An expectile computation cookbook

Abstract

A substantial body of work in the last 15 years has shown that expectiles constitute an excellent candidate for becoming a standard tool in probabilistic and statistical modeling. Surprisingly, the question of how expectiles may be efficiently calculated has been left largely untouched. We fill this gap by, first, providing a general outlook on the computation of expectiles that relies on the knowledge of analytic expressions of the underlying distribution function and mean residual life function. We distinguish between discrete distributions, for which an exact calculation is always feasible, and continuous distributions, where a Newton–Raphson approximation algorithm can be implemented and a list of exceptional distributions whose expectiles are in analytic form can be given. When the distribution function and/or the mean residual life is difficult to compute, Monte-Carlo algorithms are introduced, based on an exact calculation of sample expectiles and on the use of control variates to improve computational efficiency. We discuss the relevance of our findings to statistical practice and provide numerical evidence of the performance of the considered methods.

Appendices

Proofs of the theoretical results

Proof of Theorem 2.1

Clearly

$$\begin{aligned}{} & {} \forall i\in I, \ \forall x\in [a_i,a_{i+1}), \ \varphi (x)\\{} & {} \quad =\int _{{\mathbb {R}}} (t-x)\mathbb {1}_{\{ t>x \}} \, \mu (\textrm{d}t) =\psi (a_i)-x\overline{F}(a_i) \end{aligned}$$

where $\psi (x)=\int _{{\mathbb {R}}} t \mathbb {1}_{\{ t>x \}} \, \mu (\textrm{d}t)$ and $\overline{F}(x)=1-F(x)$ with $F(x)=\mu ((-\infty ,x])$. Consequently, for any $i\in I$ and $x\in [a_i,a_{i+1})$,

$$\begin{aligned} (1-\tau ) g_{\tau }(x)&= -x\{ (2\tau -1) \overline{F}(a_i)+1-\tau \} \\&\quad + (2\tau -1) \psi (a_i) + (1-\tau )m \\&= -x( \tau \overline{F}(a_i)+(1-\tau )F(a_i) )\\&\quad + \tau \psi (a_i) + (1-\tau ) (m-\psi (a_i)) \end{aligned}$$

where $m=\int _{{\mathbb {R}}} x \, \mu (\textrm{d}x)=\sum _{k\in I} p_k a_k$. In other words, the function $g_{\tau }$ is continuous, piecewise linear and decreasing, and tends to $+\infty$ (resp. $-\infty$) as $x\rightarrow -\infty$ (resp. $x\rightarrow +\infty$). It follows that there is a unique index $i=i(\tau )$ such that the two inequalities

$$\begin{aligned}&\tau \psi (a_i) + (1-\tau ) (m-\psi (a_i)) \\&\quad \ge a_i( \tau \overline{F}(a_i)+(1-\tau )F(a_i) ) \text{ and } \\&\tau \psi (a_{i+1}) + (1-\tau ) (m-\psi (a_{i+1})) \\&\quad < a_{i+1}( \tau \overline{F}(a_{i+1})+(1-\tau )F(a_{i+1}) ) \end{aligned}$$

hold. With this index i, the expectile $\xi _{\tau }$ is the unique root of the linear function $g_{\tau }$ on the interval $[a_i,a_{i+1})$, namely:

$$\begin{aligned} \xi _{\tau }=\frac{\tau \psi (a_i) + (1-\tau ) (m-\psi (a_i))}{\tau \overline{F}(a_i)+(1-\tau )F(a_i)}. \end{aligned}$$

(A1)

Now $\psi (a_i)=\sum _{k>i} p_k a_k$ and $\overline{F}(a_i)=\sum _{k>i} p_k$, so that, for any i,

$$\begin{aligned}{} & {} \tau \psi (a_i) + (1-\tau ) (m-\psi (a_i)) \\{} & {} \qquad - a_i( \tau \overline{F}(a_i)+(1-\tau )F(a_i) ) \\{} & {} \quad = \tau \left( \sum _{k<i} p_k(a_i-a_k) \right. \\{} & {} \qquad \left. + \sum _{k>i} p_k(a_k-a_i) \right) - \sum _{k<i} p_k(a_i-a_k). \end{aligned}$$

The above pair of inequalities is therefore equivalent to

$$\begin{aligned}{} & {} \frac{\sum _{k<i} p_k (a_i-a_k)}{\sum _{k<i} p_k (a_i-a_k) + \sum _{k>i} p_k (a_k-a_i)} \\{} & {} \quad \le \tau< \frac{\sum _{k<i+1} p_k (a_{i+1}-a_k)}{\sum _{k<i+1} p_k (a_{i+1}-a_k) + \sum _{k>i+1} p_k (a_k-a_{i+1})} \end{aligned}$$

as announced. The two identities involving $\xi _{\tau }$ follow immediately from (A1). $\square$

Proof of Theorem 2.2

Define a continuous function $G=G_{\tau }$ by setting

$$\begin{aligned} G(x)=x-\frac{g_{\tau }(x)}{g_{\tau }'(x)}, \text{ so } \text{ that } x_{n+1}=G(x_n). \end{aligned}$$

Recall that $g_{\tau }:x\mapsto \frac{2\tau -1}{1-\tau }\varphi (x) + m-x$ is convex. It is also clear that the derivative $g_{\tau }'$ is negative. In particular, $g_{\tau }$ is decreasing, and since $\xi _{\tau }$ is the unique solution of the equation $g_{\tau }(x)=0$, one has $g_{\tau }(x_0)>0$ for any $x_0<\xi _{\tau }$. Then

$$\begin{aligned}&\forall x_0<\xi _{\tau }, \ G(x_0)-x_0 =-\frac{g_{\tau }(x_0)}{g_{\tau }'(x_0)}>0 \\&\quad \text{ and } \ G(x_0)-\xi _{\tau } {=}\left\{ \frac{\xi _{\tau }-x_0}{g_{\tau }(\xi _{\tau }){-}g_{\tau }(x_0)} {-} \frac{1}{g_{\tau }'(x_0)} \right\} g_{\tau }(x_0) \le 0 \end{aligned}$$

using the convexity property of $g_{\tau }$. It follows that for any $x_0<\xi _{\tau }$, $x_0<G(x_0)\le \xi _{\tau }$, and therefore, for any starting point $x_0<\xi _{\tau }$, the Newton–Raphson sequence of iterates $(x_n)$ is nondecreasing and bounded and hence convergent. The limit must be a root of $g_{\tau }$ by taking limits in the equation $x_{n+1}=G(x_n)$, meaning that $(x_n)$ converges to $\xi _{\tau }$.

