The distance between the spaces spanned by the column of two matrices.

Description

Calculate the distance between spaces spanned by the column of two matrices. The distance is between 0 and 1. If the two spaces are the same, the distance is 0. if the two spaces are orthogonal, the distance is 1.

Usage

Distance(Z1, Z2)

Arguments

Details

Define

\mathcal{D}(\bold{Q}_1,\bold{Q}_2)=\left(1-\frac{1}{\max{(q_1,q_2)}}\text{Tr}\left(\bold{Q}_1\bold{Q}_1^\top\bold{Q}_2\bold{Q}_2^\top\right)\right)^{1/2}.

By the definition of \mathcal{D}(\bold{Q}_1,\bold{Q}_2), we can easily see that 0 \leq \mathcal{D}(\bold{Q}_1,\bold{Q}_2)\leq 1, which measures the distance between the column spaces spanned by two orthogonal matrices \bold{Q}_1 and \bold{Q}_2, i.e., \text{span}(\bold{Q}_1) and \text{span}(\bold{Q}_2). In particular, \text{span}(\bold{Q}_1) and \text{span}(\bold{Q}_2) are the same when \mathcal{D}(\bold{Q}_1,\bold{Q}_2)=0, while \text{span}(\bold{Q}_1) and \text{span}(\bold{Q}_2) are orthogonal when \mathcal{D}(\bold{Q}_1,\bold{Q}_2)=1. The Gram-Schmidt orthogonalization can be used to make \bold{Q}_1 and \bold{Q}_2 column-orthogonal matrices.

Value

Output a number between 0 and 1.

Author(s)

Yong He, Ran Zhao, Wen-Xin Zhou.

References

He, Y., Zhao, R., & Zhou, W. X. (2023). Iterative Least Squares Algorithm for Large-dimensional Matrix Factor Model by Random Projection. <arXiv:2301.00360>.

Examples

set.seed(1111)
A=matrix(rnorm(10),5,2)
B=matrix(rnorm(15),5,3)
Distance(A,B)

Iterative Alternating Least Square Estimation for Large-dimensional Matrix Factor Model

Description

This function is designed to fit the matrix factor model using the Iterative Least Squares (IALS) method, rather than Principal Component Analysis (PCA)-based methods. In detail, in the first step, we propose to estimate the latent factor matrices by projecting the matrix observations with two deterministic weight matrices, chosen to diversify away the idiosyncratic components. In the second step, we update the row/column loading matrices by minimizing the squared loss function under the identifiability condition. The estimators of the loading matrices are then treated as the new weight matrices, and the algorithm iteratively performs these two steps until a convergence criterion is reached.

Usage

IALS(X, W1 = NULL, W2 = NULL, m1, m2, max_iter = 100, ep = 1e-06)

Arguments

Details

Assume we have two weight matrices \bold{W}_i of dimension p_{i} \times m_{i} for i=1,2, as substitutes for \bold{R} and \bold{C} respectively. Then it is straightforward to estimate \bold{F}_t simply by

\hat{\bold{F}}_{t}=\frac{1}{p_{1}p_{2}}\bold{W}_1^\top\bold{X}_t\bold{W}_2.

Given \hat{\bold{F}}_t and \bold{W}_1, we can derive that

\hat{\bold{R}}=\sqrt{p_{1}}\left(\sum_{t=1}^{T}\bold{X}_t\bold{W}_2\hat{\bold{F}}_t^\top\right)\left[\left(\sum_{t=1}^{T}\hat{\bold{F}}_t\bold{W}_2^\top\bold{X}_t^\top\right)\left(\sum_{t=1}^{T}\bold{X}_t\bold{W}_2\hat{\bold{F}}_t^\top\right)\right]^{-1/2}.

Similarly, we get the following estimator of the column factor loading matrix

\hat{\bold{C}}=\sqrt{p_{2}}\left(\sum_{t=1}^{T}\bold{X}_t^\top\hat{\bold{R}}\hat{\bold{F}}_t\right)\left[\left(\sum_{t=1}^{T}\hat{\bold{F}}_t^\top\hat{\bold{R}}^\top\bold{X}_t\right)\left(\sum_{t=1}^{T}\bold{X}_t^\top\hat{\bold{R}}\hat{\bold{F}}_t\right)\right]^{-1/2}.

