Interval Estimation Under The Uniform Distribution U(a,b)
QIQING YU
Department of Mathematical Sciences
State University of New York
Binghamton, NY 13902
USA
Abstract: - In this short note, we consider interval estimation for the parameters under the uniform distribution
U(a, b). We study two approaches: (1) based on a Wald-type statistic, (2) based on a pivotal statistic. We show
that the first approach in its common form is not valid and we propose a modified version of the first approach. It
turns out it is equivalent to the confidence interval with the shortest length.
Key-Words: - Maximum likelihood estimator, Confidence Intervals, Pivotal Statistic, Wald-type Statistic
Received: April 20, 2021. Revised: January 7, 2022. Accepted: January 25, 2022. Published: February 24, 2022.
1 Introduction.
The uniform distribution U(a, b)is a common dis-
tribution and has been studied extensively (see, for
example, Kuipers and Niederreiter (2012), Stephens
(2017) and Claessen. et al. (2015), among others.
The maximum likelihood estimators (MLEs) of its pa-
rameters have explicit expressions. How to construct
a confidence interval (CI) under U(a, b)is a typical
content in a basic statistics course. For example, in
the textbook by Casella and Bergera (2002), it is ex-
plained that if the data are from U(0, b)then the exact
CI for bcan be constructed using a pivotal statistic. If
the random sample is from U(a, b)when both aand b
are parameters, then a,band θare parameters, where
θ=b−a. Under this assumption, we shall show that
there does not exist an exact CI. We shall discuss how
to construct approximate CIs.
2 Theory.
Let W1, ..., Wnbe i.i.d. from W∼U(a, b), with
the cumulative distribution function (cdf) FW(·). Let
(ˆa, ˆ
b, ˆ
θ)be the MLE of (a, b, θ), where ˆa=W(1) =
miniWi,ˆ
b=W(n)=maxiWiand ˆ
θ=ˆ
b−ˆa. Recall
that P(ˆ
b≤t) = P(W(n)≤t) = P(Wi≤t, ∀i) =
(FW(t))nand P(W(1) > t) = P(Wi> t, ∀i) =
(SW(t))n, where SW= 1 −FW. Thus the distribu-
tion of the MLE of (a, b, θ)is well understood. It is
easy to verify that (ˆa, ˆ
b, ˆ
θ)is consistent.
There are two possible approaches in construct-
ing CI’s for γ∈ {a, b, θ}: (1) base on the Wald-type
statistic ˆγ−γ
ˆσˆγ,e.g.,[ˆγ−1.96ˆσˆγ, , ˆγ+1.96ˆσˆγ], (2) based
a pivotal statistic T=W−a
b−a, where T∼U(0,1).
The first approach relies on the mean and variance.
Recall the cdfs and density functions:
FT(n)(t) = tn, fT(n)(t) = ntn−1, FT(1) = 1 −ST(1) ,
ST(1) (t) = (1 −t)n, fT(1) (t) = n(1 −t)n−1,
for t∈[0,1]; and for t, s ∈(0,1),
fT(1),T(n)(t, s) = n!fT(t)(FT(s)−FT(t))n−2fT(s)
1!(n−2)!1! .
Based on W=θT +a, it is easy to derive
σ2
ˆ
b=θ2σ2
T(n)=σ2
ˆa=θ2n
(n+ 1)2(n+ 2)
and σ2
ˆ
θ=θ2σ2
T(n)−T(1) =2(n−1)θ2
(n+ 2)(n+ 1)2.(1)
The proofs are also given in Appendix.
3 The Main Results.
We shall consider constructing the CI for a,bor θun-
der the assumption that W∼U(a, b). For simplicity,
we only discuss the case of a 95% CI (or (1−α)100%
CI’s, with α= 0.05). For general (1 −α)100% CI’s,
just replace 0.05 by α.
