Asymmetric Squared Loss and Conditional Expectiles

Why Use Asymmetric Squared Loss?

Squared-error loss treats an overprediction and an underprediction of the same size equally. In some prediction problems, however, the two directions have different consequences. A decision-maker may want an estimate that gives greater weight to observations above the prediction than to equally distant observations below it. Asymmetric squared loss expresses that preference while retaining a smooth, quadratic penalty on either side.

The minimizer of expected asymmetric squared loss is called an expectile. Its defining balance is between weighted deviations above and below the proposed value. This resembles the balance used to characterize a quantile, but the weights here multiply the sizes of deviations, not just the probabilities of being on either side.

Definition: Fix an asymmetry level $\tau\in(0,1)$. For a response $y\in\mathbb{R}$ and action $a\in\mathbb{R}$, define the asymmetric squared loss $$ L_\tau(y,a)= \begin{cases} \tau(y-a)^2, & y\geq a,\\ (1-\tau)(y-a)^2, & y<a. \end{cases} $$ Equivalently, $L_\tau(y,a)=\tau(y-a)_+^2+(1-\tau)(a-y)_+^2$, where $u_+=\max(u,0)$. The $\tau$-expectile of a distribution with finite second moment is the value of $a$ minimizing $E[L_\tau(Y,a)]$.

The parameter convention matters: $\tau$ weights squared deviations from observations above the action, while $1-\tau$ weights deviations from observations below it. When $\tau$ is large, underpredicting high observations receives greater penalty, and the resulting expectile generally moves upward. At $\tau=1/2$, both sides receive equal weight and the minimizer is the mean.

Deriving the First-Order Condition

Write $R_\tau(a)=E[L_\tau(Y,a)]$ for the population risk. Assume $E[Y^2]<\infty$. For fixed $y$, the loss is differentiable in $a$, including at $a=y$, where both one-sided derivatives are zero. Its derivative is

\frac{\partial}{\partial a}L_\tau(y,a) =-2\tau(y-a)_+ + 2(1-\tau)(a-y)_+.

Differentiating inside the expectation is justified locally: on any bounded interval $|a|\leq M$, the absolute derivative is at most $2(|Y|+M)$, an integrable random variable because a finite second moment implies a finite first moment. The dominated convergence theorem therefore gives

R_\tau'(a) =-2\tau E[(Y-a)_+]+2(1-\tau)E[(a-Y)_+].

Consequently, any minimizer satisfies the weighted-deviation equation below. The first expectation is the average positive deviation above $a$; the second is the average positive deviation below $a$.

Theorem: If $E[Y^2]<\infty$ and $0<\tau<1$, the risk $R_\tau(a)=E[L_\tau(Y,a)]$ has a unique minimizer. It is the unique solution $e_\tau$ of $$ \tau E[(Y-e_\tau)_+]=(1-\tau)E[(e_\tau-Y)_+]. $$ This solution is the $\tau$-expectile of $Y$.

Proof. The risk is finite and continuous in $a$. Let $c=\min(\tau,1-\tau)>0$. Since $L_\tau(Y,a)\geq c(Y-a)^2$,

R_\tau(a)\geq cE[(Y-a)^2] =c\bigl(\operatorname{Var}(Y)+(E[Y]-a)^2\bigr),

so $R_\tau(a)$ tends to infinity as $|a|$ tends to infinity. Continuity and this coercivity imply that a global minimizer exists. The derivative formula shows that $R_\tau'(a)$ is continuous: each positive-part term is continuous in $a$ and changes by at most the change in $a$, so its expectation is continuous.

