相關系數及其在fMRI資料中的應用
- 背景知識
-
- 皮爾遜相關系數(Pearson correlation coefficient)
- 肯德爾和諧系數(Kendall's W)
- fMRI資料
- 功能連接配接
- 局部一緻性(Regional Homogeneity)(此章節有誤)
背景知識
皮爾遜相關系數(Pearson correlation coefficient)
ρ i j = ρ ( X i , X j ) = c o v ( X i , X j ) D ( X i ) D ( X j ) \begin{aligned} \rho_{ij} &= \rho \left( \mathbf{X}_i, \mathbf{X}_j \right) \\ &= \frac{cov \left( \mathbf{X}_i, \mathbf{X}_j \right)} {\sqrt{D(\mathbf{X}_i)} \sqrt{D(\mathbf{X}_j)}} \\ \end{aligned} ρij=ρ(Xi,Xj)=D(Xi)
D(Xj)
cov(Xi,Xj)
肯德爾和諧系數(Kendall’s W)
Kendall’s W解決的原始問題是在不同人對多個事物進行評分時, 如何對這些人評分結果的一緻性進行評價.
假設存在 n n n件物體, 并由 m m m個不同的人進行評分或排序, 且第 j j j個人對第 i i i件物體的評分為 r i j r_{ij} rij(在該問題中, 每個評分實際表示的是排序, 為1到n之間的整數且不重複), 則這些評分可表示為
R = { r i j } n × m \mathbf{R}=\{ r_{ij} \}_{n \times m} R={rij}n×m
物體 i i i的總分為
R i = ∑ j = 1 m r i j R_i=\sum_{j=1}^m r_{ij} Ri=j=1∑mrij
所有物體的平均得分為
R ˉ = 1 n ∑ i = 1 n R i = 1 n ∑ i = 1 n ∑ j = 1 m r i j = 1 n ∑ j = 1 m n ( n + 1 ) 2 = m ( n + 1 ) 2 \begin{aligned} \bar{R} &=\frac{1}{n} \sum_{i=1}^n R_i \\ &= \frac{1}{n} \sum_{i=1}^n \sum_{j=1}^m r_{ij} \\ &= \frac{1}{n} \sum_{j=1}^m \frac{n(n+1)}{2} \\ &= \frac{m(n+1)}{2} \end{aligned} Rˉ=n1i=1∑nRi=n1i=1∑nj=1∑mrij=n1j=1∑m2n(n+1)=2m(n+1)
所有物體得分的方差之和為 S = ∑ i = 1 n ( R i − R ˉ ) 2 S=\sum_{i=1}^n \left( R_i -\bar{R} \right)^2 S=∑i=1n(Ri−Rˉ)2, 則Kendall’s W定義為
W = 12 S m 2 ( n 3 − n ) W=\frac{12S}{m^2 (n^3-n)} W=m2(n3−n)12S
其中 m 2 ( n 3 − n ) 12 \frac{m^2 (n^3-n)}{12} 12m2(n3−n)表示 S S S可能出現的最大值, 當且僅當所有人對同一個物體的評分一緻時(即評分一緻)得到, 而最小值則在 R i = R ˉ R_i = \bar{R} Ri=Rˉ時得到.
max S = ∑ i = 1 n ( m i − R ˉ ) 2 = ∑ i = 1 n [ m 2 i 2 − m 2 i ( n + 1 ) + m 2 ( n + 1 ) 2 4 ] = m 2 n ( n + 1 ) ⋅ 2 n + 1 6 − m 2 n ( n + 1 ) ⋅ n + 1 4 = m 2 n ( n + 1 ) ⋅ n − 1 12 = m 2 ( n 3 − n ) 12 \begin{aligned} \max{S} &= \sum_{i=1}^n \left( mi-\bar{R} \right)^2 \\ &= \sum_{i=1}^n \left[ m^2 i^2 - m^2 i (n+1) + \frac{m^2 (n+1)^2}{4} \right] \\ &= m^2 n (n+1) \cdot \frac{2n+1}{6} - m^2 n (n+1) \cdot \frac{n+1}{4} \\ &= m^2 n (n+1) \cdot \frac{n-1}{12} \\ &= \frac{m^2(n^3-n)}{12} \end{aligned} maxS=i=1∑n(mi−Rˉ)2=i=1∑n[m2i2−m2i(n+1)+4m2(n+1)2]=m2n(n+1)⋅62n+1−m2n(n+1)⋅4n+1=m2n(n+1)⋅12n−1=12m2(n3−n)
使得 W W W的範圍縮小至 [ 0 , 1 ] \left[0, 1\right] [0,1].