3.1. CIs for b:First consider the pivotal method.
Since T=W−a
b−a∼U(0,1),Tis a pivotal statis-
tic. For t∈[0,0.05],FT(n)(t) = tn, letting (u, v) =
(t1/n,(0.95 + t)1/n)yields
0.95 = P(u≤T(n)≤v) = P(u≤W(n)−a
b−a≤v)
=P(1
v≤b−a
W(n)−a≤1
u)
=P(W(n)−a
v+a≤b≤W(n)−a
u+a).
If ais given, a 95% CI for bis
[W(n)−a
v+a, W(n)−a
u+a],(2)
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DOI: 10.37394/23206.2022.21.10
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where (u, v) = (t1/n,(0.95 + t)1/n). There
are 3 typical cases: (u, v) = (0,0.951/n), or
(0.0251/n,0.9751/n), or (0.051/n,1), with lengthes:
W(n)(1
0−0.95
−1
n,0.025
−1
n−0.975
−1
n,0.05
−1
n−1)
≈W(n)(∞,0.037,0.030) if n= 100.
Thus the best choice among these three 95% CIs for
bis (W(n),W(n)−a
0.051/n+a)if ais known. Actually, it
is the shortest 95% CI, which is given by (u, v) =
(0.051/n,1), as the length of the CI in Eq. (2) is
(W(n)−a)[(0.95 + t)−1/n−t−1/n]and
((0.95 + t)−1/n−t−1/n)′
t
=−1
n(0.95+t)
−1
n−1−−1
nt
−1
n−1<0for t∈[0,0.05].
If ais unknown, estimating aby W(1) yields an ap-
proximate 95% CI
[W(n),W(n)−W(1)
0.051/n+W(1)],(3)
as P(a < W(1) ≤a+δ) = P(T(1) ≤δ
θ)
= 1 −(1 −δ
θ)n→1∀δ∈(0, θ/2), and thus
P(W(n)≤b≤W(n)−W(1)
0.051/n+W(1))≈0.95 if nis
large. The length of the CI is (ˆ
b−ˆa)(201/n−1).
Wald-type statistic may lead to another possible
95% CI ˆ
b±1.96ˆσˆ
b, with its length ≈4ˆσˆ
b. By Eq.
(1) and Eq. (3), if nis large then the ratio of these
two lengths is
(ˆ
b−a)(0.05−1/n−1)
4ˆσˆ
b
=(ˆ
b−a)(0.05−1/n−1)
4ˆ
θ√n
n+2 /(n+1)
≈0.05−1/n−1
4/n<0.8,
thus ˆ
b±1.96ˆσˆ
bis not as good as the CI in Eq. (3).
Moreover, this approach is based on the belief that
P(W(n)−b
ˆσW(n)≤t)≈Φ(t)∀t, where Φis the cdf of
N(0,1). However, if t > 0then
0.95 ≥P(W(n)−b
σW(n)≤t)
=P(W(n)≤tσW(n)+b)
= (P(T≤tσW(n)+b−a
b−a))n
= (P(T≤tσW(n)
b−a+ 1))n
= (1)n, as T∼U(0,1).
It leads to a contradiction: 0.95 ≥1. Thus ˆ
b±2ˆσˆ
b
is not a CI. But we can make use of the Wald-type
statistic as follows. Choose t < 0such that
0.05 = P(W(n)−b
σW(n)≤t)
=P(W(n)≤tσW(n)+b)
= (P(W≤tσW(n)+b))n
= (P(T≤tσW(n)+b−a
b−a))n
= (t
θ√n
n+2
n+1
θ+1)n(≈(t
n+1)n≈et).
0.95 = P(W(n)−b
σW(n)
> t) (t≈ln0.05 = −ln20)
=P(b < W(n)−σW(n)t)
=P(W(n)< b < W(n)−σW(n)t)
≈P(W(n)< b < W(n)+σW(n)ln20).
