Sharp bounds on moments of random variables

Abstract

We give new inequalities involving the expectation and variance of a random variable defined on a finite interval and having an essentially bounded probability density function. Our inequalities sharpen previous inequalities by N.S. Barnett and S.S. Dragomir. We prove our bounds are optimal, and we explicitly give the cases in which equality occurs.

1 Introduction

There has been considerable interest in the study of absolutely continuous random variables X that take values in a compact interval $[ a,b ]$ and whose probability density functions ρ are essentially bounded. See, for example, [1–6], and [7].

One result of particular importance, introduced by Barnett and Dragomir in [5], and discussed in [1–3, 5, 7], and elsewhere, gives good general bounds involving the mean and variance of X:

Theorem 1

([5, Theorem 1(ii)])

If X is an absolutely continuous random variable whose pdf ρ satisfies $m\leq \rho ( x ) \leq M$ for $\mathcal{L}^{1}$ a.e. $x\in [ a,b ]$ and $\int _{a}^{b}\rho ( x ) \,dx =1$, then

$$ \frac{m ( b-a ) ^{3}}{6}\leq \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] - \sigma ^{2} ( X ) \leq \frac{M ( b-a ) ^{3}}{6} $$

(1)

and

$$ \biggl\vert \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] - \sigma ^{2} ( X ) - \frac{ ( b-a ) ^{3}}{6} \biggr\vert \leq \frac{\sqrt{5} ( b-a ) ^{3} ( M-m ) }{60}. $$

(2)

In this paper we establish, for the first time, optimal lower and upper bounds for

$$ \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) $$

for such random variables X, in particular sharpening the key results (1) and (2). Moreover, we show that the distributions yielding the optimal bounds are unique, and we give them explicitly.

Since pdfs that differ on sets having Lebesgue measure zero correspond to identically distributed random variables, without loss of generality we will identify such pdfs. In what follows, $\mathcal{L}^{1} ( E ) = \vert E \vert$ denotes the Lebesgue measure of the set E.

2 Results

Theorem 2

Suppose X is an absolutely continuous random variable whose pdf ρ satisfies $0\leq m\leq \rho ( x ) \leq M<\infty$ for $\mathcal{L}^{1}$ a.e. $x\in [ a,b ]$ and $\int _{a}^{b}\rho ( x ) \,dx =1$.

• (1) If $m=M$, then

  $$ \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) = \frac{ ( b-a ) ^{2}}{6}. $$

• (2) If $m< M$, then

  $$\begin{aligned} & \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) \\ &\quad \geq \frac{1}{6}m ( b-a ) ^{3}+\frac{1}{12} \bigl( 2 ( b-a ) ( M-m ) +M ( b-a ) -1 \bigr) \frac{ ( 1-m ( b-a ) ) ^{2}}{ ( M-m ) ^{2}}. \end{aligned}$$

  This lower bound is sharp, and we have equality if and only if the function

  $$ \rho ( x ) = \textstyle\begin{cases} M, & \textit{if }x\in {}[ a,a+\frac{G}{2})\cup (b-\frac{G}{2},b], \\ m, & \textit{if }x\in [ a+\frac{G}{2}, b-\frac{G}{2} ], \end{cases}$$

  with

  $$ G=\frac{1-m ( b-a ) }{M-m}, $$

  is a pdf of X.

• (3) If $m< M$, then

  $$\begin{aligned} & \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) \\ &\quad \leq \frac{1}{4} ( b-a ) ^{2}-\frac{1}{12} \biggl( m ( b-a ) ^{3}+ \frac{ ( 1-m ( b-a ) ) ^{3}}{ ( M-m ) ^{2}} \biggr) . \end{aligned}$$

  This upper bound is sharp, and we have equality if and only if the function

  $$ \rho ( x ) = \textstyle\begin{cases} M, & \textit{if }x\in ( ( \frac{1}{2}a+\frac{1}{2}b ) -\frac{G}{2}, ( \frac{1}{2}a+\frac{1}{2}b ) +\frac{G}{2} ), \\ m, & \textit{if }x\in [ a, ( \frac{1}{2}a+\frac{1}{2}b ) -\frac{G}{2} ] \cup [ ( \frac{1}{2}a+\frac{1}{2}b ) +\frac{G}{2}, b ], \end{cases}$$

  with G defined as in part (2), is a pdf of X.

