A note on the sum of uniform random variables
Aniello Buonocore, Enrica Pirozzi, Luigia Caputo
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Aniello Buonocore, Enrica Pirozzi, Luigia Caputo. A note on the sum of uniform random
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Accepted Manuscript
A note on the sum of uniform random variables
Aniello Buonocore, Enrica Pirozzi, Luigia Caputo
PII:
DOI:
Reference:
S0167-7152(09)00244-2
10.1016/j.spl.2009.06.020
STAPRO 5460
To appear in:
Statistics and Probability Letters
Received date: 29 June 2009
Accepted date: 30 June 2009
Please cite this article as: Buonocore, A., Pirozzi, E., Caputo, L., A note on the sum of uniform
random variables. Statistics and Probability Letters (2009), doi:10.1016/j.spl.2009.06.020
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IPT
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A Note on the Sum of Uniform Random Variables
2
Aniello Buonocore∗,a , Enrica Pirozzia , Luigia Caputob
a Dipartimento
di Matematica e Applicazioni, Università di Napoli Federico II
Via Cintia, 80126 Napoli, Italy
b Dipartimento di Matematica, Università di Torino
Via Carlo Alberto 10, 10123 Torino, Italy
3
4
5
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7
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1
Abstract
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1. Introduction
The problem of calculating the distribution of the sum Sn of n uniform random variables has been the
object of considerable attention even in recent times. The motivation can be ascribed to various reasons
such as the necessity of handling data drawn from measurements characterized by different level of precision (Bradley and Gupta, 2002), or questions appearing in change point analysis (Sadooghi-Alvandi et al.,
2009), or, more in general, the need of aggregating scaled values with differing numbers of significant figures
(Potuschak and Müller, 2009). It appears that this problem has been taken up first in Olds (1952), where
by somewhat obscure procedures formulas for the probability density function of Sn and its distribution
function are derived. An accurate bibliography of articles published in the last century is found in Bradley
and Gupta (2002), where the authors also obtain the probability density function of Sn by non probabilistic
arguments, namely via a complicated analytical inversion of the characteristic function. Such a procedure
was successively and successfully simplified in Potuschak and Müller (2009), where again no trace of probabilistic arguments is present. An attempt to achieve the same results by a simpler procedure appears
in Sadooghi-Alvandi et al. (2009) where a given function is assumed to be the unknown probability density
function, the proof of the correctness of such an ansatz being that its Laplace transform coincides with the
moment generating function of Sn . Quite differently, the present note includes a novel proof of the above
cited results (Proposition 2.1). This is based on an inductive procedure, suitably adapted to our general
instance, used by Feller (1966) for the case of identically distributed variables, that further pinpoints the
usefulness of induction procedures in the probability context. (See also Hardy et al. (1978) for some more
illuminating examples.) In the case of identically distributed random variables, some results concerning
certain probabilities and means of random variables related to Sn are obtained (Lemma 3.1, Theorem 3.1,
corollaries 3.1 and 3.2, Proposition 3.4), as well as certain recurrence relations that are reminiscent of those
holding for Stirling numbers (Propositions 3.5, 3.6, 3.7).
TE
10
Key words: induction, recursive formulas
2000 MSC: 60G50, 60E99
EP
9
AC
C
8
DM
AN
An inductive procedure is used to obtain distributions and probability densities for the sum Sn of independent, non equally uniform random variables. Some known results are then shown to follow immediately as
special cases. Under the assumption of equally uniform random variables some new formulas are obtained
for probabilities and means related to Sn . Finally, some new recursive formulas involving distributions are
derived.
∗ Correspondig
author; Tel. +39-081-675684; Fax +39-081-675665
Email addresses: [email protected] (Aniello Buonocore), [email protected] (Enrica Pirozzi),
[email protected] (Luigia Caputo)
Preprint submitted to Statistics and Probability Letters
June 29, 2009
33
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ACCEPTED MANUSCRIPT
2. The general case
FXn (x) =
CR
PnLet {Xn }n∈N denote a sequence of uniform distributed independent random variables and denote Sn =
i=1 Xi . Without loss of generality we assume that Xn ∼ U (0, an ) with an positive real numbers. By
adopting a suitably modified procedure due to Feller (1966) we shall obtain the probability density function
fn (x) and the distribution function Fn (x) of Sn for all n ∈ N. The starting point is to write
x+ − (x − an )+
,
an
(1)
∀n ∈ N, ∀x ∈ R,
US
where (x − c)+ = max{x − c, 0}, ∀c ∈ R. Next we shall make use of
Z x
n
n−1
1
(x − c)+ , ∀n ∈ N, ∀c ∈ R+ .