fMRI資料
功能磁共振成像資料的基本元素為體素的時間序列 X i = [ x i 1 , x i 2 , ⋯   , x i T ] T \mathbf{X}_i=\left[x_{i1}, x_{i2}, \cdots, x_{iT} \right]^T Xi=[xi1,xi2,⋯,xiT]T, 則一個被試的全腦資料可表示為
X = [ X 1 , X 2 , ⋯   , X R ] T = [ x 11 x 12 ⋯ x 1 T x 21 x 22 ⋯ x 2 T ⋮ ⋮ ⋱ ⋮ x R 1 x R 2 ⋯ x R T ] R × T \begin{aligned} \mathbf{X} &=\left[\mathbf{X}_1, \mathbf{X}_2, \cdots, \mathbf{X}_{R} \right]^T \\ &= \left[ \begin{matrix} x_{11} & x_{12} & \cdots & x_{1T} \\ x_{21} & x_{22} & \cdots & x_{2T} \\ \vdots & \vdots & \ddots & \vdots \\ x_{R1} & x_{R2} & \cdots & x_{RT} \end{matrix} \right]_{R \times T} \end{aligned} X=[X1,X2,⋯,XR]T=⎣⎢⎢⎢⎡x11x21⋮xR1x12x22⋮xR2⋯⋯⋱⋯x1Tx2T⋮xRT⎦⎥⎥⎥⎤R×T
其中 R = W × H × D R=W \times H \times D R=W×H×D表示體素的數量, W , H , D W, H, D W,H,D分别表示腦影像資料在三個次元上的大小, T T T表示時間序列的長度, X r ∈ R T \mathbf{X}_r \in \mathbb{R}^T Xr∈RT表示第 R R R個體素的時間序列.
根據AAL(Anatominal Atlas Label)模版, 全腦可分為116個腦區, 即将所有體素劃分至116個集合中
R O I i = { j 1 , j 2 , ⋯   , j R i } i = 1 , 2 , ⋯   , 116 ROI_i=\{ j_1, j_2, \cdots, j_{R_i} \} \quad i=1,2,\cdots,116 ROIi={j1,j2,⋯,jRi}i=1,2,⋯,116
其中, R i R_i Ri表示第i個腦區包含的體素數量.
功能連接配接
功能連接配接的計算步驟如下
-
計算各腦區的平均時間序列
X ˉ i = 1 R i ∑ j ∈ R O I i X j \mathbf{\bar{X}}_i = \frac{1}{R_i} \sum_{j \in ROI_i} \mathbf{X}_j Xˉi=Ri1j∈ROIi∑Xj
-
計算兩個腦區之間的Pearson相關系數
ρ i j = ρ ( X ˉ i , X ˉ j ) = c o v ( X ˉ i , X ˉ j ) D ( X ˉ i ) D ( X ˉ j ) \begin{aligned} \rho_{ij} &= \rho \left( \bar{\mathbf{X}}_i, \bar{\mathbf{X}}_j \right) \\ &= \frac{cov \left( \bar{\mathbf{X}}_i, \bar{\mathbf{X}}_j \right)} {\sqrt{D(\bar{\mathbf{X}}_i)} \sqrt{D(\bar{\mathbf{X}}_j)}} \\ \end{aligned} ρij=ρ(Xˉi,Xˉj)=D(Xˉi)
D(Xˉj)
cov(Xˉi,Xˉj)
