The ICA BookICA 学习书 Yuda Bi

Part VI · Chapter 15第六部分 · 第 15 章

Group ICA, Dual Regression and IVA组 ICA、双回归与独立向量分析

Multi-subject ICA: temporal concatenation, back-reconstruction, dual regression, spatially constrained ICA, independent vector analysis, network connectivity and multimodal fusion.

多被试 ICA:时间拼接、反重建、双回归、空间约束 ICA、独立向量分析、网络连接以及多模态融合。

15.1 From one subject to many从单个被试到多个被试

Most fMRI studies compare networks between people: patients and controls, young and old, before and after treatment. ICA of each subject separately returns components in arbitrary order, sign and scale (Section 1.3第 1.3 节), and sometimes with different splits of the same network, so the components must be matched across subjects before anything can be compared. Matching by spatial correlation or by clustering is possible but error-prone, especially at higher model orders. Group methods build the correspondence into the model instead. They differ in what they assume to be shared: the maps (temporal concatenation, Section 15.2第 15.2 节); the time courses (spatial concatenation, Svensén, Kruggel & Benali, 2002, which suits experiments in which all subjects follow the same stimulus timing, but not rest); both maps and time courses up to subject-specific loadings (tensor PICA; Beckmann & Smith, 2005); or neither, with corresponding components linked only by statistical dependence (independent vector analysis, Section 15.6第 15.6 节). Calhoun, Liu and Adalı (2009) review these approaches. This chapter develops the ones most used in practice, then turns to what is computed from the components: connectivity between networks and links between imaging modalities.

大多数 fMRI 研究要在人与人之间比较网络:患者与对照、年轻人与老年人、治疗前与治疗后。对每个被试单独做 ICA,得到的成分顺序、符号与尺度都是任意的(Section 1.3第 1.3 节),同一个网络有时还会以不同的方式被拆分,因此在做任何比较之前,必须先把成分在被试之间对应起来。按空间相关或聚类来匹配是可行的,但容易出错,模型阶数较高时尤其如此。组方法则把对应关系直接写进模型。它们的区别在于假设哪些东西是共享的:共享空间图(时间拼接,Section 15.2第 15.2 节);共享时间过程(空间拼接,Svensén, Kruggel & Benali, 2002,适合所有被试遵循相同刺激时序的实验,不适合静息态);空间图和时间过程都共享、只差被试特有的载荷(张量 PICA;Beckmann & Smith, 2005);或者都不共享,相互对应的成分只靠统计依赖联系起来(独立向量分析,Section 15.6第 15.6 节)。Calhoun、Liu 与 Adalı(2009)综述了这些方法。本章先讨论实践中最常用的几种,再转向由成分计算的量:网络之间的连接,以及不同成像模态之间的联系。

15.2 Temporal concatenation时间拼接

Let Xi∈RT×V\bX_i \in \R^{T\times V}, i=1,…,Mi = 1, \dots, M, be the preprocessed data of MM subjects, registered to a common template and centered over voxels. Group ICA by temporal concatenation (Calhoun, Adalı, Pearlson & Pekar, 2001) assumes that the subjects share the same maps but have their own time courses, Xi=AiS\bX_i = \bA_i\bS. Stacking the subjects in time gives an ordinary spatial ICA model ((14.1)) with MTMT "sensors":

设 Xi∈RT×V\bX_i \in \R^{T\times V}(i=1,…,Mi = 1, \dots, M)是 MM 个被试经过预处理、配准到同一模板并在体素方向中心化后的数据。时间拼接的组 ICA(Calhoun, Adalı, Pearlson & Pekar, 2001)假设各被试共享相同的空间图,但各有自己的时间过程,即 Xi=AiS\bX_i = \bA_i\bS。把各被试在时间方向上堆叠起来,就得到一个普通的空间 ICA 模型((14.1)),其“传感器”个数为 MTMT:

(X1⋮XM)=(A1⋮AM)S,(X1⋮XM)∈RMT×V,S∈Rn×V.(15.1)\begin{pmatrix}\bX_1\\ \vdots\\ \bX_M\end{pmatrix} = \begin{pmatrix}\bA_1\\ \vdots\\ \bA_M\end{pmatrix}\bS, \qquad \begin{pmatrix}\bX_1\\ \vdots\\ \bX_M\end{pmatrix} \in \R^{MT\times V}, \quad \bS \in \R^{n\times V}.\tag{15.1}

The stacked matrix is large, and its PCA reduction is computed in two stages. Each subject's data are first reduced to qq dimensions, Yi=FiXi∈Rq×V\bY_i = \bF_i\bX_i \in \R^{q\times V}, where the rows of Fi∈Rq×T\bF_i \in \R^{q\times T} are the qq leading eigenvectors of 1VXiXi⊤\frac1V\bX_i\bX_i^\top. The reduced data are stacked into Y∈RMq×V\bY \in \R^{Mq\times V} and reduced again to nn dimensions, where the rows of G∈Rn×Mq\bG \in \R^{n\times Mq} are the nn leading eigenvectors of 1VYY⊤\frac1V\bY\bY^\top. Partition G=[G1,…,GM]\bG = [\bG_1, \dots, \bG_M] into blocks Gi∈Rn×q\bG_i \in \R^{n\times q}, so that GY=∑iGiYi\bG\bY = \sum_i \bG_i\bY_i. Spatial ICA of the n×Vn\times V matrix GY\bG\bY ((14.2)) gives the group maps S^\hat\bS and an invertible n×nn\times n mixing matrix B\bB with

堆叠后的矩阵很大,它的 PCA 降维分两步进行。先把每个被试的数据降到 qq 维:Yi=FiXi∈Rq×V\bY_i = \bF_i\bX_i \in \R^{q\times V},其中 Fi∈Rq×T\bF_i \in \R^{q\times T} 的各行是 1VXiXi⊤\frac1V\bX_i\bX_i^\top 的前 qq 个特征向量。再把降维后的数据堆叠成 Y∈RMq×V\bY \in \R^{Mq\times V},并降到 nn 维,其中 G∈Rn×Mq\bG \in \R^{n\times Mq} 的各行是 1VYY⊤\frac1V\bY\bY^\top 的前 nn 个特征向量。把 G\bG 分块为 G=[G1,…,GM]\bG = [\bG_1, \dots, \bG_M],Gi∈Rn×q\bG_i \in \R^{n\times q},于是 GY=∑iGiYi\bG\bY = \sum_i \bG_i\bY_i。对 n×Vn\times V 矩阵 GY\bG\bY 做空间 ICA((14.2)),得到组空间图 S^\hat\bS 与可逆的 n×nn\times n 混合矩阵 B\bB,满足

GY=BS^.(15.2)\bG\bY = \bB\hat\bS.\tag{15.2}

The subject-level step removes noise specific to each subject and keeps memory requirements manageable; the group-level step keeps the dimensions that the subjects share. Here the reducing matrices have orthonormal rows, FiFi⊤=I\bF_i\bF_i^\top = \bI and GG⊤=I\bG\bG^\top = \bI, as in the derivation of Erhardt et al. (2011). Implementations such as GIFT also whiten at each stage, dividing the rows of Fi\bF_i and of G\bG by the square roots of the corresponding eigenvalues. At the group level this only moves a diagonal factor into B\bB; at the subject level it changes how the back-reconstructed maps relate to the time courses (Remark 15.3注 15.3). With T=100T = 100, q=5q = 5, M=6M = 6 and n=3n = 3, as in the example of Section 15.4第 15.4 节, Y\bY is 30×V30\times V and GY\bG\bY is 3×V3\times V.

被试水平的降维去除每个被试特有的噪声,并使内存需求保持在可控范围内;组水平的降维保留各被试共有的维度。这里降维矩阵的行是单位正交的:FiFi⊤=I\bF_i\bF_i^\top = \bI,GG⊤=I\bG\bG^\top = \bI,与 Erhardt 等(2011)的推导一致。GIFT 等软件还在每一步做白化,即把 Fi\bF_i 与 G\bG 的各行除以相应特征值的平方根。组水平的白化只是把一个对角因子并入 B\bB;被试水平的白化则会改变反重建得到的空间图与时间过程之间的关系(Remark 15.3注 15.3)。取 T=100T = 100,q=5q = 5,M=6M = 6,n=3n = 3(与Section 15.4第 15.4 节的例子相同),则 Y\bY 为 30×V30\times V,GY\bG\bY 为 3×V3\times V。

15.3 Back-reconstruction反重建

Group maps are only half the answer: comparisons between subjects need each subject's own time courses and maps. Back-reconstruction obtains them by running the reductions backwards. If the group PCA loses little, Y≈G⊤GY=G⊤BS^\bY \approx \bG^\top\bG\bY = \bG^\top\bB\hat\bS, and the rows that belong to subject ii give Yi≈Gi⊤BS^\bY_i \approx \bG_i^\top\bB\hat\bS; if the subject PCA loses little, Xi≈Fi⊤Yi\bX_i \approx \bF_i^\top\bY_i. This suggests the subject time courses (T×nT\times n) and maps (n×Vn\times V)

