Unit 11, lesson 84

Extreme value theory and tail models

How the maximum domain of attraction and the Pickands-Balkema-de Haan theorem lead to generalized Pareto tails, and how the tail index, high quantiles, expected shortfall and threshold stability follow.

  • Video 84
  • Attribution and tail risk measures

A desk with four years of daily losses, about a thousand observations, is asked for the loss level exceeded on one day in a thousand. The empirical quantile at that level rests on one or two observations, and a level further out rests on none. Extreme value theory supplies a family of tail laws, the generalized Pareto distributions, that excesses over a high threshold approximately follow under a precise hypothesis on the loss distribution. This lesson states that hypothesis, derives quantiles and expected shortfall from an assumed generalized Pareto tail, shows why they do not depend on the threshold within that model, and separates the exact algebra from the uncertainty of using it on data.

Key terms

Maximum domain of attraction

Definition Let L1,L2,…L_1, L_2, \dots be independent losses, positive for adverse outcomes, with common CDF FF under the physical measure P\mathbb{P}, and let xF=sup⁡{x:F(x)<1}≤∞x_F = \sup\{x : F(x) < 1\} \le \infty be the right endpoint of FF. Extreme value theory studies the limit laws of the largest losses, such as the maximum Mn=max⁡(L1,…,Ln)M_n = \max(L_1, \dots, L_n), whose CDF is P(Mn≤x)=F(x)n\mathbb{P}(M_n \le x) = F(x)^n because all nn losses must be at most xx. The distribution FF is in the maximum domain of attraction of the generalized extreme value law GξG_{\xi} with shape ξ∈R\xi \in \mathbb{R}, written F∈MDA⁡(Gξ)F \in \operatorname{MDA}(G_{\xi}), if there are constants an>0a_n > 0 and bn∈Rb_n \in \mathbb{R} with

P(Mn−bnan≤x)=F(anx+bn)n→Gξ(x)\begin{aligned} &\mathbb{P}\Bigl(\frac{M_n - b_n}{a_n} \le x\Bigr) \\[0.4em] &\quad = F(a_n x + b_n)^n \\[0.4em] &\quad \to G_{\xi}(x) \end{aligned}

for every real xx. The generalized extreme value CDF is

Gξ(x)=exp⁡(−(1+ξx)−1/ξ)\begin{aligned} &G_{\xi}(x) \\[0.4em] &= \exp\bigl(-(1 + \xi x)^{-1/\xi}\bigr) \end{aligned}

where 1+ξx>01 + \xi x > 0 and ξ≠0\xi \ne 0, and its limit as ξ→0\xi \to 0 is the Gumbel CDF G0(x)=exp⁡(−e−x)G_0(x) = \exp(-e^{-x}) for every real xx. Off its support, Gξ(x)G_{\xi}(x) equals 00 for x≤−1/ξx \le -1/\xi when ξ>0\xi > 0 and 11 for x≥−1/ξx \ge -1/\xi when ξ<0\xi < 0, and it is continuous everywhere. By the Fisher-Tippett-Gnedenko theorem, which is stated here without proof, every nondegenerate limit of normalized maxima has the form Gξ((x−μ)/s)G_{\xi}((x - \mu)/s) for some ξ\xi, μ\mu and s>0s > 0, and the location μ\mu and scale ss can be absorbed into ana_n and bnb_n. A domain with ξ>0\xi > 0 forces xF=∞x_F = \infty, one with ξ<0\xi < 0 forces xF<∞x_F < \infty, and ξ=0\xi = 0 allows either. The normalization is part of the condition, since MnM_n itself increases to xFx_F. Not every distribution has a domain of attraction, and the Poisson and geometric distributions, for example, belong to none.

In plain words The largest loss in a sample can only grow as the sample grows, but after shifting and rescaling, its distribution can settle on a generalized extreme value law, of one of three kinds according to the sign of ξ\xi. The shape parameter ξ\xi records how heavy the tail of a single loss is. Belonging to a domain of attraction is a hypothesis about FF, not something every loss distribution satisfies.

