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Licensed Unlicensed Requires Authentication Published by De Gruyter December 4, 2022

A bivariate extension of the Omega distribution for two-dimensional proportional data

  • Ömer Özbilen and Alі İ. Genç EMAIL logo
From the journal Mathematica Slovaca

Abstract

When data generating mechanism generates two correlated data sets both defined on the unit interval, a bivariate probabilistic distribution defined on the unit square is needed for modelling the data. For this purpose, we give a Marshall-Olkin type bivariate extension of an omega distribution in this paper. This is in fact a bivariate unit-exponentiated-half-logistic distribution. We study its mathematical properties in detail. The distribution contains neither an exponential term nor any special function which complicates the computations. Maximum likelihood estimation method and its large sample inference are considered for model parameters. Alternatively, we also propose an expectation- maximization algorithm to compute the estimates. To see the performances of the proposed estimators and validate the theoretical results obtained for estimation, we present the results of a simulation study. Data fitting demonstrations show its applicability in modelling random proportions.

Acknowledgement

The authors would like to thank the Area Editor (Gejza Wimmer, Ph.D.) and the two anonymous referees for their comments.

  1. (Communicated by Gejza Wimmer )

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Appendix A. Proofs of Theorems

Proof

Proof of Theorem 2.1.

SY1,Y2(y1,y2)=P(Y1>y1,Y2>y2)=P(min(X1,X3)>y1,min(X2,X3)>y2)=P(X1>y1,X3>y1,X2>y2,X3>y2)=P(X1>y1,X2>y2,X3>max(y1,y2))=P(X1>y1)P(X2>y2)P(X3>max(y1,y2)).

Proof

Proof of Theorem 2.3. We have FY1,Y2 (y1, y2) = SY1, Y2(y1, y2)+FY1(y1)+FY2(y)−1, where FY1 and FY2 denote the marginal CDF’s of Y1 and Y2. Using this formula, one can obtain the result in the theorem.□

Proof

Proof of Theorem 2.4. For the absolutely continuous part of the joint PDF, we take the second order partial derivatives of the joint CDF. For the singular part, we use the following fact:

01y11f1(y1,y2)dy2dy1+01y21f2(y1,y2)dy1dy2+01f0(y)dy=1.

Since

01y11f1(y1,y2)dy2dy1=λ1i=13λiand01y21f2(y1,y2)dy1dy2=λ2i=13λi,

we have

01f0(y)dy=λ3i=13λi.

Since

λ3i=13λi01g(y;i=13λi,θ)dy=λ3i=13λi,

we have

f0(y)=λ3i=13λig(y;i=13λi,θ)=2θλ3yθ1(1yθ)i=13λi1(1+yθ)i=13λi+1.

Proof

Proof of Theorem 2.5. If y1 < y2, then

fY1|Y2(y1|y2)=f(y1,y2)fY2(y2)=g(y1;λ1,θ)g(y2;λ2+λ3,θ)g(y2;λ2+λ3,θ)=g(y1;λ1,θ).

The other two cases can be shown similarly.□

Proof

Proof of Theorem 2.6. If y1 < y2, then

SY1|Y2(y1|y2)=P(Y1>y1|Y2>y2)=S(y1,y2)SY2(y2)=1y1θ1+y1θλ1.

The other two cases can be shown similarly.□

Received: 2021-06-01
Accepted: 2021-09-20
Published Online: 2022-12-04
Published in Print: 2022-12-16

© 2022 Mathematical Institute Slovak Academy of Sciences

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