组空间图只是答案的一半:被试之间的比较需要每个被试自己的时间过程与空间图。反重建通过把降维步骤倒推回去来得到它们。若组水平 PCA 损失很少,则 Y≈G⊤GY=G⊤BS^\bY \approx \bG^\top\bG\bY = \bG^\top\bB\hat\bS,取出属于被试 ii 的那些行,得到 Yi≈Gi⊤BS^\bY_i \approx \bG_i^\top\bB\hat\bS;若被试水平 PCA 损失很少,则 Xi≈Fi⊤Yi\bX_i \approx \bF_i^\top\bY_i。由此得到被试的时间过程(T×nT\times n)与空间图(n×Vn\times V):

Aˇi=Fi⊤Gi⊤B,Sˇi=(Gi⊤B)+Yi.(15.3)\check\bA_i = \bF_i^\top\bG_i^\top\bB, \qquad \check\bS_i = \bigl(\bG_i^\top\bB\bigr)^+\bY_i.\tag{15.3}

This is the back-reconstruction of Calhoun et al. (2001), written in the present notation; Erhardt et al. (2011) call it GICA1. Because Fi⊤\bF_i^\top has orthonormal columns, Aˇi+=(Gi⊤B)+Fi\check\bA_i^+ = (\bG_i^\top\bB)^+\bF_i, so Sˇi=Aˇi+Xi\check\bS_i = \check\bA_i^+\bX_i: the subject maps are the least-squares regression of the subject's data on the subject's time courses. This identity relies on the orthonormal rows of Fi\bF_i; Remark 15.3注 15.3 describes what changes when the subject-level reduction is whitened. Erhardt et al. (2011) compared several back-reconstruction schemes on simulated and task data. One of them, GICA3, which they introduced, produces subject maps that add up to the group maps, with subject time courses and maps that reproduce each subject's data about as accurately as the PCA reduction allows; they recommended GICA3 together with subject-specific PCA, and GICA3 is the back-reconstruction recommended in the GIFT toolbox. They also observed that GICA1 approximates dual regression, the method of the next section. Proposition 15.2命题 15.2 makes this exact.

这就是 Calhoun 等(2001)的反重建方法,这里用本章的记号写出;Erhardt 等(2011)称之为 GICA1。由于 Fi⊤\bF_i^\top 的列单位正交,Aˇi+=(Gi⊤B)+Fi\check\bA_i^+ = (\bG_i^\top\bB)^+\bF_i,所以 Sˇi=Aˇi+Xi\check\bS_i = \check\bA_i^+\bX_i:被试的空间图就是该被试数据对其时间过程的最小二乘回归。这一恒等式依赖于 Fi\bF_i 的行单位正交;被试水平的降维做白化时有何变化,见Remark 15.3注 15.3。Erhardt 等(2011)在仿真数据和任务数据上比较了几种反重建方案。其中由他们提出的 GICA3 给出的被试空间图之和等于组空间图,而且被试的时间过程与空间图重现被试数据的精度,大致达到 PCA 降维所允许的程度;他们推荐将 GICA3 与被试特异的 PCA 配合使用,GIFT 工具箱也把 GICA3 作为推荐的反重建方法。他们还指出 GICA1 近似于下一节的双回归。Proposition 15.2命题 15.2把这一近似关系精确化。

15.4 Dual regression双回归

Dual regression (Beckmann et al., 2009; Filippini et al., 2009), called spatio-temporal regression (STR) by Erhardt et al. (2011), skips the PCA bookkeeping and uses only a set of group maps S^\hat\bS, from group ICA or from elsewhere, together with the subject's own data. Stage 1 is a spatial regression: each image of subject ii is regressed on the group maps, which gives the subject's time courses. Stage 2 is a temporal regression: each voxel's time series is regressed on these time courses, which gives the subject's maps. Stage 1 solves min⁡A∥Xi−AS^∥F\min_{\bA}\|\bX_i - \bA\hat\bS\|_F and stage 2 solves min⁡S∥Xi−A^iS∥F\min_{\bS}\|\bX_i - \hat\bA_i\bS\|_F; in closed form,

双回归(Beckmann et al., 2009; Filippini et al., 2009),Erhardt 等(2011)称之为时空回归(spatio-temporal regression, STR),它不需要 PCA 的记账,只用一组组空间图 S^\hat\bS(来自组 ICA 或其他途径)和被试自己的数据。第一步是空间回归:把被试 ii 的每一幅图像对组空间图回归,得到该被试的时间过程。第二步是时间回归:把每个体素的时间序列对这些时间过程回归,得到该被试的空间图。第一步求解 min⁡A∥Xi−AS^∥F\min_{\bA}\|\bX_i - \bA\hat\bS\|_F,第二步求解 min⁡S∥Xi−A^iS∥F\min_{\bS}\|\bX_i - \hat\bA_i\bS\|_F,其闭式解为

A^i=XiS^⊤(S^S^⊤)−1=XiS^+,S^i=(A^i⊤A^i)−1A^i⊤Xi=A^i+Xi.(15.4)\hat\bA_i = \bX_i\hat\bS^\top\bigl(\hat\bS\hat\bS^\top\bigr)^{-1} = \bX_i\hat\bS^+, \qquad \hat\bS_i = \bigl(\hat\bA_i^\top\hat\bA_i\bigr)^{-1}\hat\bA_i^\top\bX_i = \hat\bA_i^+\bX_i.\tag{15.4}

In the FSL implementation the time courses can be variance-normalized before stage 2. Replacing A^i\hat\bA_i by A^iDi−1\hat\bA_i\bD_i^{-1}, with Di\bD_i the diagonal matrix of their standard deviations, multiplies the rows of S^i\hat\bS_i by these standard deviations (Exercise 15.2习题 15.2), so the maps then reflect both the shape and the amplitude of each network (Nickerson et al., 2017). The subject maps are finally compared across subjects voxel by voxel, for instance with permutation tests. The closed forms show what the subject maps contain.

在 FSL 的实现中,可以在第二步之前把时间过程方差归一化。把 A^i\hat\bA_i 换成 A^iDi−1\hat\bA_i\bD_i^{-1}(Di\bD_i 是由各时间过程标准差构成的对角矩阵),相当于把 S^i\hat\bS_i 的各行乘以这些标准差(Exercise 15.2习题 15.2),于是空间图同时反映每个网络的形状和幅度(Nickerson et al., 2017)。最后在体素水平上比较各被试的空间图,例如用置换检验。闭式解揭示了被试空间图里到底包含什么。

Proposition 15.1 (dual regression in closed form)命题 15.1(双回归的闭式表示)

Let S^\hat\bS have full row rank and A^i\hat\bA_i full column rank, and let PS^=S^+S^\bP_{\hat\bS} = \hat\bS^+\hat\bS be the V×VV\times V orthogonal projector onto the row space of S^\hat\bS. Then the stage-2 maps of (15.4) are

设 S^\hat\bS 行满秩,A^i\hat\bA_i 列满秩,PS^=S^+S^\bP_{\hat\bS} = \hat\bS^+\hat\bS 是到 S^\hat\bS 行空间上的 V×VV\times V 正交投影矩阵。则 (15.4) 第二步得到的空间图为

S^i=S^+A^i+Xi (I−PS^).\hat\bS_i = \hat\bS + \hat\bA_i^+\bX_i\,(\bI - \bP_{\hat\bS}).

Proof.证明

By stage 1, XiPS^=XiS^+S^=A^iS^\bX_i\bP_{\hat\bS} = \bX_i\hat\bS^+\hat\bS = \hat\bA_i\hat\bS, so A^i+XiPS^=S^\hat\bA_i^+\bX_i\bP_{\hat\bS} = \hat\bS, because A^i+A^i=I\hat\bA_i^+\hat\bA_i = \bI. Split Xi=XiPS^+Xi(I−PS^)\bX_i = \bX_i\bP_{\hat\bS} + \bX_i(\bI - \bP_{\hat\bS}) in S^i=A^i+Xi\hat\bS_i = \hat\bA_i^+\bX_i.

由第一步,XiPS^=XiS^+S^=A^iS^\bX_i\bP_{\hat\bS} = \bX_i\hat\bS^+\hat\bS = \hat\bA_i\hat\bS;又 A^i+A^i=I\hat\bA_i^+\hat\bA_i = \bI,所以 A^i+XiPS^=S^\hat\bA_i^+\bX_i\bP_{\hat\bS} = \hat\bS。在 S^i=A^i+Xi\hat\bS_i = \hat\bA_i^+\bX_i 中把 Xi\bX_i 拆成 XiPS^+Xi(I−PS^)\bX_i\bP_{\hat\bS} + \bX_i(\bI - \bP_{\hat\bS}) 即得。

Without variance normalization, then, every subject map equals the group map plus a correction built only from the part of the subject's data that lies outside the span of the group maps, regressed on time courses that were themselves estimated with the group maps. The two stages rest on different assumptions, common maps in stage 1 and subject-specific maps in stage 2, a point stressed by Erhardt et al. (2011). For group maps from spatial ICA, S^S^⊤=VI\hat\bS\hat\bS^\top = V\bI (Exercise 14.5习题 14.5), so stage 1 reduces to A^i=1VXiS^⊤\hat\bA_i = \frac1V\bX_i\hat\bS^\top and each time course is computed independently of the other maps; stage 2 is different, since the time courses are correlated and every map depends on all of them. The relation to back-reconstruction is the following.