A Taylor expansion of $g_{\tau }$ on the interval $[x_n,\xi _{\tau }]\subset [x_0,\xi _{\tau }]$ (on which $g_{\tau }$ is twice continuously differentiable) with remainder in integral form entails

$$\begin{aligned} 0= & {} g_{\tau }(\xi _{\tau }) =g_{\tau }(x_n)+(\xi _{\tau }-x_n)g_{\tau }'(x_n)\\{} & {} +\int _{x_n}^{\xi _{\tau }} (\xi _{\tau }-u) g_{\tau }''(u) \, \textrm{d}u. \end{aligned}$$

This is readily rewritten as

$$\begin{aligned} x_{n+1}-\xi _{\tau } = \frac{1}{g_{\tau }'(x_n)} \int _{x_n}^{\xi _{\tau }} (\xi _{\tau }-u) g_{\tau }''(u) \, \textrm{d}u. \end{aligned}$$

Then

$$\begin{aligned}&|x_{n+1}-\xi _{\tau }| \le \frac{\max _{[x_n,\xi _{\tau }]} g_{\tau }''}{2|g_{\tau }'(x_n)|} (\xi _{\tau }-x_n)^2 \\&\quad = \frac{1}{2} \frac{(2\tau -1)\max _{[x_n,\xi _{\tau }]} f}{1-\tau +(2\tau -1) \overline{F}(x_n)} |x_n-\xi _{\tau }|^2 \\&\quad \le \frac{1}{2} \frac{(2\tau -1)\max _{[x_n,\xi _{\tau }]} f}{1-\tau +(2\tau -1) \overline{F}(\xi _{\tau })} |x_n-\xi _{\tau }|^2 \\&\quad \le \frac{1}{2} \frac{(2\tau -1)\max _{[x_k,\xi _{\tau }]} f}{1-\tau +(2\tau -1) \overline{F}(\xi _{\tau })} |x_n-\xi _{\tau }|^2 \end{aligned}$$

for any $k<n$. The proof is complete. $\square$

Proof of Theorem 2.3

With the notation of the proof of Theorem 2.2, the function G is twice continuously differentiable in a neighborhood of $\xi _{\tau }$ with

$$\begin{aligned} G'(x)= & {} \frac{g_{\tau }(x) g_{\tau }''(x)}{\{ g_{\tau }'(x) \}^2}, \\ G''(x)= & {} \frac{g_{\tau }''(x)}{g_{\tau }'(x)} + \frac{g_{\tau }(x)}{\{ g_{\tau }'(x) \}^2} \left( g_{\tau }^{(3)}(x) - 2\frac{\{ g_{\tau }''(x) \}^2}{g_{\tau }'(x)} \right) . \end{aligned}$$

In particular, $G'(\xi _{\tau })=0$ and

$$\begin{aligned} G''(\xi _{\tau })= & {} \frac{g_{\tau }''(\xi _{\tau })}{g_{\tau }'(\xi _{\tau })}\\= & {} -\frac{(2\tau -1)f(\xi _{\tau })}{(2\tau -1)\overline{F}(\xi _{\tau })+1-\tau }. \end{aligned}$$

Use a Taylor expansion to get, for any n,

$$\begin{aligned} x_{n+1}-\xi _{\tau }=G(x_n)-G(\xi _{\tau })=\frac{1}{2} (x_n-\xi _{\tau })^2 G''(c_n) \end{aligned}$$

for some $c_n\in \left]x_n,\xi _{\tau }\right[$. It follows from Theorem 2.2 that $c_n\rightarrow \xi _{\tau }$ and therefore, using the continuity of $G''$,

$$\begin{aligned}{} & {} \lim _{n\rightarrow \infty } \frac{\xi _{\tau }-x_{n+1}}{(x_n-\xi _{\tau })^2}\\{} & {} \quad = -\frac{1}{2} G''(\xi _{\tau })\\{} & {} \quad =\frac{1}{2} \times \frac{(2\tau -1)f(\xi _{\tau })}{(2\tau -1)\overline{F}(\xi _{\tau })+1-\tau } \end{aligned}$$

as required. $\square$

Proof of Theorem 2.4

From the last chain of inequalities in the proof of Theorem 2.2, we get, for any n,

$$\begin{aligned} |x_{n+1}-\xi _{\tau }|\le & {} \frac{1}{2} \frac{(2\tau -1)\max _{[x_0,\xi _{\tau }]} f}{1-\tau +(2\tau -1) \overline{F}(\xi _{\tau })} |x_n-\xi _{\tau }|^2 \\\le & {} \frac{1}{2} \frac{(2\tau -1)\max _{[(1-\varepsilon )\xi _{\tau },\xi _{\tau }]} f}{1-\tau +(2\tau -1) \overline{F}(\xi _{\tau })} |x_n-\xi _{\tau }|^2. \end{aligned}$$

Take then $c>0$ sufficiently large and write

$$\begin{aligned} \overline{F}(x)= & {} x^{-1/\gamma } \left\{ c^{1/\gamma } \overline{F}(c) \exp \left( \int _c^x \eta (t) \frac{\textrm{d}t}{t} \right) \right\} \\{} & {} \text{ with } \eta (t) = \frac{1}{\gamma } - \frac{t f(t)}{\overline{F}(t)}. \end{aligned}$$

The function $\eta$ is continuous on $[c,\infty )$ and converges to 0 at infinity. By the representation theorem for regularly varying functions (see Theorem B.1.6 p.365 in de Haan and Ferreira 2006), $\overline{F}$ is regularly varying with index $-1/\gamma$, and then the von Mises condition ensures that f is regularly varying with index $-1/\gamma -1$. By the uniform convergence theorem for regularly varying functions (see Theorem B.1.4 p.363 in de Haan and Ferreira 2006),

$$\begin{aligned} \frac{\xi _{\tau }}{\overline{F}(\xi _{\tau })} \max _{[(1-\varepsilon )\xi _{\tau },\xi _{\tau }]} f= & {} \frac{\xi _{\tau } f(\xi _{\tau })}{\overline{F}(\xi _{\tau })} \max _{x\in [1-\varepsilon ,1]} \frac{f(\xi _{\tau } x)}{f(\xi _{\tau })}\\{} & {} \rightarrow \frac{(1-\varepsilon )^{-1/\gamma -1}}{\gamma } \text{ as } \tau \uparrow 1. \end{aligned}$$

