Expectiles as Coherent Tail Risk Measures

Why Use an Expectile?

A tail risk measure should respond to the size of losses above its reported threshold, not only to the point where losses begin to enter the tail. An expectile does this by balancing the probability-weighted distances above and below a candidate value. Its balance gives larger losses more influence than smaller ones when the chosen level exceeds one-half.

Unlike a quantile, which is defined by a probability condition, an expectile is defined by an asymmetric loss function or, equivalently, a balance of expected positive and negative deviations. This tutorial develops that definition for loss random variables and shows why levels from one-half up to, but not including, one give coherent risk measures. Assume losses are integrable; the squared-error interpretation additionally requires a finite second moment.

Definition: For $0<\tau<1$, the expectile $e_\tau(X)$ of a loss $X$ is the unique real number $m$ satisfying $$ \tau E[(X-m)_+] = (1-\tau)E[(m-X)_+], $$ where $z_+=\max(z,0)$. The left side is the level-weighted expected excess loss above $m$; the right side is the complementary-level-weighted expected shortfall below $m$.

The parameter $\tau$ controls the asymmetry. At $\tau=1/2$, the two weights are equal, and the expectile is $E[X]$. At higher levels, an excess above $m$ receives more weight relative to a shortfall below it. The expectile is not generally a quantile: it need not leave a specified probability at or below itself.

The Asymmetric Squared-Error Interpretation

If $E[X^2]$ is finite, the same expectile minimizes an asymmetric squared-error objective:

L_\tau(m) = \tau E[((X-m)_+)^2] + (1-\tau)E[((m-X)_+)^2].

For values of $X$ above $m$, the squared deviation receives weight $\tau$; below $m$, it receives weight $1-\tau$. Differentiating with respect to $m$ gives

L_\tau'(m) = -2\tau E[(X-m)_+] + 2(1-\tau)E[(m-X)_+].

Thus the first-order condition $L_\tau'(m)=0$ is exactly the expectile balance equation. The objective is strictly convex, so its minimizer is unique. The balance definition itself only requires a finite first moment; the squared-error minimization argument requires a finite second moment.

For an integrable $X$, the balance also has a unique solution for every $0<\tau<1$. As $m$ increases, the left side decreases and the right side increases; their difference is continuous and strictly decreasing. It tends to a positive value without bound as $m$ tends downward without bound, and to a negative value without bound as $m$ tends upward without bound. Hence there is exactly one point of balance.

Worked Example: Solving a Two-Point Expectile

A hypothetical annual loss $X$, measured in thousands of dollars, is $0$ with probability $0.80$ and $10$ with probability $0.20$. Find $e_{0.80}(X)$. The specified distribution is the entire model; no interest or mortality assumptions are involved.

Method. Since the expectile must lie between the two possible loss values, set $0\le m\le10$. The positive excess occurs only when $X=10$; the shortfall occurs only when $X=0$. Substitute these outcomes into the balance:

$$ 0.80(0.20)(10-m) = (1-0.80)(0.80)(m-0). $$

The coefficients on the two sides are both $0.16$, so $10-m=m$, and therefore $m=5$. Check the balance: the weighted expected excess is $0.80(0.20)(10-5)=0.80$, and the weighted expected shortfall is $0.20(0.80)(5)=0.80$. Thus $e_{0.80}(X)=5$ thousand dollars.

Common mistake avoided. The ordinary mean is $0.80(0)+0.20(10)=2$, but that is not the $0.80$-expectile. Substituting $m=2$ into the balance gives a left side of $1.28$ and a right side of $0.32$, which are unequal. The balance, not the mean, determines the expectile.

Affine Transformations

Expectiles respect additions of certain cash amounts and nonnegative scaling. These properties are useful both in calculations and in establishing coherence. For a constant $b$ and $\lambda>0$, applying the balance equation to $Y=\lambda X+b$ gives

(\lambda X+b-m)_+ = \lambda\left(X-\frac{m-b}{\lambda}\right)_+, \qquad (m-\lambda X-b)_+ = \lambda\left(\frac{m-b}{\lambda}-X\right)_+.

The common factor $\lambda$ cancels from the balance. The unique solution is therefore $m=b+\lambda e_\tau(X)$. For $\lambda=0$, $Y$ is the constant $b$, whose expectile is $b$, so the same formula holds.