可见,不做方差归一化时,每个被试的空间图都等于组空间图加上一个修正项;修正项只由被试数据中落在组空间图张成空间之外的部分构成,并且是对用组空间图估计出的时间过程回归得到的。两步回归依据的假设不同:第一步假设空间图是共同的,第二步假设空间图因被试而异,Erhardt 等(2011)强调过这一点。若组空间图来自空间 ICA,则 S^S^⊤=VI\hat\bS\hat\bS^\top = V\bI(Exercise 14.5习题 14.5),第一步化为 A^i=1VXiS^⊤\hat\bA_i = \frac1V\bX_i\hat\bS^\top,每条时间过程的计算与其他空间图无关;第二步则不同,因为各时间过程相关,每张空间图都依赖于全部时间过程。与反重建的关系如下。

Proposition 15.2 (GICA1 and dual regression)命题 15.2(GICA1 与双回归)

In the setting of Section 15.2第 15.2 节, with the orthonormal PCA reducing matrices Fi\bF_i and G\bG of that section and group maps S^\hat\bS of full row rank, the dual-regression time courses (15.4) and the GICA1 time courses (15.3) satisfy

在Section 15.2第 15.2 节的设定下,降维矩阵 Fi\bF_i 与 G\bG 是该节中行单位正交的 PCA 矩阵,组空间图 S^\hat\bS 行满秩,则双回归的时间过程 (15.4) 与 GICA1 的时间过程 (15.3) 满足

A^i=Aˇi+(I−Fi⊤Fi) XiS^+.\hat\bA_i = \check\bA_i + (\bI - \bF_i^\top\bF_i)\,\bX_i\hat\bS^+.

In particular, if the subject-level reduction keeps all of subject ii's data, Fi⊤FiXi=Xi\bF_i^\top\bF_i\bX_i = \bX_i, the two methods give the same time courses and the same maps.

特别地,若被试水平的降维保留了被试 ii 的全部数据,即 Fi⊤FiXi=Xi\bF_i^\top\bF_i\bX_i = \bX_i,则两种方法给出相同的时间过程和相同的空间图。

Proof.证明

Write Y=G⊤GY+Δ\bY = \bG^\top\bG\bY + \boldsymbol\Delta with Δ=(I−G⊤G)Y\boldsymbol\Delta = (\bI - \bG^\top\bG)\bY. The rows of G\bG are eigenvectors of YY⊤\bY\bY^\top, say YY⊤G⊤=G⊤Λ\bY\bY^\top\bG^\top = \bG^\top\bLambda, so ΔY⊤G⊤=(I−G⊤G)G⊤Λ=0\boldsymbol\Delta\bY^\top\bG^\top = (\bI - \bG^\top\bG)\bG^\top\bLambda = \bzero. Since S^=B−1GY\hat\bS = \bB^{-1}\bG\bY by (15.2), this gives ΔS^⊤=0\boldsymbol\Delta\hat\bS^\top = \bzero and hence ΔS^+=0\boldsymbol\Delta\hat\bS^+ = \bzero. Taking the rows of subject ii, Yi=Gi⊤BS^+Δi\bY_i = \bG_i^\top\bB\hat\bS + \boldsymbol\Delta_i, so YiS^+=Gi⊤B\bY_i\hat\bS^+ = \bG_i^\top\bB. Finally Xi=Fi⊤Yi+(I−Fi⊤Fi)Xi\bX_i = \bF_i^\top\bY_i + (\bI - \bF_i^\top\bF_i)\bX_i; multiplying on the right by S^+\hat\bS^+ and using Fi⊤Gi⊤B=Aˇi\bF_i^\top\bG_i^\top\bB = \check\bA_i gives the stated relation. When the last term vanishes, the time courses agree, and so do the maps, since both methods obtain them by regressing Xi\bX_i on the time courses.

写 Y=G⊤GY+Δ\bY = \bG^\top\bG\bY + \boldsymbol\Delta,其中 Δ=(I−G⊤G)Y\boldsymbol\Delta = (\bI - \bG^\top\bG)\bY。G\bG 的各行是 YY⊤\bY\bY^\top 的特征向量,设 YY⊤G⊤=G⊤Λ\bY\bY^\top\bG^\top = \bG^\top\bLambda,则 ΔY⊤G⊤=(I−G⊤G)G⊤Λ=0\boldsymbol\Delta\bY^\top\bG^\top = (\bI - \bG^\top\bG)\bG^\top\bLambda = \bzero。由 (15.2),S^=B−1GY\hat\bS = \bB^{-1}\bG\bY,因此 ΔS^⊤=0\boldsymbol\Delta\hat\bS^\top = \bzero,进而 ΔS^+=0\boldsymbol\Delta\hat\bS^+ = \bzero。取出属于被试 ii 的各行,Yi=Gi⊤BS^+Δi\bY_i = \bG_i^\top\bB\hat\bS + \boldsymbol\Delta_i,所以 YiS^+=Gi⊤B\bY_i\hat\bS^+ = \bG_i^\top\bB。最后,Xi=Fi⊤Yi+(I−Fi⊤Fi)Xi\bX_i = \bF_i^\top\bY_i + (\bI - \bF_i^\top\bF_i)\bX_i;右乘 S^+\hat\bS^+,并利用 Fi⊤Gi⊤B=Aˇi\bF_i^\top\bG_i^\top\bB = \check\bA_i,即得所述关系。当最后一项为零时,两者的时间过程相同;空间图也相同,因为两种方法都是把 Xi\bX_i 对时间过程回归得到空间图的。

The difference is the part of the subject's data that its own PCA discarded, regressed on the group maps: dual regression uses all of the subject's data, GICA1 only what survived the subject-level reduction. The group-level reduction causes no difference, because what it discards is orthogonal, across voxels, to the group maps. The following script runs the whole pipeline on six synthetic subjects whose three maps are the same blobs shifted at random by up to four voxels, with subject-specific random-walk time courses.

两者之差是被试自身 PCA 舍弃的那部分数据对组空间图的回归:双回归使用了被试的全部数据,GICA1 只用了经被试水平降维后保留下来的部分。组水平的降维不造成差别,因为它舍弃的部分在体素方向上与组空间图正交。下面的脚本在六个合成被试上运行整个流程:每个被试的三张空间图是同样的斑块,只是随机平移了至多四个体素,时间过程是各被试特有的随机游走。

import numpy as np

def fastica(Z, rng, iters=200):                       # symmetric FastICA, g = tanh
    U = np.linalg.qr(rng.standard_normal((len(Z), len(Z))))[0]
    for _ in range(iters):
        Y = U @ Z
        U = np.tanh(Y) @ Z.T / Z.shape[1] - np.diag((1 - np.tanh(Y)**2).mean(1)) @ U
        u, _, vt = np.linalg.svd(U); U = u @ vt
    return U

def pca_rows(X, k):                                   # k leading eigenvectors, as rows
    d, E = np.linalg.eigh(X @ X.T / X.shape[1])
    return E[:, ::-1][:, :k].T, d[::-1][:k]

rng = np.random.default_rng(0)
side, T, M, q, n = 30, 100, 6, 5, 3                   # grid, scans, subjects, PCA dims
yy, xx = np.mgrid[0:side, 0:side]
blob = lambda cy, cx, r: np.exp(-((yy - cy)**2 + (xx - cx)**2) / (2 * r**2)).ravel()
Xs, Ss = [], []
for i in range(M):                                    # blobs shifted at random per subject
    S = np.vstack([blob(cy + rng.integers(-4, 5), cx + rng.integers(-4, 5), r)
                   for cy, cx, r in [(8, 8, 3), (20, 22, 4), (8, 22, 3)]])
    A = np.cumsum(rng.standard_normal((T, n)), 0)    # subject-specific time courses
    X = (A - A.mean(0)) / A.std(0) @ S + 0.3 * rng.standard_normal((T, side**2))
    Xs.append(X - X.mean(1, keepdims=True)); Ss.append(S)

Fs = [pca_rows(X, q)[0] for X in Xs]                  # subject PCA: q x T each
Y = np.vstack([F @ X for F, X in zip(Fs, Xs)])        # temporal concatenation: Mq x V
G, lam = pca_rows(Y, n)                               # group PCA: n x Mq
Z = np.diag(lam ** -0.5) @ G @ Y
U = fastica(Z, rng)
S_g, B = U @ Z, np.diag(lam ** 0.5) @ U.T             # group maps; G Y = B S_g

def match(S_est, S_true):                             # mean |corr| of best-matching rows
    return np.abs(np.corrcoef(S_est, S_true)[:n, n:]).max(0).mean()
r0, r1, r2 = [], [], []
for X, F, Gi, S in zip(Xs, Fs, np.split(G, M, axis=1), Ss):
    S_1 = np.linalg.pinv(Gi.T @ B) @ F @ X            # GICA1 back-reconstruction
    A_dr = X @ np.linalg.pinv(S_g)                    # dual regression, stage 1
    S_dr = np.linalg.pinv(A_dr) @ X                   # dual regression, stage 2
    r0.append(match(S_g, S)); r1.append(match(S_1, S)); r2.append(match(S_dr, S))
print(Y.shape, S_g.shape)
print(np.round([np.mean(r0), np.mean(r1), np.mean(r2)], 3))

The first line, (30, 900) (3, 900), gives the sizes of the concatenated data Y\bY (Mq×VMq\times V) and of the group maps. The second gives mean correlations between estimated and true subject maps: 0.7730.773 if the group maps themselves are used for every subject, 0.9280.928 for GICA1 and 0.9300.930 for dual regression. Both recover much of the spatial variability between subjects, and they nearly agree, as Proposition 15.2命题 15.2 leads one to expect when the subject PCA keeps most of the signal.