Moreover (see Bellini et al. 2014; Daouia et al. 2018)

$$\begin{aligned} \frac{\overline{F}(\xi _{\tau })}{1-\tau } \rightarrow \frac{1}{\gamma } - 1 \text{ as } \tau \uparrow 1. \end{aligned}$$

Conclude from these two convergences that

$$\begin{aligned}{} & {} \frac{1}{2} \frac{(2\tau -1)\max _{[(1-\varepsilon )\xi _{\tau },\xi _{\tau }]} f}{1-\tau +(2\tau -1) \overline{F}(\xi _{\tau })}\\{} & {} \quad = \frac{1}{\xi _{\tau }} \left( (1-\varepsilon )^{-1/\gamma -1} \frac{1/\gamma -1}{2} + {\text {o}}(1) \right) \text{ as } \tau \uparrow 1. \end{aligned}$$

The proof is complete. $\square$

Proof of Theorem 3.1

We first prove the asymptotic normality statement, and for this, it is enough to show that $\widehat{\Sigma }_{12,n}/\widehat{\sigma }_n^2\rightarrow {\Sigma _{12}/\sigma ^2=\Sigma _{12}/\Sigma _{22}}$ in probability. By the law of large numbers, $\widehat{\sigma }_n^2\rightarrow \sigma ^2$ in probability, so it suffices to show that $\widehat{\Sigma }_{12,n}\rightarrow \Sigma _{12}$ in probability. Finally, since $\widehat{\xi }_{\tau ,n}\rightarrow \xi _{\tau }$ in probability (for example by Theorem 2 in Holzmann and Klar 2016), it is enough to prove that $\widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n})\rightarrow \overline{F}(\xi _{\tau })$ and $\widehat{\varphi }_n^{(2)}(\widehat{\xi }_{\tau ,n})\rightarrow \varphi ^{(2)}(\xi _{\tau })$ in probability.

Fix $\varepsilon >0$. Then $\widehat{\xi }_{\tau ,n}\in [\xi _{\tau }-\varepsilon ,\xi _{\tau }+\varepsilon ]$ with arbitrarily high probability as $n\rightarrow \infty$. Then clearly

$$\begin{aligned}{} & {} \left| \widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n})-\widehat{\overline{F}}_n(\xi _{\tau }) \right| \le \frac{1}{n} \sum _{i=1}^n | \mathbb {1}_{\{ X_i>\widehat{\xi }_{\tau ,n} \}} - \mathbb {1}_{\{ X_i>\xi _{\tau } \}} | \\{} & {} \quad \le \frac{1}{n} \sum _{i=1}^n \mathbb {1}_{\{ X_i\in [\xi _{\tau }-\varepsilon ,\xi _{\tau }+\varepsilon ] \}} \end{aligned}$$

with arbitrarily high probability as $n\rightarrow \infty$. The upper bound converges in probability to $\mu ([\xi _{\tau }-\varepsilon ,\xi _{\tau }+\varepsilon ])$, by the law of large numbers, which is arbitrarily small as $\varepsilon \downarrow 0$. Conclude that $\widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n}) = (\widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n})-\widehat{\overline{F}}_n(\xi _{\tau }))+\widehat{\overline{F}}_n(\xi _{\tau })\rightarrow \overline{F}(\xi _{\tau })$ in probability, by the law of large numbers again. We now prove that $\widehat{\varphi }_n(\widehat{\xi }_{\tau ,n})=\widehat{\varphi }_n^{(1)}(\widehat{\xi }_{\tau ,n})\rightarrow \varphi ^{(1)}(\xi _{\tau })=\varphi (\xi _{\tau })$ before turning to the convergence of $\widehat{\varphi }_n^{(2)}(\widehat{\xi }_{\tau ,n})$. Write

$$\begin{aligned}{} & {} (X-x) \mathbb {1}_{\{ X>x \}} - (X-x') \mathbb {1}_{\{ X>x' \}} \\{} & {} \quad =(x'-x) \mathbb {1}_{\{ X>x \}} + (X-x') ( \mathbb {1}_{\{ X>x \}} - \mathbb {1}_{\{ X>x' \}} ) \end{aligned}$$

to obtain

$$\begin{aligned}&\left| \widehat{\varphi }_n^{(1)}(\widehat{\xi }_{\tau ,n})-\widehat{\varphi }_n^{(1)}(\xi _{\tau }) \right| \\&\quad \le |\widehat{\xi }_{\tau ,n}-\xi _{\tau }| \widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n}) \\&\qquad + \frac{1}{n} \sum _{i=1}^n |X_i-\xi _{\tau }| | \mathbb {1}_{\{ X_i>\widehat{\xi }_{\tau ,n} \}} - \mathbb {1}_{\{ X_i>\xi _{\tau } \}} | \\&\quad \le \varepsilon \left( \widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n}) + \frac{1}{n} \sum _{i=1}^n \mathbb {1}_{\{ X_i\in [\xi _{\tau }-\varepsilon ,\xi _{\tau }+\varepsilon ] \}} \right) \le 2\varepsilon \end{aligned}$$

with arbitrarily high probability as $n\rightarrow \infty$. The upper bound is arbitrarily small as $\varepsilon \downarrow 0$, so that the law of large numbers yields $\widehat{\varphi }_n^{(1)}(\widehat{\xi }_{\tau ,n})\rightarrow \varphi ^{(1)}(\xi _{\tau })$ in probability. Finally

