Volume 2008, Article ID 386715,9pages doi:10.1155/2008/386715
Research Article
Some Equivalent Forms of the
Arithematic-Geometric Mean Inequality in
Probability: A Survey
Cheh-Chih Yeh,1 Hung-Wen Yeh,2and Wenyaw Chan3
1Department of Information Management, Lunghwa University of Science and Technology,
Kueishan Taoyuan, Taoyuan County 33306, Taiwan
2Department of Biostatistics, University of Kansas, Kansas City, KS 66160, USA
3Division of Biostatistics, University of Texas-Health Science Center at Houston, Houston,
TX 77030, USA
Correspondence should be addressed to Cheh-Chih Yeh,[email protected]
Received 5 December 2007; Revised 10 April 2008; Accepted 24 June 2008
Recommended by Jewgeni Dshalalow
We link some equivalent forms of the arithmetic-geometric mean inequality in probability and mathematical statistics.
Copyrightq2008 Cheh-Chih Yeh et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
The arithmetic-geometric mean inequalityin short, AG inequalityhas been widely used in mathematics and in its applications. A large number of its equivalent forms have also been developed in several areas of mathematics. For probability and mathematical statistics, the equivalent forms of the AG inequality have not been linked together in a formal way. The purpose of this paper is to prove that the AG inequality is equivalent to some other renowned inequalities by using probabilistic arguments. Among such inequalities are those of Jensen, H ¨older, Cauchy, Minkowski, and Lyapunov, to name just a few.
2. The equivalent forms
LetXbe a random variable, we define
Er|X|:
⎧ ⎨ ⎩
E|X|r1/r
, ifr / 0,
expEln|X|, ifr 0, 2.1
Throughout this paper, letnbe a positive integer and we consider only the random variables which have finite expected values.
In order to establish our main results, we need the following lemma which is due to Infantozzi1,2, Marshall and Olkin3, Page 457, and Maligranda4,5. For other related results, we refer to6–19.
Lemma 2.1. The following inequalities are equivalent.
E1AG inequality:EX≥eElnX, whereXis a nonnegative random variable.
E2aq11aq22· · ·anqn ≤a1q1a2q2· · ·anqnifai∈0,∞andqi ∈0,1fori1,2, . . . , n withni1qi 1. The arithmetic-geometric mean inequality is usually applied in a simple version of E2withqi1/nfor eachi1,2, . . . , n.
E3aαb1−α ≤αa 1−αbif0 < α < 1anda, b >0, and the opposite inequality holds if
α >1orα <0.
E4 y1α<1αyif0< α <1andy >−1, and the opposite inequality holds ifα >1or
α <0andy >−1. E5
n
i1a
p ib
q i ≤
n
i1aip
n
i1biq forai, bi∈0,∞, i1,2, . . . , nifp >0, q > 0with pq≥1, and the opposite inequality holds ifpq <0withpq≤1.
E6
n
i1aibip1/p ≥
n
i1a
p i1/p
n
i1b
p
i1/p ifp ≤ 1 andai, bi ∈ 0,∞fori
1,2, . . . , n, and the opposite inequality holds ifp≥1. E7
n
i1aibis
t−r ≤ n i1aibri
t−sn i1aibti
s−r
if ai, bi ∈ 0,∞fori 1,2, . . . , nand r < s < t.
E8LetΩ,B, μbe a measure space. Iffi:Ω→0,∞is finitelyμ-integrable,i1,2, . . . , n and letqi≥0,
n
i1qi1. ThenΠni1f
qi
i is finitely integrable and
n i1f
qi
i dμ≤ ni1
fqi
i dμ. E9Ifa≥b ≥candf :Ω→Risμ-integrable, whereΩ,B, μis a probability space, then
|f|bdμa−c≤|f|cdμa−b|f|adμb−c.
E10Artin’s theorem. LetKbe an open convex subset ofRandf :K×a, b→0,∞satisfy
afx, yis Borel-measurable inyfor each fixedx, blogfx, yis convex inxfor each fixedy.
Ifμis a measure on the Borel subsets ofa, bsuch thatfx,·isμ-integrable for eachx∈K,then gx:logbafx, ydμyis a convex function onK.
E11 Jensen’s inequality. LetΩ be a probability space andX be a random variable taking
values in the open convex setA⊂Rwith finite expectationEX. Iff :A→Ris convex, thenEfX≥ fEX.
Proof. The proof of the equivalent relations ofE2,E3,E4, . . . ,E7can be found in1,2,
4,5.
The proof of the equivalent relations ofE1, E2, E8, E9, E10, andE11can be
found in3.
Theorem 2.2. The following inequalities are equivalent.
H0E|XY| ≤Ep|X|Eq|Y|ifX, Y are random variables and1/p1/q1withp >1 andq >1.