(y − c)+
dy =
n
−∞
(2)
DM
AN
In addition we note that, by convolution, probability density functions and distribution functions are related
as follows:
Z an+1
Fn (x) − Fn (x − an+1 )
fn (x − y)fXn+1 (y) dy =
fn+1 (x) =
, ∀n ∈ N, ∀x ∈ R.
(3)
an+1
0
Claim 2.1. One has
F1 (x) =
and
f2 (x) =
34
TE
2
(x+ )2 − [(x − a1 )+ ]2 − [(x − a2 )+ ]2 + {[x − (a1 + a2 )]+ }
,
2a1 a2
n
f3 (x) = (x+ )2
−
+
AC
C
−
(5)
2
∀x ∈ R
EP
and
37
∀x ∈ R.
Proof. It follows from (1) written for S1 ≡ X1 , and from (3).
F2 (x) =
36
(4)
∀x ∈ R
x+ − (x − a1 )+ − (x − a2 )+ + [x − (a1 + a2 )]+
,
a1 a2
Claim 2.2. One has
35
x+ − (x − a1 )+
,
a1
2
2 2 (x − a1 )+ − (x − a2 )+ − (x − a3 )+ +
o2 n
o2 n
o2
n
[x − (a1 + a2 )]+ + [x − (a1 + a3 )]+ + [x − (a2 + a3 )]+
o2 o
n
+
(2a1 a2 a3 )−1 , ∀x ∈ R.
[x − (a1 + a2 + a3 )]
Proof. Eq. (6) follows from (5) and (2). From (6) and (3) one then obtains Eq. (7).
(6)
(7)
2
Claims 2.1 and 2.2 lead us to infer a possible general forms of the distribution function of Sn and of the
probability density function of Sn+1 , as specified in the following Proposition.
Proposition 2.1. The distribution function Fn (x) of Sn and the probability density function fn+1 (x) of
Sn+1 are given by, respectively:
)
(
n
n
n
n
on
n
X
X
X
X
1
+
+ n
ν
,
[x − (aj1 + aj2 + · · · + ajν )]
···
Fn (x) =
(x ) +
(−1)
n! An
ν=1
j =j
+1
j =1 j =j +1
1
2
1
ν
2
ν−1
∀n ∈ N, ∀x ∈ R
(8)
)
(
n+1
n+1
n+1
on
X n
X
X n+1
X
1
+
+ n
ν
,
fn+1 (x) =
[x − (aj1 + aj2 + · · · + ajν )]
···
(x ) +
(−1)
n! An+1
ν=1
j =j
+1
j =1 j =j +1
1
2
1
ν
ν−1
CR
and
IPT
ACCEPTED MANUSCRIPT
∀n ∈ N, ∀x ∈ R.
(9)
ν=1
40
41
Eq. (10) identifies with Eq. (9) written for n = r since the curly brackets contains all and only all the
following terms:
1. [(x)+ ]r ;
r
r
r
2. [(x − a1 )+ ] , [(x − a2 )+ ] , . . . , [(x − ar+1 )+ ] ;
3. for 1 < ν ≤ r
(−1)ν
r
X
r
X
j1 =1 j2 =j1 +1
−(−1)ν−1
r
X
···
r
X
j1 =1 j2 =j1 +1
r+1
X
r
X
r+1
X
j1 =1 j2 =j1 +1
jν =jν−1 +1
···
···
EP
≡ (−1)ν
r+1
X
jν =jν−1 +1
43
44
3. A special case
+
[x − (aj1 + aj2 + · · · + ajν )]
n
or
+
+ or
x − (aj1 + aj2 + · · · + ajν−1 + ar+1 )
or
n
+
;
[x − (aj1 + aj2 + · · · + ajν )]
This complete the induction.
2
AC
C
45
r
X
n
jν−1 =jν−2 +1
or
n
+
.
4. (−1)r+1 [x − (a1 + a2 + · · · + ar+1 ]
42
jν =jν−1 +1
j1 =1 j2 =j1 +1
TE
38
39
DM
AN
US
Proof. We proceed by induction. Claims 2.1 and 2.2 show that Eqs. (8) and (9) hold for n = 1 and n = 2.
