Group Testing for Longitudinal Data

Abstract

We consider how to test for group differences of shapes given longitudinal data. In particular, we are interested in differences of longitudinal models of each group’s subjects. We introduce a generalization of principal geodesic analysis to the tangent bundle of a shape space. This allows the estimation of the variance and principal directions of the distribution of trajectories that summarize shape variations within the longitudinal data. Each trajectory is parameterized as a point in the tangent bundle. To study statistical differences in two distributions of trajectories, we generalize the Bhattacharyya distance in Euclidean space to the tangent bundle. This not only allows to take second-order statistics into account, but also serves as our test-statistic during permutation testing. Our method is validated on both synthetic and real data, and the experimental results indicate improved statistical power in identifying group differences. In fact, our study sheds new light on group differences in longitudinal corpus callosum shapes of subjects with dementia versus normal controls.

Appendices

Appendix

A Properties of the Generalized Bhattacharyya Distance

Non-negativity. In the first term of Eq. (4), $D_M^{\mathcal {TM}}$ is the generalized squared-Mahalanobis distance which is non-negative; consequently, the first term in Eq. (4) is non-negative. Furthermore, the determinant of a covariance matrix in the second term is also non-negative, since it is the product of all non-negative eigenvalues. Besides, it is easy to demonstrate that $(|\varSigma _1|+|\varSigma _2|) / (2\sqrt{|\varSigma _1||\varSigma _2|}) \ge 1$, indicating the second term is non-negative. Hence, $D_B^{\mathcal {TM}}(D_1, D_2) \ge 0$.

Identity of Indiscernibles. If $D_1 = D_2$, i.e., $\mu _1 = \mu _2$ and $\varSigma _1 = \varSigma _2$, we see that (1) ${{\mathrm{Log}}}_{\mu _1} \mu _2$ and ${{\mathrm{Log}}}_{\mu _2} \mu _1$ are zero tangent vectors, and (2) $|\varSigma _1| = |\varSigma _2|$. Hence, $D_M^{\mathcal {TM}}(\mu _1, D_2) = D_M^{\mathcal {TM}}(\mu _2, D_1) = 0$, i.e., the first term of Eq. (4) is 0; also, the second term is 0. Now, if $D_1 = D_2$ then $D_B^{\mathcal {TM}}(D_1, D_2) = 0$. On the other hand, assuming $D_B^{\mathcal {TM}}(D_1, D_2) = 0$, we can only obtain $\mu _1 = \mu _2$ and $|\varSigma _1| = |\varSigma _2|$, because of the non-negativity properties of the two terms in Eq. (4). But, we cannot draw the conclusion that the two covariance matrices are equal. Therefore, if $D_1=D_2$ then $D_B^{\mathcal {TM}}(D_1, D_2) = 0$, but it is possible that $D_B^{\mathcal {TM}}(D_1, D_2) =0$ for some $D_1 \not = D_2$, if $\mu _1 = \mu _2$ and $|\varSigma _1| = |\varSigma _2|$.

Symmetry. Because both terms of Eq. (4) are symmetric, the sum of them is also symmetric, i.e., $D_B^{\mathcal {TM}}(D_1, D_2) = D_B^{\mathcal {TM}}(D_2, D_1)$.

Triangle Inequality. Since, Eq. (1) in $\mathbb {R}^n$ does not satisfy the triangle inequality, our generalized variant will not satisfy it either.