Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 4, Article 86, 2004
SOME INEQUALITIES BETWEEN MOMENTS OF PROBABILITY
DISTRIBUTIONS
R. SHARMA, R.G. SHANDIL, S. DEVI, AND M. DUTTA
D EPARTMENT OF M ATHEMATICS
H IMACHAL P RADESH U NIVERSITY
S UMMER H ILL , S HIMLA -171005, I NDIA
shandil_rg1@rediffmail.com
Received 23 February, 2004; accepted 19 July, 2004
Communicated by T. Mills
A BSTRACT. In this paper inequalities between univariate moments are obtained when the random variate, discrete or continuous, takes values on a finite interval. Further some inequalities
are given for the moments of bivariate distributions.
Key words and phrases: Random variate, Finite interval, Power means, Moments.
2000 Mathematics Subject Classification. 60E15, 26D15.
1. I NTRODUCTION
The rth order moment µ0r of a continuous random variate which takes values on the interval
[a, b] with pdf φ(x) is defined as
Z b
0
(1.1)
µr =
xr φ(x)dx.
a
For a random variate which takes a discrete set of finite values xi (i = 1, 2,. . ., n) with corresponding probabilities pi (i = 1, 2,. . ., n), we define
n
X
(1.2)
µ0r =
pi xri .
i=1
The power mean of order r is defined as
(1.3)
1/r
Mr = (µ0r )
for r 6= 0,
and
(1.4)
1/r
Mr = lim (µ0r )
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
040-04
r→0
for r = 0.
2
R. S HARMA , R.G. S HANDIL , S. D EVI ,
AND
M. D UTTA
It may be noted here that M−1 , M0 and M1 respectively define harmonic mean, geometric mean
and arithmetic mean.
Kapur [1] has reported the following bound for µ0r when µ0s is prescribed, r > s, and the
random variate, discrete or continuous, takes values in the interval [a, b] with a ≥ 0,
(br − ar ) µ0s + ar bs − as br
.
b s − as
Inequality (1.5) gives the condition which the given moment values must necessarily satisfy
in order to be the moments of a probability distribution in the given range [a, b]. Kapur [1]
was motivated by the consideration of maximizing the entropy function subject to certain constraints. But before maximizing the entropy function one has to see whether the given moment
values are consistent or not i.e whether there is any probability distribution which corresponds
to the given values of moments. If there is no such distribution then the efforts of finding out
the maximum entropy probability distribution will not produce any result and hence we should
not proceed to apply Lagrange’s or any other method to find the maximum entropy probability
distribution, [2].
Here we try to obtain a generalization of inequality (1.5) for the case where r and s can
assume any real value. This shall help us in deducing bounds between power means. This will
also provide us with an alternate proof of inequality (1.5) and enable us to tighten it when the
random variate takes a finite set of discrete values x1 , x2 ,. . ., xn .
In addition some inequalities between the moments of bivariate distributions are also obtained.
(1.5)
r/s
(µ0s )
≤ µ0r ≤
2. S OME E LEMENTARY I NEQUALITIES
We prove the following theorems:
Theorem 2.1. If r is a positive real number and s is any non zero real number with r > s then
for a ≤ x ≤ b; with a > 0, we have
(br − ar ) xs + ar bs − as br
,
b s − as
and for x lying outside (a, b) we have
(2.1)
xr ≤
(br − ar ) xs + ar bs − as br
.
b s − as
If r is a negative real number with r > s then inequality (2.1) holds for x lying outside (a, b)
and inequality (2.2) holds for a ≤ x ≤ b.
(2.2)
xr ≥
Proof. Consider the following function f (x) for positive real values of x:
br − ar s as br − ar bs
(2.3)
f (x) = x − s
x +
,
b − as
b s − as
where r and s are real numbers such that r > s and s 6= 0. The function f (x) is continuous in
the interval [a, b] with a > 0. Then f 0 (x) is given by
r
b − ar
0
s−1
r−s
(2.4)
f (x) = x
rx − s s
.
b − as
r
f 0 (x) vanishes at x = 0 and c, where
(2.5)
1
r
r−s
s b − ar
c=
.
r b s − as
J. Inequal. Pure and Appl. Math., 5(4) Art. 86, 2004
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I NEQUALITIES B ETWEEN M OMENTS OF P ROBABILITY D ISTRIBUTIONS
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By Rolle’s theorem we have that c lies in the interval (a, b).
If r is a positive real number and s is a negative real number with r > s then f 0 (x) ≤ 0 iff
x ≤ c. This means that f (x) decreases in the interval (0, c) and increases in the interval (c, ∞).