由于 ρ ( X ˉ i , X ˉ j ) = ρ ( α ^ i X i , α ^ j X j ) \rho \left( \bar{\mathbf{X}}_i, \bar{\mathbf{X}}_j \right)= \rho \left(\boldsymbol{\hat{\alpha}}_i \mathbf{X}_i, \boldsymbol{\hat{\alpha}}_j \mathbf{X}_j \right) ρ(Xˉi,Xˉj)=ρ(α^iXi,α^jXj), 其中 α ^ i \boldsymbol{\hat{\alpha}}_i α^i為長度為 R i R_i Ri, 所有元素為 1 R i \frac{1}{R_i} Ri1的向量. 是以有
ρ i j = ρ ( X ˉ i , X ˉ j ) = ρ ( α i X i , α j X j ) = α ^ i T Σ X i X j α ^ j α ^ i T Σ X i X i α ^ i α ^ j T Σ X j X j α ^ j = α i T Σ X i X j α j α i T Σ X i X i α i α j T Σ X j X j α j \begin{aligned} \rho_{ij} &= \rho \left( \bar{\mathbf{X}}_i, \bar{\mathbf{X}}_j \right) \\ &= \rho \left( \boldsymbol{\alpha}_i \mathbf{X}_i, \boldsymbol{\alpha}_j \mathbf{X}_j \right) \\ &= \frac{\boldsymbol{\hat{\alpha}}_i^T \Sigma_{\mathbf{X}_i \mathbf{X}_j} \boldsymbol{\hat{\alpha}}_j} {\sqrt{\boldsymbol{\hat{\alpha}}_i^T \Sigma_{\mathbf{X}_i \mathbf{X}_i} \boldsymbol{\hat{\alpha}}_i} \sqrt{\boldsymbol{\hat{\alpha}}_j^T \Sigma_{\mathbf{X}_j \mathbf{X}_j} \boldsymbol{\hat{\alpha}}_j}} \\ &= \frac{\boldsymbol{\alpha}_i^T \Sigma_{\mathbf{X}_i \mathbf{X}_j} \boldsymbol{\alpha}_j} {\sqrt{\boldsymbol{\alpha}_i^T \Sigma_{\mathbf{X}_i \mathbf{X}_i} \boldsymbol{\alpha}_i} \sqrt{\boldsymbol{\alpha}_j^T \Sigma_{\mathbf{X}_j \mathbf{X}_j} \boldsymbol{\alpha}_j}} \\ \end{aligned} ρij=ρ(Xˉi,Xˉj)=ρ(αiXi,αjXj)=α^iTΣXiXiα^i
α^jTΣXjXjα^j
α^iTΣXiXjα^j=αiTΣXiXiαi
αjTΣXjXjαj
αiTΣXiXjαj
其中 Σ X i X j \Sigma_{\mathbf{X}_i \mathbf{X}_j} ΣXiXj表示随機向量 X i \mathbf{X}_i Xi與 X j \mathbf{X}_j Xj的協方差矩陣, α i \boldsymbol{\alpha}_i αi與 α j \boldsymbol{\alpha}_j αj分别表示所有元素為1, 長度為 R i , R j R_i, R_j Ri,Rj的向量.
