Charta is on Google Play: practise on the go. Get the app
Charta

Examine the role the generalized Pareto (GP) distribution plays when POT is used.

FRM® Part II · Market Risk Measurement and Management · Parametric Approaches (II): Extreme Value · FRM2.MR.3.e

The idea

When POT is used, the generalized Pareto (GP) distribution models the excess x ≥ 0 over the threshold u, with shape ξ and scale β. A shape of zero makes the excess exponential, and a positive shape means a fat tail. Given that a loss exceeds the threshold, the excess has distribution function G = 1 − (1 + ξ × x ÷ β)^(−1/ξ). With n observations, Nu exceedances of the threshold and confidence c, the scaled tail probability is q = n ÷ Nu × (1 − c). The value at risk is the threshold plus a scaled quantile of the excess, VaR = u + (β ÷ ξ) × (q^(−ξ) − 1). Expected shortfall needs ξ < 1, because otherwise the mean is infinite. Its formula is ES = (VaR + β − ξ × u) ÷ (1 − ξ). Typical errors are leaving out the term n ÷ Nu that converts the tail probability into the exceedance sample, forgetting to add u back, and using the VaR formula for ES. The GP is what turns a handful of exceedances into an extrapolated extreme quantile.

Formulas

Generalised Pareto distribution of the excess

G = 1 − (1 + ξ × x ÷ β)^(−1 ÷ ξ)

ξ
the shape parameter, positive for a fat tail
x
the excess over the threshold, at least 0
β
the scale parameter

Conditional on exceeding the threshold. As ξ tends to 0 the excess becomes exponential.

POT estimate of VaR

VaR = u + (β ÷ ξ) × [(n ÷ Nu × (1 − c))^(−ξ) − 1]

u
the threshold
β
the GP scale
ξ
the GP shape
n
the total number of observations
Nu
the number of exceedances
c
the confidence level

The n ÷ Nu term converts the tail probability into the exceedance sample. Leaving it out, or forgetting to add u back, are the usual errors.

POT estimate of expected shortfall

ES = (VaR + β − ξ × u) ÷ (1 − ξ)

VaR
the POT VaR at the same confidence level
β
the GP scale
ξ
the GP shape, below 1
u
the threshold

Needs ξ < 1, or the mean excess and ES are infinite. Dropping the ξ × u term overstates ES.

POT tail probability

P(L > x) = (Nu ÷ n) × [1 + ξ × (x − u) ÷ β]^(−1 ÷ ξ)

L
the loss
x
the loss level, above u
Nu
the number of exceedances
n
the total number of observations
ξ
the GP shape
u
the threshold
β
the GP scale

x must exceed u, and the bracket uses the excess x − u. Without Nu ÷ n the result is only the probability conditional on exceeding u.

A practice question

From the question bank, with the reason each wrong option is wrong.

In the POT method, which distribution is used to model the excesses over a high threshold?

  1. The generalized extreme value distributionWhy not: GEV models block maxima, not threshold excesses.
  2. The generalized Pareto distribution - correct
  3. The normal distributionWhy not: The normal is thin-tailed and is not the limit law of threshold excesses.
  4. The Gumbel distributionWhy not: Gumbel is one GEV type (xi = 0) for block maxima, not the family for excesses.

By the Pickands-Balkema-de Haan theorem, the excesses over a sufficiently high threshold converge to a generalized Pareto distribution for a wide class of parent distributions.

A harder one

In percent of portfolio value, a POT fit uses n = 1,000 daily losses, Nu = 50 exceedances, threshold u = 2.0, GP scale beta = 1.0 and shape xi = 0.25. What is the 99 percent VaR, closest to?

  1. 2.50 percentWhy not: Omits the division by xi: u + beta x [(n / Nu x (1 - c))^(-xi) - 1].
  2. 1.98 percentWhy not: Forgets to add the threshold u back to the GP quantile.
  3. 10.65 percentWhy not: Applies 1 - c directly to the excess, omitting n / Nu, so the tail probability is not converted to the exceedance scale.
  4. 3.98 percent - correct

VaR = u + (beta / xi) x [(n / Nu x (1 - c))^(-xi) - 1] = 2 + 4 x [(20 x 0.01)^(-0.25) - 1] = 2 + 4 x (1.4953 - 1) = 3.98 percent.

The hardest tier

A POT model for daily losses (percent) uses n = 2,500, Nu = 125, threshold u = 1.5, scale beta = 0.9 and shape xi = 0.3. The 99.9 percent VaR is u + (beta / xi) x [(n / Nu x (1 - c))^(-xi) - 1] and the ES is (VaR + beta - xi x u) / (1 - xi). What is the 99.9 percent expected shortfall, closest to?

  1. 8.2 percentWhy not: Reports the VaR itself, not the expected shortfall.
  2. 12.4 percent - correct
  3. 13.0 percentWhy not: Leaves out the xi x u term in the ES numerator.
  4. 11.7 percentWhy not: Applies only the 1 / (1 - xi) scaling to the VaR and drops (beta - xi x u).

VaR = 1.5 + (0.9 / 0.3) x [(20 x 0.001)^(-0.3) - 1] = 1.5 + 3 x (3.2336 - 1) = 8.20 percent. ES = (8.20 + 0.9 - 0.3 x 1.5) / 0.7 = 12.36 percent.

In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.

Start free →

More in Parametric Approaches (II): Extreme Value