Worked Example: Scaling and Adding a Certain Amount

A hypothetical loss $X$ is $1$ with probability $0.75$ and $9$ with probability $0.25$. At level $\tau=0.75$, find the expectile of $Y=2X+3$, where the values are in thousands of dollars.

Method. First solve for $e_{0.75}(X)$. The solution lies between 1 and 9, so the balance is

$$ 0.75(0.25)(9-m) = 0.25(0.75)(m-1). $$

Both coefficients equal $0.1875$. Hence $9-m=m-1$, so $2m=10$ and $e_{0.75}(X)=5$. The affine property now gives

e_{0.75}(Y)=2e_{0.75}(X)+3=2(5)+3=13.

Check directly: $Y$ is $5$ with probability $0.75$ and $21$ with probability $0.25$. At $m=13$, the weighted excess is $0.75(0.25)(21-13)=1.5$, and the weighted shortfall is $0.25(0.75)(13-5)=1.5$. Thus the answer is $13$ thousand dollars.

Common mistake avoided. Averaging the transformed outcomes gives $0.75(5)+0.25(21)=9$, not the expectile. Another error is to calculate $2(5)+3$ as $12$; the correct arithmetic is $10+3=13$.

Why Levels at Least One-Half Are Coherent

A risk measure $\rho$ on losses is coherent when it satisfies four properties: monotonicity, translation invariance, positive homogeneity, and subadditivity. For expectiles, the first three can be seen from how the balance equation responds to ordering, constant additions, and positive scaling. To prove all four together, use a dual representation.

For $1/2\le\tau<1$, set $c=\tau/(1-\tau)$, so $c\ge1$. On a probability space carrying the loss, let $\mathcal D_\tau$ be the set of random variables $Z$ such that $Z\ge0$, $E[Z]=1$, and $\operatorname{ess\,sup}Z\le c\operatorname{ess\,inf}Z$. These are normalized probability weights whose largest value is at most $c$ times their smallest value. Then

Theorem: For an integrable loss $X$ and $1/2\le\tau<1$, $$ e_\tau(X)=\sup_{Z\in\mathcal D_\tau}E[ZX]. $$ This representation holds for losses on the same underlying probability space; the weights $Z$ can emphasize adverse outcomes, subject to the stated bound on their relative size.

Justification. Let $m=e_\tau(X)$, $A=E[(X-m)_+]$, and $B=E[(m-X)_+]$. The balance says $\tau A=(1-\tau)B$, or $cA=B$. For any $Z\in\mathcal D_\tau$, write $a=\operatorname{ess\,inf}Z$ and $b=\operatorname{ess\,sup}Z$. Then

E[Z(X-m)] \le bA-aB \le caA-aB =0.

Since $E[Z]=1$, this implies $E[ZX]\le m$. To attain $m$, assign unnormalized weight $c$ where $X>m$, weight $1$ where $X<m$, and, for example, weight $1$ where $X=m$. Call this weight $Z_0$. It has maximum-to-minimum ratio at most $c$, and

$$ E[Z_0(X-m)]=cA-B=0. $$

Normalize it by setting $Z=Z_0/E[Z_0]$. The ratio bound is unchanged, $E[Z]=1$, and $E[ZX]=m$. Thus the supremum is exactly the expectile.

Each coherence property now follows from the representation. If $X\le Y$ in every scenario, then $E[ZX]\le E[ZY]$ for each nonnegative $Z$, so $e_\tau(X)\le e_\tau(Y)$ (monotonicity). For a constant $b$, $E[Z(X+b)]=E[ZX]+b$, so $e_\tau(X+b)=e_\tau(X)+b$ (translation invariance). For $\lambda\ge0$, taking the supremum gives $e_\tau(\lambda X)=\lambda e_\tau(X)$ (positive homogeneity).

Finally, for losses $X$ and $Y$ on the same probability space,

e_\tau(X+Y) = \sup_{Z\in\mathcal D_\tau}E[Z(X+Y)] \le \sup_{Z\in\mathcal D_\tau}E[ZX] + \sup_{Z\in\mathcal D_\tau}E[ZY] = e_\tau(X)+e_\tau(Y).

This is subadditivity. The inequality allows the weighting that maximizes the combined portfolio to differ from the weighting that maximizes either component alone. It does not require independence. If only marginal distributions are known, however, a numerical portfolio expectile generally also requires an assumption about how the losses occur together.

