We introduce and characterize a new family of distributions, Marshall-Olkin discrete uniform distribution. The natures of hazard rate, entropy, and distribution of minimum of sequence of i.i.d. random variables are derived. First order autoregressive (AR (1)) model with this distribution for marginals is considered. The maximum likelihood estimates for the parameters are found out. Also, the goodness of the distribution is tested with real data.
1. Introduction
Marshall and Olkin [1] introduced a new method for adding a parameter to a family of distributions with application to the exponential and Weibull families. Jose and Alice [2] discussed Marshall-Olkin family of distributions and their applications in time series modeling and reliability. Jose and Krishna [3] have developed Marshall-Olkin extended uniform distribution. These works and most of the references there in, deal with continuous distribution. Not much work is seen in the discrete case. The reason behind this may be that it is difficult to obtain compact mathematical expressions for moments, reliability, and estimation in the discrete set up.
If F¯(x) is the survival function of a distribution, then, by Marshall-Olkin method, we get another survival function G¯(x), by adding a new parameter θ to it. That is,
(1)G¯(x,θ)=θF¯(x)(1-(1-θ)F¯(x)),-∞<x<∞,θ>0.
Then the corresponding distribution function is G(x,θ)=1-G¯(x,θ)=F(x)/(1-(1-θ)F¯(x)).
Now its probability mass function (p.m.f.) is
(2)g(x,θ)=G(x,θ)-G(x-1,θ)=θf(x)[1-(1-θ)F¯(x)][1-(1-θ)F¯(x-1)],
where f(x) is the p.m.f. corresponding to F(x).
The hazard rate of X is
(3)γG(x)=g(x)G¯(x).
We consider a new family of distributions by adding an additional parameter by using the Marshall-Olkin scheme (Marshall and Olkin [1] in Section 2). The nature of hazard rate, entropy and expectation are derived. In the third section, distributions of minimum sequence of i.i.d. random variables are found out. An AR (1) model with new Marshall-Olkin discrete uniform distribution is discussed in the next section. In the fifth section, the maximum likelihood estimate (m.l.e.) for the parameters is found out and, in the last section, the goodness of the distribution is tested with a real life data.
2. Marshall-Olkin Discrete Uniform Distribution
Let X be a discrete random variable following uniform distribution with p.m.f. f(x)=1/a, x=1,2,3,…,a. We have the distribution function F(x)=x/a and the survival function F¯(x)=(a-x)/a. By Marshall-Olkin method, we can form another survival function of Marshall-Olkin discrete uniform (MODU) distribution by substituting F¯(x)=(a-x)/a in G¯(x,θ), in (1), and we get
(4)G¯(x,θ)=θ(a-x)[aθ+(1-θ)x],0<θ<∞.
We write X~MODU(a,θ) for a random variable (r.v.) with G¯(x,θ) given in (4).
2.1. Stability Property of the New Family
If we apply the same method again into the new family, by adding a new parameter “α” (α>0) to the reliability function, we get G¯1(x)=(αθ)(a-x)/[a-(1-αθ)(a-x)], which is the reliability function of the new Marshall-Olkin family with parameters (a,αθ).
2.2. Probability Mass Function
From (2), consider the probability mass function p.m.f. (5)g(x,θ)=θf(x)[1-(1-θ)F¯(x)][1-(1-θ)F¯(x-1)],
where f(x) is the p.m.f. corresponding to F(x). That is,
(6)g(x,θ)=aθ{[aθ+(1-θ)x][aθ+(1-θ)(x-1)]}.
Now we plot the p.m.f. of X~MODU for different values of a and θ (Figure 1).
P.m.f. of X~MODU.
Remark 1.
If X follows MODU (a,θ), then g(x,a,θ)=g(a-x+1,a,1/θ), x=1,2,…,a, θ>0.
Now, our next interest is to check whether (6) characterizes MODU. If X satisfies the condition g(x,a,θ)=g(a-x+1,a,1/θ), x=1,2,…,a, θ>0, this should be true for θ=1 also. For θ=1, it gives the parent distribution itself before reparameterization. Thus, Remark 1 implies that P(x)=P(a-x+1), x=1,2,3,…,a, where P(x) is the p.m.f. of discrete uniform distribution with parameter “a”. That is, P(1)=P(a),P(2)=P(a-1),…,P(a)=P(1). This relation need not characterize a discrete uniform distribution, when a>2. For example, let a=3; then we get P(1)=P(3). Since P(1)+P(2)+P(3)=1, we get P(2)=1-P(1)-P(3). That is P(2)=1-2P(1). Let P(1)=0.1=P(3) imply P(2)=0.8, which is not discrete uniform. Thus we have the following.
