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Bivariate Poisson models with varying offsets: an application to the paired mitochondrial DNA dataset

  • Pei-Fang Su EMAIL logo , Yu-Lin Mau , Yan Guo , Chung-I Li , Qi Liu , John D. Boice and Yu Shyr

Abstract

To assess the effect of chemotherapy on mitochondrial genome mutations in cancer survivors and their offspring, a study sequenced the full mitochondrial genome and determined the mitochondrial DNA heteroplasmic (mtDNA) mutation rate. To build a model for counts of heteroplasmic mutations in mothers and their offspring, bivariate Poisson regression was used to examine the relationship between mutation count and clinical information while accounting for the paired correlation. However, if the sequencing depth is not adequate, a limited fraction of the mtDNA will be available for variant calling. The classical bivariate Poisson regression model treats the offset term as equal within pairs; thus, it cannot be applied directly. In this research, we propose an extended bivariate Poisson regression model that has a more general offset term to adjust the length of the accessible genome for each observation. We evaluate the performance of the proposed method with comprehensive simulations, and the results show that the regression model provides unbiased parameter estimations. The use of the model is also demonstrated using the paired mtDNA dataset.

Acknowledgment

The authors thank the editor, the associate deitor and referees for their valuable comments nad suggestions. The work was partly supported by a Ministry of Science and Technology grant, MOST 105-2118-M-006-007-MY2 and the Mathematics Division of the National Center for Theoretical Sciences in Taiwan.

Appendix I : Recursive relation

We refine the notation BPO(Yi1 = yi1, Yi2 = yi2) equal to P(yi1, yi2). Because the offset ti0, ti1, ti2 vary, the recursive relationships proposed by Kawamura (1985) need to be re-driven. In this appendix, we would like to prove

{yi1P(yi1,yi2)=ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21)yi2P(yi1,yi2)=ti2θ2P(yi1,yi21)+ti0θ0P(yi11,yi21)

for every integers yi1, yi2 = 1, 2, …, n. First, we prove this relationship

yi1P(yi1,yi2)=ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21).

Let’s consider two cases: the first case is yi1yi2 and second case is yi2yi1. Considering yi1yi2,

P(yi1,yi2)=k=0yi1P(Xi1=k)P(Xi1=yi1k)P(Xi2=yi2k)=P(Xi0=0)P(Xi1=yi1)P(Xi2=yi2)+P(Xi0=1)P(Xi1=yi11)P(Xi2=yi21)++P(Xi0=yi11)P(Xi1=1)P(Xi2=yi2yi1+1)+P(Xi0=yi1)P(Xi1=0)P(Xi2=yi2yi1).

To calculate yi1P(yi1, yi2), the above formula becomes

yi1P(yi1,yi2)=yi1P(Xi1=0)P(Xi1=yi1)P(Xi2=yi2)+(yi11)P(Xi0=1)P(Xi1=yi11)P(Xi2=yi21)++1×P(Xi0=yi11)P(Xi1=1)P(Xi2=yi2yi1+1)+1×P(Xi0=1)P(Xi1=yi11)P(Xi2=yi21)++(yi11)P(Xi0=yi11)P(Xi1=1)P(Xi2=yi2yi1+1)+yi1P(Xi0=yi1)P(Xi1=0)P(Xi2=yi2yi1).

Therefore, yi1P(yi1, yi2) can be rearrange as

k=0(yi11)yi2ti1θ1P(Xi0=k)P(Xi1=yi1k1)P(Xi2=yi2k)+k=0(yi11)(yi21)ti0θ0P(Xi0=k)P(Xi1=yi1k1)P(Xi2=yi2k1)=ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21).

Similarly, for the case of yi2yi1, we use the same method to get the result

yi1P(yi1,yi2)=ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21).

Therefore, the second relation can be derived in the same way.

Appendix II: Solve E-step by recursive relationship

We know Xi1 and Xi0 are independent, and Yi1 = Xi0 + Xi1. Then the conditional expectation

E(Xi1|Yi1,Yi2)=E(Yi1Xi0|Yi1,Yi2)=yi1E(Xi0|Yi1,Yi2),E(Xi2|Yi1,Yi2)=E(Yi2Xi0|Yi1,Yi2)=yi2E(Xi0|Yi1,Yi2).

As long as we derive the formual of E(Xi0|Yi1,Yi2), we can straightforward get E(Xi1|Yi1,Yi2) and E(Xi2|Yi1,Yi2). Because the conditional expectation

(6)E((Xi0+Xi1)P(yi1,yi2)|Yi1,Yi2)=E(Xi0P(yi1,yi2)|Yi1,Yi2)+E(Xi1P(yi1,yi2)|Yi1,Yi2)=P(yi1,yi2)E(Xi0|Yi1,Yi2)+P(yi1,yi2)E(Xi1|Yi1,Yi2).

Using the recursive relationship, the formula can be rewritten as

(7)E((Xi0+Xi1)P(yi1,yi2)|Yi1,Yi2)=E(Yi1P(yi1,yi2)|Yi1,Yi2)=E(ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21)|Yi1,Yi2)=ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21)

Comparing (6) and (7), we know

P(yi1,yi2)E(Xi0|Yi1,Yi2)+P(yi1,yi2)E(Xi1|Yi1,Yi2)=ti1θ1P(yi11,yi2)+ti0θ0P(yi11,yi21)
E(Xi1|Yi1,Yi2)+E(Xi0|Yi1,Yi2)=ti1θ1P(yi11,yi2)P(yi1,yi2)+ti0θ0P(yi11,yi21)P(yi1,yi2).

Similarly,

E(Xi2|Yi1,Yi2)+E(Xi0|Yi1,Yi2)=ti2θ2P(yi1,yi21)P(yi1,yi2)+ti0θ0P(yi11,yi21)P(yi1,yi2).

Then, we can get the results that

E(Xi0|Yi1,Yi2)=ti0θ0P(yi11,yi21)P(yi1,yi2),E(Xi1|Yi1,Yi2)=ti1θ1P(yi11,yi2)P(yi1,yi2),E(Xi2|Yi1,Yi2)=ti2θ2P(yi1,yi21)P(yi1,yi2).

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Supplemental Material:

The online version of this article (DOI: 10.1515/sagmb-2016-0040) offers supplementary material, available to authorized users.


Published Online: 2017-3-1
Published in Print: 2017-3-1

©2017 Walter de Gruyter GmbH, Berlin/Boston

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