第一行 (30, 900) (3, 900) 给出拼接后数据 Y\bY(Mq×VMq\times V)与组空间图的尺寸。第二行是估计的被试空间图与真实被试空间图之间的平均相关系数:直接把组空间图用于每个被试时为 0.7730.773,GICA1 为 0.9280.928,双回归为 0.9300.930。两种方法都恢复了被试之间空间变异的大部分,而且结果几乎一致;当被试水平的 PCA 保留了大部分信号时,这正是Proposition 15.2命题 15.2所预期的。

Remark 15.3 (whitening at the subject level)注 15.3(被试水平的白化)

Implementations such as GIFT whiten the subject-level reduction, so that the group stage works on Yi=Λi−1/2FiXi\bY_i = \bLambda_i^{-1/2}\bF_i\bX_i, where Λi\bLambda_i is the diagonal matrix of the eigenvalues that belong to the rows of Fi\bF_i; back-reconstruction undoes the whitening with Fi⊤Λi1/2\bF_i^\top\bLambda_i^{1/2}. The GICA1 time courses become Aˇi=Fi⊤Λi1/2Gi⊤B\check\bA_i = \bF_i^\top\bLambda_i^{1/2}\bG_i^\top\bB, and the maps of (15.3) become

GIFT 等软件对被试水平的降维做白化,组水平的步骤处理的是 Yi=Λi−1/2FiXi\bY_i = \bLambda_i^{-1/2}\bF_i\bX_i,其中 Λi\bLambda_i 是由与 Fi\bF_i 各行对应的特征值构成的对角矩阵;反重建时用 Fi⊤Λi1/2\bF_i^\top\bLambda_i^{1/2} 撤销白化。于是 GICA1 的时间过程变为 Aˇi=Fi⊤Λi1/2Gi⊤B\check\bA_i = \bF_i^\top\bLambda_i^{1/2}\bG_i^\top\bB,而 (15.3) 中的空间图变为

Sˇi=(Gi⊤B)+Λi−1/2FiXi.\check\bS_i = \bigl(\bG_i^\top\bB\bigr)^+\bLambda_i^{-1/2}\bF_i\bX_i.

The proof of Proposition 15.2命题 15.2 carries over, with Fi⊤Λi1/2Yi\bF_i^\top\bLambda_i^{1/2}\bY_i in place of Fi⊤Yi\bF_i^\top\bY_i in its last step, so the relation between the time courses holds as stated, Fi⊤Fi\bF_i^\top\bF_i being the orthogonal projector onto the subspace retained for subject ii. The maps behave differently. Sˇi\check\bS_i is a least-squares fit in the whitened coordinates of that subspace, whereas Aˇi+Xi\check\bA_i^+\bX_i is the least-squares fit in the original coordinates; the two agree when q=nq = n and Gi⊤B\bG_i^\top\bB is invertible, but in general not when q>nq > n. The identity Sˇi=Aˇi+Xi\check\bS_i = \check\bA_i^+\bX_i of Section 15.3第 15.3 节 and the equality of the maps in Proposition 15.2命题 15.2 therefore require an unwhitened subject-level reduction. Rerun with this whitening, the example above still satisfies the time-course relation up to rounding error, but the GICA1 and dual-regression maps now differ, on average over subjects, by 6% in relative Frobenius norm for q=5q = 5 (4% without whitening) and by 38% for q=T=100q = T = 100, where nothing is discarded (none without whitening).

Proposition 15.2命题 15.2的证明仍然适用,只需在最后一步用 Fi⊤Λi1/2Yi\bF_i^\top\bLambda_i^{1/2}\bY_i 代替 Fi⊤Yi\bF_i^\top\bY_i;因此时间过程之间的关系照样成立,其中 Fi⊤Fi\bF_i^\top\bF_i 是到被试 ii 所保留子空间的正交投影。空间图则不同。Sˇi\check\bS_i 是在该子空间的白化坐标下的最小二乘拟合,而 Aˇi+Xi\check\bA_i^+\bX_i 是在原坐标下的最小二乘拟合;当 q=nq = n 且 Gi⊤B\bG_i^\top\bB 可逆时二者一致,但 q>nq > n 时一般不同。因此,Section 15.3第 15.3 节中的恒等式 Sˇi=Aˇi+Xi\check\bS_i = \check\bA_i^+\bX_i 以及Proposition 15.2命题 15.2中空间图相同的结论,都要求被试水平的降维不做白化。对上面的例子加上这种白化后重新运行,时间过程之间的关系在舍入误差范围内仍然成立,但 GICA1 与双回归的空间图按 Frobenius 范数计的相对差别(对被试取平均)在 q=5q = 5 时为 6%(不白化时为 4%),在什么都不舍弃的 q=T=100q = T = 100 时为 38%(不白化时为零)。

15.5 Spatially constrained ICA and NeuroMark空间约束 ICA 与 NeuroMark

Back-reconstruction and dual regression obtain subject components by regression and never ask them to be independent. Spatially constrained ICA instead runs ICA on each subject's data and uses reference maps to guide it. In the semi-blind spatial ICA of Lin et al. (2010), each component maximizes a non-Gaussianity contrast subject to a constraint that keeps it close to a spatial reference, such as a mask defined by an atlas. In group-information-guided ICA (GIG-ICA; Du & Fan, 2013), the references are the group maps, and each subject component is found by a multi-objective optimization that balances its independence against its similarity to the corresponding group component. Schematically, with z\bz the whitened data of subject ii at a random voxel and sˉl\bar s_l the value of reference map ll there, the unit vector wl\bw_l that extracts component ll solves

反重建与双回归都通过回归得到被试的成分,从不要求它们相互独立。空间约束 ICA 则对每个被试的数据做 ICA,并用参考空间图来引导它。在 Lin 等(2010)的半盲空间 ICA 中,每个成分在保持与某个空间参考(例如由图谱定义的掩模)足够接近的约束下,最大化一个非高斯性对比函数。在组信息引导 ICA(GIG-ICA;Du & Fan, 2013)中,参考就是组空间图,每个被试成分通过一个多目标优化求得,在独立性与它和对应组成分的相似性之间取得平衡。示意地说,记 z\bz 为被试 ii 在随机体素处的白化数据,sˉl\bar s_l 为参考图 ll 在该处的取值,则提取成分 ll 的单位向量 wl\bw_l 求解

max⁡∥wl∥=1 ( J(wl⊤z),  E[sˉl wl⊤z]),\max_{\|\bw_l\| = 1}\ \Bigl(\, J\bigl(\bw_l^\top\bz\bigr),\ \ \E\bigl[\bar s_l\,\bw_l^\top\bz\bigr] \Bigr),

two objectives combined, for example, by a weighted sum, where JJ is a negentropy approximation (Section 6.7第 6.7 节) and the reference is standardized. Because every subject is decomposed on its own, the order and meaning of the components are fixed by the references, and new subjects can be processed one at a time without rerunning a group analysis.

两个目标可以通过加权和等方式合并,其中 JJ 是负熵的近似(Section 6.7第 6.7 节),参考图已标准化。由于每个被试单独分解,成分的顺序与含义由参考图固定,新被试可以逐个处理,无需重新做组分析。

NeuroMark (Du et al., 2020) turns this into a pipeline for comparisons across studies. Group ICA of two large healthy-control data sets, from the Human Connectome Project and the Genomics Superstruct Project with 1,828 subjects in total, yielded 53 intrinsic networks that replicated across the two data sets; they are grouped into seven functional domains: subcortical, auditory, sensorimotor, visual, cognitive control, default mode and cerebellar. The networks of each new subject are estimated by spatially constrained ICA with these templates as references, which keeps components and connectivity measures comparable across data sets and disorders. Du et al. applied the pipeline to several brain disorders, including schizophrenia, autism, mild cognitive impairment and Alzheimer's disease, bipolar disorder and major depression.