3 Proofs

We first record a useful lemma, proven using a clever algebraic manipulation by Barnett and Dragomir in [5].

Lemma 3

([5, (2.5)])

Suppose X is as stated in our theorem. Then,

$$ \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) = \int _{a}^{b} ( b-x ) ( x-a ) \rho ( x ) \,dx . $$

Proof of (1)

In this case, X is a continuous uniform random variable on $[ a,b ]$, and so its expected value and variance are $E ( X ) = ( a+b ) /2$ and $\sigma ^{2} ( X ) = ( b-a ) ^{2}/12$. Substituting these values yields the result immediately.

Proof of (2)

Suppose $m< M$. Let $c=b-a$, and define

$$\begin{aligned}& f ( x ) = x ( c-x ) \quad \text{for all }x\in [ 0,c ] , \\& \tau _{a} ( x ) = x+a\quad \text{for all real }x, \\& \phi ( x ) = \frac{x-m}{M-m}\quad \text{for all }x\in [ m,M ] . \end{aligned}$$

The translation operator $\tau _{a}$ is invertible, with $\tau _{a}^{-1} ( x ) =x-a$ for all real x. Since $1/ ( M-m ) >0$, the affine transformation $\phi : [ m,M ] \rightarrow [ 0,1 ]$ is invertible with $\phi ^{-1} ( x ) = ( M-m ) x+m$ for all $x\in [ 0,1 ]$. Define

$$\begin{aligned} &\mathcal{C}_{\rho } = \biggl\{ \rho :m\leq \rho ( x ) \leq M\text{ for }\mathcal{L}^{1}\text{ a.e. }x\in [ a,b ] , \text{and }\int _{a}^{b}\rho ( x ) \,dx =1 \biggr\} , \\ &\mathcal{C}_{h} = \biggl\{ h:m\leq h ( x ) \leq M \text{ for }\mathcal{L}^{1}\text{ a.e. }x\in [ 0,c ] ,\text{and }\int _{0}^{c}h ( x ) \,dx =1 \biggr\} , \\ &\mathcal{C}_{g} = \biggl\{ g:0\leq g ( x ) \leq 1 \text{ for }\mathcal{L}^{1}\text{ a.e. }x\in [ 0,c ] ,\text{and }\int _{0}^{c}g ( x ) \,dx =G \biggr\} . \end{aligned}$$

We say that a function ρ, h, or g is admissible provided it is an element of $\mathcal{C}_{\rho }$, $\mathcal{C}_{h}$, or $\mathcal{C}_{g}$, respectively. Whenever $\rho _{\ast }$, $h_{\ast }$, and $g_{\ast }$ are admissible, define

$$\begin{aligned}& I_{\rho _{\ast }} = \int _{a}^{b} ( b-x ) ( x-a ) \rho _{\ast } ( x ) \,dx , \\& I_{h_{\ast }} = \int _{0}^{c} ( c-x ) x\cdot h_{\ast } ( x ) \,dx = \int _{0}^{c}f ( x ) \cdot h_{\ast } ( x ) \,dx , \\& I_{g_{\ast }} = \int _{0}^{c} ( c-x ) x\cdot g_{\ast } ( x ) \,dx = \int _{0}^{c}f ( x ) \cdot g_{\ast } ( x ) \,dx . \end{aligned}$$

If $\rho \in \mathcal{C}_{\rho }$, $h=\rho \circ \tau _{a}$, and $g=\phi \circ h$, then $m\leq h ( x ) \leq M$ for $\mathcal{L}^{1}$ a.e. $x\in [ 0,c ]$, $0\leq g ( x ) \leq 1$ for $\mathcal{L}^{1}$ a.e. $x\in [ 0,c ]$, $\rho =h\circ \tau _{a}^{-1}$, $h=\phi ^{-1}\circ g$, and

$$\begin{aligned} 1 & = \int _{a}^{b}\rho ( x ) \,dx = \int _{0}^{c}h ( x ) \,dx = \int _{0}^{c} \bigl[ ( M-m ) g ( x ) +m \bigr] \,dx \\ & = ( M-m ) \int _{0}^{c}g ( x ) \,dx +mc, \end{aligned}$$

from which it follows that $h\in \mathcal{C}_{h}$ and $g\in \mathcal{C}_{g}$.