Theorem 2.
A random variable X follows MODU(2,θ), x=1,2 if and only if its p.m.f. satisfies
(7)g(x,a,θ)=g(a-x+1,a,1θ).
Proof.
Suppose that g(x,a,θ)=g(a-x+1,a,1/θ) is true for Marshall-Olkin (MO) scheme, then its parent distribution is discrete uniform on {1,2}. That is,
(8)g(x,2,θ)=g(2-x+1,2,1θ)=12,
that is,
(9)g(1,2,θ)=g(2,2,1θ).
We know the p.m.f. g(x,a,θ)=θf(x)/[1-(1-θ)F¯(x-1)][1-(1-θ)F¯(x)]. That is,
(10)g(1,2,θ)=θP(1)[1-(1-θ)F¯(0)][1-(1-θ)F¯(1)]=θP(1)[1-(1-θ)][1-(1-θ)F¯(1)]=12,
by (8); that is,
(11)2θP(1)=[1-(1-θ)][1-(1-θ)F¯(1)].(12)g(2,2,1θ)=P(2)θ[1-(1-1/θ)F¯(1)][1-(1-1/θ)F¯(2)]=P(2)θ[1-(1-1/θ)F¯(1)]=P(2)θ[1-((θ-1)/θ)F¯(1)].
That is, g(2,2,1/θ)=P(2)/[θ-(θ-1)F¯(1)]=1/2, by (8).
That is, 2P(2)=[θ-(θ-1)F¯(1)] or 2P(2)-θ=-(θ-1)F¯(1). We get,
(13)F¯(1)=(2P(2)-θ)1-θ.
By substituting this in (11),
(14)2θP(1)=[1-(1-θ)][1-(1-θ)((2P(2)-θ)(1-θ))]=[1-(1-θ)][1-(2P(2)-θ)],2θP(1)=θ[1-(2P(2)-θ)].
That is, 2P(1)=1-2P(2)+θ or 2(P(1)+P(2))=1+θ. So 2=1+θ, since P(1)+P(2)=1.
That is, θ=1. But, when θ=1, it corresponds to discrete uniform distribution on {1,2}. The converse is straight forward.
Remark 3.
It is clear that MODU distribution does not possess additive property.
Remark 4.
MODU is not infinitely divisible (i.d.), since its support is on {1,2,3,…,a}, a finite range. It is not log convex, since the class of log convex distribution forms a subclass of the class of i.d. distributions.
Corollary 5.
If X~MODU(a,θ), then F¯(x)≤G¯(x,θ)≤θF¯(x), if θ≤1, and θF¯(x)<G¯(x,θ)<F¯(x), if 0<θ<1.
Proof.
Let X~MODU(a,θ), F¯(x)=(a-x)/a and G¯(x,θ)=θ(a-x)/[aθ+(1-θ)x]. So, F¯(x)≤G¯(x,θ) implies that (a-x)/a≤θ(a-x)/[aθ+(1-θ)x]. That is 1/a-θ/[aθ+(1-θ)x]≤0, aθ+(1-θ)x-aθ≤0, or (1-θ)x≤0. This happens only when θ≥1. Now assume that G¯(x,θ)≤θF¯(x). That is, θ(a-x)/[aθ+(1-θ)x]≤θ(a-x)/a. So (1-θ)a-(1-θ)x≤0; that is, (a-x)(1-θ)≤0, which is true only when θ≥1, since (a-x)≥0. Hence, the proof is completed.
2.3. Expectation, Standard Deviation, and Entropy of MODU Distribution
We numerically compute the expectation and standard deviation (Tables 1 and 2) and entropy (Table 3) of the MODU distribution with different a and θ, since compact expressions are not available for calculating the same. We have
(15)Shannon’sEntropy=-∑Ip(xi)logp(xi),i∈I={1,2,…}=-∑aθ[aθ+(1-θ)xi][aθ+(1-θ)(xi-1)]×logaθ[aθ+(1-θ)xi][aθ+(1-θ)(xi-1)].