Afterwards, we sequentially update \bold{F}, \bold{R} and \bold{C}. In simulation, the iterative procedure is terminated either when a pre-specified maximum iteration number (\text{maxiter}=100) is reached or when

\sum_{t=1}^{T}\|\hat{\bold{S}}^{(s+1)}_t-\hat{\bold{S}}^{(s)}_t\|_{F} \leq \epsilon \cdot Tp_1p_2,

where \hat{\bold{S}}^{(s)}_t is the common component estimated at the s-th step, \epsilon is a small constant (10^{-6}) given in advance.

Value

The return value is a list. In this list, it contains the following:

Author(s)

Yong He, Ran Zhao, Wen-Xin Zhou.

References

He, Y., Zhao, R., & Zhou, W. X. (2023). Iterative Alternating Least Square Estimation for Large-dimensional Matrix Factor Model. <arXiv:2301.00360>.

Examples

set.seed(11111)
T=20;p1=20;p2=20
k1=3;k2=3

R=matrix(runif(p1*k1,min=-1,max=1),p1,k1)
C=matrix(runif(p2*k2,min=-1,max=1),p2,k2)

X=E=array(0,c(T,p1,p2))
F=array(0,c(T,k1,k2))

for(t in 1:T){
  F[t,,]=matrix(rnorm(k1*k2),k1,k2)
  E[t,,]=matrix(rnorm(p1*p2),p1,p2)
}
for(t in 1:T){
X[t,,]=R%*%F[t,,]%*%t(C)+E[t,,]
}

#Estimating the matrix factor model using the default initial values
fit1 = IALS(X, W1 = NULL, W2 = NULL,k1, k2, max_iter = 100, ep = 1e-06) 
Distance(fit1$R,R);Distance(fit1$C,C)

#Estimating the matrix factor model using one-step iteration
fit2 = IALS(X, W1 = NULL , W2 = NULL, k1, k2, max_iter = 1, ep = 1e-06) 
Distance(fit2$R,R);Distance(fit2$C,C)

Estimating the Pair of Factor Numbers via Eigenvalue Ratios Corresponding to IALS

Description

The function is to estimate the pair of factor numbers via eigenvalue ratios corresponding to IALS method.

Usage

KIALS(X, W1 = NULL, W2 = NULL, kmax, max_iter = 100, ep = 1e-06)

Arguments

Details

In detail, we first set k_{\max} is a predetermined upper bound for k_1,k_2 and thus by IALS method, we can obtain the estimate of \bold{F}_t, denote as \hat{\bold{F}}_t, which is of dimension k_{\max}\times k_{\max}. Then the dimensions k_1 and k_2 are further determined as follows:

\hat{k}_{1}=\arg\max_{j \leq k_{\max}}\frac{\lambda_{j}\left(\dfrac{1}{T}\sum_{t=1}^{T}\hat{\bold{F}}_t\hat{\bold{F}}_t^\top\right)}{\lambda_{j+1}\left(\frac{1}{T}\sum_{t=1}^{T}\hat{\bold{F}}_t\hat{\bold{F}}_t^\top\right)},

\hat{k}_{2}=\arg\max_{j \leq k_{\max}}\frac{\lambda_{j}\left(\dfrac{1}{T}\sum_{t=1}^{T}\hat{\bold{F}}_t^\top\hat{\bold{F}}_t\right)}{\lambda_{j+1}\left(\frac{1}{T}\sum_{t=1}^{T}\hat{\bold{F}}_t^\top\hat{\bold{F}}_t\right)}.

Value

Author(s)

Yong He, Ran Zhao, Wen-Xin Zhou.

References

He, Y., Zhao, R., & Zhou, W. X. (2023). Iterative Alternating Least Square Estimation for Large-dimensional Matrix Factor Model. <arXiv:2301.00360>.

Examples

set.seed(11111)
T=20;p1=20;p2=20
k1=3;k2=3

R=matrix(runif(p1*k1,min=-1,max=1),p1,k1)
C=matrix(runif(p2*k2,min=-1,max=1),p2,k2)

X=E=array(0,c(T,p1,p2))
F=array(0,c(T,k1,k2))

for(t in 1:T){
  F[t,,]=matrix(rnorm(k1*k2),k1,k2)
  E[t,,]=matrix(rnorm(p1*p2),p1,p2)
}

for(t in 1:T){
X[t,,]=R%*%F[t,,]%*%t(C)+E[t,,]
}

kmax=8
K=KIALS(X, W1 = NULL, W2 = NULL, kmax, max_iter = 100, ep = 1e-06);K