Thus an approximate 95% CI for bis
(W(n), W(n)+ ˆσW(n)ln20), with
length ≈ˆ
b−ˆa
nln20 = (ˆ
b−ˆa)ln201/n,
as σ2
ˆ
b=(b−a)2n
(n+1)2(n+2) by Eq. (1). It is of interest to
compare its length to the length of the CI in (3):
(ˆ
b−ˆa)(0.05−1/n−1)
= (ˆ
b−ˆa)(201/n−1)
≈(ˆ
b−ˆa)ln201/n, as
ln201/n
=lnx
=lnx−ln1
≈(lnx)′|x=1(x−1)
=x−1(with x= 201/n).
Thus these two approximate CI’s have the same
length asymptotically.
Remark. In general, an approximate (1 −α)100%
CI for bis
[W(n), W(n)−ˆσW(n)lnα]Wald-type method
[W(n),ˆ
θ
α1/n+W(1)]pivotal method.
3.2. CI for a:W(1) −a
θ=T(1) is a pivatol statistic and
P(T(1) > t) = (1−t)nif t∈[0,1]. Let (u, v)satisfy
0.95 = P(u≤W(1)−a
θ≤v) (= (1−u)n−(1−v)n)
0.95 = P(W(1) −uθ ≥a≥W(1) −vθ),then it leads
to an approximate 95% CI for a,e.g, let
((1 −u)n,(1 −v)n)
= (0.95,0),(0.975,0.025),(1,0.05), then (u, v) =
(1 −0.95 1
n,1) or (1 −0.975 1
n,1−0.025 1
n), or
(0,1−0.05 1
n).
Then their length
=θ(0.95 1
n,0.975 1
n−0.025 1
n,1−0.05 1
n). It can be
shown that the shortest 95% CI for ais [ˆa−vθ, ˆa]
if θis given, where v= 1 −0.051/n; otherwise, an
approximate 95% CI is [W(1) −vˆ
θ, W(1)].
Moreover, a 95% CI based on Wald-type statistic
ˆa−a
ˆσˆais [ˆa−tˆσˆa,ˆa], where t≈n(1 −0.051/n)and
ˆσˆ
θ≈ˆ
θ/n. The reason is as follows.
0.95 = P(W(1)−a
σW(1) ≤t)
=P(W(1) −tσW(1) ≤a)
=P(W(1) −tσW(1) ≤a≤W(1))
=P(W(1) ≤tσW(1) +a)
= 1 −P(W(1) > tσW(1) +a)
= 1 −P(T(1) >tσW(1) +a−a
θ)
= 1 −P(T(1) >tσW(1)
θ)
= 1 −(1 −tσW(1)
θ)n
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DOI: 10.37394/23206.2022.21.10
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= 1 −(1 −tθ√n
n+2
θ(n+1) )n.=>
0.051/n= 1 −t√n
n+2
(n+1) =>
t= (1 −0.051/n)√n
n+2
(n+1) ≈n(1 −0.051/n).
Moreover, σW(1) =θ√n
n+2
(n+1) ≈θ/n.
Remark. Both approaches lead to approximately the
same CI, as expected.
3.3. CI for θ:Let 0.95 = 1 −P(T(n)≤v) = 1 −vn,
i.e. v= 0.051/n. Then
P(T(n)=W(n)−a
θ> v) = P(W(n)−a
v> θ)
=P(W(n)−a
v> θ > ˆ
b−ˆa).
Hence, an approximate 95% CI for θis (ˆ
θ, ˆ
θ
0.051/n],
where ˆ
θ=W(n)−W(1), Or in general, (ˆ
θ, ˆ
θ
α1/n].
On the other hand, in order to study Wald-type
approach, we need to find the distribution of ˆ
θ
(=W(n)−W(1)). Let fbe the density of (T(n), T(1)).
G(t)def
=P(T(n)−T(1) ≤t)
= 1(0 ≤x−y≤t)f(x, y)dxdy
=t
0x
0f(x, y)dydx +1
tx
x−tf(x, y)dydx (t∈
[0,1]).