NeuroMark(Du et al., 2020)把这一思路做成了可用于跨研究比较的流程。对两个大型健康对照数据集(来自人类连接组计划与基因组超结构计划,共 1,828 名被试)做组 ICA,得到 53 个在两个数据集之间可复现的内在网络,它们分属七个功能域:皮层下、听觉、感觉运动、视觉、认知控制、默认模式与小脑。每个新被试的网络都以这些模板为参考,用空间约束 ICA 估计,从而使成分与连接指标在不同数据集和不同疾病之间保持可比。Du 等把这一流程用于多种脑疾病,包括精神分裂症、自闭症、轻度认知障碍与阿尔茨海默病、双相障碍和重度抑郁。

15.6 Independent vector analysis独立向量分析

Temporal concatenation assumes common maps; dual regression and constrained ICA estimate subject maps around group references. Independent vector analysis (IVA; Kim, Eltoft & Lee, 2006) treats the subjects symmetrically and links them only through statistical dependence. There are KK data sets, here the subjects after their own PCA reduction, each with its own mixing matrix:

时间拼接假设空间图是共同的;双回归与约束 ICA 围绕组参考来估计被试的空间图。独立向量分析(independent vector analysis, IVA;Kim, Eltoft & Lee, 2006)则对称地对待各被试,只通过统计依赖把它们联系起来。设有 KK 个数据集(这里是经过各自 PCA 降维的被试),每个数据集有自己的混合矩阵:

x[k]=A[k]s[k],k=1,…,K,si=(si[1],…,si[K])⊤,i=1,…,n.(15.5)\bx^{[k]} = \bA^{[k]}\bs^{[k]}, \quad k = 1, \dots, K, \qquad \bs_i = \bigl(s_i^{[1]}, \dots, s_i^{[K]}\bigr)^\top, \quad i = 1, \dots, n.\tag{15.5}

The vector si\bs_i collects the ii-th source of every data set and is called the ii-th source component vector (SCV). The model assumes that the SCVs s1,…,sn\bs_1, \dots, \bs_n are mutually independent, while the components within an SCV may be, and are expected to be, dependent: the default-mode map of one subject resembles that of another. For spatial ICA of fMRI the samples are again voxels, so si[k]s_i^{[k]} is the value of map ii of subject kk at a random voxel.

向量 si\bs_i 汇集了各数据集的第 ii 个源,称为第 ii 个源成分向量(source component vector, SCV)。模型假设各 SCV s1,…,sn\bs_1, \dots, \bs_n 相互独立,而同一 SCV 内部的分量可以相互依赖,而且通常预期是相互依赖的:一个被试的默认模式图与另一个被试的相似。对 fMRI 的空间 ICA 而言,样本仍然是体素,所以 si[k]s_i^{[k]} 是被试 kk 的第 ii 张空间图在随机体素处的取值。

IVA estimates unmixing matrices W[1],…,W[K]\bW^{[1]}, \dots, \bW^{[K]}, with outputs y[k]=W[k]x[k]\by^{[k]} = \bW^{[k]}\bx^{[k]} and estimated SCVs yi=(yi[1],…,yi[K])⊤\by_i = (y_i^{[1]}, \dots, y_i^{[K]})^\top, by minimizing the mutual information among the estimated SCVs,

IVA 估计解混矩阵 W[1],…,W[K]\bW^{[1]}, \dots, \bW^{[K]},输出为 y[k]=W[k]x[k]\by^{[k]} = \bW^{[k]}\bx^{[k]},估计的 SCV 为 yi=(yi[1],…,yi[K])⊤\by_i = (y_i^{[1]}, \dots, y_i^{[K]})^\top,所用的准则是最小化各估计 SCV 之间的互信息

I(y1;… ;yn)=∑i=1nH(yi)−H(y[1],…,y[K]).I(\by_1; \dots; \by_n) = \sum_{i=1}^n H(\by_i) - H\bigl(\by^{[1]}, \dots, \by^{[K]}\bigr).

The transformation is block diagonal, which gives the first identity below; the second splits the entropy of an estimated SCV into its marginal entropies minus the mutual information I(yi)I(\by_i) among its components:

变换是分块对角的,由此得到下面第一个等式;第二个等式把一个估计 SCV 的熵分解为各边缘熵之和减去其分量之间的互信息 I(yi)I(\by_i):

H(y[1],…,y[K])=H(x[1],…,x[K])+∑k=1Klog⁡∣det⁡W[k]∣,H(yi)=∑k=1KH(yi[k])−I(yi).\begin{aligned} H\bigl(\by^{[1]}, \dots, \by^{[K]}\bigr) &= H\bigl(\bx^{[1]}, \dots, \bx^{[K]}\bigr) + \sum_{k=1}^K \log\bigl|\det\bW^{[k]}\bigr|, \\ H(\by_i) &= \sum_{k=1}^K H\bigl(y_i^{[k]}\bigr) - I(\by_i). \end{aligned}

Hence the cost is

因此代价函数为

IIVA=∑i=1n[∑k=1KH(yi[k])−I(yi)]−∑k=1Klog⁡∣det⁡W[k]∣−H(x[1],…,x[K])(15.6)\mathcal{I}_{\mathrm{IVA}} = \sum_{i=1}^n \Bigl[\sum_{k=1}^K H\bigl(y_i^{[k]}\bigr) - I(\by_i)\Bigr] - \sum_{k=1}^K \log\bigl|\det\bW^{[k]}\bigr| - H\bigl(\bx^{[1]}, \dots, \bx^{[K]}\bigr)\tag{15.6}

(Anderson, Adalı & Li, 2012; Adalı, Anderson & Fu, 2014). Without the terms I(yi)I(\by_i), the cost would be, up to a constant, the sum of KK separate ICA costs (6.4). These terms reward dependence within each estimated SCV, and they are what aligns the components across data sets. With two data sets separated correctly but in different orders, an estimated SCV would pair sources from two different SCVs, which are independent, so its I(yi)I(\by_i) would vanish; with a common order I(yi)=I(si)>0I(\by_i) = I(\bs_i) > 0, which lowers the cost (Exercise 15.3习题 15.3). IVA thus solves the correspondence problem as part of the estimation.

(Anderson, Adalı & Li, 2012; Adalı, Anderson & Fu, 2014)。若去掉 I(yi)I(\by_i) 项,代价函数就是(相差一个常数)KK 个各自独立的 ICA 代价 (6.4) 之和。这些项奖励每个估计 SCV 内部的依赖,正是它们把各数据集的成分对齐。若两个数据集都已正确分离但顺序不同,则估计的 SCV 会把两个不同 SCV 中的源配在一起,而它们是独立的,所以其 I(yi)I(\by_i) 为零;顺序一致时 I(yi)=I(si)>0I(\by_i) = I(\bs_i) > 0,代价更低(Exercise 15.3习题 15.3)。这样,IVA 在估计过程中顺带解决了对应问题。

Algorithms differ in the density they assume for the SCVs. The first IVA algorithms (Kim, Eltoft & Lee, 2006) used a super-Gaussian multivariate density of Laplacian type. IVA-G (Anderson, Adalı & Li, 2012) models each SCV as multivariate Gaussian and so uses only second-order statistics, the correlations of corresponding components across data sets. Even Gaussian sources can then be separated, provided their correlation structures across data sets differ. Two Gaussian SCVs whose covariance matrices satisfy Σm=DΣlD\bSigma_m = \bD\bSigma_l\bD for a diagonal D\bD cannot be separated (Exercise 15.5习题 15.5), and Anderson, Fu, Phlypo and Adalı (2014) showed that obstructions of this kind are essentially the only ones. The situation parallels SOBI (Theorem 9.6定理 9.6), with data sets in place of time lags.

各种算法的区别在于对 SCV 假设的密度。最早的 IVA 算法(Kim, Eltoft & Lee, 2006)采用拉普拉斯型的超高斯多元密度。IVA-G(Anderson, Adalı & Li, 2012)把每个 SCV 建模为多元高斯,因而只利用二阶统计量,即对应成分在各数据集之间的相关。这样一来,即使是高斯源也能分离,只要它们在各数据集之间的相关结构不同。若两个高斯 SCV 的协方差矩阵满足 Σm=DΣlD\bSigma_m = \bD\bSigma_l\bD(D\bD 为对角矩阵),它们就无法分离(Exercise 15.5习题 15.5);Anderson、Fu、Phlypo 与 Adalı(2014)证明,这类障碍本质上是唯一的障碍。这与 SOBI 的情形(Theorem 9.6定理 9.6)相似,只是用数据集代替了时间延迟。

Lee et al. (2008) applied IVA to group fMRI. Michael et al. (2014) compared group ICA with back-reconstruction and IVA on simulated data with spatially varying networks. As the variability of the maps across subjects grew, IVA stayed close to the ground truth, whereas group ICA produced subject maps that were partly composites of other maps. They also noted that group ICA remains preferable when group-level maps are the goal, and that IVA is more demanding in memory and computation.

