Expected shortfall and coherent risk

Definition Let $L$ be an integrable loss, so that $\mathbb{E}[\lvert L \rvert] < \infty$, with larger positive values worse, and let $\alpha \in (0, 1)$. With the lower quantiles $q_u(L) = \operatorname{VaR}_u(L)$ of Value at risk and quantile estimation, the expected shortfall at level $\alpha$ is

It is a finite amount in currency units, as the next card shows. Because $u \mapsto q_u$ is nondecreasing, $\operatorname{ES}_{\alpha}(L) \ge \operatorname{VaR}_{\alpha}(L)$, and because $\int_0^1 q_u\,du = \mathbb{E}[L]$, the average over the top part of the levels is at least the overall average, $\operatorname{ES}_{\alpha}(L) \ge \mathbb{E}[L]$. When $L$ has no atom at $q_{\alpha}$, the tail $\{L > q_{\alpha}\}$ has probability exactly $1 - \alpha$ and $\operatorname{ES}_{\alpha}(L) = \mathbb{E}[L \mid L > q_{\alpha}] = \mathbb{E}[L \mid L \ge q_{\alpha}]$. With an atom at the quantile both conditional means can differ from ES, because ES always averages exactly $1 - \alpha$ of probability. Some texts call this quantity conditional value at risk or average value at risk.

Definition The quantile integral can be evaluated through the quantile representation of $L$. If $U$ is uniform on $(0, 1)$, then $q_U(L)$ has the law of $L$, because $q_u \le \ell$ holds exactly when $u \le F_L(\ell)$. Let $q = q_{\alpha}(L)$ and $p = \mathbb{P}(L > q)$. For $\alpha < u \le F_L(q)$ the quantile equals $q$, and $q_u > q$ exactly when $u > F_L(q)$, so

The last term is finite because $L$ is integrable. The weight $F_L(q) - \alpha = 1 - \alpha - p$ is the part of the atom at $q$ that the tail needs, and it lies between $0$ and $\mathbb{P}(L = q)$ because $\mathbb{P}(L < q) \le \alpha \le F_L(q)$. Dividing by $1 - \alpha$ and using $\mathbb{E}[(L - q)^{+}] = \mathbb{E}[L\,\mathbf{1}_{\{L > q\}}] - qp$ gives

Definition For an integrable loss $L$, a level $\alpha \in (0, 1)$ and a real threshold $z$, put

Then $G$ is finite and convex, its minimizers are exactly the $z$ with

a closed interval that contains $q_{\alpha}(L)$, and

For the proof, the map $z \mapsto (\ell - z)^{+}$ is convex for each $\ell$, so $G$ is convex. The one-sided difference quotients of $(\ell - z)^{+}$ in $z$ are bounded by $1$ in absolute value, so dominated convergence moves the one-sided derivatives inside the expectation,

A point minimizes a convex function exactly when $G'_{-}(z) \le 0 \le G'_{+}(z)$, which is the displayed quantile condition. The lower quantile satisfies it, and the previous card gives $G(q_{\alpha}) = \operatorname{ES}_{\alpha}(L)$. The minimizer need not be unique, since the condition holds on a whole interval when $F_L$ equals $\alpha$ across a gap between loss values, but every minimizer gives the same value. For $n$ equally weighted scenarios whose losses $L_i(w)$ are affine in the portfolio weights $w$, the problem

subject to $u_i \ge L_i(w) - z$ and $u_i \ge 0$ for every scenario and to linear constraints on $w$, is a linear program. At an optimum each $u_i$ equals $(L_i(w) - z)^{+}$, so its optimal value, when it exists, is the smallest scenario ES over the allowed portfolios, found without sorting the losses of every trial portfolio.