G′(t) = t
0f(t, y)dy−t
0f(t, y)dy+1
tf(x, x−t)dx
=1
tn(n−1)tn−2dx =n(n−1)tn−2(1 −t) =>
G(t) = t
0n(n−1)xn−2(1 −x)dx
=n(n−1)[xn−1
n−1−xn
n]|t
0=>
G(t) = ntn−1−(n−1)tn, t ∈[0,1].(4)
Let vbe determined by 0.05 = P(ˆ
θ−θ
σˆ
θ≤v). Then
0.05 = P(ˆ
θ−θ
σˆ
θ≤v) = P(ˆ
θ−vσˆ
θ≤θ)
=P(W(n)−W(1) ≤vσˆ
θ+θ)
=P(T(n)−T(1) ≤vσˆ
θ+θ
θ)
=P(T(n)−T(1) ≤vθ2(n−1)
n+2
n+1 +θ
θ)
=P(T(n)−T(1) ≤
v2(n−1)
n+2
n+1 + 1)
=G(v2(n−1)
n+2
n+1 ) (= 0.05);
0.95 = P(ˆ
θ−vσˆ
θ> θ)
=P(ˆ
θ−vσˆ
θ> θ ≥ˆ
θ)
(= 1 −G(v2(n−1)
n+2
n+1 )).
Thus [ˆ
θ, ˆ
θ−vˆσˆ
θ]is an approximate 95% CI for θ,
where vis specified by G(v2(n−1)
n+2
n+1 )=0.05 and
G(t) = ntn−1−(n−1)tnby Eq. (4).
4 Summary.
The confidence intervals for the parameters under
U(a, b)are not of the typical form of [ˆ
ψ−u, ˆ
ψ+u],
but are of the form either ˆ
ψ, ˆ
ψ+u], or ˆ
ψ−u, ˆ
ψ], where
ψ∈ {a, b, θ}.
Appendix
E(T(1)) = 1
n+1 ,
V(T(1)) = n
(n+1)2(n+2) ,
E(T(n)) = n
n+1 .
V(T(n)) = n
(n+1)2(n+2) ,
E(W(1)) = b
n+1 +an
n+1 .
E(W(n)) = bn
n+1 +a
n+1 .
V(W(1)) = V(W(n)) = θ2n
(n+1)2(n+2) ,
Let Z=W(n)−W(1),
then E(Z) = (b−a)(n−1)/(n+ 1).
σ2
Z=2n
(n+1)2(n+2) θ2
−2[E(W(n)W(1))−E(W(n))E(W(1))].
fW(i),W(j)(x, y) =
n!(FW(x))i−1fW(x)(FW(y)−FW(x))j−i−1fW(y)(SW(y))n−j
(i−1)!1!(j−i−1)!1!(n−j)! ,
x < y,i < j,
fW(1),W(n)(x, y) = n!fW(x)(FW(y)−FW(x))n−2fW(y)
(n−2)!
=n(n−1)(y−x)n−2
θ2,a < x < y < b.
σ2
ˆ
θ=σ2
W(n)−W(1)
=2nθ2
(n+1)2(n+2) −2θ2
(n+2)(n+1)2=2(n−1)θ2
(n+2)(n+1)2.
References:
[1] Casella, G. and Bergera, R. L. (2002). Statistical
Inference (2nd ed.) Duxbury. U.S.
[2] Kuipers, L., & Niederreiter, H. (2012). Uniform
distribution of sequences. Courier Corporation.
[3] Stephens, M. A. (2017). Tests for the uniform dis-
tribution. In Goodness-of-fit techniques (pp. 331-
366). Routledge.
[4] Claessen, K., Duregard, J., & Palka, M. H.
(2015). Generating constrained random data with
uniform distribution. Journal of functional pro-
gramming, 25.
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DOI: 10.37394/23206.2022.21.10
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E-ISSN: 2224-2880
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