Journal of Probability

Journal of Probability / 2014 / Article

Research Article | Open Access

Volume 2014 |Article ID 979312 | 10 pages | https://doi.org/10.1155/2014/979312

Marshall-Olkin Discrete Uniform Distribution

Academic Editor: Serkan Eryílmaz
Received24 Nov 2013
Revised11 Feb 2014
Accepted20 Feb 2014
Published07 Apr 2014

Abstract

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 is the survival function of a distribution, then, by Marshall-Olkin method, we get another survival function , by adding a new parameter to it. That is, Then the corresponding distribution function is .

Now its probability mass function (p.m.f.) is where is the p.m.f. corresponding to .

The hazard rate of is

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 be a discrete random variable following uniform distribution with p.m.f. , . We have the distribution function and the survival function . By Marshall-Olkin method, we can form another survival function of Marshall-Olkin discrete uniform (MODU) distribution by substituting in , in (1), and we get

We write for a random variable (r.v.) with 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 “ to the reliability function, we get , which is the reliability function of the new Marshall-Olkin family with parameters .

2.2. Probability Mass Function

From (2), consider the probability mass function p.m.f. where is the p.m.f. corresponding to . That is,

Now we plot the p.m.f. of for different values of and (Figure 1).

Remark 1. If follows , then , , .

Now, our next interest is to check whether (6) characterizes MODU. If satisfies the condition , , , this should be true for also. For , it gives the parent distribution itself before reparameterization. Thus, Remark 1 implies that , , where is the p.m.f. of discrete uniform distribution with parameter “”. That is, . This relation need not characterize a discrete uniform distribution, when . For example, let ; then we get . Since , we get . That is . Let imply , which is not discrete uniform. Thus we have the following.

Theorem 2. A random variable follows , if and only if its p.m.f. satisfies

Proof. Suppose that is true for Marshall-Olkin (MO) scheme, then its parent distribution is discrete uniform on . That is, that is, We know the p.m.f. . That is, by (8); that is, That is, , by (8).
That is, or . We get, By substituting this in (11), That is, or . So , since .
That is, . But, when , it corresponds to discrete uniform distribution on . 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 , 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 , then , if , and , if .

Proof. Let , and . So, implies that . That is , , or . This happens only when . Now assume that . That is, . So ; that is, , which is true only when , since . 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 and , since compact expressions are not available for calculating the same. We have



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



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



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 , then the is decreasing with increasing value of when and the increasing with increasing value of when .

From Remark 1 and Tables 1 and 2, we have the following.

Remark 7. The for is equal to for .

From Remark 1, Table 3, and Figure 2, the following is observed.

Remark 8. The entropy of is equal to the entropy of .

Let ; then the mean, median, and mode of the distribution (in Figure 1 and Tables 1 and 2) for different and are computed in Table 4.


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 , since mode < median < mean and the distribution is negatively skewed, when , since mode > median > mean. Also, it is to be noted that the distribution is unimodal; that is, when , the mode = 1 and, when , the mode .
Therefore, the MODU distribution can be applied for the data showing positive skewness when and the data showing negative skewness when .

2.4. Hazard Function

The hazard function of is That is,

Theorem 10. Let ; then the distribution is with Increasing Failure Rate (IFR) when , Decreasing Failure Rate (DFR) when , and constant FR when .

Proof. From (17), the hazard function of is .
Then . If , the distribution is with IFR.
That is, if or if , the distribution is with IFR.
Similarly, if , the distribution is with DFR, and if , 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 , and DFR, when .


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

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 be a sequence of i.i.d. random variables with common survival function . Let be a geometric random variable independent of   such that , ; , and . Let . Then is distributed as MODU distribution if and only if   follows discrete uniform distribution with .

Proof. Proof follows as in the same lines as given in Satheesh et al. [46] for a similar characterization of minimum of sequence of i.i.d. random variables.
The survival function of is   =     =    =    =    =  . Substituting , the survival function of the discrete uniform distribution; that is, which is the survival function of . 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: where is a sequences of i.i.d. random variables following a discrete uniform distribution with parameter independent of . Then the process is stationary and is marginally distributed with marginal distribution.

Theorem 12. In an AR (1) process with structure (20) is stationary Markovian with marginal if and only if is distributed as uniform with parameter .

Proof. From the structure (20), it follows that Under stationary equilibrium, this implies If we take , then,
, which is the survival function of . We can also show that follows uniform distribution with parameter . By retracing the steps we can easily show the converse.

5. Maximum Likelihood Estimates (m.l.e.) for the Parameters of MODU

Let be independent random samples from . Then, from the likelihood function of the distribution, we can write We calculated the m.l.e. of and numerically (using Mathematica) as the solution of these two nonlinear equations. But, here, the maximum of the range of observation is . So m.l.e. of the parameter is the largest value of the observation. Then, from (23), we can find m.l.e. of .

Algorithm 13. Consider the following.(1)Through simulation, 2500 random samples were generated from inverse function of distribution function for some given value of the parameters and .(2)Then we iterate the m.l.e. for the parameters from the equations (23) and (24) (Table 6).(3)For accuracy, we repeat the same calculation ten times with same values of and .(4)The mean and the standard error (SE) of these estimates are calculated.(5)Then it is calculated with different values of the parameter and the same is found with an increase in sample size 5000 also.(6)The mean estimated values for the parameters are tabulated (Table 7).


Mean SE

1 2500 20200.1 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


2500 5000
Mean SE Mean SE

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 “” and 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 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).


Month
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


Month May June July Aug Sept Oct Nov Dec Jan Feb

m.l.e. 2.0347 2.0364 1.9247 2.0432 2.0347 1.9368 2.0743 1.9995 2.0039 1.995


Sample size Mean SE

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).


May August October February

m.l.e. of 2.0347 2.0432 1.9368 1.995
statistics 14.800 9.4000 13.200 10.400
value 0.0632 0.3096 0.1052 0.2381


Month m.l.e. of 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 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).


May August October February

m.l.e. of 2.0347 2.04320 1.93680 1.99500
statistics 9.4668 6.45238 1.80953 6.95238
value 0.1489 0.37446 0.93640 0.325270

Result. From the 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 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).


Month
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


Month May June July Aug Sept Oct Nov Dec Jan Feb

m.l.e. 0.3511 0.3544 0.3536 0.3802 0.3619 0.3522 0.3720 0.3707 0.3745 0.3594


Sample size Mean SE

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.


May September February

m.l.e. of 0.3511 0.3630 0.3594
m.l.e. of 20 19 18
statistics 23.60 25.20 35.60
value 0.0027 0.0014 0.00002


Month m.l.e. of 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 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).


May September February

m.l.e. of 0.3511 0.3630 0.3594
20 19 18
statistics 9.80007 7.23453 7.79048
value 0.1333 0.2997 0.2538

Result. From the 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 satisfies the relation , , . The failure rate of is IFR, when , DFR, when , and constant, when . Like discrete uniform distribution, the probability function degenerates when a is large. In the case of , the standard deviation is decreasing with the increasing value of , when , and is increasing with increasing value of , when . For and , the is equal in both cases. It is also observed that entropy. The MODU distribution is positively skewed, when , since, here, mode < median < mean and the distribution is negatively skewed, when , since then mode > median > mean. It is to be noted that the distribution is unimodal; that is, when , the mode and, when , mode . 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.

References

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Copyright © 2014 E. Sandhya and C. B. Prasanth. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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