Example Let the daily loss be exponential with mean 1010 currency units, so F(x)=1−e−x/10F(x) = 1 - e^{-x/10} for x≥0x \ge 0. With an=10a_n = 10 and bn=10ln⁡nb_n = 10 \ln n, for x≥−ln⁡nx \ge -\ln n,

F(10x+10ln⁡n)n=(1−e−xn)n→exp⁡(−e−x),\begin{aligned} &F(10x + 10\ln n)^n \\[0.4em] &\quad = \Bigl(1 - \frac{e^{-x}}{n}\Bigr)^{n} \\[0.4em] &\quad \to \exp(-e^{-x}), \end{aligned}

so F∈MDA⁡(G0)F \in \operatorname{MDA}(G_0). For n=250n = 250 the Gumbel approximation puts the median of the largest of 250250 daily losses at 10(ln⁡250−ln⁡ln⁡2)≈58.8810(\ln 250 - \ln \ln 2) \approx 58.88, while the exact median, which solves F(m)250=0.5F(m)^{250} = 0.5, is about 58.8958.89. Over all xx the two CDFs differ by at most about 0.00110.0011. A loss with survival function 1−F(x)=x−41 - F(x) = x^{-4} for x≥1x \ge 1 is instead in MDA⁡(G1/4)\operatorname{MDA}(G_{1/4}), with an=n1/4/4a_n = n^{1/4}/4 and bn=n1/4b_n = n^{1/4}, and a loss uniform on a bounded interval is in MDA⁡(G−1)\operatorname{MDA}(G_{-1}).

Threshold excesses

Definition Fix a threshold u<xFu < x_F, so that the exceedance probability pu=P(L>u)=1−F(u)p_u = \mathbb{P}(L > u) = 1 - F(u) is positive. The excess over uu is Y=L−uY = L - u on the event {L>u}\{L > u\}, and its CDF under the conditional probability given L>uL > u is the excess distribution function

Fu(y)=P(L−u≤y∣L>u)=F(u+y)−F(u)1−F(u)\begin{aligned} &F_u(y) \\[0.4em] &= \mathbb{P}(L - u \le y \mid L > u) \\[0.4em] &= \frac{F(u + y) - F(u)}{1 - F(u)} \end{aligned}

for 0≤y<xF−u0 \le y < x_F - u, with Fu(y)=1F_u(y) = 1 for y≥xF−uy \ge x_F - u when xFx_F is finite. Since the event {L>u+y}\{L > u + y\} lies inside {L>u}\{L > u\}, the definition of conditional probability gives the exact factorization

P(L>u+y)=pu(1−Fu(y))\begin{aligned} &\mathbb{P}(L > u + y) \\[0.4em] &\quad = p_u \bigl(1 - F_u(y)\bigr) \end{aligned}

for every y≥0y \ge 0. When E[max⁡(L,0)]<∞\mathbb{E}[\max(L, 0)] < \infty, the mean excess e(u)=E[L−u∣L>u]e(u) = \mathbb{E}[L - u \mid L > u] is finite. In a sample, the excesses are the amounts by which observed losses pass uu, and the fraction of observations above uu estimates pup_u. Fitting a tail to them is beyond this lesson.

In plain words The tail beyond a threshold splits into two parts, how often the threshold is crossed and how far past it a crossing goes. The probability of a loss beyond u+yu + y is exactly the probability of crossing uu times the conditional probability of then going more than yy past it.

Example For an exponential loss with mean 2020 and the threshold u=60u = 60, the exceedance probability is pu=e−3≈0.0498p_u = e^{-3} \approx 0.0498. The exponential law is memoryless, so Fu(y)=1−e−y/20F_u(y) = 1 - e^{-y/20} for every threshold u≥0u \ge 0, and the excess is again exponential with mean 2020. The factorization gives P(L>100)=e−3⋅e−2=e−5≈0.00674\mathbb{P}(L > 100) = e^{-3} \cdot e^{-2} = e^{-5} \approx 0.00674.