Lee 等(2008)把 IVA 用于组 fMRI 分析。Michael 等(2014)在空间网络存在变异的仿真数据上比较了带反重建的组 ICA 与 IVA:随着空间图在被试之间的变异增大,IVA 的估计仍接近真实值,而组 ICA 得到的被试空间图部分地混入了其他空间图。他们也指出,若目标是组水平的空间图,组 ICA 仍更合适,而且 IVA 对内存与计算的要求更高。

15.7 Functional network connectivity功能网络连接

Spatial ICA leaves the time courses unconstrained, so their correlations carry information about how networks interact. Jafri et al. (2008) introduced functional network connectivity (FNC), the correlation among the time courses of spatially independent components, and applied it to schizophrenia. For subject ii, let A~i\tilde\bA_i be the time courses of (15.4), or of any back-reconstruction, after centering in time and usually detrending and low-pass filtering. The static FNC is the n×nn\times n correlation matrix

空间 ICA 对时间过程不加约束,所以时间过程之间的相关携带了网络之间相互作用的信息。Jafri 等(2008)提出了功能网络连接(functional network connectivity, FNC),即空间独立成分的时间过程之间的相关,并把它用于精神分裂症研究。对被试 ii,记 A~i\tilde\bA_i 为 (15.4)(或任何一种反重建)得到的时间过程,经过时间上的中心化,通常还做了去趋势与低通滤波。静态 FNC 是 n×nn\times n 相关矩阵

Ri=diag⁡(Ci)−1/2 Ci diag⁡(Ci)−1/2,Ci=A~i⊤A~i,(15.7)\bR_i = \diag(\bC_i)^{-1/2}\,\bC_i\,\diag(\bC_i)^{-1/2}, \qquad \bC_i = \tilde\bA_i^\top\tilde\bA_i,\tag{15.7}

where diag⁡(Ci)\diag(\bC_i) is the diagonal matrix with the same diagonal as Ci\bC_i. The entries of Ri\bR_i are usually Fisher-transformed, z=tanh⁡−1(r)z = \tanh^{-1}(r), before statistics across subjects, because the transformed values have approximately constant variance. Connectivity also changes within a scan. Dynamic FNC computes the correlation matrix (15.7) from the time points inside a window that slides along the scan, often with tapered weights, and studies how the matrices evolve. Allen et al. (2014) used tapered windows of about 44 seconds and clustered the windowed matrices of all subjects with k-means into a few recurring connectivity states, whose frequency and transitions can be compared between groups. Short windows give noisy estimates (Exercise 15.4习题 15.4): the window length trades temporal resolution against precision, and fluctuations in windowed correlations must be distinguished from the sampling variability that stationary data would also show.

其中 diag⁡(Ci)\diag(\bC_i) 是与 Ci\bC_i 对角元相同的对角矩阵。在做跨被试统计之前,Ri\bR_i 的元素通常要做 Fisher 变换 z=tanh⁡−1(r)z = \tanh^{-1}(r),因为变换后的值方差近似恒定。连接在一次扫描之内也会变化。动态 FNC 用沿扫描滑动的窗口内的时间点计算相关矩阵 (15.7)(常用渐变加权的窗),研究这些矩阵如何演化。Allen 等(2014)使用约 44 秒的渐变窗,并用 k 均值把所有被试的窗口相关矩阵聚成少数几个反复出现的连接状态,进而在组间比较各状态出现的频率与相互转换。窗口越短,估计越嘈杂(Exercise 15.4习题 15.4):窗长需要在时间分辨率与精度之间权衡,而且窗口相关的起伏必须与平稳数据同样会表现出的抽样波动区分开来。

15.8 Multimodal fusion多模态融合

ICA can also link different kinds of data from the same subjects. Suppose each of MM subjects contributes one feature map from each of two modalities, for instance a gray-matter map from structural MRI and a contrast map from a task fMRI experiment, and stack them into X1∈RM×V1\bX_1 \in \R^{M\times V_1} and X2∈RM×V2\bX_2 \in \R^{M\times V_2}, one row per subject. Joint ICA (Calhoun et al., 2006) assumes that the modalities share the mixing matrix:

ICA 还能把同一批被试的不同类型数据联系起来。设 MM 个被试中每人在两种模态下各提供一张特征图,例如结构 MRI 的灰质图与任务 fMRI 实验的对比图,把它们按每个被试一行堆叠成 X1∈RM×V1\bX_1 \in \R^{M\times V_1} 与 X2∈RM×V2\bX_2 \in \R^{M\times V_2}。联合 ICA(joint ICA;Calhoun et al., 2006)假设两种模态共享混合矩阵:

[X1, X2]=A [S1, S2],A∈RM×n.(15.8)\bigl[\bX_1,\ \bX_2\bigr] = \bA\,\bigl[\bS_1,\ \bS_2\bigr], \qquad \bA \in \R^{M\times n}.\tag{15.8}

Each joint source, a row of [S1,S2][\bS_1, \bS_2], is a pair of linked maps, one per modality, and the corresponding column of A\bA gives its expression in each subject, the subject loadings. The samples are the V1+V2V_1 + V_2 features, and independence is imposed on the joint sources. Differences between groups in the loadings, such as between patients and controls, point to linked changes in both modalities. Joint ICA (15.8) forces the modalities to share their loadings exactly. Parallel ICA (Liu et al., 2009) relaxes this. It runs a separate ICA for each modality, X1=A1S1\bX_1 = \bA_1\bS_1 and X2=A2S2\bX_2 = \bA_2\bS_2, and adds to the combined Infomax objective a term that rewards correlation between the loadings of a component of one modality and those of a component of the other, so that linked components emerge where the data support them; Liu et al. used it to relate fMRI components to genetic (SNP) components. In both models the link between modalities runs through the subject dimension, the one thing the data sets have in common.

每个联合源([S1,S2][\bS_1, \bS_2] 的一行)是一对相互关联的图,每种模态一张;A\bA 中对应的列给出它在每个被试中的表达强度,称为被试载荷。样本是 V1+V2V_1 + V_2 个特征,独立性施加在联合源上。载荷在组间(例如患者与对照之间)的差异,意味着两种模态中存在相互关联的变化。联合 ICA (15.8) 强制两种模态的载荷完全相同。并行 ICA(parallel ICA;Liu et al., 2009)放宽了这一点:它对每种模态分别做 ICA,X1=A1S1\bX_1 = \bA_1\bS_1,X2=A2S2\bX_2 = \bA_2\bS_2,并在合并的 Infomax 目标函数上加一项,奖励一种模态某个成分的载荷与另一种模态某个成分的载荷之间的相关,使相互关联的成分只在数据支持时才出现;Liu 等用它把 fMRI 成分与遗传(SNP)成分联系起来。在两种模型中,模态之间的联系都通过被试这一维度建立,这是各数据集唯一的共同之处。

Key points

本章要点

  • Group ICA by temporal concatenation assumes common maps and subject-specific time courses; a two-stage PCA, first per subject and then of the stacked reduced data, makes the MT×VMT\times V problem tractable.
  • Back-reconstruction (GICA1) inverts the reductions; dual regression regresses each subject's data on the group maps and then on the resulting time courses. Both have closed forms. Their time courses differ only by the regression, on the group maps, of the part of the subject's data that the subject-level PCA discards, whether or not the reductions are whitened; if that PCA discards nothing, the time courses coincide, and so do the maps unless the subject-level reduction is whitened.
  • Without variance normalization, dual-regression maps equal the group maps plus a correction built from the part of the data outside the span of the group maps; with it, they also reflect network amplitude.
  • Spatially constrained ICA (GIG-ICA, NeuroMark) estimates each subject's components by ICA guided by reference maps, which fixes their order and meaning across subjects and studies.
  • IVA links data sets through dependence within source component vectors; the within-SCV mutual information aligns the components and, with second-order statistics, can separate Gaussian sources whose cross-dataset correlations differ.
  • FNC and dynamic FNC analyse correlations among component time courses; joint and parallel ICA link imaging modalities through subject loadings.
  • 时间拼接的组 ICA 假设空间图共同、时间过程因被试而异;两步 PCA(先对每个被试,再对堆叠后的降维数据)使 MT×VMT\times V 规模的问题变得可行。
  • 反重建(GICA1)把降维步骤倒推回去;双回归先把每个被试的数据对组空间图回归,再对所得时间过程回归。两者都有闭式解。无论降维是否白化,它们的时间过程之差都只是被试水平 PCA 所舍弃的那部分数据对组空间图的回归;若该 PCA 不舍弃任何数据,则时间过程一致,而且只要被试水平的降维不做白化,空间图也一致。
  • 不做方差归一化时,双回归的空间图等于组空间图加上一个修正项,修正项来自数据中落在组空间图张成空间之外的部分;做方差归一化后,空间图还反映网络的幅度。
  • 空间约束 ICA(GIG-ICA、NeuroMark)在参考图引导下对每个被试做 ICA,使成分的顺序与含义在被试之间、研究之间保持一致。
  • IVA 通过源成分向量内部的依赖把各数据集联系起来;SCV 内部的互信息把成分对齐,而且只用二阶统计量就能分离跨数据集相关结构不同的高斯源。
  • FNC 与动态 FNC 分析成分时间过程之间的相关;联合 ICA 与并行 ICA 通过被试载荷把不同成像模态联系起来。

15.9 Exercises习题

Exercise 15.1习题 15.1 basic基础

A study has M=50M = 50 subjects with T=200T = 200 volumes of V=50,000V = 50{,}000 voxels. Each subject is reduced to q=30q = 30 dimensions and the group to n=20n = 20. Give the sizes of the stacked data in (15.1), of Fi\bF_i, Y\bY, G\bG, Gi\bG_i and B\bB, and of the back-reconstructed Aˇi\check\bA_i and Sˇi\check\bS_i. Compare the covariance matrices whose eigenvectors are needed with and without the two-stage reduction.