H1E|Z||X|h|Y|k ≤ E|Z||X|hE|Z||Y|kifX, Y, Zare random variables andhk 1
withh >0andk >0.
H2E|Z||X|h|Y|k ≥ E|Z||X|hE|Z||Y|kifX, Y, Zare random variables andhk 1
H∗
2E|X|h|Y|k≥E|X|
hE|Y|kifX, Y are random variables andhk1withhk <0, that is,E|XY| ≥Ep|X|Eq|Y|if1/p1/q1with0< p <1.
H3E|X|h|Y|k ≤ E|X|hE|Y|k ifX, Y are random variables andhk ≤ 1withh >0
andk >0.
H4E|X|h|Y|k≥E|X|hE|Y|kifX, Yare random variables andhk≥1withhk <0.
L1 E|Z||X|st−r ≤E|Z||X|ts−rE|Z||X|rt−sifX, Zare random variables andr < s < t.
L2 E|Z||X|st−r ≥E|Z||X|ts−rE|Z||X|rt−sifX, Zare random variables ands < r < t.
L3 E|X|r1/r ≤ E|X|s1/s if X is a random variable and r ≤ s, that is, E|X|r1/r is
nondecreasing onr.
L4(see [10,18])E|X|p≥E|X|p, whereXis a random variable ifp≥1orp≤0, and the
opposite inequality holds if0≤p≤1.
R1 E|X|rp/E|Y|rq ≤ E|X|p/|Y|qr ifX, Y are random variables andp ≥ qr with
p >0, q >0, r >0.
R2 E|X|rp/E|Y|rq ≤ E|X|p/|Y|qr ifX, Y are random variables andp ≥ qr with
p <0, q <0, r >0.
R3 E|X|rp/E|Y|rq ≥ E|X|p/|Y|qr ifX, Y are random variables andp ≤ qr with
p >0, q <0, r >0.
R4 E|X|p/E|Y|p−1≤E|X|p/|Y|p−1ifX, Yare random variables andp≥1.
R5 E|X|p/E|Y|p−1≤E|X|p/|Y|p−1ifX, Yare random variables andp <0.
R6 E|X|p/E|Y|p−1≥E|X|p/|Y|p−1ifX, Yare random variables and0< p <1.
C1 Cauchy-Bunyakovski and Schwarz’s (CBS) inequality: E|XY|2 ≤ E|X|2E|Y|2if
X, Yare random variables. C∗
1 E|XY Z|
2≤E|Z||X|2E|Z||Y|2ifX, Y, Zare random variables.
C2 EZ|X|s|Y|1−sEZ|X|1−s|Y|s ≤ E|Z||X|E|Z||Y| if X, Y, Z are random
variables ands∈0,1(the inequality is reversed ifs >1ors <0).
C3 E|Z||X|pr|Y|p−rE|Z||X|p−r|Y|pr ≤ E|Z||X|ps|Y|p−sE|Z||X|p−s|Y|ps
for anyp∈RifX, Y, Zare random variables and|r| ≤ |s|.
C4 E|Z||X|r|Y|sE|Z||X|s|Y|r ≤ E|Z||X|u|Y|vE|Z||X|v|Y|uif X, Y, Z are
random variables and either0≤v≤s≤r≤u, rsuvor0≤u≤r≤s≤v, rsuv. C5 E|Z||X|rE|Z||X|−r≤ E|Z||X|sE|Z||X|−sifX, Y, Zare random variables and
|r| ≤ |s|.
C6 E|Z||X|p−s|Y|sE|Z||X|s|Y|p−s ≤ E|Z||X|p−r|Y|rE|Z||X|r|Y|p−r if X, Y, Z
are random variables and eitherp/2≤s≤r≤por0≤r ≤s≤p/2.
C7 E|Z||X|2−s|Y|sE|Z||X|s|Y|2−s ≤ E|Z||X|2−r|Y|rE|Z||X|r|Y|2−r if X, Y, Z
are random variables and either0≤r≤s≤1or1≤s≤r≤2.
C8 E|Z||X|ks|Y|l−tE|Z||X|k−s|Y|ltincreases with|s|ifX, Y, Zare random variables
andk/ls/t.
MMinkowski’s inequality:Ep|XY| ≤Ep|X|Ep|Y|ifX, Yare random variables,p≥1, and the opposite inequality holds ifp≤1.
TTriangle inequality:Ep|X−Y| ≤Ep|X−Z|Ep|Z−Y|ifX, Y, Zare random variables, p≥1,and the opposite inequality holds ifp≤1.
J1G2G−11EY≤EG2G−11YifYis a random variable,G1andG2are two continuous and
strictly increasing functions such thatG2G−11is convex.