Expectation and standard deviation (sd) with different “a” and “θ” (θ<1) for X~MODU(a,θ).
a
θ=0.133
θ=0.25
θ=0.5
θ=0.75
E(X)
sd(X)
E(X)
sd(X)
E(X)
sd(X)
E(X)
sd(X)
10
2.597426
2.186733
3.358938
2.519965
4.375428
2.778938
5.026709
2.855936
12
2.995745
2.638189
3.919444
3.033456
5.145940
3.341060
5.930269
3.432611
15
3.597637
3.312588
4.762709
3.801608
6.302744
4.182802
7.286014
4.296371
20
4.606958
4.43258
6.171538
5.079024
8.232135
5.583787
9.546128
5.734303
25
5.620115
5.550024
7.582418
6.354713
10.16236
6.983619
11.80657
7.171237
30
6.635235
6.666144
8.994330
7.629531
12.09300
8.382876
14.06717
8.607673
50
10.70370
11.12519
14.64612
12.72530
19.81722
13.97760
23.11022
14.35143
75
15.79535
16.69483
21.71398
19.09236
29.47374
20.96928
34.41451
21.52963
100
20.88903
22.26307
28.78287
25.45854
39.13069
27.96038
45.71897
28.70733
Expectation and standard deviation with different “a” and “θ” (θ>1) for X~MODU(a,θ).
a
θ=1.33
θ=2
θ=4
θ=7.5
E(X)
sd(X)
E(x)
sd(x)
E(x)
sd(x)
E(x)
sd(x)
10
5.97329
2.855936
6.62457
2.77893
7.64106
2.51996
8.40257
2.18673
12
7.06973
3.432611
7.85406
3.34106
9.08055
3.03345
10.0042
2.63818
15
8.71398
4.296371
9.69725
4.18280
11.2372
3.80160
12.4023
3.31258
20
11.4538
5.734303
12.7678
5.58378
14.8284
5.07902
16.3930
4.43258
25
14.1934
7.171237
15.8376
6.98361
18.4175
6.35471
20.3798
5.55002
30
16.9328
8.607673
18.9070
8.38287
22.0056
7.62953
24.3647
6.66614
50
27.8897
14.35143
31.1827
13.9776
36.3538
12.7253
40.2963
11.1251
75
41.5854
21.52963
46.5262
20.9692
54.2860
19.0923
60.2046
16.6948
100
55.2810
28.70733
61.86931
27.96038
72.21713
25.45854
80.11097
22.26307
Entropy with different “a” and “θ” for X~MODU(a,θ).
a
θ=0.1
θ=0.133
θ=0.25
θ=0.5
θ=0.75
θ=1.33
θ=2
θ=4
θ=7.5
θ=10
10
1.5310
1.6923
1.9985
2.2241
2.2889
2.2889
2.2241
1.9985
1.6923
1.5310
15
1.9142
2.0846
2.4004
2.6290
2.6943
2.6943
2.6290
2.4004
2.0846
1.9142
20
2.1933
2.3674
2.6868
2.9165
2.9819
2.9819
2.9165
2.6868
2.3674
2.1933
30
2.5923
2.7692
3.0914
3.3218
3.3874
3.3874
3.3218
3.0914
2.7692
2.5923
50
3.0997
3.2782
3.6017
3.8326
3.8982
3.8982
3.8326
3.6017
3.2782
3.0997
100
3.7913
3.9705
4.2947
4.5257
4.5913
4.5913
4.5257
4.2947
3.9705
3.7913
Remark 6.
If X~MODU (a,θ), then the sd(X) is decreasing with increasing value of θ when θ>1 and the sd(X) increasing with increasing value of θ when θ<1.
From Remark 1 and Tables 1 and 2, we have the following.
Remark 7.
The sd(X) for X~MODU (a,θ) is equal to sd(X) for X~MODU (a,1/θ).
From Remark 1, Table 3, and Figure 2, the following is observed.
Entropy with different “a” and “θ” for X~MODU (a,θ).
Remark 8.
The entropy of X~MODU (a,θ) is equal to the entropy of X~MODU (a,1/θ).
Let X~MODU (a,θ); then the mean, median, and mode of the distribution (in Figure 1 and Tables 1 and 2) for different a and θ are computed in Table 4.