Pickands-Balkema-de Haan theorem

Definition For a shape ξ∈R\xi \in \mathbb{R} and a scale β>0\beta > 0, let Hξ,βH_{\xi,\beta} be the generalized Pareto CDF of the next card, extended by 11 beyond its endpoint when ξ<0\xi < 0. The Pickands-Balkema-de Haan theorem, stated here without proof, says that F∈MDA⁡(Gξ)F \in \operatorname{MDA}(G_{\xi}) if and only if there is a positive measurable function β(u)\beta(u) with

sup⁡0≤y<xF−u∣Fu(y)−Hξ,β(u)(y)∣→0\begin{aligned} &\sup_{0 \le y < x_F - u} \bigl\lvert F_u(y) \\[0.4em] &\quad - H_{\xi,\beta(u)}(y) \bigr\rvert \to 0 \end{aligned}

as u↑xFu \uparrow x_F. The shape is the same ξ\xi as in the domain of attraction. The convergence is of an absolute difference of CDFs, uniformly over the possible excesses. It gives no rate, it does not make the relative error small for a tail probability that is itself small, and it does not say that any finite threshold has an exactly generalized Pareto excess law. The theorem concerns one distribution FF. Applying it to a sample also needs independent, identically distributed losses, or a separate treatment of clustering and changing volatility.

In plain words Under the extreme value hypothesis, excesses over a threshold look more and more like a generalized Pareto variable as the threshold rises toward the end of the distribution. The approximation is measured in probability units, so it can be good overall and still be poor far out in the tail, where every probability is tiny.

Example Let a standardized loss be standard normal, which is in MDA⁡(G0)\operatorname{MDA}(G_0), and take β(u)=(1−Φ(u))/φ(u)\beta(u) = (1 - \Phi(u))/\varphi(u), one valid choice for this case. At u=2u = 2 it gives β≈0.4214\beta \approx 0.4214, and the largest gap between FuF_u and the exponential CDF with mean β\beta is about 0.0380.038, near y≈0.75y \approx 0.75. At u=3u = 3 the gap is about 0.0220.022. Yet at u=2u = 2 the conditional probability of an excess beyond 22, a loss beyond 44, is about 0.001390.00139 exactly and about 0.008680.00868 under the exponential approximation, more than six times larger.

Generalized Pareto distribution

Definition The generalized Pareto distribution with shape ξ∈R\xi \in \mathbb{R} and scale β>0\beta > 0 is the law of a random variable YY with survival function

P(Y>y)=(1+ξyβ)−1/ξ\begin{aligned} &\mathbb{P}(Y > y) \\[0.4em] &\quad = \Bigl(1 + \frac{\xi y}{\beta}\Bigr)^{-1/\xi} \end{aligned}

for ξ≠0\xi \ne 0 and P(Y>y)=e−y/β\mathbb{P}(Y > y) = e^{-y/\beta} for ξ=0\xi = 0, at every y≥0y \ge 0 with 1+ξy/β>01 + \xi y/\beta > 0, and with survival 00 beyond that range. The scale β\beta is in loss units, here currency units, and the shape ξ\xi is dimensionless. The case ξ=0\xi = 0 is the limit of the others, because the logarithm of the survival, −ξ−1ln⁡(1+ξy/β)-\xi^{-1}\ln(1 + \xi y/\beta), tends to −y/β-y/\beta. For ξ>0\xi > 0 the support is [0,∞)[0, \infty) and the survival decays like the power y−1/ξy^{-1/\xi}, for ξ=0\xi = 0 it decays exponentially, and for ξ<0\xi < 0 the support is [0,−β/ξ][0, -\beta/\xi]. For ξ>0\xi > 0 the number 1/ξ1/\xi is the tail index, and some texts give that name to ξ\xi itself. The moment E[Yk]\mathbb{E}[Y^k] of order k>0k > 0 is finite exactly when ξ<1/k\xi < 1/k, and every moment is finite when ξ≤0\xi \le 0. In particular