If $g\in \mathcal{C}_{g}$, $h=\phi ^{-1}\circ g$, and $\rho =h\circ \tau _{a}^{-1}$, then $m\leq h ( x ) \leq M$ for $\mathcal{L}^{1}$ a.e. $x\in [ 0,c ]$, $m\leq \rho ( x ) \leq M$ for $\mathcal{L}^{1}$ a.e. $x\in [ a,b ]$, $g=\phi \circ h$, $h=\rho \circ \tau _{a}$, and

$$ \frac{1-mc}{M-m}=G= \int _{0}^{c}g ( x ) \,dx = \int _{0}^{c} \frac{h ( x ) -m}{M-m}\,dx = \frac{1}{M-m} \biggl( \int _{0}^{c}h ( x ) \,dx -mc \biggr) , $$

from which it follows that

$$ 1= \int _{0}^{c}h ( x ) \,dx = \int _{a}^{b}\rho ( x ) \,dx , $$

and hence, $h\in \mathcal{C}_{h}$ and $\rho \in \mathcal{C}_{\rho }$.

We say that a function ρ, h, or g is a minimizer provided it is an element of $\mathcal{C}_{\rho }$, $\mathcal{C}_{h}$, or $\mathcal{C}_{g}$, respectively, and provided it results in the lowest possible value for $I_{\rho _{\ast }}$, $I_{h_{\ast }}$, or $I_{g_{\ast }}$, respectively, among all admissible functions $\rho _{\ast }$, $h_{\ast }$, or $g_{\ast }$.

If ρ is a minimizer and $h=\rho \circ \tau _{a}$, then h is admissible, as shown above, and

$$ I_{h}= \int _{0}^{c} ( c-x ) x\cdot h ( x ) \,dx = \int _{0}^{c} ( c-x ) x\cdot \rho ( x+a ) \,dx = \int _{a}^{b} ( b-u ) ( u-a ) \cdot \rho ( u ) \,du =I_{\rho }, $$

and so h is a minimizer. If, additionally, $g=\phi \circ h$, then g is admissible, as shown above, and we have

$$\begin{aligned} I_{g} & = \int _{0}^{c} ( c-x ) x\cdot g ( x ) \,dx \\ & = \int _{0}^{c} ( c-x ) x\cdot \frac{h ( x ) -m}{M-m}\,dx \\ & = \frac{1}{M-m} \biggl( I_{h}-\frac{1}{6}mc^{3} \biggr) . \end{aligned}$$

Thus, g is a minimizer as well, as $I_{g}$ is a strictly increasing affine function of $I_{h}$.

Similarly, if g is a minimizer and $h=\phi ^{-1}\circ g$, then h is admissible, as shown above, and we have

$$\begin{aligned} I_{h} & = \int _{0}^{c} ( c-x ) x\cdot h ( x ) \,dx \\ & = \int _{0}^{c} ( c-x ) x\cdot \bigl( ( M-m ) g ( x ) +m \bigr) \,dx \\ & = ( M-m ) I_{g}+\frac{1}{6}mc^{3}. \end{aligned}$$

Then, h is a minimizer, as $I_{h}$ is a strictly increasing affine function of $I_{g}$. If, additionally, $\rho =h\circ \tau _{a}^{-1}$, then ρ is admissible, as shown above, and we have

$$ I_{\rho }= \int _{a}^{b} ( b-x ) ( x-a ) \rho ( x ) \,dx = \int _{0}^{c} ( c-u ) u\cdot h ( u ) \,du =I_{h}, $$

and so ρ is a minimizer.

By Lemma 3, X minimizes

$$ \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) $$

if and only if ρ is admissible and minimizes $I_{\rho }$. By the argument above, that occurs if and only if the function g defined by $g=\phi \circ h$, where $h=\rho \circ \tau _{a}$, minimizes $I_{g}$.

