On a moment inequality for laws on (−∞,+∞)

Statistics and Probability Letters 79 (2009) 2334–2337

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On a moment inequality for laws on (−∞, +∞) Slavko Simic ∗ Mathematical Institute SANU, Kneza Mihaila 36, 11000 Belgrade, Serbia

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Article history: Received 20 May 2009 Accepted 10 August 2009 Available online 18 August 2009

abstract We give a positive solution to the conjectured moment inequality for probability distributions with support on R. A considerable improvement of the classical Jensen’s inequality follows. © 2009 Elsevier B.V. All rights reserved.

MSC: 60E15

1. Introduction The aim of this paper is to obtain a new moment inequality which is valid for all probability distributions with support on

(−∞, ∞) and also to make some additional remarks on the results from Simic (2008). In the sequel we implicitly assume

that all stated moments actually exist. The two fundamental moment inequalities for probability distributions with support on R+ (cf. Rossberg et al., 1985) are the following: 1. Jensen’s inequality For each m ∈ N,

(EX )m ≤ EX m .

(1)

2. Lyapunov’s inequality For m, n, p ∈ N, m < p < n,

(EX p )n−m ≤ (EX m )n−p (EX n )p−m . Note that for non-degenerate laws there is a strict inequality in (1) and (2) (cf. Hardy et al., 1978; Mitrinovic, 1970). The above inequalities admit an extension to the laws with support on R. Proposition 0. For a non-degenerate probability distribution with support on (−∞, +∞) and m, n, p ∈ N, we have (i)

(EX )2m < EX 2m ; (ii)

(EX p )2n−2m < (EX 2m )2n−p (EX 2n )p−2m , for each p, 2m < p < 2n.

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Tel.: +381 63 277718. E-mail address: [email protected].

0167-7152/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2009.08.003

(2)

S. Simic / Statistics and Probability Letters 79 (2009) 2334–2337

2335

Indeed, since |EX | ≤ E |X |, we get

(EX )2m = (|EX |)2m ≤ (E |X |)2m < E |X |2m = EX 2m , and by (2),

(EX p )2n−2m = (|EX p |)2n−2m ≤ (E |X |p )2n−2m < (E |X |2m )2n−p (E |X |2n )p−2m = (EX 2m )2n−p (EX 2n )p−2m . Note that if m or n is an odd number, then the inequalities (1) and (2) do not hold for an arbitrary distribution with support on R. We proved in Simic (2008) the following inequality which is a mixture of Jensen’s and Lyapunov’s inequalities. Theorem A. For any probability distribution F with support on (0, ∞), r , s, t ∈ N and 1 < r < s < t, we have

(EX s − (EX )s )t −r ≤ C (r , s, t )(EX r − (EX )r )t −s (EX t − (EX )t )s−r ,

(3)

with s t −r 2 t −s t s−r




C ( r , s, t ) =

r 2







.

(4)

2

An interesting question is whether the relation (3) can be valid for distributions with support on (−∞, +∞). Our main contribution here is to give a necessary and sufficient conditions for orders r , s, t such that this assertion is fulfilled. 2. Results We prove firstly the following Proposition 1. In the moment inequality (3), the constant C (r , s, t ) is best possible. Denote by µk , µk := E (X − EX )k the central moments of a distribution F . A considerable improvement of Jensen’s inequality in terms of µ2 , µ3 is given in the next Proposition 2. For any law F with support on (0, ∞) and with the variance σX2 (=µ2 > 0), we have EX m ≥ (EX )m +

m  2

EX +

µ3 3µ2

m−2

σX2 ,

m ≥ 2.

(5)

Our main result is the positive answer to the conjecture arising in Simic (2008) which is contained in the following Proposition 3. The inequality (3) is valid for any non-degenerate law F with support on (−∞, +∞) if and only if r and t are positive even numbers and s is any integer between them. Therefore, for arbitrary m, n, p ∈ N the inequality

(EX p − (EX )p )2n−2m ≤ C (2m, p, 2n)(EX 2m − (EX )2m )2n−p (EX 2n − (EX )2n )p−2m ,

(6)

holds whenever 2m < p < 2n. An improved variant of Jensen’s inequality in this case is the next Proposition 4. For an arbitrary non-degenerate law F with support on (−∞, +∞), the inequality



EX 2n ≥ (EX )2n + n(2n − 1) (EX )2 +

2µ3 3µ2

EX +

µ4 6µ2

 n −1

σX2 ,

(7)

holds for each n ∈ N. 3. Proofs Proof of Proposition 1. Denote ds := EX s −(EX )s , s ∈ N. Note that for a non-degenerate law F we have d2 = EX 2 −(EX )2 = σX2 > 0.

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S. Simic / Statistics and Probability Letters 79 (2009) 2334–2337

To prove the assertion from Proposition 1., we make a shift X → X + x, x > 0 and consider ds (x) = E (X + x)s − (E (X + x))s =

Z

(t + x)s dF (t ) −

Z

s Z  s Z (t + x)dF (t ) = (x + t )s dF (t ) − x + tdF (t ) .

Developing in x, we get ds (x) =

s   X s

k

k=0

EX k xs−k −

s   X s

k

k=0

(EX )k xs−k =

s   X s

k

k=2

dk xs−k .

For sufficiently large x, it follows that ds (x) ∼ d2

s

xs−2

2

(x → ∞).

Hence

(ds (x))t −r ∼ (dr (x))t −s (dt (x))s−r d2

t −r s xs−2 2 s−r t −s t xr −2 d2 2 xt −2




d2

r 2
















→ C (r , s, t ) (x → ∞).