Mean, median, and mode of the MODU distribution with different a and θ.
a
θ
Mean
Median
Mode
10
0.2
3.06
2
1
0.5
4.40
3
1
2
6.60
7
10
5
7.90
9
10
20
0.2
5.57
4
1
0.5
8.23
7
1
2
12.76
14
20
5
15.40
17
20
100
0.2
25.70
17
1
0.5
39.10
34
1
2
61.80
67
100
5
75.20
84
100
From Figure 1 and from Table 4, the following is clear.
Remark 9.
The MODU distribution is positively skewed, when θ<1, since mode < median < mean and the distribution is negatively skewed, when θ>1, since mode > median > mean. Also, it is to be noted that the distribution is unimodal; that is, when θ<1, the mode = 1 and, when θ>1, the mode =a.
Therefore, the MODU distribution can be applied for the data showing positive skewness when θ>1 and the data showing negative skewness when θ<1.
2.4. Hazard Function
The hazard function of X~MODU (a,θ) is
(16)γG(x)=g(x)G¯(x)=θ(a-x+1)/[aθ+(1-θ)(x-1)]θ(a-x)/[aθ+(1-θ)x]-1.
That is,
(17)γG(x)=a(a-x)(aθ+(1-θ)(x-1)).
Theorem 10.
Let X~MODU(a,θ); then the distribution is with Increasing Failure Rate (IFR) when θ>(a-2x)/(2a-2x), Decreasing Failure Rate (DFR) when θ<(a-2x)/(2a-2x), and constant FR when θ=(a-2x)/(2a-2x).
Proof.
From (17), the hazard function of X~MODU(a,θ) is γG(x)=a/{(a-x)(aθ+(1-θ)(x-1))}.
Then γG(x+1)=a/(a-x-1)(aθ+(1-θ)x). If γG(x+1)-γG(x)>0, the distribution is with IFR.
That is, if γG(x+1)/γG(x)>1 or
(18)(a-x)(aθ+(1-θ)(x-1))(a-x-1)(aθ+(1-θ)x)>1,(a-x)(aθ+(1-θ)(x-1))-(a-x-1)(aθ+(1-θ)x)>0,(-a+2aθ+2x-2xθ)>0,θ(2a-2x)>a-2xor
if θ>(a-2x)/(2a-2x), the distribution is with IFR.
Similarly, if θ<(a-2x)/(2a-2x), the distribution is with DFR, and if θ=(a-2x)/(2a-2x), the distribution is with a constant FR.
We illustrate these results graphically (Figure 3) and numerically (Table 5).
From Figure 3, also, it is clear that the failure rate is IFR, when θ>(a-2x)/(2a-2x), and DFR, when θ<(a-2x)/(2a-2x).
Hazard function γG(x) of X~MODU(a,θ) with a=10.
X
θ
0.75
0.5
0.2
0.1
2
5
7.5
1
0.1481481
0.2222222
0.5555556
1.111111
0.05555556
0.02222222
0.01481481
2
0.1612903
0.2272727
0.4464286
0.6578947
0.06578947
0.02717391
0.01824818
3
0.1785714
0.2380952
0.3968254
0.5102041
0.07936508
0.03401361
0.02304147
4
0.2020202
0.2564103
0.3787879
0.4504505
0.09803922
0.04385965
0.03003003
5
0.2352941
0.2857143
0.3846154
0.4347826
0.1250000
0.05882353
0.04081633
6
0.2857143
0.3333333
0.4166667
0.4545455
0.1666667
0.08333333
0.05882353
7
0.3703704
0.4166667
0.4901961
0.5208333
0.2380952
0.1282051
0.09259259
8
0.5405405
0.5882353
0.6578947
0.6849315
0.3846154
0.2272727
0.1694915
9
1.052632
1.111111
1.190476
1.219512
0.8333333
0.5555556
0.4347826
Hazard function γG(x) of X~MODU(a,θ).
3. MODU Distribution as the Distribution of Minimum of a Sequence of i.i.d. Random Variables
The following theorem gives a characterization of minimum of a sequence of i.i.d. random variables following MODU distribution.
Theorem 11.
Let {Xi,i≥1} be a sequence of i.i.d. random variables with common survival function F¯(x). Let N be a geometric random variable independent of {Xi,i≥1} such that P(N=n)=pqn-1, n=1,2,3,…; 0<p=θ<1, and q=1-p. Let UN=min1≤i≤n(Xi). Then {UN} is distributed as MODU distribution if and only if {Xi} follows discrete uniform distribution with x=1,2,3,…,a.