We will now use the Bathtub Principle [8, Theorem 1.14] with Ω, μ, f, G, $\mathcal{C}$, and I there replaced by $[ 0,c ]$, the Lebesgue measure, f, G, $\mathcal{C}_{g}$, and $I_{g}$, respectively, to find a function g that minimizes $I_{g}$ and to prove that it is unique among minimizers of $I_{g}$. It will then follow from our reasoning above that the function ρ defined by $\rho =h\circ \tau _{a}^{-1}$, where $h=\phi ^{-1}\circ g$, uniquely minimizes $I_{\rho }$, and hence solves our problem. Let

$$\begin{aligned} s & = \sup \bigl\{ t: \bigl\vert \bigl\{ x\in [ 0,c ] :f ( x ) < t \bigr\} \bigr\vert \leq G \bigr\} \\ & = \sup \biggl\{ t: \biggl\vert \biggl\{ x\in [ 0,c ] : \frac{1}{4}c^{2}- \biggl( x-\frac{c}{2} \biggr) ^{2}< t \biggr\} \biggr\vert \leq G \biggr\} \\ & = \sup \biggl\{ t: \biggl\vert \biggl\{ x\in [ 0,c ] :x> \frac{c}{2}+\sqrt{\frac{1}{4}c^{2}-t}\text{ or }x< \frac{c}{2}-\sqrt{\frac{1}{4}c^{2}-t} \biggr\} \biggr\vert \leq G \biggr\} \\ & = \sup \bigl\{ t:c-\sqrt{c^{2}-4t}\leq G \bigr\} \\ & = \sup \biggl\{ t:t\leq \frac{1}{4} \bigl( c^{2}- ( c-G ) ^{2} \bigr) \biggr\} \\ & = \frac{1}{4} \bigl( c^{2}- ( c-G ) ^{2} \bigr) . \end{aligned}$$

For this s, we have $\vert \{ x\in [ 0,c ] :f ( x ) =s \} \vert =0$ and

$$\begin{aligned} & \bigl\vert \bigl\{ x\in [ 0,c ] :f ( x ) < s \bigr\} \bigr\vert \\ &\quad = \biggl\vert \biggl\{ x\in [ 0,c ] :\frac{1}{4}c^{2}- \biggl( x-\frac{c}{2} \biggr) ^{2}< \frac{1}{4} \bigl( c^{2}- ( c-G ) ^{2} \bigr) \biggr\} \biggr\vert \\ &\quad = \biggl\vert \biggl\{ x\in [ 0,c ] : \biggl( x- \frac{c}{2} \biggr) ^{2}>\frac{1}{4} ( c-G ) ^{2} \biggr\} \biggr\vert \\ &\quad = \biggl\vert \biggl[0, \frac{G}{2}\biggl)\cup \biggr(c-\frac{G}{2}, c\biggr] \biggr\vert \\ &\quad = G. \end{aligned}$$

Since $\vert \{ x\in [ 0,c ] :f ( x ) < s \} \vert =G$, by the Bathtub Principle [8, Theorem 1.14] there is a unique $g\in \mathcal{C}_{g}$ that minimizes the integral $I_{g}$, and it is given by $g ( x ) =\chi _{ \{ f< s \} } ( x )$.

Thus,

$$ g ( x ) = \textstyle\begin{cases} 1, & \text{if }f ( x ) < s, \\ 0, & \text{if }f ( x ) \geq s\end{cases}\displaystyle = \textstyle\begin{cases} 1, & \text{if }x\in {}[ 0,\frac{G}{2})\cup (c-\frac{G}{2}, c], \\ 0, & \text{if }x\in [ \frac{G}{2}, c-\frac{G}{2} ] .\end{cases}$$

We then calculate

$$ h ( x ) = ( M-m ) g ( x ) +m= \textstyle\begin{cases} M, & \text{if }x\in {}[ 0,\frac{G}{2})\cup (c-\frac{G}{2}, c], \\ m, & \text{if }x\in [ \frac{G}{2}, c-\frac{G}{2} ] \end{cases}$$

and

$$\begin{aligned} \rho ( x ) & = h ( x-a ) = \textstyle\begin{cases} M, & \text{if }x\in {}[ a, a+\frac{G}{2})\cup (a+c-\frac{G}{2}, a+c], \\ m, & \text{if }x\in [ a+\frac{G}{2}, a+c-\frac{G}{2} ] \end{cases}\displaystyle \\ & = \textstyle\begin{cases} M, & \text{if }x\in {}[ a, a+\frac{G}{2})\cup (b-\frac{G}{2}, b], \\ m, & \text{if }x\in [ a+\frac{G}{2}, b-\frac{G}{2} ] .\end{cases}\displaystyle \end{aligned}$$

