On the independence of integer and fractional parts

On the independence of integer and fractional parts

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Steutel, F. W., & Thiemann, J. G. F. (1987). On the independence of integer and fractional parts. (Memorandum COSOR; Vol. 8729). Technische Universiteit Eindhoven.

Document status and date: Published: 01/01/1987

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EINDHOVEN UNIVERSITY OF TECHNOLOGY

Department of Mathematics and Computing Science

Memorandum COSOR 87-29

On the independence of integer and fractional parts

by

F.W. Steutel and J.G.F. Thiemann

Eindhoven, Netherlands October 1987

On the independence of integer and fractional parts

by F.W. Steutel and J.G.F. Thiemann Eindhoven University of Technology,

Department of Mathematics and Computing Science, P.O. Box 513,

5600 MB Eindhoven, The Netherlands.

ABSTRACT

We use the independence of the integer and fractional parts of exponentially distributed random variables to obtain expressions for the order statistics from a geometric distribution. As our main result we show that a strong form of this independence characterizes the exponential distribution.

1.

Introduction

It is easily verified that for an exponentially distributed random variable the integer part and the fractional part are independent. We formulate this property as a lemma.

Lemma 1. Let Y have an exponential distribution, let [Yl denote the largest integer not

exced-ing Y (the integer part) and let {Y} := Y - [Y] (the fractional part). Then

[YJ and {Y} are independent .

It is also easily verified that there is the following (well-known?) relation between exponential

and geometric random variables; this relation was pointed out to me long ago by L. de Haan

(1978) in the context of problems 56 and 57 in Statistica Neerlandica, where nobody had used it (see Jagers (1978)).

2

-Lemma 2. If X has a geometric distribution, Le., if

P(X =k)=p(1-pi' (k =0,1,'" ;O<p

< 1) ,

then

d

X = [YJ ,

where Y has an exponential distribution with density

fy(y)

=

'A.e-'A.y

(y > 0; 'A.

=

-log(1 -

p» .

In section 2 of this note we return to order statistics of geometric random variables, and in

sec-tion 3 we prove that a random variable Y has an exponential distribution if and only if [cYJ

and {cY} are independent for every c

>

O.

2. Order statistic from the geometric distribution

It is well known and not hard to prove that the order statistics Y l;n S; Y 2;n S; • . . S; Yn;n from an exponential distribution satisfy (cf. Feller (1971), p. 19)

d Ie-I Y . n-}

YIe;n

=

L --.

(k

=

1,2, ... ,n) ,

j=O n-j

(1)

where Y l' . . . , Y_n are iid exponential random variables. This is most easily seen as follows:

waiting for YIe;n to finish we first wait for Y l;n' after which the remaining Y's start anew by

the lack-of-memory property, so

d Ie-I

YIe;n

=

L

Y l;n-j ,

j=O

d

with the Y l;n-j independent and (cf. (1» Y l;n-j

=

_{Yn- j} I (n - j) , Since the geometric distribution has a similar lack-of-memory property:

(2)

P (X ~ k

+

m I X ~ m)

=

P (X ~ k). it is tempting to conjecture that a relation similar to (2) would hold for geometric order statistics. It is easily seen. however, that this is not so. Instead of (2) we have the following result.

Theorem 1. Let X";lI (k

=

1, 2, ...• n) denote the order statistics from a geometric distribu-tion on {O,l.2 •... }. Then

3 -d k-l k-1 Y . _n-J Xk;n

=

I:X1;n-j

+

[I: {--. }] ,

j=O j=O n-} (3) d

where Xj and Yj are related by Xj

=

_{[Yj ]} as in Lemma 2, and all variables in the right-hand

side of (3) are independent.

d

Proof. Using Lemmas 1 and 2, and the fact that X l;m

=

[Y 11m], one easily has

d d d k-l y . k-l y . k-l Y .

Xk;n

=

[Y]k;n = [Yk;n]

=

[I:

n-~]

=

[I: [

n-~]

+

I: {

n-~}] =

j=O n-) j=O n-) j=O n-}

k-l y . k-l y . d k-1 k-l Y .

=

I: [

n-~]

+

[I: {

n-J.}]

=

I:

X l;n-j

+

[I:

{~}]

.

j=O n-) j=O n-) j=O j=O n-j

where all random variables in the last expression are independent.