$$\begin{aligned} (X-x)^2 - (X-x')^2 = 2(x'-x)(X-x)-(x'-x)^2 \end{aligned}$$

so that, with arbitrarily high probability as $n\rightarrow \infty$,

$$\begin{aligned}&\left| \widehat{\varphi }_n^{(2)}(\widehat{\xi }_{\tau ,n})-\widehat{\varphi }_n^{(2)}(\xi _{\tau }) \right| \\&\quad \le |\widehat{\xi }_{\tau ,n}-\xi _{\tau }|^2\widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n}) + 2 |\widehat{\xi }_{\tau ,n}-\xi _{\tau }| \widehat{\varphi }_n^{(1)}(\widehat{\xi }_{\tau ,n})\\&\qquad + \frac{1}{n} \sum _{i=1}^n (X_i-\xi _{\tau })^2 | \mathbb {1}_{\{ X_i>\widehat{\xi }_{\tau ,n} \}} - \mathbb {1}_{\{ X_i>\xi _{\tau } \}} | \\&\quad \le \varepsilon \left( \varepsilon \widehat{\overline{F}}_n(\widehat{\xi }_{\tau ,n}) + 2 \widehat{\varphi }_n^{(1)}(\widehat{\xi }_{\tau ,n}) + \varepsilon \right. \\&\qquad \left. \times \frac{1}{n} \sum _{i=1}^n \mathbb {1}_{\{ X_i\in [\xi _{\tau }-\varepsilon ,\xi _{\tau }+\varepsilon ] \}} \right) \\&\quad \le 2\varepsilon \left( \varepsilon + \widehat{\varphi }_n^{(1)}(\widehat{\xi }_{\tau ,n}) \right) . \end{aligned}$$

Conclude in a similar fashion that $\widehat{\varphi }_n^{(2)}(\widehat{\xi }_{\tau ,n})\rightarrow \varphi ^{(2)}(\xi _{\tau })$ in probability, as required. The fact that $\widetilde{\xi }_{\tau ,n}$ has the lowest asymptotic variance among all asymptotically unbiased linear combinations of $\widehat{\xi }_{\tau ,n}$ and $\overline{X}_n-m$ is then obvious since

$$\begin{aligned} \widehat{\xi }_{\tau ,n}- \frac{\Sigma _{12}}{\sigma ^2} (\overline{X}_n-m) = \widetilde{\xi }_{\tau ,n} + {\text {o}}_{{\mathbb {P}}}(1/\sqrt{n}) \end{aligned}$$

has this property.

It remains to prove the assertions about the variance reduction factor $1-R(\tau ,\mu )$. This function is clearly zero at $\tau =1/2$. Note also that

$$\begin{aligned} \varphi ^{(2)}(x)={\mathbb {E}}((X-x)^2 \mathbb {1}_{\{ X>x \}})=2\int _x^{\infty } (t-x) \overline{F}(t) \, \textrm{d}t. \end{aligned}$$

As a result, and since $\tau \mapsto \xi _{\tau }$ is continuously differentiable on $I=(\tau _1,\tau _2)$ (see Proposition 1(iii) in Holzmann and Klar 2016), the function $\tau \mapsto 1-R(\tau ,\mu )$ is continuously differentiable on this interval. It remains to prove the statements about monotonicity. Set

$$\begin{aligned} u(\tau )&=u(\tau ,\mu ) =(1-\tau ) {\mathbb {E}}((X-\xi _{\tau })^2) \\&\qquad + (2 \tau -1) \varphi ^{(2)}(\xi _{\tau }) \\ \text{ and } v(\tau )&=v(\tau ,\mu ) =(1-\tau )^2 {\mathbb {E}}((X-\xi _{\tau })^2) \\&\qquad + (2 \tau -1) \varphi ^{(2)}(\xi _{\tau }) \end{aligned}$$

so that $R(\tau ,\mu )=(u(\tau ))^2/(\sigma ^2 v(\tau ))$ and therefore

$$\begin{aligned} \frac{\partial R}{\partial \tau }(\tau ,\mu ) = \frac{u(\tau )}{\sigma ^2 (v(\tau ))^2} (2 u'(\tau ) v(\tau ) - v'(\tau ) u(\tau )). \end{aligned}$$

Writing $u(\tau )=(1-\tau ) {\mathbb {E}}((X-\xi _{\tau })^2 \mathbb {1}_{\{ X<\xi _{\tau } \}}) + \tau {\mathbb {E}}((X-\xi _{\tau })^2 \mathbb {1}_{\{ X>\xi _{\tau } \}})$ yields in particular that $u(\tau )>0$ for any $\tau$, meaning that the partial derivative $\frac{\partial R}{\partial \tau }(\tau ,\mu )$ has the same sign as $2 u'(\tau ) v(\tau ) - v'(\tau ) u(\tau )$. Now

$$\begin{aligned} u'(\tau )&=2\varphi ^{(2)}(\xi _{\tau })-{\mathbb {E}}((X-\xi _{\tau })^2)\\&\quad -2\frac{\textrm{d}\xi _{\tau }}{\textrm{d}\tau } ( (2\tau -1)\varphi (\xi _{\tau })+(1-\tau )(m-\xi _{\tau }) ) \\&= 2\varphi ^{(2)}(\xi _{\tau })-{\mathbb {E}}((X-\xi _{\tau })^2) \text{(using } \text{(3)) } \\ \text{ and } v'(\tau )&=2\varphi ^{(2)}(\xi _{\tau })-2(1-\tau ){\mathbb {E}}((X-\xi _{\tau })^2)\\&\quad -2\frac{\textrm{d}\xi _{\tau }}{\textrm{d}\tau } ( (2\tau -1)\varphi (\xi _{\tau })+(1-\tau )^2(m-\xi _{\tau }) ) \\&=2\varphi ^{(2)}(\xi _{\tau })-2(1-\tau ){\mathbb {E}}((X-\xi _{\tau })^2)\\&\quad +2\tau (1-\tau )\frac{\textrm{d}\xi _{\tau }}{\textrm{d}\tau } (m-\xi _{\tau }) \text{(from }~(3) \text{ again). } \end{aligned}$$

Straightforward calculations yield

$$\begin{aligned}{} & {} 2 u'(\tau ) v(\tau ) - v'(\tau ) u(\tau ) \\{} & {} \quad =2\left( (1-2\tau ) {\mathbb {E}}((X-\xi _{\tau })^2 \mathbb {1}_{\{ X<\xi _{\tau } \}}) {\mathbb {E}}((X-\xi _{\tau })^2 \mathbb {1}_{\{ X>\xi _{\tau } \}})\right. \\{} & {} \qquad \left. + \tau (1-\tau ) u(\tau ) \frac{\textrm{d}\xi _{\tau }}{\textrm{d}\tau } (m-\xi _{\tau }) \right) . \end{aligned}$$