This ρ is the desired minimizer. Finally, we will calculate the minimum value of $I_{\rho }$,

$$\begin{aligned} I_{\rho } & = I_{h} \\ & = \int _{0}^{c}f ( x ) h ( x ) \,dx = \int _{0}^{G/2}Mf ( x ) \,dx + \int _{G/2}^{c-G/2}mf ( x ) \,dx + \int _{c-G/2}^{c}Mf ( x ) \,dx \\ & = \frac{1}{24}MG^{2} ( 3c-G ) +\frac{1}{12}m \bigl( 2c^{3}-G^{2} ( 3c-G ) \bigr) +\frac{1}{24}MG^{2} ( 3c-G ) \\ & = \frac{1}{12}G^{2} ( 3c-G ) ( M-m ) + \frac{1}{6}mc^{3}. \end{aligned}$$

We have

$$ G ( M-m ) =1-mc $$

and also

$$\begin{aligned} G ( 3c-G ) & = \frac{1-mc}{M-m} \biggl( 3c-\frac{1-mc}{M-m} \biggr) \\ & = \frac{1}{ ( M-m ) ^{2}} ( 1-mc ) ( 3Mc-2mc-1 ) . \end{aligned}$$

Substituting these expressions, we obtain

$$\begin{aligned} & \int _{0}^{c}f ( x ) h ( x ) \,dx \\ &\quad = \frac{1}{6}mc^{3}+\frac{1}{12} ( 1-mc ) \frac{1}{ ( M-m ) ^{2}} ( 1-mc ) ( 3Mc-2mc-1 ) \\ &\quad = \frac{1}{6}mc^{3}+\frac{1}{12} \bigl( 2c ( M-m ) +Mc-1 \bigr) \frac{ ( 1-mc ) ^{2}}{ ( M-m ) ^{2}} \\ &\quad = \frac{1}{6}m ( b-a ) ^{3}+\frac{1}{12} \bigl( 2 ( b-a ) ( M-m ) +M ( b-a ) -1 \bigr) \frac{ ( 1-m ( b-a ) ) ^{2}}{ ( M-m ) ^{2}}, \end{aligned}$$

as asserted in (2).

Proof of (3)

Suppose $m< M$. Let c, G, f, $\tau _{a}$, ϕ, $\mathcal{C}_{p}$, $\mathcal{C}_{h}$, $\mathcal{C}_{g}$, $I_{\rho _{\ast }}$, $I_{h_{\ast }}$, and $I_{g_{\ast }}$ be as in the proof of (2). We say that a function ρ, h, or g is a maximizer provided it is an element of $\mathcal{C}_{\rho }$, $\mathcal{C}_{h}$, or $\mathcal{C}_{g}$, respectively, and provided it results in the highest possible value for $I_{\rho _{\ast }}$, $I_{h_{\ast }}$, or $I_{g_{\ast }}$, respectively, among all admissible functions $\rho _{\ast }$, $h_{\ast }$, or $g_{\ast }$.

If ρ is a maximizer and $h=\rho \circ \tau _{a}$, then h is admissible and $I_{h}=I_{\rho }$, as shown above, and so h is a maximizer. If, additionally, $g=\phi \circ h$, then g is admissible and

$$ I_{g}=\frac{1}{M-m} \biggl( I_{h}- \frac{1}{6}mc^{3} \biggr) , $$

as shown above, and so g is a maximizer, as $I_{g}$ is a strictly increasing affine function of $I_{h}$.

Similarly, if g is a maximizer and $h=\phi ^{-1}\circ g$, then h is admissible and

$$ I_{h}= ( M-m ) I_{g}+\frac{1}{6}mc^{3}, $$

as shown above, and so h is a maximizer, as $I_{h}$ is a strictly increasing affine function of $I_{g}$. If, additionally, $\rho =h\circ \tau _{a}^{-1}$, then ρ is admissible and $I_{\rho }=I_{h}$, as shown above, and so ρ is a maximizer.

By Lemma 3, X maximizes

$$ \bigl[ b-E ( X ) \bigr] \bigl[ E ( X ) -a \bigr] -\sigma ^{2} ( X ) $$

if and only if ρ is admissible and maximizes $I_{\rho }$. By the argument above, that occurs if and only if the function g defined by $g=\phi \circ h$, where $h=\rho \circ \tau _{a}$, maximizes $I_{g}$.