Therefore we conclude that the constant C (r , s, t ) in the moment inequality (3) is best possible.



Proof of Proposition 2. Since EX 2 − (EX )2 = µ2 = σX2 > 0, putting in (3): r = 2, s = 3, t = m, we obtain EX m − (EX )m ≥

 m   EX 3 − (EX )3 m−2 3µ2

2

σX2 .

But EX 3 − (EX )3 = µ3 + 3µ2 EX and the result follows. Proof of Proposition 3. We show firstly that for odd r or t, the inequality (3) cannot be valid for arbitrary probability distributions with support on R. If s is an even number, this is evidently true for the uniform discrete distributions with masses at the points of the form {±k}. If s is an odd number, consider the system of (linear in p) equations n X

n X

pi = 1;

1

n X

pi xi = 0;

1

2k1 +1

pi xi

1

= 0;

n X

2k2 +1

pi xi

= a > 0;

k1 6= k2 ∈ N.

(8)

1

For n > 4, by elementary algebra theorems, there exist a positive p = {pi }n1 and a real x = {xi }n1 such that (8) is fulfilled. This means that there exist discrete distributions with support on R such that EX = 0;

EX r ∨t = 0;

EX s > 0.

It is evident that in this case the inequality (3) does not hold. Now we shall prove that (3) is valid for any probability law with support on R whenever r and t are even numbers and s is a number between them i.e. we prove the inequality (6). For this cause, the following assertion is of crucial importance. Lemma 1. Denote λm := (EX m − (EX )m )/

m 2




, m ∈ N. Then the inequality

(λm+n )2 ≤ λ2m λ2n ,

(9)

holds for arbitrary laws with support on (−∞, ∞) and m, n ∈ N. Proof. We give a short and elegant proof applying the method originated from Simic (2007). Let f : R → R be given by f (x) := u2

x2m

xm+n

x2n

m 2

m+n 2

2n 2


+ 2uv


+ v2  

where u, v are real parameters. Since f 00 (x) = 2(u2 x2m−2 + 2uv xm+n−2 + v 2 x2n−2 ) = 2(uxm−1 + v xn−1 )2 ≥ 0, we conclude that f is a convex function on R. Therefore Jensen’s inequality (cf. Rossberg et al., 1985) gives Ef (X ) ≥ f (EX ),

S. Simic / Statistics and Probability Letters 79 (2009) 2334–2337

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that is u2 λ2m + 2uvλm+n + v 2 λ2n ≥ 0.

(10)

Since u, v are arbitrary real numbers, the relation (10) is possible only if

(λm+n )2 − λ2m λ2n ≤ 0, as required. We are able now to prove the inequality (6). Assume firstly that p is an even number p = 2k, 2m < 2k < 2n. From the above lemma we get λ22k ≤ λ2k−2 λ2k+2 λ

λ

−2 ≥ λ2k2k+2 . i.e. 2k λ2k Hence

λ2l−2 λ2k−2 ≥ , λ2k λ2l

l = k + 1, . . . , n.

(11)

Similarly, from the inequality (9) it follows that

λ2k λ2l ≥ , λ2k−2 λ2l−2

l = m + 1, . . . , k.

(12)

Multiplying the inequalities from (11) and (12), we get



λ2k−2 λ2k

n−k

λ2k λ2k−2

k−m

≥

 n  Y λ2l−2 λ2l

k+1

=

λ2k , λ2n

and



≥

 k  Y λ2l λ2k = . λ2l−2 λ2m m+1

Hence



λ2k λ2n

k−m

 ≤

λ2k−2 λ2k

(n−k)(k−m)

 ;

λ2k λ2m

 n −k

 ≤

λ2k λ2k−2

(k−m)(n−k)

.

Multiplying those inequalities we obtain a version of the inequality (6) with p = 2k, 2m < 2k < 2n i.e.,

(λ2k )n−m ≤ (λ2m )n−k (λ2n )k−m .

(13)

If p is an odd number p = 2k + 1, 2m < 2k + 1 < 2n, from (9) we get (λ2k+1 ) ≤ λ2k λ2k+2 . Hence by (13), 2

(λ2k+1 )2(n−m) ≤ (λ2k )n−m (λ2k+2 )n−m ≤ (λ2m )n−k (λ2n )k−m (λ2m )n−k−1 (λ2n )k+1−m , i.e.

(λ2k+1 )2n−2m ≤ (λ2m )2n−(2k+1) (λ2n )2k+1−2m , and this is exactly the inequality (6) with p = 2k + 1. The proof is complete.  Proof of Proposition 4. Putting in (13) k = 2, m = 1 and recalling the definition of λk , we get EX

2n

− (EX )

2n

 ≥

2n 2



EX 4 − (EX )4 6σX2

n−1

σX2 .

Since σX2 = µ2 and EX 4 − (EX )4 = µ4 + 4µ3 EX + 6µ2 (EX )2 , the assertion from Proposition 4. follows. References Hardy, G.H., Littlewood, J.E., Polya, G., 1978. Inequalities. Cambridge University Press, Cambridge. Mitrinovic, D.S., 1970. Analytic Inequalities. Springer-Verlag, Berlin. Rossberg, H.J., Jesiak, B., Siegel, G., 1985. Analytic Methods of Probability Theory. Akademie-Verlag, Berlin. Simic, S., 2008. On a new moments inequality. Statist. Probab. Lett. 78 (16), 2671–2678. Simic, S., 2007. On logarithmic convexity for differences of power means. J. Inequal. Appl. 8. Article ID 37359.