This quantity is positive on (0, 1/2) and negative on (1/2, 1) because $\tau \mapsto \xi _{\tau }$ is strictly increasing (see Proposition 1(ii) in Holzmann and Klar 2016) and $\xi _{1/2}=m$. The proof is complete. $\square$

Proof of Theorem 3.2

Let , where c is to be found so as to minimize the relative asymptotic variance of . Write

Applying Proposition 3.1 entails that the desired value of c satisfies

$$\begin{aligned} c \frac{q_{\alpha _n}}{\xi _{\tau _n}} {\rightarrow } -\frac{V_{12}}{V_{22}}= & {} - \left( \left\{ \min \left( \frac{\lambda }{1/\gamma -1}, 1 \right) \right\} ^{1-\gamma } \right. \\{} & {} \left. + \lambda \left\{ \min \left( \frac{\lambda }{1/\gamma -1}, 1 \right) \right\} ^{-\gamma } - \lambda \right) { \text{ as } n\rightarrow \infty .} \end{aligned}$$

Apply Proposition 1(i) in Daouia et al. (2020) to obtain

$$\begin{aligned} \frac{q_{\alpha _n}}{\xi _{\tau _n}} = \frac{q_{\alpha _n}}{q_{\tau _n}} \frac{q_{\tau _n}}{\xi _{\tau _n}} \rightarrow \lambda ^{-\gamma } (1/\gamma -1)^{\gamma } \text{ as } n\rightarrow \infty , \end{aligned}$$

leading to a choice of c minimizing the relative asymptotic variance of as

$$\begin{aligned} c= & {} - \left( \frac{\lambda }{1/\gamma -1} \right) ^{\gamma } \left( \left\{ \min \left( \frac{\lambda }{1/\gamma -1}, 1 \right) \right\} ^{1-\gamma } \right. \\{} & {} \left. + \lambda \left\{ \min \left( \frac{\lambda }{1/\gamma -1}, 1 \right) \right\} ^{-\gamma } - \lambda \right) . \end{aligned}$$

The statement on the asymptotic normality of with this choice of c is then immediate. Finding the maximum of $\lambda \mapsto C(\gamma ,\lambda )$ is done by noting that this function is decreasing past $1/\gamma -1$, and

$$\begin{aligned}{} & {} \forall \lambda \in (0,1/\gamma -1), \ C(\gamma ,\lambda )\\{} & {} \qquad =\frac{1-2\gamma }{2\gamma } \left( \frac{\lambda ^{1/2-\gamma }}{\gamma (1/\gamma -1)^{1-\gamma }} - \sqrt{\lambda } \right) ^2. \end{aligned}$$

Maximizing this function over $(0,1/\gamma -1)$ is straightforward and leads to the value $\lambda ^{\star }$ specified in the statement of Theorem 3.2. The last asymptotic normality result follows by plugging $\lambda ^{\star }$ into $C(\gamma ,\lambda )$. $\square$

Detailed calculations related to the examples

Distribution supported on a set with three elements (Example 2.2)

Let $\mu$ be the probability distribution on a set $\{a,b,c\}$ with $a<b<c$ characterized by $\mu (\{b\})=p$ and $\mu (\{c\})=q$, with $p,q>0$ and $p+q<1$. Then, from Corollary 2.1,

$$\begin{aligned} \xi _{\tau }= & {} \frac{\tau (pb+qc) + (1-\tau )(1-p-q)a}{(2\tau -1)(p+q)+1-\tau }, \\{} & {} \text{ for } \tau \le 1-\frac{q(c-b)}{(1-p)(b-a)+q(a+c-2b)}, \end{aligned}$$

and

$$\begin{aligned} \xi _{\tau } = \frac{\tau qc + (1-\tau )\{(1-p-q)a+pb\}}{(2\tau -1)q+1-\tau } \text{ otherwise }. \end{aligned}$$

In particular, for the distribution $\mu$ on $\{0,1,2\}$ with $\mu (\{1\})=p$ and $\mu (\{2\})=q$,

$$\begin{aligned} \xi _{\tau } = \left\{ \begin{array}{ll} \dfrac{\tau (p+2q)}{(2\tau -1)(p+q)+1-\tau } &{} \text{ for } \tau \le 1-\dfrac{q}{1-p}, \\ \dfrac{2\tau q + (1-\tau )p}{(2\tau -1)q+1-\tau } &{} \text{ otherwise. } \end{array} \right. \end{aligned}$$

Taking $p=p_1(1-p_2)+p_2(1-p_1)$ and $q=p_1 p_2$, for $p_1,p_2\in (0,1)$, yields the expectile of the sum of two independent random variables having Bernoulli distributions with parameters $p_1$ and $p_2$:

$$\begin{aligned} \xi _{\tau } = \left\{ \begin{array}{ll} \dfrac{\tau (p_1+p_2)}{(2\tau -1)(p_1+p_2-p_1p_2)+1-\tau }\\ \qquad \qquad \qquad \qquad \text{ for } \tau \le \dfrac{1-p_1-p_2+p_1p_2}{1-p_1-p_2+2p_1p_2}, \\ \dfrac{(2\tau -1) 2p_1p_2 + (1-\tau )(p_1+p_2)}{(2\tau -1)p_1p_2+1-\tau } \text{ otherwise. } \end{array} \right. \end{aligned}$$

Uniform distribution on $\{1,\ldots ,n\}$ (Example 2.3)

Fix $n\ge 2$. For the uniform distribution on $\{1,\ldots ,n\}$, solving the inequalities of Corollary 2.2 is equivalent to finding the unique index $i\in \{1,\ldots ,n-1\}$ such that

$$\begin{aligned}{} & {} \frac{i(i-1)}{i(i-1)+(n-i)(n-i+1)} \le \tau \\{} & {} \quad < \frac{i(i+1)}{i(i+1)+(n-i)(n-i-1)}. \end{aligned}$$

This is equivalent to finding the unique solution (which we already know to exist, by Corollary 2.2) to the inequalities $P_{\tau }(i+1)<0\le P_{\tau }(i)$ for $i\in \{1,\ldots ,n-1\}$, where $P_{\tau }$ is the polynomial

$$\begin{aligned} P_{\tau }(x)=(2\tau -1)x^2 - \{ 2\tau (n+1)-1 \} x + \tau n(n+1). \end{aligned}$$