Define

$$ F ( x ) =\frac{c^{2}}{4}-f ( x ) = \biggl( x- \frac{c}{2} \biggr) ^{2}\quad \text{for all }x\in [ 0,c ] . $$

Whenever $g_{\ast }$ is admissible, define

$$ I_{g_{\ast }}^{\prime }= \int _{0}^{c}F ( x ) \cdot g_{ \ast } ( x ) \,dx . $$

For each admissible $g_{\ast }$, we have

$$ I_{g_{\ast }}+I_{g_{\ast }}^{\prime }= \int _{0}^{c}f ( x ) \cdot g_{\ast } ( x ) \,dx + \int _{0}^{c}F ( x ) \cdot g_{\ast } ( x ) \,dx = \int _{0}^{c}\frac{c^{2}}{4} \cdot g_{\ast } ( x ) \,dx =\frac{c^{2}}{4}G. $$

Since this sum is constant, g maximizes $I_{g}$, as needed, if and only if g minimizes $I_{g}^{\prime }$.

We will now use the Bathtub Principle [8, Theorem 1.14] with Ω, μ, f, G, $\mathcal{C}$, and I there replaced by $[ 0,c ]$, the Lebesgue measure, F, G, $\mathcal{C}_{g}$, and $I_{g}^{\prime }$, respectively, to find a function g that minimizes $I_{g}^{\prime }$ and to prove that it is unique among minimizers of $I_{g}$. It will then follow from our reasoning above that the function ρ defined by $\rho =h\circ \tau _{a}^{-1}$, where $h=\phi ^{-1}\circ g$, uniquely maximizes $I_{\rho }$, and hence solves our problem. Let

$$\begin{aligned} s & = \sup \bigl\{ t: \bigl\vert \bigl\{ x\in [ 0,c ] :F ( x ) < t \bigr\} \bigr\vert \leq G \bigr\} \\ & = \sup \biggl\{ t: \biggl\vert \biggl\{ x\in [ 0,c ] : \biggl( x- \frac{c}{2} \biggr) ^{2}< t \biggr\} \biggr\vert \leq G \biggr\} \\ & = \sup \biggl\{ t: \biggl\vert \biggl( \frac{c}{2}-\sqrt{t}, \frac{c}{2}+\sqrt{t} \biggr) \biggr\vert \leq G \biggr\} \\ & = \sup \{ t:2\sqrt{t}\leq G \} \\ & = \frac{G^{2}}{4}. \end{aligned}$$

For this s, we have $\vert \{ x\in [ 0,c ] :F ( x ) =s \} \vert =0$ and

$$\begin{aligned} & \bigl\vert \bigl\{ x\in [ 0,c ] :F ( x ) < s \bigr\} \bigr\vert \\ &\quad = \biggl\vert \biggl\{ x\in [ 0,c ] : \biggl( x- \frac{c}{2} \biggr) ^{2}< \frac{G^{2}}{4} \biggr\} \biggr\vert \\ &\quad = \biggl\vert \biggl( \frac{c}{2}-\frac{G}{2}, \frac{c}{2}+ \frac{G}{2} \biggr) \biggr\vert \\ &\quad = G. \end{aligned}$$

Since $\vert \{ x\in [ 0,c ] :F ( x ) < s \} \vert =G$, by the Bathtub Principle [8, Theorem 1.14] there is a unique $g\in \mathcal{C}_{g}$ that minimizes the integral $I_{g}^{\prime }$, and it is given by $g ( x ) =\chi _{ \{ F< s \} } ( x )$.

