability reaches zero there. These distinctions explain why the shape parameter changes the kinds of extreme excesses a model permits.

If a question concerns an original observation $X$ and a threshold $u$, the model answers the conditional question

P(X>u+y\mid X>u)=P(Y>y\mid X>u).

To obtain the unconditional probability $P(X>u+y)$, one also needs the probability of crossing $u$:

P(X>u+y)=P(X>u)\,P(X>u+y\mid X>u).

The GPD excess distribution supplies the second factor; it does not specify the first factor on its own.

Common Mistake: Ignoring the Support

A frequent error is to substitute any proposed $y$ into the algebraic survival formula without first checking whether $y$ is in the support. For instance, with $\sigma=8$ and $\xi=-0.25$, substituting $y=40$ gives $1+\xi y/\sigma=1-1.25=-0.25$. Raising this negative number to the fourth power produces a positive number, but it is not a valid survival probability from the GPD formula. The formula's required condition, $1+\xi y/\sigma>0$, fails. The endpoint is 32, so $P(Y>40\mid X>u)=0$.

Another mistake is to treat $P(Y>y\mid X>u)$ as the probability that a new observation exceeds $u+y$ without conditioning. The correct unconditional probability is the conditional tail probability multiplied by $P(X>u)$. Keeping the support and the conditioning explicit prevents both errors.

Key takeaway: The GPD uses $\sigma>0$ to set the scale and $\xi$ to determine tail type and support. Check the support first, then use the survival or quantile formula; interpret every excess probability as conditional on crossing the threshold.

Check Your Understanding

Use the support condition as well as the formulas when answering.

For $\sigma=5$ and $\xi=-0.2$, what is the finite endpoint? Is an excess of 24 in the support?
For $\sigma=4$, $\xi=0$, calculate $P(Y>8\mid X>u)$.
Write the formula for the $p$-quantile when $\xi\ne0$, and state the corresponding formula when $\xi=0$.
If $P(X>u)=0.03$ and $P(Y>10\mid X>u)=0.20$, what is $P(X>u+10)$?
In words, distinguish the tail behavior for positive shape from the support behavior for negative shape.

Shape	Support for \(Y\)	Tail behavior
\(\xi>0\)	\(y\ge0\), unbounded above	Power-decreasing survival tail
\(\xi=0\)	\(y\ge0\), unbounded above	Exponential survival tail
\(\xi<0\)	\(0\le y\le-\sigma/\xi\)	Finite upper endpoint