14.13 Expected Shortfall and coherent risk measures

14.13.1Definitions#

This is the sign convention of M02 · 2.8 restated for losses: there, with return $r$, $\text{VaR}_c = -q_{1-c}(r)$. ES is the average of the quantiles beyond $\alpha$. When $F_L$ is continuous, $\operatorname{ES}_\alpha(L) = \mathbb E[L \mid L \ge \operatorname{VaR}_\alpha(L)]$, which is the version in M02. When $L$ has atoms (option books, defaultable bonds, any discrete scenario set), the conditional-expectation version can fail to be subadditive, and (14.10) is the correct definition. Other names in the literature: CVaR, average VaR, tail VaR.

Subadditivity is the axiom that matters for a desk. It says merging two books cannot create risk, so the sum of the desk-level numbers is a conservative bound for the firm, and limits allocated to desks add up. A measure that violates it can be gamed: split a book in two and report less risk.

14.13.2VaR is not subadditive#

VaR satisfies axioms 1–3 because quantiles are monotone and commute with increasing affine maps. It fails 4.

Proofcounterexample

Two independent bonds each lose 100 with probability 0.04 and 0 otherwise. Take $\alpha = 0.95$. For each bond $F_{L_i}(0) = 0.96 \ge 0.95$, so $\operatorname{VaR}_{0.95}(L_i) = 0$. For the sum, $P(L_1 + L_2 = 0) = 0.96^2 = 0.9216 < 0.95$ and $P(L_1 + L_2 \le 100) = 1 - 0.04^2 = 0.9984 \ge 0.95$, so $\operatorname{VaR}_{0.95}(L_1 + L_2) = 100 > 0 + 0$.∎

The trading version is the deep out-of-the-money short option: a position that loses a lot with probability below $1-\alpha$ has a VaR equal to its usual small gain, with a negative sign. Two desks each selling such options report negative VaR; the firm holding both does not (Problem 14.18). VaR rewards exactly the negatively skewed exposure that M09 · 9.9 warns about. VaR is subadditive on linear portfolios of jointly elliptical returns (for example multivariate normal or multivariate $t$), which is why delta-normal VaR rarely shows the problem; see McNeil, Frey and Embrechts, Quantitative Risk Management (rev. ed., 2015), ch. 8. Option books are not linear and returns are not elliptical in a crash.

In the bond example, the ES at 95% of each bond is the average of the top 5% of its quantiles: 4% of the mass at 100 and 1% at 0, so $\operatorname{ES} = (0.04 \times 100 + 0.01 \times 0)/0.05 = 80$. For the sum, the top 5% contains the 0.16% atom at 200 and 4.84% of the 7.68% atom at 100: $\operatorname{ES} = (0.0016 \times 200 + 0.0484 \times 100)/0.05 = 103.2 \le 160$.

14.13.3ES is coherent#

Proof

If $U$ is uniform on $(0,1)$, then $q_U$ has the distribution of $L$, so $\mathbb E[(L - c)^+] = \int_0^1 (q_u - c)^+\,du$. Call the bracket in (14.11) $f(c)$.

At $c = q_\alpha$: $q_u \le q_\alpha$ for $u < \alpha$ and $q_u \ge q_\alpha$ for $u > \alpha$, because the quantile function is non-decreasing. Hence

$$f(q_\alpha) = q_\alpha + \frac{1}{1-\alpha}\int_\alpha^1 (q_u - q_\alpha)\,du = \frac{1}{1-\alpha}\int_\alpha^1 q_u\,du = \operatorname{ES}_\alpha(L).$$

For any other $c$, drop the non-negative integral over $(0,\alpha)$ and use $x^+ \ge x$ on $(\alpha,1)$:

$$f(c) \ge c + \frac{1}{1-\alpha}\int_\alpha^1 (q_u - c)\,du = \frac{1}{1-\alpha}\int_\alpha^1 q_u\,du = \operatorname{ES}_\alpha(L). \;\;∎$$

Proof

Monotonicity: $L_1 \le L_2$ a.s. implies $(L_1 - c)^+ \le (L_2 - c)^+$, so $f_{L_1}(c) \le f_{L_2}(c)$ for every $c$; take the minimum over $c$. Translation invariance: $f_{L+a}(c + a) = f_L(c) + a$, and $c \mapsto c + a$ is a bijection. Positive homogeneity: for $\lambda > 0$, $f_{\lambda L}(\lambda c) = \lambda f_L(c)$; for $\lambda = 0$ both sides are 0.

