Inequality for the variance of an asymmetric loss

Let $\hat{y}$ be a predicted value of an observed value $y$. In this paper, we make the assumptions (I) and (II):

1. (I)

   The prediction error $z:=\hat{y}-y$ is the realized value of a random variable $Z$, whose probability density function $f(z)$ satisfies $f(x)=f(-x)$ for $x\in\mathbb{R}$ and $f(x)\geq f(y)$ for $0\leq x\leq y$.

2. (II)

   Let $k_{1}$, $k_{2}\in\mathbb{R}_{>0}$. If there is a mismatch between $y$ and $\hat{y}$, then we suffer a loss

   $\displaystyle L(z):=\begin{cases}k_{1}z,&z\geq 0,\\ -k_{2}z,&z<0.\end{cases}$

Under the assumptions (I) and (II), we solve the minimization problem for the expected value of $L(Z+c)$:

$$C=\arg{\min_{c}}\{\operatorname{{E}}[L(Z+c)]\}.$$

In addition, we give the following theorem.

Theorem 1 is obtained by the following lemma.

These results are a generalization of the results of [5]. The paper [5] made the assumptions (I’) and (II):

1. (I’)

   The prediction error $z:=\hat{y}-y$ is the realized value of a random variable $Z$, whose probability density function is a generalized Gaussian distribution function (see, e.g., [1], [2], and [3]) with mean zero

   $\displaystyle f(z):=\frac{1}{2ab\Gamma(a)}\exp{\left(-\left\lvert\frac{z}{b}\right\rvert^{\frac{1}{a}}\right)},$

   where $\Gamma(a)$ is the gamma function and $a,b>0$.

Assumption (I) is weaker than (I’). Thus, we assume a more general situation than in [5]. In [5], under the assumptions (I’) and (II), the minimization problem for the expected value of $L(Z+c)$ is solved and the inequality $\operatorname{{V}}[L(Z+C)]\leq\operatorname{{V}}[L(Z)]$ is obtained. This inequality is derived from the following inequality: For $a,x>0$, we have

$\displaystyle x^{a}\gamma(a,x)^{2}-x^{a}\Gamma(a)^{2}+2\gamma(a,x)\Gamma(2a,x)>0,$

(1)

where

$\displaystyle\Gamma(a):=\int_{0}^{+\infty}t^{a-1}e^{-t}dt,\quad\Gamma(a,x)$

$\displaystyle:=\int_{x}^{+\infty}t^{a-1}e^{-t}dt,\quad\gamma(a,x):=\int_{0}^{x}t^{a-1}e^{-t}dt.$

Inequality (1) is the special case of Lemma 2 that $f(z)$ is a generalized Gaussian distribution function.

Assumptions (I) and (II) have a background in the procurement from an electricity market. Suppose that we purchase electricity $\hat{y}$ from an market, based on a forecast of the electricity $y$ that will be needed. This situation makes the assumption (I). If $\hat{y}-y>0$, then there is a waste of procurement fee proportional to $\hat{y}-y$. If $y-\hat{y}>0$, then we are charged with a penalty proportional to $y-\hat{y}$. This situation makes the assumption (II). For details, see [4].

For $c\in\mathbb{R}$, let $\operatorname{{sgn}}(c):=1\>(c\geq 0)$; $-1\>(c<0)$. From $\int_{0}^{\infty}f(z)dz=\frac{1}{2}$, the expected value of $L(Z+c)$ and $L(Z+c)^{2}$ are as follows: For any $c\in\mathbb{R}$,

	$\displaystyle\operatorname{{E}}[L(Z+c)]$	$\displaystyle=(k_{1}+k_{2})\int_{|c|}^{\infty}zf(z)dz+\frac{c(k_{1}-k_{2})}{2}+|c|(k_{1}+k_{2})\int_{0}^{|c|}f(z)dz,$
	$\displaystyle\operatorname{{E}}[L(Z+c)^{2}]$	$\displaystyle=(k_{1}^{2}+k_{2}^{2})\int_{0}^{\infty}z^{2}f(z)dz+\operatorname{{sgn}}(c)(k_{1}^{2}-k_{2}^{2})\int_{0}^{|c|}z^{2}f(z)dz$
		$\displaystyle\qquad+2c(k_{1}^{2}-k_{2}^{2})\int_{|c|}^{\infty}zf(z)dz+\frac{c^{2}(k_{1}^{2}+k_{2}^{2})}{2}+c|c|(k_{1}^{2}-k_{2}^{2})\int_{0}^{|c|}f(z)dz.$