Worked Example: Combining Two Independent Losses

Two hypothetical annual losses $X$ and $Y$, measured in thousands of dollars, are independent. Each is $0$ with probability $0.75$ and $8$ with probability $0.25$. Find the separate and combined $0.75$-expectiles and check subadditivity. Independence is the stated joint-model assumption.

Separate risks. For either component, the balance for a value $m$ between 0 and 8 is

$$ 0.75(0.25)(8-m)=0.25(0.75)m. $$

The coefficients are equal, so $8-m=m$, giving $e_{0.75}(X)=e_{0.75}(Y)=4$. Their sum is $4+4=8$.

Combined risk. By independence, $X+Y$ is 0 with probability $0.75^2=0.5625$, 8 with probability $2(0.75)(0.25)=0.3750$, and 16 with probability $0.25^2=0.0625$. Its expectile lies between 0 and 8. For a candidate $m$ in that interval, the expected excess is

$$ 0.375(8-m)+0.0625(16-m) =3+1-0.4375m =4-0.4375m. $$

The expected shortfall is $0.5625m$. Substitution into the balance gives

0.75(4-0.4375m)=0.25(0.5625m), \qquad 3-0.328125m=0.140625m, \qquad m=6.4.

Check: the expected excess is $0.375(8-6.4)+0.0625(16-6.4)=0.6+0.6=1.2$, and its weighted value is $0.75(1.2)=0.9$. The weighted shortfall is $0.25(0.5625)(6.4)=0.9$. Thus $e_{0.75}(X+Y)=6.4\le8=e_{0.75}(X)+e_{0.75}(Y)$.

Common mistake avoided. The combined distribution has probability $0.375$ at 8, not $0.25$: either component can be the one with loss 8. Omitting one of those two cases gives the wrong portfolio balance.

Interpretation and Exam Traps

Expectiles combine severity and probability: both the probability of a loss outcome and its distance from the candidate value enter the balance. This makes the reported value sensitive to the sizes of losses, unlike a quantile that is determined by a cumulative probability condition. At $\tau=1/2$, the expectile reduces to the mean. At higher levels, it emphasizes the upper side of the loss distribution.

The coherence result is specifically for $1/2\le\tau<1$. The dual proof depends on $c=\tau/(1-\tau)\ge1$; for levels below one-half, the same argument does not establish coherence. At $\tau=1$, the balance definition no longer has two positive weights, so that endpoint is excluded. Expectiles are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus as a choice of tail risk measure and as an application of the coherence axioms.

Do not swap the weights. The weight on excess losses is $\tau$; the weight on shortfalls is $1-\tau$.
Do not confuse an expectile with a quantile. A quantile solves a probability condition; an expectile solves a weighted-deviation balance.
Do not assume additivity. Coherence gives subadditivity, not equality. The dependence structure affects a portfolio calculation.
Check the candidate’s interval. For a two-point loss, the expectile should lie between the possible loss values. Verify the final value by substituting it back into both sides of the balance.

Key takeaway: The expectile $e_\tau(X)$ balances the $\tau$-weighted excess above it against the $(1-\tau)$-weighted shortfall below it. For $1/2\le\tau<1$, its dual representation as a supremum of weighted expected losses proves monotonicity, translation invariance, positive homogeneity, and subadditivity.

Check Your Understanding

For each problem, use the stated loss distribution and level, and verify any calculated expectile by its balance equation.

A hypothetical loss is 3 with probability 0.80 and 15 with probability 0.20. Find its $0.80$-expectile. (Answer: 9)
A hypothetical loss $X$ is 2 with probability 0.75 and 10 with probability 0.25. Find $e_{0.75}(3X-4)$, using the affine property and checking that the transformed loss values are consistent with your answer. (Answer: 14)
A loss is 0 or 10, each with probability 0.50. Find its $0.60$-expectile by writing both sides of the balance explicitly. (Answer: 6)
Two independent hypothetical losses each equal 0 with probability 0.80 and 10 with probability 0.20. Find the $0.75$-expectile of their sum. The sum takes values 0, 10, and 20; first calculate all three probabilities, then solve the balance. (Answer: $300/43$, approximately 6.9767)
Suppose $X\le Y$ in every scenario on a common probability space, and both losses are integrable. Use the dual representation for a level $0.75$ to state the ordering of their expectiles and identify the coherence property that justifies it. (Answer: $e_{0.75}(X)\le e_{0.75}(Y)$; monotonicity)

What You'll Learn