For uniqueness, the derivative of the loss for a fixed $y$ is $$ -2\tau(y-a)_+ + 2(1-\tau)(a-y)_+. $$ As a function of $a$, this is strictly increasing: its slope is $2\tau>0$ for $a<y$, and $2(1-\tau)>0$ for $a>y$, with the derivative continuous at $y$. Therefore, for $a_1<a_2$, the derivative at $a_2$ is strictly greater than at $a_1$ for every $y$. Taking expectations shows $R_\tau'(a_2)>R_\tau'(a_1)$. Thus $R_\tau'$ is strictly increasing. A differentiable coercive risk has a minimizer, and at any minimizer its derivative is zero. Strict increase permits at most one zero, so the minimizer is unique. Setting the derivative equal to zero gives the displayed weighted-deviation equation. $\square$

Conditional Expectiles

In prediction, the action can depend on features $X$. Given $X$, the relevant target is the minimizer of conditional risk, not necessarily the minimizer of the marginal risk. Assume $E[Y^2]<\infty$; then $E[Y^2\mid X]<\infty$ almost surely. Conditional expectations of the positive-part terms below are therefore finite almost surely.

Definition: A conditional $\tau$-expectile of $Y$ given $X$ is an $X$-measurable random variable $e_\tau(X)$ that, almost surely, uniquely minimizes $E[L_\tau(Y,a)\mid X]$ over actions $a\in\mathbb{R}$. Equivalently, it is the unique solution, almost surely, of $$ \tau E[(Y-e_\tau(X))_+\mid X] =(1-\tau)E[(e_\tau(X)-Y)_+\mid X]. $$

To see why the same condition applies conditionally, fix a conditional distribution given $X$, outside a null set. The preceding derivative calculation applies to that distribution: its second moment is finite, and the derivative is dominated locally by an integrable multiple of $1+|Y|$. Its conditional risk is coercive and has a unique minimizer by the same argument. The zero-derivative equation is precisely the conditional balance displayed above. The minimizer can be taken $X$-measurable: the conditional positive-part expectations, as functions of a fixed action, have measurable versions; the unique root can be located by testing the sign of their difference at rational actions. Thus the equation identifies the pointwise conditional-risk minimizer rather than merely a marginal summary.

Worked Example: A Discrete Distribution

Suppose $Y$ equals $0$, $4$, and $10$ with probabilities $0.3$, $0.5$, and $0.2$, respectively. Find its $\tau=0.7$ expectile.

The candidate lies between $4$ and $10$, so the only positive deviation above it comes from $Y=10$. For a candidate $a$ in this interval,

E[(Y-a)_+]=0.2(10-a), \qquad E[(a-Y)_+]=0.3a+0.5(a-4)=0.8a-2.

The balance equation is $0.7[0.2(10-a)]=0.3(0.8a-2)$. Expanding both sides gives $1.4-0.14a=0.24a-0.6$, hence $2=0.38a$ and $a=100/19\approx5.263$. This value is indeed between $4$ and $10$, as assumed.

Check the balance directly. The weighted upper deviation is $0.7\cdot0.2(10-100/19)=0.7\cdot(18/19)=63/95$. The weighted lower deviation is $0.3[0.3(100/19)+0.5(100/19-4)] =0.3(42/19)=63/95$. The two sides agree, verifying the result.

Worked Example: A Bernoulli Response

Let $Y$ be Bernoulli with $P(Y=1)=p$, and take $0<p<1$. The expectile lies between $0$ and $1$. For $0<a<1$, the only positive deviation above $a$ occurs when $Y=1$, while the only positive deviation below $a$ occurs when $Y=0$. Thus

E[(Y-a)_+]=p(1-a), \qquad E[(a-Y)_+]=(1-p)a.

The balance equation yields $\tau p(1-a)=(1-\tau)(1-p)a$, or $\tau p=a[\tau p+(1-\tau)(1-p)]$. Therefore,

e_\tau=\frac{\tau p}{\tau p+(1-\tau)(1-p)}.

For $p=0.4$ and $\tau=0.75$, this is $0.75(0.4)/[0.75(0.4)+0.25(0.6)]=0.3/0.45=2/3$. The upper weighted deviation is $0.75(0.4)(1/3)=0.1$; the lower weighted deviation is $0.25(0.6)(2/3)=0.1$. In contrast, the median of this Bernoulli distribution is $0$, since more than half its probability is at zero. The expectile and median need not coincide.