Proof.
Proof follows as in the same lines as given in Satheesh et al. [4–6] for a similar characterization of minimum of sequence of i.i.d. random variables.
The survival function of UN is G¯(x) = P(UN>x) = ∑1∞[F¯(x)]nP(N=n) = ∑1∞[F¯(x)]nθ(1-θ)n = θ∑1∞[F¯(x)(1-θ)]n = θF¯(x)/[1-(1-θ)F¯(x)]. Substituting F¯(x)=(a-x)/a, the survival function of the discrete uniform distribution; that is,
(19)G¯(x)=θ[(a-x)/a][1-(1-θ)((a-x)/a)]=θ(a-x)[a-(1-θ)(a-x)]=θ(a-x)[aθ+(1-θ)x],
which is the survival function of MODU(a,θ). By retracing the steps we can easily show the converse.
4. AR (1) Model with MODU Distribution as Innovating Distribution
Arnold and Robertson [7], Chernick [8], and Satheesh et al. [9] studied some properties of the AR (1) models. As Satheesh et al. [9] discussed, construct a first order autoregressive minification process with the following structure:
(20)Xn={ϵnwith probabilityθmin(Xn-1,ϵn)with probability(1-θ),0≤θ≤1,
where {ϵn} is a sequences of i.i.d. random variables following a discrete uniform distribution with parameter a independent of {Xn-1,Xn-2,…}. Then the process is stationary and is marginally distributed with MODU(a,θ) marginal distribution.
Theorem 12.
In an AR (1) process with structure (20) {Yi,n} is stationary Markovian with MODU(a,θ) marginal if and only if {ϵn} is distributed as uniform with parameter a,x=1,2,3,…,a.
Proof.
From the structure (20), it follows that
(21)F¯Xn(x)=θF¯ϵn(x)+(1-θ)F¯Xn-1(x)F¯ϵn(x).
Under stationary equilibrium, this implies
(22)F¯X(x)=θF¯ϵn(x)[1-(1-θ)F¯ϵn(x)].
If we take F¯ϵn(x)=(a-x)/a, then,
F¯X(x)=θ(a-x)/[a-(1-θ)(a-x)], which is the survival function of MODU(a,θ). We can also show that F¯ϵn(x) follows uniform distribution with parameter a. By retracing the steps we can easily show the converse.
5. Maximum Likelihood Estimates (m.l.e.) for the Parameters of MODU<inline-formula>
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M245">
<mml:mo mathvariant="bold">(</mml:mo>
<mml:mi>a</mml:mi>
<mml:mo mathvariant="bold">,</mml:mo>
<mml:mi>θ</mml:mi>
<mml:mo mathvariant="bold">)</mml:mo></mml:math>
</inline-formula>
Let x1,x2,…,xn be independent random samples from MODU(a,θ). Then, from the likelihood function of the distribution, we can write
(23)(nθ)-∑(a-xi)[aθ+(1-θ)xi]-∑(a-xi+1)[aθ+(1-θ)(xi-1)]=0,(24)(na)-θ∑1[aθ+(1-θ)xi]-θ∑1[aθ+(1-θ)(xi-1)]=0.
We calculated the m.l.e. of a and θ numerically (using Mathematica) as the solution of these two nonlinear equations. But, here, the maximum of the range of observation is a. So m.l.e. of the parameter a(a^) is the largest value of the observation. Then, from (23), we can find m.l.e. of θ.
Algorithm 13.
Consider the following.
Through simulation, 2500 random samples were generated from inverse function of distribution function G(x) for some given value of the parameters θ and a.
Then we iterate the m.l.e. for the parameters from the equations (23) and (24) (Table 6).
For accuracy, we repeat the same calculation ten times with same values of θ and a.
The mean and the standard error (SE) of these estimates are calculated.
Then it is calculated with different values of the parameter θ and the same is found with an increase in sample size 5000 also.
The mean estimated values for the parameters are tabulated (Table 7).
Initial observations.
N
a
a^
θ
θ^
Mean θ^
SE(θ^)
1
2500
20
20
0.1
0.104118
0.1016693
0.1016693
2
0.111474
3
0.095902
4
0.101358
5
0.098971
6
0.095781
7
0.101448
8
0.098904
9
0.103342
10
0.105396
Mean of m.l.e. of θ from 10 repetitions. (a=20,a^=max(x)=20, sample size =n).