Corollary. For the probability generating function of Xk;n one has

k-l 1-(1-p )n-j

PXk;tt(z)

=

IT

1-(1 )n-_j • Qn,k(z) ,

j=O -p Z

where Qn,k is a probability generating function and a polynomial of degree k - 1.

o

(4)

Remark 1. For the order statistics of exponentially distributed random variables mentioned

ear-lier we have for the Laplace transfonn

_ k-l 'J..(n-j)

C\>Yk;n (s) -

IT

'J..( '}i;

j=O n-j S

here the extra factor occurring in (4) is missing.

Remark 2. A special case of this Corollary occurs in Jagers (1979). The explicit fonn of Qn,k

seems intractable.

Remark 3. From Theorem 1 it follows that

k-l k-l

I:

X 1;n-j S Xk;n S

I:

X1;n-j

+

n - 1 •

j=O j=O

(5)

where the inequalities should be read in distribution (i.e. in the sense of stochastic ordering);

the inequality on the right in (5) holds, in fact, with probability one on the probability space

4 -3. A characterization of the exponential distribution

Let Y be a random variable with distribution function F. For [Y] and {Y} (cf. Lemma 1) to be

independent it is necessary and sufficient that (F(n

+

y) - F«n»/(F(n

+

1) - F(n» is

independent of n E N for y E (0,1]. There are many F's with this property, but if F is

required to have a continuous density

f

then it follows that

f

must satisfy

for some (X, i.e. that

f

is "globally exponential",

The object of this section is to prove the following theorem,

Theorem 2. If a random variable Y is non-constant and such that [eY] and {eY} are

indepen-dent for every e > 0, then Y has an exponential distribution on (0,00) or on (--00,0).

We first prove three lemmas. The first lemma is a direct consequence of the independence of

[eYJ and leY}.

Lemma 3. Let Y satisfy the conditions of Theorem 2 and let k E

.z,

a,b E [0,1], e > 0 be

such that

Peke S;; Y < (k + l)e) > 0 and peek +a)c < Y < (k +b)c)

=

0 .

Then, for each 1 E Z,

P((l +a)e

<

Y

<

(l +b)e)

=

O.

Proof: P([Y/e] = k) = P(ke S;; Y < (k + l)c)

>

0, so

P«l+a)c <Y

<

(l+b)c)=P([Ylc]=1)P(a

<

{Ylc} <b)=

= P([Y/e] = l)P([Ylc] = krl peek +a)c

<

Y

<

(k +b)e)

=

O.

Lemma 4. Let Y satisfy the conditions of Theorem 2. Then

i) P (Y < 0)

=

1 or P (Y ~ 0)

=

1 ,

ii) P ( I Y I < 'Y)

>

0 and P ( I Y I

> 'Y)

>

0 for each 'Y > 0 .

Proof.

i) Suppose P(Y < 0) < 1 and P(Y ~ 0) < 1. Then P(Y ~ 0) > 0 and P(Y < 0) > 0 and therefore, for a large enough. we have

P(OS Y

<

a) >

t

P(OS Y) and P(-«S Y < 0) >

t

P(Y

<

0) From this it follows that

p({L} <.11 [L]=-I)= P(-2aS Y <-a) = 2a 2 2a P (-2a S Y < 0)

=

1 _ P (-a S Y < 0) < 1 _ 1I2P (Y < 0) =.1 ;

P(-2a S Y < 0) P(Y < 0) 2

P ({ L } <.1 1 [L]

=

0)

=

P (0 S Y < a) > 112 P (0 S Y) =.1

2a 2 2a P(O S Y < 2a) P(O S Y) 2

The inequality of these conditional probabilities contradicts the independence of

[:a.]

and

{ :a. },

and so (i) holds.

ii) We consider the case P(Y ~ 0) = 1; the case P(Y < 0) = 1 can be handled in a similar way. Again, we argue by contradiction.

First suppose that P (Y < y)

=

0 for some y > O. Then

3 := sup {y > 0 I P (0 S Y < y)

=

O} < ex> and P (0 S Y < 3)

=

0 .

Also, for each a > 3. we have P (0 S Y < a) > 0 and hence. by Lemma 3,

P(a < Y < a+3)

=

O.

So, for all a> 3 we have P(a < Y < a

+

3) = 0, and therefore P (Y > 3) =

o.