This polynomial has discriminant $4\tau (1-\tau )(n+1)(n-1)+1>0$ and then (for $\tau \ne 1/2$) two real roots $x_{\tau ,-}$ and $x_{\tau ,+}$ defined as

$$\begin{aligned} x_{\tau ,\pm } = \frac{2\tau (n+1)-1 \pm \sqrt{4\tau (1-\tau )(n+1)(n-1)+1}}{2(2\tau -1)}. \end{aligned}$$

A straightforward calculation yields $P_{\tau }(1)=\tau n(n-1)>0$ and $P_{\tau }(n)=-(1-\tau )n(n-1)<0$. It follows that when $\tau >1/2$ (resp. $\tau <1/2$), only the lowest (resp. largest) of the two roots $x_{\tau ,-}$ and $x_{\tau ,+}$ belongs to the interval [1, n]. Conclude that, in both cases, $P_{\tau }(i+1)<0\le P_{\tau }(i) \Leftrightarrow i\le x_{\tau ,-}<i+1 \Leftrightarrow i=\lfloor x_{\tau ,-} \rfloor$. With this index i,

$$\begin{aligned} \xi _{\tau }=\frac{\tau n(n+1) - (2\tau -1) i(i+1)}{2\tau n-2(2\tau -1)i}. \end{aligned}$$

Consequently

$$\begin{aligned} \xi _{\tau } = \left\{ \begin{array}{l} \dfrac{n(n+1)}{2} \text{ when } \tau =1/2, \\ \dfrac{\tau n(n+1) - (2\tau -1) \lfloor x_{\tau } \rfloor (\lfloor x_{\tau } \rfloor +1)}{2\tau n-2(2\tau -1)\lfloor x_{\tau } \rfloor } \text { otherwise, } \end{array} \right. \end{aligned}$$

with

$$\begin{aligned} x_{\tau }=\dfrac{2\tau (n+1)-1 - \sqrt{4\tau (1-\tau )(n+1)(n-1)+1}}{2(2\tau -1)}. \end{aligned}$$

Geometric distribution (Example 2.4)

For the geometric distribution with success probability $p\in (0,1)$, namely, $\mu (\{k\})=p(1-p)^{k-1}$ for any positive integer k, the inequalities of Theorem 2.1 read as

$$\begin{aligned} \frac{(1-p)^i - (1-pi)}{2(1-p)^i - (1-pi)}\le \tau <\frac{(1-p)^{i+1} - (1-p(i+1))}{2(1-p)^{i+1} - (1-p(i+1))}. \end{aligned}$$

Solving these inequalities is equivalent to finding the index $i\ge 1$ such that $h_{\tau }(i+1)<0\le h_{\tau }(i)$, where

$$\begin{aligned} h_{\tau }(x)=(2\tau -1)(1-p)^x - (1-\tau )(px-1). \end{aligned}$$

Straightforward calculations reveal that the unique root $x_{\tau }$ of $h_{\tau }$ over $[1,+\infty )$ satisfies the equation

$$\begin{aligned}{} & {} -\log (1-p)\left( x_{\tau }-\frac{1}{p} \right) \exp \left( -\log (1-p)\left( x_{\tau }-\frac{1}{p} \right) \right) \\{} & {} \quad =\frac{2\tau -1}{1-\tau } \left( -\frac{\log (1-p)}{p} (1-p)^{1/p} \right) . \end{aligned}$$

This is a transcendental equation (unless $\tau \ne 1/2$, for which $x_{1/2}=1/p$). Nevertheless, since by construction

$$\begin{aligned} -\log (1-p)\left( x_{\tau }-\frac{1}{p} \right) \ge -\log (1-p)\left( 1-\frac{1}{p} \right) >-1 \end{aligned}$$

for any $p\in (0,1)$, and

$$\begin{aligned}{} & {} \frac{2\tau -1}{1-\tau } \left( -\frac{\log (1-p)}{p} (1-p)^{1/p} \right) \\{} & {} \quad = \frac{2\tau -1}{1-\tau } \{ - (1-p)^{1/p} \log ((1-p)^{1/p}) \} > -e^{-1} \end{aligned}$$

for any $\tau ,p\in (0,1)$, one may express $x_{\tau }$ using the main branch of Lambert’s W function, that is

$$\begin{aligned} x_{\tau } = \frac{1}{p} - \frac{1}{\log (1-p)} W\left( -\frac{(1-p)^{1/p}\log (1-p)}{p} \frac{2\tau -1}{1-\tau } \right) \end{aligned}$$

where, for $x>0$, W(x) is the unique (positive) solution to the equation $w e^w=x$. The main branch of the Lambert function is available numerically in R using (for instance) the gsl package (Hankin et al. 2023), acting as a wrapper for the GNU Scientific Library.

Note further that when $\tau >1/2$, the function $h_{\tau }$ is obviously decreasing, so the inequalities $h_{\tau }(i+1)<0\le h_{\tau }(i)$ are equivalent to $i\le x_{\tau }<i+1$, i.e. $i=\lfloor x_{\tau } \rfloor$. When $\tau <1/2$, it is readily shown that $h_{\tau }'$ is decreasing and $h_{\tau }'(1)=-(1-2\tau ) (1-p)\log (1-p)-(1-\tau )p<0$ for any $p\in (0,1)$, so again $h_{\tau }$ is decreasing and $h_{\tau }(i+1)<0\le h_{\tau }(i) \Leftrightarrow i=\lfloor x_{\tau } \rfloor$. Conclude that

$$\begin{aligned} \xi _{\tau }= & {} \frac{(2\tau -1)(1-p)^{\lfloor x_{\tau } \rfloor } (1+p\lfloor x_{\tau } \rfloor )+1-\tau }{p\{ (2\tau -1)(1-p)^{\lfloor x_{\tau } \rfloor }+1-\tau \}} \text{ with } \\ x_{\tau }= & {} \frac{1}{p} - \frac{1}{\log (1-p)} W\left( -\frac{(1-p)^{1/p}\log (1-p)}{p} \frac{2\tau -1}{1-\tau } \right) . \end{aligned}$$

Cardano and Ferrari formulae (relevant to Sect. 2.3.1)