E[Y]=β1−ξ,ξ<1,Var⁡(Y)=β2(1−ξ)2(1−2ξ)\begin{aligned} \mathbb{E}[Y] &= \frac{\beta}{1 - \xi}, \quad \xi < 1, \\[0.4em] \operatorname{Var}(Y) &= \frac{\beta^2}{(1 - \xi)^2(1 - 2\xi)} \end{aligned}

where the variance needs ξ<12\xi < \tfrac12. More generally, a result of Gnedenko stated here says that F∈MDA⁡(Gξ)F \in \operatorname{MDA}(G_{\xi}) with ξ>0\xi > 0 exactly when 1−F(x)=x−1/ξℓ(x)1 - F(x) = x^{-1/\xi}\ell(x) for a slowly varying ℓ\ell, meaning ℓ(tx)/ℓ(x)→1\ell(tx)/\ell(x) \to 1 as x→∞x \to \infty for every t>0t > 0. Then the moments of max⁡(L,0)\max(L, 0) of order below 1/ξ1/\xi are finite and those of order above 1/ξ1/\xi are infinite.

In plain words The scale sets the size of typical excesses, and the shape sets how quickly very large excesses become rare. A larger shape means a heavier tail and a smaller tail index. Once the shape reaches 1/k1/k, the moment of order kk is infinite, although every quantile is still finite.

Example With β=20\beta = 20 and ξ=0.2\xi = 0.2, the tail index is 55, the mean excess is 20/0.8=2520/0.8 = 25 currency units and the variance is 400/(0.64⋅0.6)≈1041.7400/(0.64 \cdot 0.6) \approx 1041.7, a standard deviation of about 32.2732.27. Moments of order below 55 are finite and the others infinite. With ξ=0.4\xi = 0.4 the tail index is 2.52.5, the mean is about 33.3333.33 and the variance about 5555.65555.6, but the third moment is infinite. With ξ=−0.2\xi = -0.2 an excess cannot pass 20/0.2=10020/0.2 = 100, so above a threshold of 100100 the model caps the loss at 200200. Whether a real loss is bounded depends on the contract and the size of the position, so a finite endpoint is an economic assumption as well as a mathematical one.

High quantile extrapolation

Definition Assume a tail model for a loss LL with CDF FF, in which the exceedance probability pu=P(L>u)p_u = \mathbb{P}(L > u) is positive and the excess law over uu is exactly Hξ,βH_{\xi,\beta}, with nothing assumed below uu. At a confidence level α∈(0,1)\alpha \in (0, 1), value at risk is the lower quantile and expected shortfall is the quantile integral,

VaR⁡α=inf⁡{x:F(x)≥α},ES⁡α=11−α∫α1VaR⁡s ds,\begin{aligned} \operatorname{VaR}_{\alpha} &= \inf\{x : F(x) \ge \alpha\}, \\[0.4em] \operatorname{ES}_{\alpha} &= \frac{1}{1 - \alpha}\int_{\alpha}^{1} \operatorname{VaR}_{s}\,ds, \end{aligned}

as in Expected shortfall and coherent risk. For x≥ux \ge u in the support, the factorization gives P(L>x)=pu(1+ξ(x−u)/β)−1/ξ\mathbb{P}(L > x) = p_u(1 + \xi(x - u)/\beta)^{-1/\xi}, which is continuous and strictly decreasing. If 1−pu<α<11 - p_u < \alpha < 1, setting it equal to 1−α1 - \alpha and solving gives qα=VaR⁡αq_{\alpha} = \operatorname{VaR}_{\alpha} as