As we also have P(O S Y < 3)

=

0, we conclude that P(Y

=

3)

=

1, i.e. Y is constant. This contradiction proves that P (Y < y) > 0 for every y > O.

Next suppose that P(Y > y) = 0 for some y> O. Then 3:= inf{y> 0 I P(Y >

1>

= O} > 0, since 3 = 0 would imply P(Y = 0) = I, i.e. Y is constant. Moreover P(Y > 3) =

o.

Now let

a

E

(~

3, 3). Then, by the definition of 3,

P(3 < Y < 2a)

=

0 and peaS Y < 2a) > 0 . So, by Lemma 3,

P(3-a < Y < a)

=

0 . (7)

Now suppose that P (0 S Y S 3 - a) > O. Then, since 3 - a < ; 3 < a, we have

P(O S Y < ; 3) > 0 and, by (7), P(3-a < Y < ; 3) = 0, hence, by Lemma 3,

5 4 5

- 6 ~ definition of O. So (P (0 S Y S 0 - a)

=

O.

All this implies P (0 S Y

<

a) = 0 for each a E

(~

0, 0), so P (0 S Y

<

0)

=

O.

Since also P(O < Y)

=

O. we have P(Y

=

0)

=

1, i.e. Y is constant So, again we obtain a

con-tradiction, and hence P (Y

>

y)

>

0 for every y

>

O.

0

Lemma

5.

Let Y satisfy the conditions of Theorem 2. then [Y] has a geometric distribution

either on {O, 1,2 ... } or on {-1,-2,-3, ... }.

Proof. We first consider the case P(Y ~ 0) = 1 (cf. Lemma 4).

Let

Pi := P([Y] = i) (i = 0,1,2, ... )

Then, for each i • j E N with i < j, we have

PiPj =P(i S Y < i+l)PU S Y <j+l)=

Y i Y i+l Y Y 1

= P ([ -;- ] = 0) P (-: S {-;-} < - . ) P ([ -;- ] = 1) P (0 S {-;-} < -;- ) =

} } } } } J J

=P(OS Y < l)PCi+jS Y <i+j+l)=pcj1j+j'

Now Po> 0 by Lemma 4 ii), So, taking i = 1. we get

hence

E1. .

Pj = Pj+l U = 2,3, ... ) ;

Po

Also by Lemma 4 ii) we have P (Y ~ 3) > O.

00 00 P

So 0

<

L

PI

=

L

(_1),= P3. which implies P3

*'

O.

1=3 k=O Po

The foregoing now gives, for 1

=

2,3,4, ... ,

PI )1-2 PI 1-2 PI 1

P3Pl

=

P3(- P2

= (-)

PsPo

= (-)

P3Po

Po Po Po

so PI =

(E1./

Po. As the last equality holds also for I = 0,1, the random variable [Y] is

Po

7

-For the case P (Y < 0)

=

1 we define

qj :=P([y] =-I-i) (i =0,1,2,"')

and prove the relations qjqj

=

qi+jqO (0 < i < j) by considering the random variable Y Ij.

The rest of the proof is similar to the proof given above. []

Proof of Theorem 2

We consider the case P (Y ;?! 0) = 1; the case P (Y < 0) can be treated in the same way.

Let n E IV. Then the random variable nY satisfies the conditions of Theorem 2 and therefore

[ny] is geometrically distributed on {O, 1,2 •... }, by Lemma 5. So p E (0, 1) exists such that, for each m e {O, 1, 2, . . . }, we have

pm = P([nY];?! m) = P(nY;?! m) = P(Y;?!.!!!.) , n

and, in particular, pI!

=

P (Y;?! 1); hence

m

P (Y ;?! .!!!.)

=

P (Y ;?! 1)-; .

n

So P (Y ;?! r)

=

P (Y;?! 1)' for each rational r E [0, (0), Since both members of the equality

are non-increasing functions of

r

on [0,(0), the equality holds for each

r

e [0,(0). Hence Y is

exponentially distributed on [0,(0). []

References

Feller, W. (1971), The theory of probability and its applications, Vol. 2, 2-nd ed., Wiley, New York.

de Haan, L. (1978), Personal communication.

Jagers, A.A. (1978), Solution of Problem 57 in Statistica Neerlandica, Vol. 33, Dr. 3.

Eindhoven University of Technology,

Department of Mathematics and Computing Science, P.O. Box 513,

5600 MB Eindhoven, The Netherlands.