θ
n
2500
5000
Mean θ^
SE(θ^)
Mean θ^
SE(θ^)
0.1
0.1016693
0.1020
0.09995393
0.0020
0.25
0.2487304
0.0020
0.2480182
0.0020
0.5
0.5020383
0.0030
0.4920417
0.0030
0.75
0.7512596
0.0023
0.7496541
0.0019
0.9
0.9015150
0.0060
0.8997414
0.0043
2
2.0007690
0.0191
1.9986450
0.0131
5
4.9308800
0.0507
5.0252140
0.0362
10
9.8801710
0.0900
9.9767230
0.0900
50
51.069840
0.4321
100.04960
0.4006
100
100.58670
1.3979
100.04960
0.4666
It is clear that the SE of m.l.e. of the parameters “a” and θ is decreasing with increase in sample size.
6. An Application of MODU Example 14.
The data was collected from the daily attendance register of the science and commerce (nonlanguage) supplementary class (20 working days in a month) of a higher secondary school in Palakkad district, Kerala. We took a sample of 50 students (out of 360) and let X be the number of days these students attended in the class for the whole year 2012-2013 (an academic year is 10 months) (see Tables 8, 9, and 10).
Observed frequency of 50 students in 10 months from science and commerce classes.
Month
x
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
May
1
1
0
3
2
4
2
2
0
2
3
2
1
2
2
4
3
8
6
2
June
0
2
1
2
4
1
1
1
2
4
2
2
1
3
2
4
5
5
3
5
July
0
0
0
2
2
3
1
3
2
3
4
1
4
2
5
5
4
2
4
3
Aug
2
1
3
1
1
1
1
2
3
1
1
4
4
2
1
6
5
3
3
5
Sep
1
2
1
3
1
2
3
3
0
2
0
1
3
4
5
4
1
6
3
5
Oct
1
0
2
0
3
0
2
3
0
6
2
4
3
1
4
4
2
8
2
3
Nov
1
0
2
3
3
0
1
3
1
4
3
2
0
2
1
6
4
4
7
3
Dec
1
3
0
1
3
3
2
0
4
0
2
1
2
5
2
4
4
5
3
5
Jan
0
0
0
2
2
3
1
3
2
3
4
1
4
2
5
5
4
2
4
3
Feb
2
1
3
1
1
1
0
2
3
1
1
4
4
2
2
6
5
3
3
5
The m.l.e. of θ (sample size = 50, m.l.e. of a, a^=20).
Month
May
June
July
Aug
Sept
Oct
Nov
Dec
Jan
Feb
m.l.e. θ^
2.0347
2.0364
1.9247
2.0432
2.0347
1.9368
2.0743
1.9995
2.0039
1.995
Mean m.l.e. of θ and mean Standard Error (SE) of θ for the samples of 10 months.
Sample size
a^=max(x)
Mean θ^
SE(θ^)
50
20
2.00832
0.04486918
We arbitrarily fix 4 months, and the m.l.e.s are computed. Initially we fit discrete uniform to the data (see Tables 11 and 12).
Goodness of fit of uniform distribution with α=0.05.
May
August
October
February
m.l.e. of θ=θ^
2.0347
2.0432
1.9368
1.995
χ2 statistics
14.800
9.4000
13.200
10.400
P value
0.0632
0.3096
0.1052
0.2381
Mean, median, and mode of the data in May, August, October, and February.
a
Month
m.l.e. of θ=θ^
Mean
Median
Mode
20
May
2.0347
12.80
14
18
August
2.0432
12.30
13
16
October
1.9368
12.86
13
18
February
1.9950
12.64
14
16
Since mode > median > mean, the data exhibits negative skewness with m.l.e. of θ>1 and it is observed that the distribution is unimodal; hence, MODU distribution is supposed to give a better fit than uniform distribution (see Table 13).
Goodness of fit of MODU distribution with α=0.05.
May
August
October
February
m.l.e. of θ=θ^
2.0347
2.04320
1.93680
1.99500
χ2 statistics
9.4668
6.45238
1.80953
6.95238
P value
0.1489
0.37446
0.93640
0.325270
Result. From the P values, it is seen that MODU distribution is better fit than uniform distribution.