To solve a real cubic polynomial equation of the form $x^3+bx^2+cx+d=0$, Cardano’s method consists in letting $p=c-b^2/3$ and $q=d+b(2b^2-9c)/27$, and in computing the discriminant $\Delta =-4p^3-27q^2$ of the so-called depressed cubic $X^3+pX+q=0$. If $\Delta \le 0$, then the unique real root of the equation is given by

$$\begin{aligned} x= \root 3 \of {\frac{-q+\sqrt{\frac{-\Delta }{27}}}{2}} + \root 3 \of {\frac{-q-\sqrt{\frac{-\Delta }{27}}}{2}} -\frac{b}{3}. \end{aligned}$$

If on the contrary $\Delta >0$, then necessarily $p<0$ and there are 3 real solutions, given by Viète’s formula:

$$\begin{aligned} x= 2 \sqrt{\frac{-p}{3}} \cos \left( \frac{1}{3} \arccos \left( \frac{3q}{2p} \sqrt{\frac{3}{-p}} \right) + \frac{2k \pi }{3} \right) - \frac{b}{3}, \end{aligned}$$

$k \in \{ 0,1,2 \}$. To solve a real quartic polynomial equation $x^4+bx^3+cx^2+dx+e=0$, Ferrari’s method first finds a root $\lambda$ to the cubic equation $8 \lambda ^3 -4c \lambda ^2+(2bd-8e) \lambda -b^2 e+4ce-d^2=0$. The four (possibly complex) solutions to the quartic equation are then

$$\begin{aligned}{} & {} \frac{\varepsilon _1 \sqrt{2 \lambda -c+\frac{b^2}{4}}}{2} - \frac{b}{4} \\{} & {} \quad + \frac{\varepsilon _2 \sqrt{-2 \lambda -c -\varepsilon _1 \left( \frac{2(d-b) \lambda }{\sqrt{2 \lambda -c+\frac{b^2}{4}}}+ b \sqrt{2 \lambda -c+\frac{b^2}{4}} \right) + \frac{b^2}{2}}}{2} \end{aligned}$$

where $\varepsilon _1, \varepsilon _2 \in \{ -1,1 \}$.

Student distribution with $\nu =4$ degrees of freedom (Example 2.13)

Consider the Student distribution with $\nu$ degrees of freedom, having probability density function

$$\begin{aligned} f(x)=\frac{\Gamma ((\nu +1)/2)}{\sqrt{\nu \pi } \Gamma (\nu /2)} \left( 1 + \frac{x^2}{\nu } \right) ^{-(\nu +1)/2}, \ x\in {\mathbb {R}}. \end{aligned}$$

Here $\Gamma$ is Euler’s Gamma function. When $\nu =4$, the probability density function simplifies to

$$\begin{aligned} f(x)=\frac{3}{8} \left( 1+\frac{x^2}{4} \right) ^{-5/2}, \ x\in {\mathbb {R}}. \end{aligned}$$

The change of variables $t=2\tan (\theta )$ combined with the trigonometric identities $\cos (3\phi )=4\cos ^3(\phi )-3\cos (\phi )$, $\sin (3\phi )=3\sin (\phi )-4\sin ^3(\phi )$ and $\sin (\arctan \theta )=\theta /\sqrt{1+\theta ^2}$ then yield, after straightforward calculations, the following closed form for the survival function:

$$\begin{aligned} \overline{F}(x)=\int _x^{\infty } f(t) \, \textrm{d}t=\frac{1}{2} - \frac{x}{8} \frac{3+x^2/2}{(1+x^2/4)^{3/2}}. \end{aligned}$$

Further straightforward calculations based on the change of variables $u=t^2$ then provide

$$\begin{aligned} \varphi (x)=\int _x^{\infty } \overline{F}(t)\,\textrm{d}t= \frac{1}{2} \left( \frac{x^2+2}{\sqrt{x^2+4}} - x \right) . \end{aligned}$$

Since the Student distribution is centered, $m=0$ and Eq. (3) is

$$\begin{aligned} \xi _{\tau }^4 +4 \xi _{\tau }^2- \frac{(2 \tau -1)^2}{\tau (1-\tau )}=0. \end{aligned}$$

This is a biquadratic equation, leading to $\xi _{\tau }^2 = -2+1/\sqrt{\tau (1-\tau )}$ because $\xi _{\tau }^2\ge 0$, and then

$$\begin{aligned} \xi _{\tau }= {\text {sign}}(2 \tau -1) \sqrt{\frac{1}{\sqrt{\tau (1-\tau )}}-2}. \end{aligned}$$

In general, the distribution function and mean residual life function of the Student distribution involve the hypergeometric function. It is not hard to see that, while the distribution function and mean residual life function can in fact be written in closed form when $\nu$ is an even integer, resulting in a polynomial equation characterizing $\xi _{\tau }$, only the cases $\nu \in \{2,4,6\}$ result in an equation of degree 4 or lower.

Fisher distribution with (4, 4) degrees of freedom (Example 2.14)

The Fisher distribution with degrees of freedom $\nu _1>0$ and $\nu _2>0$ has density function

$$\begin{aligned} f(x)=\frac{(\nu _1/\nu _2)^{\nu _1/2}}{B(\nu _1/2,\nu _2/2)} x^{\nu _1/2-1} (1+\nu _1 x/\nu _2)^{-(\nu _1+\nu _2)/2}, \end{aligned}$$

$x>0$, where B is the Beta function. In the specific case $\nu _1=\nu _2=4$, one finds $\varphi (x)=(3x+2)/(x+1)^2$ for $x>0$, and $m=2$. Equation (3) is thus equivalent to the cubic equation

$$\begin{aligned} \xi _{\tau }^3-\frac{3 \tau }{1-\tau } \xi _{\tau }-\frac{2 \tau }{1-\tau }=0. \end{aligned}$$

The discriminant of this equation is $\Delta =108 \tau ^2 (2 \tau -1)/(1-\tau )^3$. If $\tau \le 1/2$, then $\Delta \le 0$ and the unique solution is

$$\begin{aligned} \xi _{\tau }= \root 3 \of {\frac{\tau }{1-\tau }} \left( \root 3 \of {1+\sqrt{\frac{1-2\tau }{1-\tau }}} + \root 3 \of {1-\sqrt{\frac{1-2\tau }{1-\tau }}} \right) . \end{aligned}$$

If now $\tau > 1/2$, then $\Delta > 0$ and the 3 possible solutions are

$$\begin{aligned} \xi _{\tau }= 2 \sqrt{\frac{\tau }{1-\tau }} \cos \left( \frac{1}{3} \arccos \left( \sqrt{\frac{1-\tau }{\tau }} \right) + \frac{2 k \pi }{3} \right) , \end{aligned}$$

