Percentile* & 0.3% & 5% & 25% & 50% & 75% & 95% & 99.7%\
Longitudinal Training& 0.1694 & 0.4802 & 1.0765 & 1.7649 & 2.7630 & 4.8060 & 5.6826\
Cross-sectional Training& **0.1123** & **0.3024** & **0.5863** & **0.8737** & **1.2743** & **2.2659** & **2.7836**\

SGR
---

Determining the initial image, $I_0$, and the initial momentum, $m_0$, of Eq.  is computationally costly. However, in simple geodesic regression, the initial image is fixed to the *first* image of a subject’s longitudinal image set (left-most part of Fig. \[fig:1\]). Furthermore, the similarity measure $d(\cdot,\cdot)$ is chosen as the geodesic distance between images and [*approximated*]{} so that the geodesic regression problem can be solved by computing pair-wise image registrations with respect to the first image. Specifically, we define the quadratic distance $d^2$ between two images $A$ and $B$ as $$\begin{aligned}
& d^2(A, B) = \frac{1}{2} \int_0^1\lVert v^*\rVert^2_L dt, \\
\notag
& \text{where} \, v^* = \operatorname*{arg\,min}_v \frac{1}{2} \int_0^1 \lVert v\rVert^2_L dt + \frac{1}{\sigma^2} \lVert Q(1) - B\rVert_2^2, \\
\notag
& \text{s.t.} \quad Q_t + \nabla Q^Tv = 0, \text{and}~Q(0) = A\enspace. 
\label{eq:geo_dist}\end{aligned}$$ Assume we have an image $I(t_0)$ at time $t_0$ as well as two images $A(t_i)$ and $B(t_i)$. Further, assume that the 