Example 15.
The data was collected from the daily attendance register of the supplementary language class (20 working days in a month) from the same higher secondary school in Palakkad district, Kerala. We took a sample of 50 students and let X be the number of days these students attended in the same for the whole year 2012-2013 (an academic year is 10 months) (see Tables 14, 15, and 16).
Observed frequency of 50 students in 10 months from the language classes.
Month
X
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
May
5
5
7
5
5
1
1
4
1
5
2
1
0
1
1
2
1
1
1
1
June
3
9
7
4
2
4
3
1
3
1
1
1
1
2
3
0
1
0
1
3
July
7
4
5
5
2
3
6
0
4
2
2
1
2
1
0
2
1
1
1
1
Aug
10
4
5
2
1
4
2
2
4
1
0
1
0
2
3
4
1
0
3
1
Sep
7
3
10
2
3
1
5
3
3
1
2
1
3
1
1
2
0
1
1
0
Oct
8
4
4
3
5
3
2
4
1
4
3
0
3
1
1
0
1
2
0
1
Nov
9
5
7
0
3
2
2
4
1
2
2
2
2
1
2
2
0
3
1
0
Dec
8
4
3
5
7
4
2
5
0
0
3
0
1
1
1
1
4
1
0
0
Jan
6
2
7
3
4
1
7
5
3
1
2
1
2
1
1
3
0
0
0
1
Feb
5
6
7
6
5
2
2
0
3
1
0
2
2
3
1
2
2
1
0
0
The m.l.e. of θ and SE of θ (sample size = 50, m.l.e. of a, a^=max(x)=20).
Month
May
June
July
Aug
Sept
Oct
Nov
Dec
Jan
Feb
m.l.e. θ^
0.3511
0.3544
0.3536
0.3802
0.3619
0.3522
0.3720
0.3707
0.3745
0.3594
Mean m.l.e. of θ and mean SE of θ for the samples of 10 months.
Sample size
a^=max(x)
Mean θ^
SE (θ^)
50
20
0.3630059
0.0033407
We arbitrarily fix 3 months, and the m.l.e.s are computed. Fit discrete uniform to the above data. We have the results as shown in Tables 17 and 18.
Goodness of fit of uniform distribution with α=0.05.
May
September
February
m.l.e. of θ=θ^
0.3511
0.3630
0.3594
m.l.e. of a=a^
20
19
18
χ2 statistics
23.60
25.20
35.60
P value
0.0027
0.0014
0.00002
Mean median and mode of the data in May, September, and February.
a
Month
m.l.e. of θ=θ^
Mean
Median
Mode
20
May
0.3510
7.80
5
3
September
0.3619
6.78
5
1
February
0.3594
6.82
5
3
Since mode < median < mean, the data exhibits positive skewness with m.l.e. of θ<1 and it is observed that the distribution is unimodal; hence, MODU distribution is supposed to give a better fit than uniform distribution (see Table 19).
Goodness of fit of MODU distribution with α=0.05.
May
September
February
m.l.e. of θ=θ^
0.3511
0.3630
0.3594
a^
20
19
18
χ2 statistics
9.80007
7.23453
7.79048
P value
0.1333
0.2997
0.2538
Result. From the P values it is seen that, here, MODU distribution is a better fit than uniform distribution.
7. Conclusions
The p.m.f.s of MODU distribution g(x,θ) satisfies the relation g(x,θ)=g(a-x+1,1/θ), x=1,2,…,a, θ>0. The failure rate of MODU(a,θ) is IFR, when θ>(a-2x)/(2a-2x), DFR, when θ<(a-2x)/(2a-2x), and constant, when θ=(a-2x)/(2a-2x). Like discrete uniform distribution, the probability function degenerates when a is large. In the case of MODU(a,θ), the standard deviation is decreasing with the increasing value of θ, when θ>1, and is increasing with increasing value of θ, when θ<1. For X~MODU(a,θ) and X~MODU(a,1/θ), the sd(X) is equal in both cases. It is also observed that entropy(a,θ)=entropy(a,1/θ). The MODU distribution is positively skewed, when θ<1, since, here, mode < median < mean and the distribution is negatively skewed, when θ>1, since then mode > median > mean. It is to be noted that the distribution is unimodal; that is, when θ<1, the mode =1 and, when θ>1, mode =a. An application of MODU distribution is also discussed.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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