Worked Example: A Continuous Uniform Distribution

Suppose $Y$ is uniform on $[0,1]$, and $\tau=0.8$. For $a\in[0,1]$, integrating the positive deviations gives

E[(Y-a)_+]=\int_a^1(y-a)\,dy=\frac{(1-a)^2}{2}, \qquad E[(a-Y)_+]=\int_0^a(a-y)\,dy=\frac{a^2}{2}.

The balance is $0.8(1-a)^2/2=0.2a^2/2$, so $4(1-a)^2=a^2$. Since $a\in[0,1]$, taking nonnegative square roots gives $2(1-a)=a$, and hence $a=2/3$. Substitution checks the equation: the upper weighted deviation is $0.8(1/3)^2/2=2/45$, and the lower weighted deviation is $0.2(2/3)^2/2=2/45$.

Worked Example: Conditional Expectiles for Two Groups

Let $X$ take values $0$ and $1$. Suppose $Y\mid X=0$ is Bernoulli with success probability $0.25$, while $Y\mid X=1$ is Bernoulli with success probability $0.75$. For $\tau=0.6$, the Bernoulli formula gives

e_{0.6}(0)=\frac{0.6(0.25)}{0.6(0.25)+0.4(0.75)} =\frac{0.15}{0.45}=\frac{1}{3},

e_{0.6}(1)=\frac{0.6(0.75)}{0.6(0.75)+0.4(0.25)} =\frac{0.45}{0.55}=\frac{9}{11}.

These are separate minimizers of conditional risk for the two feature values. For $X=0$, the upper weighted deviation at $1/3$ is $0.6(0.25)(2/3)=0.1$, and the lower weighted deviation is $0.4(0.75)(1/3)=0.1$. For $X=1$, the corresponding values at $9/11$ are $0.6(0.75)(2/11)=9/110$ and $0.4(0.25)(9/11)=9/110$. Each group satisfies its own conditional balance.

How Expectiles Differ From Quantiles

The first-order condition makes the distinction precise. For an expectile, the balance involves the magnitudes of deviations: $\tau E[(Y-a)_+]$ and $(1-\tau)E[(a-Y)_+]$. For a quantile, the defining balance concerns probabilities on either side of the candidate, as characterized through quantile loss. Consequently, changing the distance of observations from the candidate can change an expectile even if the probabilities on either side do not change.

At $\tau=1/2$, the asymmetric loss reduces to one-half of squared error. The balance equation becomes $E[(Y-a)_+]=E[(a-Y)_+]$. Since $Y-a=(Y-a)_+-(a-Y)_+$, equality implies $E[Y-a]=0$, so the expectile is $E[Y]$. This recovers squared-error prediction by the conditional mean when the same argument is applied to the conditional distribution.

Common Mistake: Reversing the Weights

Wrong reasoning: “A large $\tau$ penalizes values below the action more, because $1-\tau$ is the weight on the lower side.” This identifies the smaller weight but ignores which side the parameter $\tau$ multiplies.
Correction: Values above the action receive weight $\tau$; values below receive $1-\tau$. Thus a large $\tau$ emphasizes underprediction of high outcomes and typically raises the expectile.
Another error: Replacing deviation expectations with probabilities above and below the action. That is not the expectile equation: each positive deviation must retain its size inside the expectation.

Key takeaway: A conditional expectile is the unique conditional-risk minimizer of asymmetric squared loss. It is characterized by equality between the $\tau$-weighted mean upper deviation and the $(1-\tau)$-weighted mean lower deviation.

Check Your Understanding

Use the loss convention and weighted-deviation equation stated above.

For an action $a$, which side of the action receives weight $\tau$, and which receives weight $1-\tau$?
State the first-order condition for the $\tau$-expectile of $Y$.
Why does a finite second moment ensure that the risk has a minimizer?
For a Bernoulli response with success probability $p$, write the expectile formula and describe how it depends on $\tau$.
Explain why an expectile equation balances deviation magnitudes rather than only probabilities.