J2EetX≥etEXfor anyt∈RifXis a random variable.
Proof. The sketch of the proof of this theorem is illustrated by the following maps of equivalent circles:
1 E3⇒H0⇔H1⇔H2⇔H2∗;
2 H1⇒L1⇒H0⇒L3⇒H3;
3 H2⇒L2⇒H2∗⇒H4;
4 L1⇒L3⇔L4, L2⇒L3⇒E1;
5 H3 ⇒ R1 ⇒ R4 ⇒ H2∗, H4 ⇒ R2 ⇒ R5 ⇒ H2, H4 ⇒ R3 ⇒
R6⇒H2;
6 H0⇔M⇔T;
7 C1⇒H0⇒H1⇒C2⇒C3⇒C4⇒C6⇒C7⇒C1⇔C1∗;
8 C4⇒C5⇒C3⇔C8;
9 E11⇒J1⇒L3, E11⇒J2⇒E1.
Now, we are in a position to give the proof of this theorem as follows.
E3⇒H0, see Casella and Berger7, page 187.
H0⇔H1is clear.
H1 ⇒H2: Ifh < 0 andk > 0, then−k/h >0 and−h/k1/k 1. This andH1
imply
E|Z||X|−h/k|Y|1/k≤E|Z||X|−h/kE|Z||Y|1/k
. 2.2
Replacing|Y|by|X|h|Y|kin the above inequality, we obtainH
2.
Similarly, we can prove the case thath >0 andk <0.
H2⇒H1is proved similarly.
H2⇔H2∗is clear.
H1 ⇒ L1. Letting|X|, |Y|, h, andkbe replaced by|X|t, |X|r, s−r/t−rand
t−s/t−rinH1, respectively, we obtainL1.
H2⇒L2is similarly proved.
L1⇒H0: Leth t−s/t−r, k s−r/t−r. Thenhk1, h >0, k >0. It
follows fromL1that
E|X||Y|E|X|t/t−s|Y|−r/s−r|X|−1/t−s|Y|1/s−rs
≤E|X|t/t−s|Y|−r/s−r|X|−1/t−s|Y|1/s−rrt−s/t−r
×E|X|t/t−s|Y|−r/s−r|X|−1/t−s|Y|1/s−rts−r/t−r
E1/h|X|E1/k|Y|.
2.3
That is,H0holds.
L2⇒H2∗is similarly proved.
H0⇒L3⇒H3. TakingY 1 inH0, we see that
which implies
E|X|r/s≤E|X|r/s, p r
s. 2.5
Replacing|X|by|X|s,
E|X|sr/s≤E|X|r. 2.6
Thus
E|X|s1/s ≤E|X|r1/r ifr > s >0,
E|X|s1/s ≥E|X|r1/r
ifr < s <0.
2.7
This provesL3.
Next, letphk. Thenh/pk/p1 and 0< p≤1. This andH0imply
E|X|h/p|Y|k/p≤E|X|h/pE|Y|k/p
. 2.8
Replacing|X|and|Y|by|X|pand|Y|pin the above inequality, respectively, and usingL
3,
we obtain
E|X|h|Y|k≤E p|X|
h Ep|Y|
k≤E|X|hE|Y|k
. 2.9
This provesH3holds.
H∗
2⇒H4is proved similarly.
L1⇒L3.aTakingZ1 andt0 inL1,
E|X|s−r ≤E|X|r−s ifr < s <0, 2.10
which implies
E|X|r1/r ≤E|X|s1/s
ifr < s <0. 2.11
bTakingZ1 andr 0 inL1,
E|X|st≤E|X|ts if 0< s < t, 2.12
which implies
E|X|s1/s≤E|X|t1/t
cTakingZ1 ands0 inL1,
1≤E|X|t−rE|X|rt ifr <0< t, 2.14
which implies
E|X|r1/r≤E|X|t1/t
if r <0< t. 2.15
dIt follows froma,b, andcthatL3holds.
Thus, we complete the proof.
L2⇒L3is similarly proved.
L3⇒L4is clear.
L4⇒L3by using the technique ofH0⇒L3.
L3⇒E1. Lettingr→0 ands1 inL3, we obtainE1.
H3⇒R1. It follows fromp≥qrandp, q, r∈0,∞thatq/pr/p≤1. This and
H3imply
E|X|q/p|Y|r/p≤E|X|q/pE|Y|r/p
. 2.16
Replacing|X|and|Y|by|Y|r and|X|p|Y|−q in the above inequality, respectively, we obtain R1.
H4⇒R2andH4⇒R3are similarly proved.
R1⇒R4, R2⇒R5andR3⇒R6follow by takingqp−1 andr1.
R4 ⇒ H2∗,R5 ⇒ H2and R6 ⇒ H0 follow by takingp h, k 1−p in
R4, R5andR6, respectively.