Let us now assume that they hold for n = r − 1 and prove that they also hold for n = r. To this purpose,
we re-write Eq. (9) for n = r − 1 and x = y and then integrate both sides over (−∞, x). By virtue of (2),
Eq. (8) with n = r then follows. To obtain Eq. (9) for n = r we make use of (3) and of the just obtained
expression of Fr (x). Hence,
(
r
r
r
r
or
n
X
X
X
X
1
[x − (aj1 + aj2 + · · · + ajν )]+ +
fr+1 (x) =
···
(x+ )r +
(−1)ν
r! Ar+1
ν=1
jν =jν−1 +1
j1 =1 j2 =j1 +1
+ r
+
(10)
− (x − ar+1 )
)
r
r
r
r
o
n
X
X
X X
r
+
.
[x − (aj1 + aj2 + · · · + ajν + ar+1 )]
···
−
(−1)ν
Let us assume that the random variables in {Xn }n∈N are identically distributed.
Proposition 3.1. When an = a > 0 for all n ∈ N then
n
n
1 X
ν n
Fn (x) =
(−1)
(x − νa)+ ,
n! an ν=0
ν
and
∀n ∈ N, ∀x ∈ R
n+1
X
n
1
ν n+1
(−1)
(x − νa)+ ,
fn+1 (x) =
n! an+1 ν=0
ν
3
∀n ∈ N, ∀x ∈ R.
(11)
(12)
Proof. Eq. (11) follows from (8) after noting that now An = an and that
aj1 + aj2 + · · · + ajν = νa
IPT
ACCEPTED MANUSCRIPT
n
n
X
j1 =1 j2 =j1 +1
46
···
n
X
jν =jν−1
on n n
n
+
=
[x − (aj1 + aj2 + · · · + ajν )]
· (x − νa)+ .
ν
+1
Eq. (12) follows from (9) by a similar argument.1
US
n
X
CR
for ν = 0, 1, . . . , n. Indeed, in the sum on ν in (8), the term in curly bracket becomes [(x − νa)+ ] , so that
2
Hereafter, for simplicity we shall take an = a = 1 for all n ∈ N. Then, from Eqs. (3) there follows
so that
fn+1 (k) = Fn (k) − Fn (k − 1),
∀n ∈ N, k ∈ {0, 1, . . . , n + 1},
whereas from Eqs. (11) and (12) one obtains
k
n
1 X
n (−1)ν
(k − ν)+ ,
n! ν=0
ν
Fn (k) =
and
∀n ∈ N, ∀x ∈ R
DM
AN
fn+1 (x) = Fn (x) − Fn (x − 1),
∀n ∈ N, k ∈ {0, 1, . . . , n}
k
X
n−1
1
ν n
fn (k) =
(−1)
,
(k − ν)+
(n − 1)! ν=0
ν
∀n ∈ N, k ∈ {0, 1, . . . , n}.
(13)
(14)
(15)
(16)
Fn (x) =
k
X
j=1
TE
Proposition 3.2. When an = a = 1 for all n ∈ N then
fn+1 (x + j − k),
∀n ∈ N, k ∈ {1, 2, . . . , n}, k − 1 ≤ x ≤ k.
(17)
EP
Proof. Starting from (13), by iteration it follows that
Fn (x)
= fn+1 (x) + Fn (x − 1) = fn+1 (x) + fn+1 (x − 1) + Fn (x − 2)
47
AC
C
= ··· =
k
X
j=1
fn+1 (x + j − k) + Fn (x − k).
Since x − k ≤ 0, one has Fn (x − k) = 0, which completes the proof.
2
Proposition 3.3. When an = a = 1 for all n ∈ N then
Z
k
Fn (x) dx = Fn+1 (k),
∀n ∈ N, k ∈ {1, 2, . . . , n}.
k−1
(18)
1 Note that Eqs. (11) and (12) obtained by us as a special case of (8) and (9) are in agreement with a result due to Feller
(1966).
4
k−1
k
X
j=1
48
k−1
j=1
=
k
fn+1 (x + j − k) dx =
k
X
j=1
k
Fn+1 (x + j − k)
k−1
CR
Proof. Making use of (17) one obtains
Z k
k Z
X
Fn (x) dx =
IPT
ACCEPTED MANUSCRIPT
[Fn+1 (j) − Fn+1 (j − 1)] = Fn+1 (k) − Fn+1 (0).
The proof is then a consequence of Fn (0) = 0 for all n ∈ N.
Pn,k = Fn (k) − Fn (k − 1) = fn+1 (k),
Lemma 3.1. When an = a = 1 for all n ∈ N then
US
Consider now the event Sn,k = {k − 1 ≤ Sn ≤ k} and let Pn,k := P (Sn,k ). From (14) it follows that
∀n ∈ N, k ∈ {1, 2, . . . , n}.