qα=u+βξ(rξ−1),r=pu1−α,\begin{aligned} q_{\alpha} &= u + \frac{\beta}{\xi}\bigl(r^{\xi} - 1\bigr), \\[0.4em] r &= \frac{p_u}{1 - \alpha}, \end{aligned}

for ξ≠0\xi \ne 0, and qα=u+βln⁡rq_{\alpha} = u + \beta \ln r for ξ=0\xi = 0, the limit of the first formula. The condition α>1−pu\alpha > 1 - p_u makes r>1r > 1 and so qα>uq_{\alpha} > u, and when ξ<0\xi < 0 it keeps qαq_{\alpha} below the endpoint u−β/ξu - \beta/\xi. Given L>qαL > q_{\alpha}, the excess over qαq_{\alpha} is again generalized Pareto with shape ξ\xi and scale β+ξ(qα−u)\beta + \xi(q_{\alpha} - u), as the threshold stability card shows. The tail is continuous, so ES⁡α=E[L∣L>qα]\operatorname{ES}_{\alpha} = \mathbb{E}[L \mid L > q_{\alpha}], and for ξ<1\xi < 1

ES⁡α=qα+e(qα),e(x)=β+ξ(x−u)1−ξ.\begin{aligned} &\operatorname{ES}_{\alpha} = q_{\alpha} + e(q_{\alpha}), \\[0.4em] &e(x) = \frac{\beta + \xi(x - u)}{1 - \xi} . \end{aligned}

For ξ≥1\xi \ge 1 the tail has an infinite mean and ES⁡α=∞\operatorname{ES}_{\alpha} = \infty. These formulas are exact under the assumed tail. Using them on data is an extrapolation, since 1−α1 - \alpha can be far smaller than any frequency the sample can measure, and the uncertainty in the estimates of ξ\xi, β\beta and pup_u usually dominates the result.

In plain words The quantile is the point beyond the threshold where the tail probability has fallen from pup_u to 1−α1 - \alpha. The expected shortfall adds the average overshoot beyond that point, and with a positive shape the overshoot grows with the quantile.

Example Take the constructed tail with u=100u = 100, pu=0.05p_u = 0.05, β=20\beta = 20 and ξ=0.2\xi = 0.2, in currency units, and α=0.99\alpha = 0.99. The tail ratio is r=0.05/0.01=5r = 0.05/0.01 = 5 and 50.2≈1.379735^{0.2} \approx 1.37973, so q0.99≈100+100⋅0.37973≈137.973q_{0.99} \approx 100 + 100 \cdot 0.37973 \approx 137.973. Putting the exact quantile back into the tail formula 0.05(1+0.2(q−100)/20)−50.05(1 + 0.2(q - 100)/20)^{-5} returns 0.010.01. The scale beyond the quantile is 20+0.2⋅37.973≈27.59520 + 0.2 \cdot 37.973 \approx 27.595, the mean excess is 27.595/0.8≈34.49327.595/0.8 \approx 34.493, and ES⁡0.99≈172.466\operatorname{ES}_{0.99} \approx 172.466. With the same uu, pup_u and β\beta, the exponential case ξ=0\xi = 0 gives q0.99=100+20ln⁡5≈132.189q_{0.99} = 100 + 20\ln 5 \approx 132.189, while ξ=0.4\xi = 0.4 gives q0.99≈145.183q_{0.99} \approx 145.183 and ES⁡0.99≈208.638\operatorname{ES}_{0.99} \approx 208.638. The shape moves the expected shortfall much more than the quantile, because it enters both the scale of the overshoot and the denominator 1−ξ1 - \xi.

