An inequality for the tails of binomial distributions

An inequality for the tails of binomial distributions

Citation for published version (APA):

van den Akker, A. G., & Steutel, F. W. (1976). An inequality for the tails of binomial distributions. (Memorandum COSOR; Vol. 7602). Technische Hogeschool Eindhoven.

Document status and date: Published: 01/01/1976

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EINDHOVEN UNIVERSITY OF TECHNOLOGY

Department of Mathematics

PROBABILITY THEORY, STATISTICS AND OPERATIONS RESEARCH GROUP

Memorandum COSOR 76-02 An inequality for the tails of

binomial distributions by

A.G. van den Akker and F.W. Steutel

Eindhoven, February 1976 The Netherlands

I. Introduction. In this note we inequali ty. Le t np and variance x 2 0

give a brief self-contained proof of the following K be a binomial random variable with expectation

n

np(l - p) (n = 1,2, ... ; P E [0,1

J),

then for all

(I ) K - np P(

I

n

I

;;;

-2x2 2 x) :0; 2e •

This inequality is not new (see

[IJ

for references). Our proof is a variant of the proof of a similar inequality given by Lorentz ([2J, p. 19, a reference which is generally overlooked). Inequality (1) is stronger than his inequality for moderate values of p (p(l-p) > 1/8).

2. Proof of the inequality.

By Chebyshev's inequality we have

(2)

for all c

-cr r

I

_{Kn - np}

I

P (

I

K - np

I

2 c) :0; e E e

n

2

°

and r 2 0. Further we have

(3)

r IK - npI

E e n

r(K -np) -r(K -np)

:0; E e n + E e n

and, for all r E

R,

(4)

r(K - np)

E e n = {per(l-p) + (I - p) e-rp n}

At this point we need an inequality that is implied by a lemma to be proved in section 3:

2

-From (2), (3), (4) and (5) we obtain

(6)

2

P(IK - npl 2 c) $ 2e- cr+nr /8,

n

where r ;,:

a

may be chosen freely. The value r

=

4c/n minimizes the right-hand side of (6), and we get

2

I

I

?-2c /n

P( K - np 2 c) ~ ~e ,

n

which ~s equivalent to (I).

3. Proof of the lemma.

We prove the following lemma, which shows that the inequality (5) is best possible. Lemma: Let f(p,r) r E 1\ \ {a}, with -2 r(l-p) -rp = r log{pe + (1 -p)e } 1 f (p,0)

=

2P

(I - p), then

for p E [O,IJ and

1 max f(p,r) =

'8 .

r,p

1 1

Proof. Apparently, we have p(Z'O) =

s.

As f(p,r) = f(J -p, - r), we may restrict ourselves to r 2 O. By differentiation with respect to p it is seen that for each fixed r the function f has a unique maximum at p = per), with per) r e - 1 - r r r (e - I)

Substituting this value of p l.n f(p,r), we have to show that for r ;::

a

(7) g(r) : = - -r r e - 1 r + log e - 1 _ 1 - r2/8 $; 0. r

We have g(r) -1- 0 for r -1- 0. Hence it is sufficient to show that g'(r) :5: 0,

or pu t ling y

=

r / (er - I), that

v' r

3

-As y'

=

ylr - y -

y2/~

this 1S equivalent to (I - Y - r/2)2 , 0, and the lemma is proved.

Remark. It is tempting to try to make the right-hand side of (5) dependent on p. The inequality (5) becomes invalid, however, when 1/8 is replaced by f(p,O)

=

p(l-p)/2. One can obtain improvements of (5) and hence of (1) for restricted values of p and r and of p and x respectively.

References.

[JJ N.L. Johnson and S. Katz, Discrete Distributions, Houghton Mifflin. Boston, Mass. 1970.

[2J G.G. Lorentz, Bernstein Polynomials, University of Toronto Press, Toronto, 1953.