Further, since c lies in the interval (a, b) and f (a) = f (b) = 0, it follows that
f (x) ≤ 0
(2.6)
for a ≤ x ≤ b,
and for x lying outside (a, b)
f (x) ≥ 0.
(2.7)
On substituting the value of f (x) from equation (2.3) in inequalities (2.6) and (2.7), we obtain
inequalities (2.1) and (2.2) respectively.
If r is a negative real number with r > s then f 0 (x) ≤ 0 iff x ≥ c. This means that f (x)
increases in the interval (0, c) and decreases in the interval (c, ∞). Since c lies in the interval
(a, b) and f (a) = f (b) = 0 it follows that inequality (2.7) holds for a ≤ x ≤ b while inequality
(2.6) holds for x lying outside (a, b) and thus we get inequalities for the case when r is negative
real number.
Theorem 2.2. For a ≤ x ≤ b with a > 0, we have
xr ≤
(2.8)
(br − ar ) log x + ar log b − br log a
,
log b − log a
and for x lying outside (a, b), we have
xr ≥
(2.9)
(br − ar ) log x + ar log b − br log a
,
log b − log a
where r is a real number.
Proof. Consider the following function f (x) defined for positive real values of x,
(2.10)
f (x) = xr −
(br − ar )
br log a − ar log b
log x +
.
log b − log a
log b − log a
The function f (x) is continuous in the interval [a, b] where a > 0. Then f 0 (x) is given by
1
b r − ar
0
r
(2.11)
f (x) =
rx −
,
x
log b − log a
and we have f 0 (x) = 0 at x = c where
b r − ar
c=
r (log b − log a)
(2.12)
r1
.
By Rolle’s Theorem we have that c lies in the interval (a, b). Also f 0 (x) ≤ 0 iff x ≤ c. This
means that f (x) decreases in the interval (0, c) and increases in the interval (c, ∞). Further,
since c lies in the interval (a, b) and f (a) = f (b) = 0 it follows that
(2.13)
f (x) ≤ 0
for a ≤ x ≤ b,
and for x lying outside (a, b) we have
(2.14)
f (x) ≥ 0.
On substituting the value of f (x) from equation (2.10) in inequalities (2.13) and (2.14), we
obtain inequalities (2.8) and (2.9) respectively.
J. Inequal. Pure and Appl. Math., 5(4) Art. 86, 2004
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4
R. S HARMA , R.G. S HANDIL , S. D EVI ,
AND
M. D UTTA
3. I NEQUALITIES B ETWEEN M OMENTS
Theorem 3.1. Let r be a positive real number and s be any non zero real number with r > s. If
a positive random variate takes values xi (i = 1, 2,. . ., n) in the interval [a, b], with a > 0, then
we have
(3.1)
µ0r ≤
(br − ar ) µ0s + ar bs − as br
,
b s − as
and
(3.2)
0
r
r
x
−
x
µs + xrj−1 xsj − xsj−1 xrj
j
j−1
0
µr ≥
,
xsj − xsj−1
where j = 2, 3,. . ., n.
If a continuous random variate takes values in the interval [a, b], with a > 0, then the upper
bound for µ0r is given by the inequality (3.1) whereas the lower bound is given by following
inequality
r/s
µ0r ≥ (µ0s )
(3.3)
.
Proof. It is seen that µ0r can be expressed in terms of µ0s in the following form :
!
"
#
n
r
r
s r
s r
r
r
r s
r s
X
x
−
x
x
x
−
x
x
x
−
x
x
x
−
x
x
α
α
α
α
α
α
β
β
β
β
β
β
(3.4) µ0r =
µ0s +
+
pi xri − s
xs +
,
s i
xsβ − xsα
xsβ − xsα
x
−
x
xsβ − xsα
α
β
i=1
where α and β take one of the values among 1, 2,. . ., n with α 6= β. Without loss of generality
we can arrange values of the variate such that a = x1 ≤ x2 ≤ · · · ≤ xn = b. If we take α = 1
and β = n then x1 ≤ xi ≤ xn for i = 1, 2, ,. . ., n. It follows from (2.1) that the last term in
equation (3.4) is negative and we conclude that the upper bound for µ0r is given by inequality
(3.1). Further if xα = xj−1 and xβ = xj , j = 2, 3,. . ., n then each xi lies outside (xj−1 , xj )
and it follows from (2.2) that the last term in equation (3.4) is positive and we conclude that the
lower bound for µ0r is given by inequality (3.2). It is also clear that equality in the inequalities
(3.1) and (3.2) holds iff n = 2.
