FCCor: Calculation of functional correlation between two...

In hadjipantelis/tPACE: Functional Data Analysis and Empirical Dynamics

Calculation of functional correlation between two simultaneously observed processes.

Description

Calculation of functional correlation between two simultaneously observed processes.

Usage

FCCor(
  x,
  y,
  Lt,
  bw = stop("bw missing"),
  kern = "epan",
  Tout = sort(unique(unlist(Lt)))
)

Arguments

x	A list of function values corresponding to the first process.
y	A list of function values corresponding to the second process.
Lt	A list of time points for both x and y.
bw	A numeric vector for bandwidth of length either 5 or 1, specifying the bandwidths for E(X), E(Y), var(X), var(Y), and cov(X, Y). If bw is a scalar then all five bandwidths are chosen to be the same.
kern	Smoothing kernel for mu and covariance; "rect", "gauss", "epan", "gausvar", "quar" (default: "gauss")
Tout	Output time points. Default to the sorted unique time points.

Details

FCCor calculate only the concurrent correlation corr(X(t), Y(t)) (note that the time points t are the same). It assumes no measurement error in the observed values.

Value

A list with the following components:

corr	A vector of the correlation corr(X(t), Y(t)) evaluated at Tout.
Tout	Same as the input Tout.
bw	The bandwidths used for E(X), E(Y), var(X), var(Y), and cov(X, Y).

Examples

set.seed(1)
n <- 200
nGridIn <- 50
sparsity <- 1:5 # must have length > 1
bw <- 0.2
kern <- 'epan'
T <- matrix(seq(0.5, 1, length.out=nGridIn))

## Corr(X(t), Y(t)) = 1/2
A <- Wiener(n, T)
B <- Wiener(n, T) 
C <- Wiener(n, T) + matrix((1:nGridIn) , n, nGridIn, byrow=TRUE)
X <- A + B
Y <- A + C
indEach <- lapply(1:n, function(x) sort(sample(nGridIn, sample(sparsity, 1))))
tAll <- lapply(1:n, function(i) T[indEach[[i]]])
Xsp <- lapply(1:n, function(i) X[i, indEach[[i]]])
Ysp <- lapply(1:n, function(i) Y[i, indEach[[i]]])

plot(T, FCCor(Xsp, Ysp, tAll, bw)[['corr']], ylim=c(-1, 1))
abline(h=0.5)

hadjipantelis/tPACE documentation built on Aug. 16, 2022, 10:45 a.m.