H0⇒MCasella and Berger7, page 188.
M⇒H0 see5: Let 1/p1/q1 withp >1 andq >1. It follows from Benoulli’s
inequalityE4that
pt|X||Y| ≤|Y|1/p−1t|X|p− |Y|p/p−1, fort >0. 2.17
This andMimply
ptE|X||Y| ≤E|Y|p/p−11/p
tE|X|p1/pp−E|Y|p/p−1, fort >0. 2.18
Hence
pE|X||Y| ≤lim
t→0inf
1
t
E|Y|p/p−11/p
tE|X|p1/pp−E|Y|p/p−1
pEp|X|Eq|Y|.
2.19
This provesH0holds.
M⇒Tfollows by replacingXandY byX−ZandZ−Y inM, respectively.
C1⇒H0. LetFx E|Y|q|X|p|Y|−qxforx∈0,1. Then, it follows fromC1that
F
x1
2
x2
2
E|Y|q|X|p|Y|−qx11/2|Y|q|X|p|Y|−qx21/2
≤E|Y|q|X|p|Y|−qx11/2E|Y|q|X|p|Y|−qx21/2
Fx1
1/2
Fx2
1/2
.
2.20
Thus, lnFis midconvex on0,1, and hence lnFis convex on0,1. Hence
lnF
r p
1−r q
≤ 1
plnFr
1
qlnF1−r. 2.21
Therefore,
F
r p
1−r q
≤F1/prF1/q1−r. 2.22
Lettingr→1−in the both sides of the above inequality,
E|XY|F
1
p
≤F1/p1F1/q0 E p|X|
Eq|Y|
. 2.23
This showsH0 see13.
H1⇒C2. First note that, as shown above,H1andH2are equivalent. It follows
fromH1that, fors∈0,1,
E|Z||X|s|Y|1−s≤E|Z||X|sE|Z||Y|1−s ,
E|Z||X|1−s|Y|s≤E|Z||X|1−sE|Z||Y|s .
2.24
These implyC2for the cases∈0,1.
Similarly, we can prove the case fors >1 ors <0 by usingH2.
C2 ⇒ C3follows by replacing s, |X|, |Y| inC2by1/21r/s if rs > 0 or
1/21−r/sifrs <0, |X|ps|Y|p−s,|X|p−s|Y|ps, respectively.
C3 ⇒ C4 follows by replacing p r, p−r, ps, p − s in C3 by r, s, u, v,
respectively.
C4 ⇒ C6follows by replacingr, s, u, vbyp−s, s, p−r, rors, p−s, r, p−r in
C4, respectively.
C6⇒C7follows by takingp2 withr≥0 inC6.
C7⇒C1follows by takings1 andr 0 inC7.
C1⇔C1∗is clear.
C4⇒C5follows by lettingrsuv0, ur, vsandY 1 inC4.
C5 ⇒ C3 follows by replacing |Z|and |X|in C5 by |Z||X||Y|p and |X||Y|−1,
C3⇒C8. Replacing|X|by|X|uand|Y|by|Y|vinC3and changing appropriately
the notation for the exponents, we obtainC8.
C8⇒C3is clear.
To complete our proof of equivalence of all inequalities in this theorem and in
Lemma 2.1, it suffices to show further the following implications.
E11⇒J1follows by takingf G2G−11inE11.
J1 ⇒ L3: LetG1Y |Y|r1, G2Y |Y|r2, wherer2/r1 > 1hencer2 > r1 > 0 or
r2 < r1 <0. Then it follows fromJ1that|EY|r2/r1 ≤ E|Y|r2/r1. SettingY |X|r1, we obtain
L3, see14, page 162.
E11⇒J2follows by takingfx etxinE11.
J2⇒E1follows by takingt1 and replacingXby lnXinJ2.
Remark 2.3. Lettingr p, sp−1h, uph, v p−1 andY Z 1 withh≥0 and
p∈RinC4, we obtain the inequality5of18:
E|X|p−1hE|X|p≤E|X|phE|X|p−1. 2.25
That is,
rp: E
|X|p−1
E|X|p 2.26
is a decreasing function of Sclove et al.18proved this property by means of the convexity offt lnE|X|t, see14. Clearly, our method is simpler than theirs.
Remark 2.4. EachHi orHi∗is called H ¨older’s inequality, eachCi orC∗iis called CBS
inequality, eachLiis called Lyapunov’s inequality, eachRiis called Radon’s inequality, each Jiis related to Jensen’s inequality.
Acknowledgments
The authors wish to thank three reviewers for their valuable suggestions that lead to substantial improvement of this paper. This work is dedicated to Professor Haruo Murakami on his 80th birthday.
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