If the value of µ0s coincides with one of xsj−1 or xsj , then from inequality (3.2) we have
r/s
µ0r ≥ (µ0s )
(3.5)
.
Also if xj−1 approaches xj we get inequality (3.5) and we conclude that for a continuous
random variate the lower bound for µ0r is given by inequality (3.5). The upper bound for µ0r can
be deduced from Theorem 2.1. Multiplying both sides of inequality (2.1) by pdf φ(x) we get,
on using the properties of definite integrals, inequality (3.1).
Theorem 3.2. Let r and s be negative real numbers with r > s. If a positive random variate
takes values xi (i = 1, 2,. . ., n) in the interval [a, b], with a > 0, we have
(3.6)
µ0r ≥
(br − ar ) µ0s + ar bs − as br
,
b s − as
and
(3.7)
0
r
r
r
s
s
r
x
−
x
j
j−1 µs + xj−1 xj − xj−1 xj
µ0r ≤
,
xsj − xsj−1
where j = 2, 3,. . ., n.
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I NEQUALITIES B ETWEEN M OMENTS OF P ROBABILITY D ISTRIBUTIONS
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If a continuous random variate takes values in the interval [a, b], with a > 0, the lower
bound for µ0r is given by inequality (3.6) whereas the upper bound for µ0r is given by following
inequality:
r/s
µ0r ≤ (µ0s )
(3.8)
.
Proof. We again consider equation (3.4). If we take α = 1 and β = n then x1 ≤ xi ≤ xn for
i = 1, 2,. . ., n. It follows from Theorem 2.1 that the last term in equation (3.4) is positive and
we conclude that the lower bound for µ0r is given by inequality (3.6). Also if xα = xj−1 and
xβ = xj , j = 2, 3,. . ., n then each xi lies outside (xj−1 , xj ). It follows from Theorem 2.1 that
the last term in equation (3.4) is negative and we conclude that the upper bound for µ0r is given
by inequality (3.7). Also if xj−1 approaches xj we get inequality (3.8). The lower bound for
µ0r can be deduced from Theorem 2.1. Multiplying both sides of inequality (2.2) by pdf φ(x) we
get, on using the properties of definite integrals, inequality (3.6).
Theorem 3.3. For a random variate which takes values xi (i = 1, 2,. . ., n) in the interval [a, b],
with a > 0, we have
(br − ar ) log M0 + ar log b − br log a
,
(3.9)
µ0r ≤
log b − log a
and
xrj − xrj−1 log M0 + xrj−1 log xj − xrj log xj−1
0
(3.10)
µr ≥
,
log xj − log xj−1
where j = 2, 3,. . .n, r is a real number and
M0 = xP1 1 xP2 2 · · · xPnn .
(3.11)
For a continuous random variate which takes values in the interval [a, b] with a > 0 the upper bound for µ0r is given by inequality (3.9) whereas the lower bound for µ0r is given by the
following inequality
µ0r ≥ (M0 )r .
(3.12)
Proof. It is seen that µ0r can be expressed in terms of log M0 in the following form:
(3.13) µ0r =
xrβ − xrα
xrα log xβ − xrβ log xα
log M0 +
log xβ − log xα
log xβ − log xα
n
X
xrβ − xrα
xrβ log xα − xrα log xβ
r
+
Pi x i −
log xi +
.
log xβ − log xα
log xβ − log xα
i=1
Without loss of generality we can arrange values of the variate such that a = x1 < x2 < · · · <
xn = b. If we take α = 1and β = n then x1 ≤ xi ≤ xn for i = 1, 2,. . ., n. It follows from
Theorem 2.2 that last term in equation (3.13) is negative and we conclude that the upper bound
for µ0r is given by inequality (3.9). Also if xα = xj−1 and xβ = xj , j = 2, 3,. . ., n then each
xi lies outside (xj−1 , xj ). It follows from Theorem 2.2 that the last term in equation (3.13) is
positive and we conclude that the lower bound for µ0r is given by inequality (3.10).
If the value of M0 coincides with one of xj−1 or xj then from inequality (3.10) we have
(3.14)
µ0r ≥ (M0 )r .
Also if xj−1 approaches xj we get inequality (3.14) and we conclude that for the continuous
random variate the lower bound for µ0r is given by inequality (3.14). The upper bound for µ0r
can be deduced from Theorem 2.2. Multiplying both sides of inequality (2.8) by pdf φ(x) we
get, on using the properties of definite integrals, inequality (3.9).