Thus,

$$ g ( x ) = \textstyle\begin{cases} 1, & \text{if }F ( x ) < s, \\ 0, & \text{if }F ( x ) \geq s\end{cases}\displaystyle = \textstyle\begin{cases} 1, & \text{if } x\in ( \frac{c}{2}-\frac{G}{2}, \frac{c}{2}+\frac{G}{2} ) , \\ 0, & \text{if } x\in [ 0, \frac{c}{2}-\frac{G}{2} ] \cup [ \frac{c}{2}+\frac{G}{2}, c ]. \end{cases}$$

We then calculate

$$ h ( x ) = ( M-m ) g ( x ) +m= \textstyle\begin{cases} M, & \text{if } x\in ( \frac{c}{2}-\frac{G}{2}, \frac{c}{2}+\frac{G}{2} ) , \\ m, & \text{if } x\in [ 0, \frac{c}{2}-\frac{G}{2} ] \cup [ \frac{c}{2}+\frac{G}{2}, c ] \end{cases}$$

and

$$ \rho ( x ) =h ( x-a ) = \textstyle\begin{cases} M, & \text{if }x\in ( ( \frac{1}{2}a+\frac{1}{2}b ) -\frac{G}{2}, ( \frac{1}{2}a+\frac{1}{2}b ) +\frac{G}{2} ) , \\ m, & \text{if }x\in [ a, ( \frac{1}{2}a+\frac{1}{2}b ) -\frac{G}{2} ] \cup [ ( \frac{1}{2}a+\frac{1}{2}b ) +\frac{G}{2}, b ]. \end{cases}$$

This ρ is the desired maximizer. Finally, we will calculate the maximum value of $I_{\rho }$,

$$\begin{aligned} I_{\rho } & = I_{h} \\ & = \int _{0}^{\frac{c}{2}-\frac{G}{2}}mf ( x ) \,dx + \int _{ \frac{c}{2}-\frac{G}{2}}^{ \frac{c}{2}+\frac{G}{2}}Mf ( x ) \,dx + \int _{ \frac{c}{2}+\frac{G}{2}}^{c}mf ( x ) \,dx \\ & = \frac{1}{24}m ( c-G ) ^{2} ( 2c+G ) + \frac{1}{12}GM \bigl( 3c^{2}-G^{2} \bigr) +\frac{1}{24}m ( c-G ) ^{2} ( 2c+G ) \\ & = \frac{1}{12}m ( c-G ) ^{2} ( 2c+G ) + \frac{1}{12}GM \bigl( 3c^{2}-G^{2} \bigr) . \end{aligned}$$

We now substitute $GM=Gm+1-mc$ into the previous expression.

$$\begin{aligned} I_{\rho } & = \frac{1}{12}m ( c-G ) ^{2} ( 2c+G ) +\frac{1}{12} ( Gm+1-mc ) \bigl( 3c^{2}-G^{2} \bigr) \\ & = \frac{1}{4}c^{2}-\frac{1}{12}mc^{3}- \frac{1}{12}G^{2}+\frac{1}{12}G^{2}mc \\ & = \frac{1}{4}c^{2}-\frac{1}{12}mc^{3}- \frac{1}{12}G^{2} ( 1-mc ) \\ & = \frac{1}{4}c^{2}-\frac{1}{12} \biggl( mc^{3}+ \frac{ ( 1-mc ) ^{3}}{ ( M-m ) ^{2}} \biggr) \\ & = \frac{1}{4} ( b-a ) ^{2}-\frac{1}{12} \biggl( m ( b-a ) ^{3}+ \frac{ ( 1-m ( b-a ) ) ^{3}}{ ( M-m ) ^{2}} \biggr) , \end{aligned}$$

as asserted in (3). □

Finally, we will comment on the choice of f and F in the proof. The proof makes essential use of the relationship between $I_{\rho }$ and $I_{g}$. $I_{h}$ is an intermediate quantity, related to $I_{\rho }$ by the translation $\tau _{a}$ and related to $I_{g}$ by the affine transformation ϕ. f is simply what results after starting with $( b-x ) ( x-a )$ in $I_{\rho }$ and translating by a to obtain $I_{h}$:

$$ I_{\rho }= \int _{a}^{b} ( b-x ) ( x-a ) \rho ( x ) \,dx = \int _{0}^{c} ( c-x ) x\cdot h ( x ) \,dx = \int _{0}^{c}f ( x ) \cdot h ( x ) \,dx =I_{h}. $$

In the proof of part (3) of the theorem, we need to maximize an integral, but the version of the Bathtub Principle that we use is stated for minimizations. It is for this reason that we apply the Bathtub Principle with the complementary function $F ( x ) =-f ( x ) +c^{2}/4$. Here, the $c^{2}/4$ constant term ensures that $F ( x )$ is nonnegative on $[ 0,c ]$. The specific choices of f and F do not affect the bounds in the theorem, which are absolute.