Therefore, the expected value and the variance of $L(Z)$ are as follows:

	$\displaystyle\operatorname{{E}}[L(Z)]$	$\displaystyle=(k_{1}+k_{2})\int_{0}^{\infty}zf(z)dz,$
	$\displaystyle\operatorname{{V}}[L(Z)]$	$\displaystyle=(k_{1}^{2}+k_{2}^{2})\int_{0}^{\infty}z^{2}f(z)dz-(k_{1}+k_{2})^{2}\left(\int_{0}^{\infty}zf(z)dz\right)^{2}.$

We determine the value $c$ that gives the minimum value of $\operatorname{{E}}[L(Z+c)]$. From

	$\displaystyle\frac{d}{dc}\operatorname{{E}}[L(Z+c)]$	$\displaystyle=\frac{k_{1}-k_{2}}{2}+\operatorname{{sgn}}(c)(k_{1}+k_{2})\int_{0}^{|c|}f(z)dz,$
	$\displaystyle\frac{d^{2}}{dc^{2}}\operatorname{{E}}[L(Z+c)]$	$\displaystyle=(k_{1}+k_{2})f(c)\geq 0,$

we can see that $\operatorname{{E}}[L(Z+c)]$ has the minimum value at the zero point of $\frac{d}{dc}\operatorname{{E}}[L(Z+c)]$. The zero point $C$ satisfies the following equation:

$$\frac{k_{1}-k_{2}}{2}+\operatorname{{sgn}}(C)(k_{1}+k_{2})\int_{0}^{|C|}f(z)dz=0.$$

From this, $C=0$ if and only if $k_{1}=k_{2}$. Also, we have

	$\displaystyle\operatorname{{E}}[L(Z+C)]$	$\displaystyle=(k_{1}+k_{2})\int_{|C|}^{\infty}zf(z)dz,$
	$\displaystyle\operatorname{{V}}[L(Z+C)]$	$\displaystyle=(k_{1}^{2}+k_{2}^{2})\int_{0}^{\infty}z^{2}f(z)dz-2(k_{1}+k_{2})^{2}\int_{0}^{|C|}f(z)dz\int_{0}^{|C|}z^{2}f(z)dz$
		$\displaystyle\qquad-4|C|(k_{1}+k_{2})^{2}\int_{0}^{|C|}f(z)dz\int_{|C|}^{\infty}zf(z)dz+\frac{C^{2}(k_{1}+k_{2})^{2}}{4}$
		$\displaystyle\qquad\qquad-(k_{1}+k_{2})^{2}\left(\int_{|C|}^{\infty}zf(z)dz\right)^{2}-C^{2}(k_{1}+k_{2})^{2}\left(\int_{0}^{|C|}f(z)dz\right)^{2}.$

Let

	$\displaystyle\beta(x)$	$\displaystyle:=-\left(\int_{0}^{\infty}zf(z)dz\right)^{2}+2\int_{0}^{x}f(z)dz\int_{0}^{x}z^{2}f(z)dz+4x\int_{0}^{x}f(z)dz\int_{x}^{\infty}zf(z)dz$
		$\displaystyle\qquad-\frac{x^{2}}{4}+\left(\int_{x}^{\infty}zf(z)dz\right)^{2}+x^{2}\left(\int_{0}^{x}f(z)dz\right)^{2}.$

Then, $\operatorname{{V}}[L(Z)]-\operatorname{{V}}[L(Z+C)]=(k_{1}+k_{2})^{2}\beta(C)$ holds. From $\beta(0)=0$ and

	$\displaystyle\frac{d}{dx}\beta(x)$	$\displaystyle=4\int_{0}^{x}f(z)dz\int_{x}^{\infty}zf(z)dz-\frac{x}{2}+2x\left(\int_{0}^{x}f(z)dz\right)^{2}$
		$\displaystyle\qquad+2f(x)\int_{0}^{x}z^{2}f(z)dz+2xf(x)\int_{x}^{\infty}zf(z)dz,$

if Lemma 2 is proved, then Theorem 1 is immediately obtained. We prove Lemma 2.