J. Inequal. Pure and Appl. Math., 5(4) Art. 86, 2004
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6
R. S HARMA , R.G. S HANDIL , S. D EVI ,
AND
M. D UTTA
4. I NEQUALITIES B ETWEEN M OMENTS OF B IVARIATE D ISTRIBUTIONS
The moments of a bivariate probability distribution are the generalizations of those of univariate one and are equally important in the theory of mathematical statistics. For a discrete
probability distribution, if pi is the probability of the occurrence of the pair of values (xi , yi )
i = 1, 2,. . ., n, the moment µ0rs about the origin is given by
µ0rs
(4.1)
=
n
X
Pi xri yis .
i=1
We obtain a bound on
µ0rs
in the following theorem:
Theorem 4.1. Let µ0rs be the moment of order r in x and of order s in y, about the origin (0, 0),
of a discrete bivariate probability distribution. The random variates x and y vary respectively
over the finite positive real intervals [a, b] and [c, d]. If µ0k m is the corresponding moment of
order k in x and m in y such that r ≥ k, s ≥ m and rm = ks then we must have by necessity,
r+s
(4.2)
(µ0k m ) k+m ≤ µ0r s ≤
(br ds − ar cs ) µ0k m + ar cs bk dm − ak cm br ds
.
bk dm − ak cm
Proof. If u, v, α and β are positive real numbers with α + β = 1 then from Hölder’s inequality
[3],
!α
!β
n
n
n
X
X
X
β
(4.3)
uαi vi ≤
ui
vi .
i=1
i=1
i=1
We make the following substitutions,
(4.4)
ui = pi xri yis , vi = pi and α =
k+m
.
r+s
This gives,
uαi viβ = pi xki yim .
(4.5)
Also,
(4.6)
n
X
!α
ui
=
n
X
i=1
! k+m
r+s
pi xri yis
,
i=1
and
n
X
(4.7)
!β
vi
= 1.
i=1
From (4.3), (4.5), (4.6) and (4.7), we get
(4.8)
µ0r s ≥ (µ0k m )
r+s
k+m
.
For a ≤ x ≤ b, c ≤ y ≤ d, r ≥ k, s ≥ m and rm = ks, inequality (4.3) will remain valid if we
,
substitute n = 2, u1 = p1 ar cs , u2 = p2 br ds , v1 = p1 , v2 = p2 , α = k+m
r+s
p1 =
bk dm − xk y m
,
bk dm − ak cm
p2 =
x k y m − ak c m
.
bk dm − ak cm
and
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I NEQUALITIES B ETWEEN M OMENTS OF P ROBABILITY D ISTRIBUTIONS
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These substitutions give
(br ds − ar cs ) xk y m + ar cs bk dm − ak cm br ds
.
bk dm − ak cm
Without loss of generality we can have that the random variate take values a = x1 < x2 <
· · · < xn = b and c = y1 < y2 < · · · < yn = d therefore a ≤ xi ≤ b and c ≤ yi ≤ d,
i = 1, 2,. . ., n. From inequality (4.9), it follows that
xr y s ≤
(4.9)
xri yis ≤
(br ds − ar cs ) xki yim + ar cs bk dm − ak cm br ds
,
bk dm − ak cm
or
n
X
(br ds − ar cs )
Pi xri yis ≤
Pn
i=1
k m
r s k m
k m r s
i=1 Pi xi yi + a c b d − a c b d
bk dm − ak cm
Pn
i=1
Pi
,
or
(br ds − ar cs ) µ0k m + ar cs bk dm − ak cm br ds
.
bk dm − ak cm
Inequality (4.2) also holds for the continuous bivariate distributions. The upper bound in inequality (4.2) is a consequence of inequality (4.9). Multiplying both sides of inequality (4.9) by
joint pdf φ(x, y) and integrating over the corresponding limits, we get the maximum value of
µ0rs where
Z bZ d
Z bZ d
r s
0
x y φ(x, y)dxdy
and
φ(x, y)dxdy = 1.