DM
AN
P (Sn+1 ≤ k, Sn,k ) = Fn+1 (k) − Fn (k − 1),
∀n ∈ N, k ∈ {1, 2, . . . , n}.
Proof. Let n ∈ N and 1 ≤ k ≤ n. Then,
P (Sn+1 ≤ k, Sn,k ) = P (Xn+1 ≤ k − Sn , Sn,k ) =
ZZ
2
(19)
(20)
fXn+1 (x)fn (y) dx dy
T
where T denotes the domain in the x-y plane defined by 0 < x < 1 and k − 1 < y < k − x. Hence, by
integration along the y-axis from k − 1 to k − x, for all x ∈ (0, 1) we obtain
Z 1
Z k−x
Z 1
Fn (k − x) dx − Fn (k − 1)
fn (y) dy =
dx
P (Sn+1 ≤ k, Sn,k ) =
=
0
k−1
0
Z
k
k−1
Fn (x) dx − Fn (k − 1).
Eq. (20) follows from (21) and (18).
50
Lemma 3.1 will be used to prove the following theorem.
TE
49
(21)
2
Theorem 3.1. When an = a = 1 for all n ∈ N then
EP
P (Sn+1 ≤ k|Sn,k ) =
k
,
n+1
∀n ∈ N, k ∈ {1, 2, . . . , n}.
(22)
Proof. Let k = 1. ∀n ∈ N, from (15) there follows Fn (1) = 1/n!, whereas Fn (0) = 0. Hence, making use
of (19) one obtains
AC
C
P (Sn+1 ≤ 1|Sn,1 ) =
P (Sn+1 ≤ 1, Sn,1 )
Fn+1 (1) − Fn (0)
n!
1
=
=
=
.
Pn,1
Fn (1) − Fn (0)
(n + 1)!
n+1
(23)
This proves (22) for k = 1. From (23) it follows that
1
≡ P (Sn+1 ≤ 1|Sn,1 ) = E [P (Xn+1 ≤ 1 − y|Sn,1 , Sn = y)] = 1 − E [Sn |Sn,1 ]
n+1
which ultimately implies
51
52
53
54
55
n
.
n+1
Since X1 , X2 , . . . , Xn are uniform iid random variables, the mean of each of them conditional on Sn,1 is
1/(n + 1). Hence, given that Sn,1 occurs, the means of S1 , S2 , . . . , Sn partition [0, 1] into n + 1 equally wide
intervals. Therefore, for 1 < k ≤ n, if Sn,k occurs, the interval that is partitioned into n + 1 equally wide
intervals is now [0, k]. This implies that Xn+1 cannot exceed k/(n + 1) to insure that Sn+1 remains below
k.
2
5
E [Sn |Sn,1 ] =
IPT
ACCEPTED MANUSCRIPT
Corollary 3.1. When an = a = 1 for all n ∈ N then
n
k,
n+1
∀n ∈ N, k ∈ {1, 2, . . . , n}.
CR
E [Sn |Sn,k ] =
Proof. Due to Theorem 3.1 one has
(24)
k
≡ P (Sn+1 ≤ k|Sn,k ) = E [P (Xn+1 ≤ k − y|Sn,k , Sn = y)] = k − E [Sn |Sn,k ]
n+1
which ultimately yields Eq. (24).
Corollary 3.2. When an = a = 1 for all n ∈ N then
Proof. Since
57
n
kPn,k ,
n+1
∀n ∈ N, k ∈ {1, 2, . . . , n}.
(25)
DM
AN
E Sn 1Sn,k =
2
US
56
E Sn 1Sn,k
,
E [Sn |Sn,k ] =
Pn,k
Eq. (25) follows from (24).
2
Proposition 3.4. When an = a = 1 for all n ∈ N then
n
n
X
n
X
kfn+1 (k) = (n + 1)
k=1
Fn+1 (k),
k=1
∀n ∈ N, k ∈ {1, 2, . . . , n}.
(26)
TE
Proof. Let n ∈ N. From (19) and (25) we obtain
n
n
k=1
k=1
X n X
n
= E [Sn ] =
E Sn 1Sn,k =
kfn+1 (k).