$k \in \{ 0,1,2 \}$. Since $\tau > 1/2$, $\arccos ( \sqrt{(1-\tau )/\tau } )\in [0,\pi /2]$, and therefore (taking the constraint $\xi _{\tau } \ge 0$ into account) $k=0$ is the only admissible solution, namely

$$\begin{aligned} \xi _{\tau }= 2 \sqrt{\frac{\tau }{1-\tau }} \cos \left( \frac{1}{3} \arccos \left( \sqrt{\frac{1-\tau }{\tau }} \right) \right) . \end{aligned}$$

Table 3 Table of expectiles of the standard Gaussian distribution, computed via the Newton–Raphson algorithm (see Example 2.12)

Table 4 Table of expectiles of the standard log-normal distribution, computed via the Newton–Raphson algorithm

Pareto distribution with extreme value index 1/4 (Example 2.15)

The Pareto distribution with extreme value index $\gamma >0$ has survival function $\overline{F}(x)=x^{-1/\gamma }$ for $x>1$. This distribution has a finite first moment when $\gamma <1$, and since $\varphi (x)=\gamma x^{1-1/\gamma }/(1-\gamma )$ for $x>1$ and $m=1/(1-\gamma )$, Eq. (3) leads to

$$\begin{aligned} (1-\gamma )(1-\tau ) \xi _{\tau }^{1/\gamma } - (1-\tau ) \xi _{\tau }^{1/\gamma -1} - \gamma (2\tau -1)=0. \end{aligned}$$

When $\gamma =1/4$, this is the quartic equation $\xi _{\tau }^4+b \xi _{\tau }^3 + c \xi _{\tau }^2 + d \xi _{\tau }+ e=0$, where $b=-4/3$, $c=0$, $d=0$ and $e=(1-2 \tau )/(3(1-\tau ))$. Ferrari’s method leads to finding $\lambda =\lambda _{\tau }$ which is a root of the cubic equation

$$\begin{aligned} \lambda _{\tau }^3 - \frac{1-2 \tau }{3(1-\tau )} \lambda _{\tau } -\frac{2}{9} \frac{1-2 \tau }{3(1-\tau )}=0. \end{aligned}$$

The discriminant of this equation is

$$\begin{aligned} \Delta = -\frac{4}{27} \frac{(1-2 \tau )^2}{(1-\tau )^2} \frac{\tau }{1-\tau } \le 0 \end{aligned}$$

for all $\tau \in (0,1)$, from which the unique solution of the equation involving $\lambda$ is

$$\begin{aligned} \lambda _{\tau }= & {} \root 3 \of {\frac{1-2 \tau +|1-2 \tau | \sqrt{\frac{\tau }{1-\tau }}}{27 (1-\tau )}} \\{} & {} + \root 3 \of {\frac{1-2 \tau -|1-2 \tau | \sqrt{\frac{\tau }{1-\tau }}}{27 (1-\tau )}}. \end{aligned}$$

Ferrari’s method yields four possible solutions. The only real-valued solution greater than 1 is obtained with $\varepsilon _1=\varepsilon _2=1$, leading to the solution

$$\begin{aligned} \xi _{\tau }= & {} \frac{1}{2} \sqrt{-2 \lambda _{\tau }-\frac{8}{3} \frac{\lambda _{\tau }}{\sqrt{2 \lambda _{\tau }+\frac{4}{9}}} + \frac{4}{3} \sqrt{2 \lambda _{\tau }+\frac{4}{9}} + \frac{8}{9}}\\{} & {} + \frac{1}{2} \sqrt{2 \lambda _{\tau }+\frac{4}{9}} + \frac{1}{3}. \end{aligned}$$

Table 5 Table of expectiles of the Student distribution with $\nu$ degrees of freedom, computed via the Newton–Raphson algorithm

Fig. 10

Quantiles (black) and expectiles (red) of the standard Gaussian distribution, as functions of $\tau \in (0,1)$. (Color figure online)

Fig. 11

Quantiles (black) and expectiles (red) of the log-normal distribution, as functions of $\tau \in (0,1)$. (Color figure online)

Fig. 12

Quantiles (black) and expectiles (red) of the Student distribution with 2 (solid curves), 4 (dashed curves) and 10 (dotted curves) degrees of freedom, as functions of $\tau \in (0,1)$. The Student distribution with 2 degrees of freedom is Koenker’s distribution (Koenker 1993), for which quantiles and expectiles are identical. (Color figure online)

Fig. 13

Quantiles (black) and expectiles (red) of the chi-squared distribution with 1 (dotted curves), 5 (dashed curves) and 10 (solid curves) degrees of freedom, as functions of $\tau \in (0,1)$. (Color figure online)

Catalog of expectile functions of continuous distributions

This section provides a catalog of expectile functions of continuous distributions, obtained either via numerical means or, in exceptional cases, in closed or analytic form. Tables 3, 4, 5 and 6 give reference values for the expectiles of the standard Gaussian, log-normal, Student and chi-squared distributions, respectively. Figures 10, 11, 12 and 13 provide graphical representations of the corresponding expectile functions over (0, 1). Table 7 gathers closed-form expressions for expectiles of certain bounded continuous distributions. Table 8 gives analytic-form expressions for expectiles of some unbounded continuous distributions. Tables 9 and 10 list closed-form expressions for expectiles of the Hall–Weiss distribution and the Pareto distribution, respectively, with particular parameters.

Table 6 Table of expectiles of the chi-squared distribution with $\nu$ degrees of freedom, computed via the Newton–Raphson algorithm

Table 7 Closed-form expressions for expectiles in exceptional cases of certain bounded continuous distributions

Table 8 Analytic-form expressions for expectiles in exceptional cases of certain unbounded continuous distributions

Table 9 Closed-form expressions for expectiles in exceptional cases of the Hall–Weiss distribution with parameters $\alpha >0$ and $\beta \ge 0$, having distribution function $F(x)=1-(x^{-\alpha } + x^{-\alpha -\beta })/2$, $x>1$

Table 10 Closed-form expressions for expectiles in exceptional cases of the Pareto distribution with extreme value index $\gamma >0$, having distribution function $F(x)=1-x^{-1/\gamma }$, $x>1$