µrs =
µ0r s ≤
a
c
a
Now consider,
RbRd
f α g β dxdy
a c
R R
α R R
β =
b d
b d
f dxdy
g dxdy
a c
a c
Z bZ
a
d
RbRd
a
d
"
≤
a
!α
f
c
Z bZ
c
c
c
g
RbRd
f dxdy
a
c
!β
dxdy
g dxdy
#
αf
βg
+ RbRd
dx dy
RbRd
f
dxdy
g
dxdy
a c
a c
= 1,
where α + β = 1 and f and g are positive functions. We therefore have
Z b Z d
α Z b Z d
β
Z bZ d
α β
(4.10)
f g dx dy ≤
f dxdy
g dxdy ,
a
c
a
a
c
c
and make the following substitutions,
f = xr y s
φ(x, y), g = φ(x, y)
and
α=
Inequality (4.10) then yields the minimum value of µ0rs .
k+m
.
r+s
5. A PPLICATIONS OF R ESULTS
On using the results derived in Section 3 and giving particular values to r and s it is possible
to derive a host of results connecting the Harmonic mean (H), Geometric mean (G), Arithmetic
mean (A) and Root mean square (R) when one of the means is given and the random variate
takes the prescribed set of positive values x1 , x2 , . . . , xn .
J. Inequal. Pure and Appl. Math., 5(4) Art. 86, 2004
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R. S HARMA , R.G. S HANDIL , S. D EVI ,
AND
M. D UTTA
If we put r = +1 and s = −1 we get inequalities between A and H, if we put r = 0 and
s = −1 we get inequalities between G and H, and so on. Root mean square R corresponds to
r = 2. In particular the following inequalities are obtained from the general result,
1
1
(5.1)
[(xj−1 + xj ) A − xj−1 xj ] 2 ≤ R ≤ [(a + b) A − ab] 2 ,
(5.2)
xj−1 xj
ab
≤H≤
,
a+b−A
xj−1 + xj − A
b
(5.3)
A−a b−A
a
1
b−a
≤G≤
A−x
xj −A
xj j−1 xj−1
1
j −xj−1
x
,
1
1
xj−1 xj (xj−1 + xj ) 2
ab (a + b) 2
2
2
2
2
(5.4) xj−1 + xj−1 xj + xj −
≤ R ≤ a + ab + b −
,
H
H
(xj−1 + xj ) −
(5.5)
(5.6)
h
x (H−xj−1 ) xj−1 (xj −H)
xj j
xj−1
log
1
j −xj−1 )
i H(x
xj x2j−1
G
xj−1
(5.7)
x
j
log xj−1
ab
log ab
a b b
log Ga
G
(5.8)
log
(5.9)
G
xj
xj−1
1
≤ G ≤ bb(H−a) aa(b−H) H(b−a) ,
2
≤R ≤
2
G b
a
log
2
b a
G
log ab
,
xj−1 xj
xj
log xj−1
xj−1
,
≤H≤
xj xj
G
log xj−1
G
xj xj−1
G
x
j
log xj−1
≤A≤
log
G b
a
b a
G
log ab
,
R2 + ab
R2 + xj−1 xj
≤A≤
,
a+b
xj−1 + xj
(5.10)
(5.11)
x2j
G
xj−1 xj
ab
≤A≤a+b− ,
H
H
a2
ab (a + b)
xj−1 xj (xj−1 + xj )
≤H≤ 2
,
2
2
+ ab + b − R
xj−1 + xj−1 xj + x2j − R2
and
(5.12)
b
R2 −a2
b2 −a2
a
b2 −R2
b2 −a2
R2 −x2
j−1
x2 −x2
j
j−1
≤ G ≤ xj
2
x2
j −R
x2 −x2
j
j−1
xj−1
,
where j = 2, 3,. . ., n.
We now deduce the result that the power mean Mr is an increasing function of r. If r is
positive and s is any real number with r > s then from inequality (3.3) we have
1
(5.13)
1
(µ0r ) r ≥ (µ0s ) s ,
or Mr ≥ Ms.
J. Inequal. Pure and Appl. Math., 5(4) Art. 86, 2004
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I NEQUALITIES B ETWEEN M OMENTS OF P ROBABILITY D ISTRIBUTIONS
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If r is a negative real number with r > s we again get inequality (5.13) from inequality (3.8).
From inequality (3.12) we have Mr ≥ M0 for r > 0, and Mr ≤ M0 for r < 0. Hence we
conclude that the power mean of order r is an increasing function of r. In particular, we get that
M−1 ≤ M0 ≤ M1 ≤ M2 .
R EFERENCES
[1] J.N. KAPUR AND A. RANI, Testing the consistency of given values of a set of moments of a
probability distribution, J. Bihar Math. Soc., 16 (1995), 51–63.
[2] J.N. KAPUR, Maximum Entropy Models in Science and Engineering, Wiley Eastern and John Wiley,
2nd Edition, 1993.
[3] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, 2nd
Edition, 1952.
J. Inequal. Pure and Appl. Math., 5(4) Art. 86, 2004
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