某研究有 M=50M = 50 名被试,每人 T=200T = 200 个体积,每个体积 V=50,000V = 50{,}000 个体素。每个被试降到 q=30q = 30 维,组水平降到 n=20n = 20 维。写出 (15.1) 中堆叠数据的尺寸,Fi\bF_i、Y\bY、G\bG、Gi\bG_i、B\bB 的尺寸,以及反重建得到的 Aˇi\check\bA_i 与 Sˇi\check\bS_i 的尺寸。比较采用与不采用两步降维时,需要求特征向量的协方差矩阵的大小。

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The stacked data are MT×V=10,000×50,000MT \times V = 10{,}000 \times 50{,}000. Then Fi\bF_i is 30×20030 \times 200, Y\bY is Mq×V=1,500×50,000Mq \times V = 1{,}500 \times 50{,}000, G\bG is 20×1,50020 \times 1{,}500, each block Gi\bG_i is 20×3020 \times 30, B\bB is 20×2020 \times 20, Aˇi\check\bA_i is 200×20200 \times 20 and Sˇi\check\bS_i is 20×50,00020 \times 50{,}000. The two-stage reduction needs the eigenvectors of 50 matrices of size 200×200200 \times 200 and of one of size 1,500×1,5001{,}500 \times 1{,}500, and never holds more than one subject's raw data together with Y\bY. A single PCA of the stacked data needs the eigenvectors of a 10,000×10,00010{,}000 \times 10{,}000 matrix computed from all 5×1085 \times 10^8 entries of the stacked data.

堆叠数据为 MT×V=10,000×50,000MT \times V = 10{,}000 \times 50{,}000。Fi\bF_i 为 30×20030 \times 200,Y\bY 为 Mq×V=1,500×50,000Mq \times V = 1{,}500 \times 50{,}000,G\bG 为 20×1,50020 \times 1{,}500,每个分块 Gi\bG_i 为 20×3020 \times 30,B\bB 为 20×2020 \times 20,Aˇi\check\bA_i 为 200×20200 \times 20,Sˇi\check\bS_i 为 20×50,00020 \times 50{,}000。两步降维只需求 50 个 200×200200 \times 200 矩阵和一个 1,500×1,5001{,}500 \times 1{,}500 矩阵的特征向量,内存中至多同时保存一个被试的原始数据和 Y\bY。若对堆叠数据直接做一次 PCA,则要从堆叠数据的全部 5×1085 \times 10^8 个元素算出一个 10,000×10,00010{,}000 \times 10{,}000 的矩阵并求其特征向量。

Exercise 15.2习题 15.2 basic基础

Let A^i\hat\bA_i have full column rank and let Di\bD_i be the diagonal matrix of the standard deviations of its columns. Show that regressing Xi\bX_i on A^iDi−1\hat\bA_i\bD_i^{-1} instead of A^i\hat\bA_i multiplies the rows of the stage-2 maps by the diagonal entries of Di\bD_i. Why does this make the maps sensitive to the amplitude of each network?

设 A^i\hat\bA_i 列满秩,Di\bD_i 是由其各列标准差构成的对角矩阵。证明:把 Xi\bX_i 对 A^iDi−1\hat\bA_i\bD_i^{-1}(而不是 A^i\hat\bA_i)回归,相当于把第二步空间图的各行乘以 Di\bD_i 的对角元。为什么这会使空间图对各网络的幅度敏感?

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Since Di\bD_i is diagonal and invertible,

由于 Di\bD_i 是可逆的对角矩阵,

(A^iDi−1)+=(Di−1A^i⊤A^iDi−1)−1Di−1A^i⊤=Di(A^i⊤A^i)−1A^i⊤=DiA^i+,\begin{aligned} (\hat\bA_i\bD_i^{-1})^+ &= \bigl(\bD_i^{-1}\hat\bA_i^\top\hat\bA_i\bD_i^{-1}\bigr)^{-1}\bD_i^{-1}\hat\bA_i^\top \\ &= \bD_i(\hat\bA_i^\top\hat\bA_i)^{-1}\hat\bA_i^\top = \bD_i\hat\bA_i^+, \end{aligned}

so the new maps are DiA^i+Xi=DiS^i\bD_i\hat\bA_i^+\bX_i = \bD_i\hat\bS_i. By Proposition 15.1命题 15.1, the unnormalized maps S^i=S^+A^i+Xi(I−PS^)\hat\bS_i = \hat\bS + \hat\bA_i^+\bX_i(\bI - \bP_{\hat\bS}) are anchored to the group maps whatever the size of the subject's fluctuations: the amplitude of network ll is carried by its time course, whose standard deviation is the ll-th diagonal entry of Di\bD_i. After normalization the time courses have unit variance, and the amplitude moves into the map: a subject whose network fluctuates twice as strongly gets a map twice as large. Group comparisons of normalized maps therefore test shape and amplitude together, comparisons of unnormalized maps mainly shape.

所以新的空间图为 DiA^i+Xi=DiS^i\bD_i\hat\bA_i^+\bX_i = \bD_i\hat\bS_i。由Proposition 15.1命题 15.1,未归一化的空间图 S^i=S^+A^i+Xi(I−PS^)\hat\bS_i = \hat\bS + \hat\bA_i^+\bX_i(\bI - \bP_{\hat\bS}) 无论被试的波动有多大都锚定在组空间图上:网络 ll 的幅度由其时间过程携带,其标准差就是 Di\bD_i 的第 ll 个对角元。归一化之后时间过程方差为一,幅度转移到空间图中:某网络波动强一倍的被试,其空间图也大一倍。因此,比较归一化后的空间图同时检验形状与幅度,比较未归一化的空间图则主要检验形状。

Exercise 15.3习题 15.3 intermediate进阶

Let K=2K = 2 and suppose that both data sets are separated exactly, but possibly in different orders: y[1]=s[1]\by^{[1]} = \bs^{[1]} and yi[2]=sπ(i)[2]y_i^{[2]} = s_{\pi(i)}^{[2]} for a permutation π\pi. Show that the IVA cost (15.6) exceeds its value for π=id\pi = \mathrm{id} by ∑i: π(i)≠iI(si)\sum_{i:\,\pi(i)\neq i} I(\bs_i), where I(si)I(\bs_i) is the mutual information between si[1]s_i^{[1]} and si[2]s_i^{[2]}.

设 K=2K = 2,两个数据集都被准确分离,但顺序可能不同:y[1]=s[1]\by^{[1]} = \bs^{[1]},yi[2]=sπ(i)[2]y_i^{[2]} = s_{\pi(i)}^{[2]},其中 π\pi 是一个置换。证明 IVA 代价函数 (15.6) 比 π=id\pi = \mathrm{id} 时的值大 ∑i: π(i)≠iI(si)\sum_{i:\,\pi(i)\neq i} I(\bs_i),其中 I(si)I(\bs_i) 是 si[1]s_i^{[1]} 与 si[2]s_i^{[2]} 之间的互信息。

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Permuting the outputs of the second data set replaces W[2]\bW^{[2]} by ΠW[2]\bPi\bW^{[2]} for a permutation matrix Π\bPi, which leaves ∣det⁡W[2]∣|\det\bW^{[2]}| unchanged, and the last term of the cost does not depend on the unmixing matrices. The marginal entropies H(yi[k])H(y_i^{[k]}) are the same numbers in a different order, so their sum is unchanged. Only the terms −I(yi)-I(\by_i) can differ. If π(i)=i\pi(i) = i, then yi=si\by_i = \bs_i and I(yi)=I(si)I(\by_i) = I(\bs_i), as for the identity. If π(i)≠i\pi(i) \neq i, then yi=(si[1],sπ(i)[2])\by_i = (s_i^{[1]}, s_{\pi(i)}^{[2]}) pairs components of two different SCVs, which are independent, so I(yi)=0I(\by_i) = 0. The cost therefore increases by ∑i: π(i)≠iI(si)≥0\sum_{i:\,\pi(i)\neq i} I(\bs_i) \ge 0, strictly whenever one of these SCVs has dependent components. The ICA terms decide how each data set is unmixed; the within-SCV terms decide how the data sets are aligned.

把第二个数据集的输出重新排序,相当于把 W[2]\bW^{[2]} 换成 ΠW[2]\bPi\bW^{[2]}(Π\bPi 为置换矩阵),这不改变 ∣det⁡W[2]∣|\det\bW^{[2]}|;代价函数的最后一项与解混矩阵无关。各边缘熵 H(yi[k])H(y_i^{[k]}) 只是换了顺序,其和不变。只有 −I(yi)-I(\by_i) 项可能不同。若 π(i)=i\pi(i) = i,则 yi=si\by_i = \bs_i,I(yi)=I(si)I(\by_i) = I(\bs_i),与恒等置换时相同。若 π(i)≠i\pi(i) \neq i,则 yi=(si[1],sπ(i)[2])\by_i = (s_i^{[1]}, s_{\pi(i)}^{[2]}) 把两个不同 SCV 的分量配在一起,它们相互独立,所以 I(yi)=0I(\by_i) = 0。因此代价增加了 ∑i: π(i)≠iI(si)≥0\sum_{i:\,\pi(i)\neq i} I(\bs_i) \ge 0;只要这些 SCV 中有一个的分量相互依赖,增加量就严格为正。ICA 项决定每个数据集如何解混,SCV 内部的互信息项决定各数据集如何对齐。

Exercise 15.4习题 15.4 intermediate进阶

Two network time courses are independent sequences of independent Gaussian values. Their sample correlation rr over a window of LL time points has mean zero and variance 1/(L−1)1/(L-1), and rL−2/1−r2r\sqrt{L-2}/\sqrt{1-r^2} has a Student tt distribution with L−2L - 2 degrees of freedom. With TR=2\mathrm{TR} = 2 s and windows of 44 s, compute LL, the standard deviation of rr and P(∣r∣>0.4)P(|r| > 0.4). Real BOLD time courses are positively autocorrelated; does this make matters better or worse?