Threshold stability

Definition Suppose the excess over uu is exactly Hξ,βH_{\xi,\beta}, and take a higher threshold v≥uv \ge u, with v−u<−β/ξv - u < -\beta/\xi when ξ<0\xi < 0. Dividing survival functions,

P(L>v+y∣L>v)=(1+ξyβ(v))−1/ξ,β(v)=β+ξ(v−u),\begin{aligned} &\mathbb{P}(L > v + y \mid L > v) \\[0.4em] &\quad = \Bigl(1 + \frac{\xi y}{\beta(v)}\Bigr)^{-1/\xi}, \\[0.4em] &\beta(v) = \beta + \xi(v - u), \end{aligned}

so the excess over vv is generalized Pareto with the same shape and a scale that is linear in the threshold, and pv=pu(1+ξ(v−u)/β)−1/ξp_v = p_u(1 + \xi(v - u)/\beta)^{-1/\xi}. For ξ=0\xi = 0 the scale stays β\beta. The tail of LL above vv is the same function in both descriptions, so the quantile formula with vv, pvp_v and β(v)\beta(v) returns the same qαq_{\alpha} for every α\alpha with 1−α<pv1 - \alpha < p_v. For ξ<1\xi < 1 the mean excess e(v)=β(v)/(1−ξ)e(v) = \beta(v)/(1 - \xi) is linear in vv with slope ξ/(1−ξ)\xi/(1 - \xi). This exact stability belongs to the assumed law. With data, a higher threshold leaves fewer excesses and noisier estimates, while a lower one supplies more excesses whose law may be further from generalized Pareto. Estimated shapes that stay roughly constant across a range of thresholds support a choice, but the stability of the assumed law does not show that estimates from sparse data are stable.

In plain words If the tail beyond 100100 is generalized Pareto, then so is the tail beyond 120120, with the same shape and a larger scale, and both descriptions give the same quantiles. Moving the threshold inside the model only rewrites the same tail.

Example In the constructed tail, raise the threshold to v=120v = 120. Then β(120)=20+0.2⋅20=24\beta(120) = 20 + 0.2 \cdot 20 = 24 and p120=0.05⋅1.2−5≈0.020094p_{120} = 0.05 \cdot 1.2^{-5} \approx 0.020094. The new tail ratio is p120/0.01≈2.0094p_{120}/0.01 \approx 2.0094, its power 0.20.2 is 50.2/1.2≈1.149775^{0.2}/1.2 \approx 1.14977, and

q0.99=120+120 (1.14977−1)≈137.973,\begin{aligned} q_{0.99} &= 120 + 120\,(1.14977 - 1) \\[0.4em] &\approx 137.973, \end{aligned}

the same quantile as before. Exactly, both thresholds give q0.99=100⋅50.2q_{0.99} = 100 \cdot 5^{0.2}. The mean excess rises from 2525 at u=100u = 100 to 24/0.8=3024/0.8 = 30 at v=120v = 120, a slope of 0.2/0.8=0.250.2/0.8 = 0.25.

Check yourself

  1. In the constructed tail with u=100u = 100, pu=0.05p_u = 0.05, β=20\beta = 20 and ξ=0.2\xi = 0.2, find the value at risk and the expected shortfall at α=0.999\alpha = 0.999, and compare the value at risk with the exponential case ξ=0\xi = 0. Why is this an extrapolation for a sample of 10001000 daily losses?
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The tail ratio is r=0.05/0.001=50r = 0.05/0.001 = 50 and 500.2≈2.1867250^{0.2} \approx 2.18672, so

q0.999≈100+100⋅1.18672≈218.672.\begin{aligned} q_{0.999} &\approx 100 + 100 \cdot 1.18672 \\[0.4em] &\approx 218.672 . \end{aligned}

The scale beyond it is 20+0.2⋅118.672≈43.73420 + 0.2 \cdot 118.672 \approx 43.734, the mean excess is 43.734/0.8≈54.66843.734/0.8 \approx 54.668, and ES⁡0.999≈273.341\operatorname{ES}_{0.999} \approx 273.341. The exponential case gives 100+20ln⁡50≈178.240100 + 20\ln 50 \approx 178.240, about 4040 currency units lower. A sample of 10001000 losses has on average only one loss beyond the 99.999.9 percent quantile and about 5050 above the threshold, so the answer comes from the assumed shape of the tail, not from losses observed at that level. A small error in ξ\xi moves it a lot.