2
n+1
(27)
Hence,
0
n
n Z
xfn (x) dx = xFn (x) −
AC
C
E [Sn ] ≡
Z
EP
Making use of (18), we are easily led to
0
n
0
Fn (x) dx = n −
n Z
X
k=1
k
k−1
Fn (x) dx = n −
n
X
Fn+1 (k).
k=1
n
n X
=
Fn+1 (k).
2
(28)
k=1
58
Eq. (26) finally follows by equating the right hand sides of (27) and (28).
59
The forthcoming recursive formulas are a consequence of Theorem 3.1.
2
Proposition 3.5. When an = a = 1 for all n ∈ N then
Fn+1 (k) = Fn (k)
k
n+1−k
+ Fn (k − 1)
,
n+1
n+1
6
∀n ∈ N, k ∈ {1, 2, . . . , n + 1}.
(29)
IPT
ACCEPTED MANUSCRIPT
Proof. Let n ∈ N and 1 ≤ k ≤ n + 1. From (19) and (20) one obtains
or
CR
P (Sn+1 ≥ k, Sn,k ) = Pn,k − P (Sn+1 ≤ k, Sn,k ) = Fn (k) − Fn (k − 1) − Fn+1 (k) + Fn (k − 1),
P (Sn+1 ≥ k, Sn,k ) = Fn (k) − Fn+1 (k).
On the other hand,
(30)
Hence, from (19) and (22) one derives
P (Sn+1 ≥ k, Sn,k ) = Fn (k) − Fn (k − 1) −
US
P (Sn+1 ≥ k, Sn,k ) = P (Sn+1 ≥ k|Sn,k ) Pn,k = [1 − P (Sn+1 ≤ k|Sn,k )] Pn,k .
k
k
Fn (k) +
Fn (k − 1).
n+1
n+1
Eq. (29) then immediately follows after equating the right hand sides of (30) and (31).
61
Note that (29) trivially holds also for k = 0, yielding 0 = 0.
Remark 3.1. Since
=
E Sn 1Sn,k
=
62
DM
AN
60
Z
k
k−1
k
xfn (x) dx = xFn (x)
k−1
−
Z
k
(31)
2
Fn (x) dx
k−1
kFn (k) − (k − 1)Fn (k − 1) − Fn+1 (k).
Eq. (25) can be alternatively obtained via (29).
Proposition 3.6. When an = a = 1 for all n ∈ N then
k
n+2−k
+ Pn,k−1
,
n+1
n+1
TE
Pn+1,k = Pn,k
∀n ∈ N, k ∈ {1, 2, . . . , n + 1}.
(32)
Proof. Let n ∈ N and 1 ≤ k ≤ n + 1. By difference of Eqs. (29) written for k and for k − 1, one obtains
63
n+1−k
1
1
k
+ Fn (k − 1)
+ Pn,k−1
− Fn (k − 2)
,
n+1
n+1
n+1
n+1
EP
Pn+1,k = Pn,k
whence (32) follows after noting that Fn (k − 1) − Fn (k − 2) = Pn,k−1 .
2
AC
C
Proposition 3.7. When an = a = 1 for all n ∈ N then
fn+1 (k) = fn (k)
k
n+1−k
+ fn (k − 1)
,
n
n
∀n ∈ N, k ∈ {1, 2, . . . , n + 1}.
(33)
Proof. Let n ∈ N and 1 ≤ k ≤ n + 1. From (19) and (32) it follows that
fn+1 (k) = Pn,k = Pn−1,k
n+1−k
k
+ Pn−1,k−1
.
n
n
64
By making again use of (19), Eq. (33) is finally obtained.
65
Acknowledgements
66
We wish to thank Professors R. Johnson and L.M. Ricciardi for helpful comments.
7
2
74
75
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73
US
71
72
DM
AN
70
Bradley, D.M., Gupta, R.C., 2002. On the distribution of the sum of n non-identically distributed uniform random variables.
Ann. Inst. Statist. Math. 54 (3), 689–700.
Feller, W., 1966. An introduction to probability theory and its applications. Vol. II. Wiley, New York.
Hardy, G.H., Littlewood, J.E., Pólya, G., 1978. Inequalities. Cambridge University Press, London.
Olds, E.G., 1952. A note on the convolution of uniform distributions. Ann. Math. Stat. 23, 282–285.
Potuschak, H., Müller, W.G., 2009. More on the distribution of the sum of uniform random variables. Stat. Papers 50, 177–183.
Sadooghi-Alvandi, S.M., Nematollahi, A.R., Habibi, R., 2009. On the distribution of the sum of independent uniform random
variables. Stat. Papers 50, 171–175.
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68
References
AC
C
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