Subadditivity: let $c_i = \operatorname{VaR}_\alpha(L_i)$. Since $(x + y)^+ \le x^+ + y^+$,

$$\begin{aligned} \operatorname{ES}_\alpha(L_1 + L_2) &\le f_{L_1 + L_2}(c_1 + c_2) = c_1 + c_2 + \frac{\mathbb E[(L_1 - c_1 + L_2 - c_2)^+]}{1-\alpha} \\ &\le f_{L_1}(c_1) + f_{L_2}(c_2) = \operatorname{ES}_\alpha(L_1) + \operatorname{ES}_\alpha(L_2). \end{aligned}$$

No assumption on the dependence between $L_1$ and $L_2$ is used. Finally, $\operatorname{ES}_\alpha$ averages $q_u \ge q_\alpha$ over $u \in (\alpha, 1)$.∎

Representation (14.11) is also the practical reason ES is used for optimisation: replacing $\mathbb E$ by an average over scenarios turns "minimise ES subject to a return target" into a linear programme in $(c, w)$ (Rockafellar and Uryasev 2000).

14.13.4ES under the Student-$t$#

Proof

$g_\nu(x) = c_\nu(1 + x^2/\nu)^{-(\nu+1)/2}$. Differentiate $h(x) = (1 + x^2/\nu)^{-(\nu-1)/2}$:

$$h'(x) = -\frac{\nu-1}{2}\cdot\frac{2x}{\nu}\Big(1 + \frac{x^2}{\nu}\Big)^{-(\nu+1)/2} = -\frac{\nu-1}{\nu}\cdot\frac{x\,g_\nu(x)}{c_\nu}.$$

Hence, for $\nu > 1$ (so that $h(x) \to 0$ as $x \to \infty$),

$$\int_{t}^\infty x\,g_\nu(x)\,dx = \frac{\nu c_\nu}{\nu - 1}\,h(t) = \frac{\nu}{\nu-1}\Big(1 + \frac{t^2}{\nu}\Big)g_\nu(t) = g_\nu(t)\,\frac{\nu + t^2}{\nu - 1}.$$

Divide by $1 - \alpha = P(T_\nu > t_\alpha)$. The continuous case of (14.10) gives the conditional mean. ES commutes with increasing affine maps by translation invariance and positive homogeneity.∎

As $\nu \to \infty$, (14.12) tends to $\phi(z_\alpha)/(1-\alpha)$, the normal formula of M02 Proposition 2.14. A numerical integral reproduces (14.12): for $\nu = 4$, $\alpha = 0.975$, both give 3.99356.

14.13.5Estimating ES on the Nifty#

Three estimators, in increasing order of structure:

1. Historical. Sort the $n$ losses and average the worst $\lceil n(1-\alpha)\rceil$. No model, but at 97.5% a 500-day window gives 13 tail observations, and the estimate cannot exceed the worst loss in the window.
2. Parametric Student-$t$. Fit $(\mu, s, \nu)$ by maximum likelihood and apply (14.12). Uses every observation to estimate the tail shape. Wrong if the tail is not $t$-like.
3. Filtered (conditional). Fit a GARCH model with $t$ innovations (M13 · 13.4), standardise the residuals, and apply 1 or 2 to the residuals scaled by tomorrow's forecast volatility. This is the right estimator for a daily risk number, because most of the variation in tail risk is variation in volatility (M02 · 2.7).