  1. A tail has u=100u = 100, pu=0.05p_u = 0.05, β=20\beta = 20 and ξ=−0.2\xi = -0.2. A colleague computes P(L>250)\mathbb{P}(L > 250) by putting y=150y = 150 into the generalized Pareto survival function and obtains a negative number. Find the mistake, the correct probability and the 9999 percent value at risk of this model.
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The survival formula holds only where 1+ξy/β>01 + \xi y/\beta > 0. Here 1−0.2⋅150/20=−0.51 - 0.2 \cdot 150/20 = -0.5, and raising it to the power −1/ξ=5-1/\xi = 5 gives −0.03125-0.03125, so the colleague reports 0.05⋅(−0.03125)=−0.00156250.05 \cdot (-0.03125) = -0.0015625. With ξ=−0.2\xi = -0.2 an excess cannot pass −β/ξ=100-\beta/\xi = 100, so the loss is at most 200200 and the correct answer is P(L>250)=0\mathbb{P}(L > 250) = 0. For the value at risk, r=5r = 5 and 5−0.2≈0.724785^{-0.2} \approx 0.72478, so

q0.99=100−100 (0.72478−1)≈127.522,\begin{aligned} q_{0.99} &= 100 - 100\,(0.72478 - 1) \\[0.4em] &\approx 127.522, \end{aligned}

which lies below the endpoint 200200, as it must. The expected shortfall is 127.522+(20−0.2⋅27.522)/1.2≈139.602127.522 + (20 - 0.2 \cdot 27.522)/1.2 \approx 139.602.

  1. Show that for 0<ξ<10 < \xi < 1 the ratio ES⁡α/VaR⁡α\operatorname{ES}_{\alpha}/\operatorname{VaR}_{\alpha} tends to 1/(1−ξ)1/(1 - \xi) as α→1\alpha \to 1 under a generalized Pareto tail. Evaluate the ratio at α=0.99\alpha = 0.99 with u=100u = 100, pu=0.05p_u = 0.05 and β=20\beta = 20, for ξ=0.2\xi = 0.2 and ξ=0.4\xi = 0.4. What happens when ξ≥1\xi \ge 1?
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Combining the two terms of the expected shortfall gives

ES⁡α=qα+β+ξ(qα−u)1−ξ=qα+β−ξu1−ξ.\begin{aligned} \operatorname{ES}_{\alpha} &= q_{\alpha} + \frac{\beta + \xi(q_{\alpha} - u)}{1 - \xi} \\[0.4em] &= \frac{q_{\alpha} + \beta - \xi u}{1 - \xi} . \end{aligned}

Dividing by qαq_{\alpha} leaves (1+(β−ξu)/qα)/(1−ξ)(1 + (\beta - \xi u)/q_{\alpha})/(1 - \xi). For ξ>0\xi > 0 the tail ratio rr grows without bound as α→1\alpha \to 1, so qα→∞q_{\alpha} \to \infty and the ratio tends to 1/(1−ξ)1/(1 - \xi). With ξ=0.2\xi = 0.2 the constant β−ξu\beta - \xi u is 00, so the ratio is exactly 1.251.25 at every level above 0.950.95, and indeed 172.466/137.973≈1.25172.466/137.973 \approx 1.25. With ξ=0.4\xi = 0.4 it is −20-20, and at α=0.99\alpha = 0.99 the ratio is 208.638/145.183≈1.437208.638/145.183 \approx 1.437, still well below the limit 1/0.6≈1.6671/0.6 \approx 1.667. When ξ≥1\xi \ge 1 the excess has an infinite mean, so the expected shortfall is infinite while every value at risk is finite. The formula with the denominator 1−ξ≤01 - \xi \le 0 then gives no meaningful number.

How this lesson connects