两条网络时间过程是相互独立的序列,各自由独立的高斯值构成。它们在 LL 个时间点的窗口内的样本相关系数 rr 均值为零、方差为 1/(L−1)1/(L-1),且 rL−2/1−r2r\sqrt{L-2}/\sqrt{1-r^2} 服从自由度为 L−2L - 2 的 Student tt 分布。取 TR=2\mathrm{TR} = 2 秒、窗长 44 秒,计算 LL、rr 的标准差以及 P(∣r∣>0.4)P(|r| > 0.4)。真实的 BOLD 时间过程存在正的自相关,这会让情况变好还是变坏?

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L=44/2=22L = 44/2 = 22, so the standard deviation of rr is 1/21≈0.221/\sqrt{21} \approx 0.22. For r=0.4r = 0.4, t=0.420/0.84≈1.95t = 0.4\sqrt{20}/\sqrt{0.84} \approx 1.95, and P(∣t20∣>1.95)≈0.065P(|t_{20}| > 1.95) \approx 0.065: in about one window in fifteen, a pair of completely unrelated networks shows ∣r∣>0.4|r| > 0.4 by chance. (A simulation with 2×1052\times 10^5 windows gives 0.0650.065 as well.) Positive autocorrelation makes neighboring samples partly redundant, so the effective number of independent samples is smaller than LL and the sampling variance of rr is larger: matters get worse, and fluctuations of windowed correlations must be compared with what stationary data with the same autocorrelation would produce.

L=44/2=22L = 44/2 = 22,所以 rr 的标准差为 1/21≈0.221/\sqrt{21} \approx 0.22。当 r=0.4r = 0.4 时,t=0.420/0.84≈1.95t = 0.4\sqrt{20}/\sqrt{0.84} \approx 1.95,而 P(∣t20∣>1.95)≈0.065P(|t_{20}| > 1.95) \approx 0.065:大约每十五个窗口中就有一个,在两个毫不相关的网络之间仅凭偶然就出现 ∣r∣>0.4|r| > 0.4。(用 2×1052\times 10^5 个窗口做仿真,结果同样是 0.0650.065。)正的自相关使相邻样本部分冗余,有效的独立样本数小于 LL,rr 的抽样方差因而更大:情况更糟。因此,窗口相关的起伏必须与具有相同自相关的平稳数据所能产生的起伏相比较。

Exercise 15.5习题 15.5 advanced硬核

In the IVA model (15.5), let two SCVs sl,sm∈RK\bs_l, \bs_m \in \R^K be zero-mean Gaussian, independent of each other and of the other SCVs, with covariance matrices Σl\bSigma_l and Σm=DΣlD\bSigma_m = \bD\bSigma_l\bD, where D=diag⁡(d1,…,dK)\bD = \diag(d_1, \dots, d_K) is invertible. For an angle θ\theta, define in each data set kk the new sources

在 IVA 模型 (15.5) 中,设两个 SCV sl,sm∈RK\bs_l, \bs_m \in \R^K 是零均值高斯向量,彼此独立,也与其他 SCV 独立,协方差矩阵分别为 Σl\bSigma_l 与 Σm=DΣlD\bSigma_m = \bD\bSigma_l\bD,其中 D=diag⁡(d1,…,dK)\bD = \diag(d_1, \dots, d_K) 可逆。对角度 θ\theta,在每个数据集 kk 中定义新的源

s~l[k]=cos⁡θ sl[k]+sin⁡θ sm[k]/dk,s~m[k]=−dksin⁡θ sl[k]+cos⁡θ sm[k].\begin{aligned} \tilde s_l^{[k]} &= \cos\theta\,s_l^{[k]} + \sin\theta\,s_m^{[k]}/d_k, \\ \tilde s_m^{[k]} &= -d_k\sin\theta\,s_l^{[k]} + \cos\theta\,s_m^{[k]}. \end{aligned}

Show that the new SCVs s~l\tilde\bs_l, s~m\tilde\bs_m are independent of each other and of the other SCVs, have covariance matrices Σl\bSigma_l and Σm\bSigma_m, and are obtained from the old sources by an invertible linear transformation within each data set. Conclude that IVA cannot separate these two sources.

证明新的 SCV s~l\tilde\bs_l、s~m\tilde\bs_m 彼此独立、也与其他 SCV 独立,协方差矩阵分别为 Σl\bSigma_l 与 Σm\bSigma_m,并且是在每个数据集内对旧源作可逆线性变换得到的。由此说明 IVA 无法分离这两个源。

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Let u=sl\bu = \bs_l and v=D−1sm\bv = \bD^{-1}\bs_m. Then u\bu and v\bv are independent zero-mean Gaussian vectors with the same covariance, Cov⁡(v)=D−1ΣmD−1=Σl\Cov(\bv) = \bD^{-1}\bSigma_m\bD^{-1} = \bSigma_l. The new SCVs are s~l=cos⁡θ u+sin⁡θ v\tilde\bs_l = \cos\theta\,\bu + \sin\theta\,\bv and s~m=D(−sin⁡θ u+cos⁡θ v)\tilde\bs_m = \bD(-\sin\theta\,\bu + \cos\theta\,\bv). They are jointly Gaussian, with

令 u=sl\bu = \bs_l,v=D−1sm\bv = \bD^{-1}\bs_m。则 u\bu 与 v\bv 是相互独立的零均值高斯向量,协方差相同:Cov⁡(v)=D−1ΣmD−1=Σl\Cov(\bv) = \bD^{-1}\bSigma_m\bD^{-1} = \bSigma_l。新的 SCV 为 s~l=cos⁡θ u+sin⁡θ v\tilde\bs_l = \cos\theta\,\bu + \sin\theta\,\bv,s~m=D(−sin⁡θ u+cos⁡θ v)\tilde\bs_m = \bD(-\sin\theta\,\bu + \cos\theta\,\bv)。它们联合高斯,且

Cov⁡(s~l)=(cos⁡2θ+sin⁡2θ) Σl=Σl,Cov⁡(s~m)=DΣlD=Σm,E[s~ls~m⊤]=(−cos⁡θsin⁡θ+sin⁡θcos⁡θ) ΣlD=0,\begin{aligned} \Cov(\tilde\bs_l) &= (\cos^2\theta + \sin^2\theta)\,\bSigma_l = \bSigma_l, \\ \Cov(\tilde\bs_m) &= \bD\bSigma_l\bD = \bSigma_m, \\ \E\bigl[\tilde\bs_l\tilde\bs_m^\top\bigr] &= (-\cos\theta\sin\theta + \sin\theta\cos\theta)\,\bSigma_l\bD = \bzero, \end{aligned}

and uncorrelated jointly Gaussian vectors are independent. They are functions of (sl,sm)(\bs_l, \bs_m) only, hence independent of the other SCVs. In data set kk, (sl[k],sm[k])(s_l^{[k]}, s_m^{[k]}) is mapped to (s~l[k],s~m[k])(\tilde s_l^{[k]}, \tilde s_m^{[k]}) by the matrix

不相关的联合高斯向量相互独立。它们只是 (sl,sm)(\bs_l, \bs_m) 的函数,因而与其他 SCV 独立。在数据集 kk 中,(sl[k],sm[k])(s_l^{[k]}, s_m^{[k]}) 经下面的矩阵变为 (s~l[k],s~m[k])(\tilde s_l^{[k]}, \tilde s_m^{[k]}):

(cos⁡θsin⁡θ/dk−dksin⁡θcos⁡θ),\begin{pmatrix} \cos\theta & \sin\theta/d_k \\ -d_k\sin\theta & \cos\theta \end{pmatrix},

whose determinant is 11. Replacing columns ll and mm of each A[k]\bA^{[k]} accordingly, using the inverse of this matrix, therefore gives a second IVA model that produces exactly the same observations, with SCVs that satisfy every assumption and have the same distributions as before. No criterion computed from the observations can prefer θ=0\theta = 0, and for θ\theta not a multiple of π/2\pi/2 the new sources are genuine mixtures of the old ones. The case D=I\bD = \bI shows that Gaussian SCVs with identical covariance across data sets are never separable, just as two Gaussian sources are not in ICA.

其行列式为 11。用这个矩阵的逆相应地替换每个 A[k]\bA^{[k]} 的第 ll、mm 列,就得到另一个 IVA 模型,它产生完全相同的观测,其 SCV 满足所有假设且分布与原来相同。任何由观测计算的准则都无法偏好 θ=0\theta = 0;而当 θ\theta 不是 π/2\pi/2 的整数倍时,新的源是旧源的真正混合。取 D=I\bD = \bI 可见,跨数据集协方差相同的高斯 SCV 永远无法分离,正如 ICA 中两个高斯源无法分离一样。

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