局部一緻性(Regional Homogeneity)(此章節有誤)
局部一緻性的計算過程如下
- 選取中心體素 X i = [ x i 1 , x i 2 , ⋯   , x i T ] T \mathbf{X}_i=[x_{i1}, x_{i2}, \cdots, x_{iT}]^T Xi=[xi1,xi2,⋯,xiT]T
-
以體素 X i \mathbf{X}_i Xi為中心選取周圍的體素(在DPARSF standard processing pipeline中, 為相鄰的 K = 27 K=27 K=27個體素)
X ~ i = [ X i 1 , X i 2 , ⋯   , X i K ] T = [ x i 1 1 x i 1 2 ⋯ x i 1 T x i 2 1 x i 2 2 ⋯ x i 1 T ⋮ ⋮ ⋱ ⋮ x i K 1 x i K 2 ⋯ x i K T ] \begin{aligned} \widetilde{\mathbf{X}}_i &=\left[ \mathbf{X}_{i_1}, \mathbf{X}_{i_2}, \cdots, \mathbf{X}_{i_{K}} \right]^T \\ &= \left[ \begin{matrix} x_{i_1 1} & x_{i_1 2} & \cdots & x_{i_1 T} \\ x_{i_2 1} & x_{i_2 2} & \cdots & x_{i_1 T} \\ \vdots & \vdots & \ddots & \vdots \\ x_{i_{K} 1} & x_{i_{K} 2} & \cdots & x_{i_{K} T} \end{matrix} \right] \end{aligned} X
i=[Xi1,Xi2,⋯,XiK]T=⎣⎢⎢⎢⎡xi11xi21⋮xiK1xi12xi22⋮xiK2⋯⋯⋱⋯xi1Txi1T⋮xiKT⎦⎥⎥⎥⎤
-
根據Kendall’s W的計算, 得到該體素的局部一緻性
(1) 計算所有時間點的和
X ~ i t S = ∑ j = 1 K x i j t t = 1 , 2 , ⋯   , T \widetilde{\mathbf{X}}_{it}^S=\sum_{j=1}^{K} x_{i_j t} \quad t=1,2, \cdots, T X
itS=j=1∑Kxijtt=1,2,⋯,T
也可以表示成為矩陣與向量的形式
X ~ i S = ∑ j = 1 K X i j = α X ~ i \begin{aligned} \widetilde{\mathbf{X}}_i^S &= \sum_{j=1}^{K} \mathbf{X}_{i_j} \\ &= \boldsymbol{\alpha} \widetilde{\mathbf{X}}_i \end{aligned} X
iS=j=1∑KXij=αX
i
其中 α = [ 1 , 1 , ⋯   , 1 ] \boldsymbol{\alpha}=[1, 1, \cdots, 1] α=[1,1,⋯,1]是維數為 1 × 27 1 \times 27 1×27的向量.
(2) 計算所有體素的平均值
X ˉ i = 1 T ∑ j = 1 K ∑ t = 1 T x i j t = 1 T ∑ t = 1 T X ~ i t S \begin{aligned} \bar{\mathbf{X}}_i &= \frac{1}{T} \sum_{j=1}^{K} \sum_{t=1}^T x_{i_j t} \\ &= \frac{1}{T} \sum_{t=1}^T \widetilde{\mathbf{X}}_{it}^S \end{aligned} Xˉi=T1j=1∑Kt=1∑Txijt=T1t=1∑TX
itS
(3) 計算Kendall’s W
W i = 12 S i K 2 ( T 3 − T ) W_i = \frac{12S_i}{K^2 (T^3 - T)} Wi=K2(T3−T)12Si
其中 S S S表示所有時間序列和的方差
S i = D ( X ~ i S ) = D ( α X ~ i ) = ∑ j = 1 K D ( X i j ) = α T Σ X ~ i X ~ i α \begin{aligned} S_i &= D(\widetilde{\mathbf{X}}_i^S ) = D(\boldsymbol{\alpha} \widetilde{\mathbf{X}}_i) \\ &= \sum_{j=1}^K D(\mathbf{X}_{i_j}) \\ &= \boldsymbol{\alpha}^T \Sigma_{\widetilde{\mathbf{X}}_i \widetilde{\mathbf{X}}_i} \boldsymbol{\alpha} \end{aligned} Si=D(X
iS)=D(αX
i)=j=1∑KD(Xij)=αTΣX
iX
iα
是以
W i = 12 K 2 ( T 3 − T ) S i = 12 K 2 ( T 3 − T ) ∑ j = 1 K D ( X i j ) = 12 K 2 ( T 3 − T ) α T Σ X ~ i X ~ i α \begin{aligned} W_i &= \frac{12}{K^2(T^3-T)} S_i \\ &= \frac{12}{K^2(T^3-T)} \sum_{j=1}^K D(\mathbf{X}_{i_j}) \\ &= \frac{12}{K^2(T^3-T)} \boldsymbol{\alpha}^T \Sigma_{\widetilde{\mathbf{X}}_i \widetilde{\mathbf{X}}_i} \boldsymbol{\alpha} \end{aligned} Wi=K2(T3−T)12Si=K2(T3−T)12j=1∑KD(Xij)=K2(T3−T)12αTΣX
iX
iα
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