Definition Let $\mathcal{Z}_{\alpha}$ be the set of random variables $Z$ on the same probability space with $0 \le Z \le 1/(1 - \alpha)$ almost surely and $\mathbb{E}[Z] = 1$. For an integrable loss $L$ and $\alpha \in (0, 1)$,

and the supremum is attained. Each $Z \in \mathcal{Z}_{\alpha}$ is the density $d\mathbb{Q}/d\mathbb{P}$ of a probability measure $\mathbb{Q}$ that multiplies probabilities by factors between $0$ and $1/(1 - \alpha)$, so ES is the largest expected loss over these reweightings. For the upper bound, take any $Z \in \mathcal{Z}_{\alpha}$ and any real $z$. Because $\mathbb{E}[Z] = 1$, $Z \ge 0$ and $Z \le 1/(1 - \alpha)$,

and the choice $z = q_{\alpha}$ gives $\mathbb{E}[ZL] \le \operatorname{ES}_{\alpha}(L)$. For attainment, let $q = q_{\alpha}$ and set $Z^{*} = 1/(1 - \alpha)$ on $\{L > q\}$, $Z^{*} = 0$ on $\{L < q\}$ and, when $\mathbb{P}(L = q) > 0$, $Z^{*} = \beta$ on $\{L = q\}$ with

The quantile inequalities $\mathbb{P}(L > q) \le 1 - \alpha \le \mathbb{P}(L \ge q)$ give $0 \le \beta \le 1/(1 - \alpha)$. Then $\mathbb{E}[Z^{*}] = 1$, and $\mathbb{E}[Z^{*}L]$ is the tail-mass formula of the quantile integral card divided by $1 - \alpha$, which is ES. Without an atom at $q$ one has $\mathbb{P}(L > q) = 1 - \alpha$, and the upper tail alone carries the weight.

Definition Let $\rho$ assign a real number to every integrable loss on a fixed probability space, with losses positive when adverse. It is a coherent risk measure if, for all integrable $L$, $L_1$ and $L_2$, every constant $c$ and every $\lambda \ge 0$,

where $L_1 \le L_2$ holds almost surely. These are monotonicity, translation equivariance, positive homogeneity and subadditivity. In the profit and loss convention $X = -L$, translation reads $\rho(X + m) = \rho(X) - m$, since a sure gain lowers risk. Expected shortfall is coherent, and each property follows from the dual envelope with one set $\mathcal{Z}_{\alpha}$ for every loss. Monotonicity holds because every $Z$ is nonnegative, translation because $\mathbb{E}[Z(L + c)] = \mathbb{E}[ZL] + c$, and positive homogeneity because $\mathbb{E}[Z\lambda L] = \lambda\,\mathbb{E}[ZL]$ with $\lambda \ge 0$. For subadditivity, every $Z \in \mathcal{Z}_{\alpha}$ gives

and the supremum over $Z$ on the left is $\operatorname{ES}_{\alpha}(L_1 + L_2)$. No independence or distributional assumption enters, because one set of weights serves every loss. VaR has the first three properties but not the fourth, as the previous lesson shows.

Definition For $n$ equally weighted losses with order statistics $L_{(1)} \le \dots \le L_{(n)}$ and $k = \lceil n\alpha \rceil$, the ES of the empirical law satisfies

which is the quantile integral of the empirical distribution, with the fractional weight $(k - n\alpha)/n$ on the cutoff observation. When $n\alpha$ is an integer it is the mean of the $n(1 - \alpha)$ largest losses. The formula is exact for the sample, but as an estimate of the ES of the loss distribution it is uncertain. About $n(1 - \alpha)$ observations determine it, they set both the cutoff and the severity of the tail, and a single new large loss can move it far. Dependence and changing volatility change its sampling behavior further. If $\mathbb{E}[L^{+}] = \infty$, the quantile integral diverges and $\operatorname{ES}_{\alpha}(L) = \infty$ at every level, although every VaR is finite. A heavy tail of gains alone does not make ES infinite, although the derivations above assumed $\mathbb{E}[\lvert L \rvert] < \infty$. Extreme value models extend a tail beyond the data, but they add assumptions about severity that the sample cannot confirm. Statistical theory for ES estimators is beyond this lesson.