Displaying 2 results from an estimated 2 matches for "epsilon2".
Did you mean:
epsilon
2007 Aug 05
1
Understanding of Johansen test.
...zero"
My understanding is following : suppose, y[t] is of order 3 and p = 1
Then Delta y[t] = A[0]*y[t-1] + epsilon[t]
Hence : Delta y1[t] = a[11]*y1[t-1] + a[12]*y1[t-1] +a[13]*y1[t-1] + epsilon1[t]
Delta y2[t] = a[12]*y1[t-1] + a[22]*y1[t-1] +a[23]*y1[t-1] + epsilon2[t]
Delta y3[t] = a[31]*y1[t-1] + a[32]*y1[t-1] +a[33]*y1[t-1] + epsilon3[t]
But is rank of A[0] is 0 then it is possible to find non-zero coef for all of above three equations such that : a[11]*y1[t-1] + a[12]*y1[t-1] +a[13]*y1[t-1] = 0
a[12]*y1[t-1] + a[22]*y1...
2010 Nov 18
0
On efficiency, Vectorize and loops
...3<-function(u,i){
sqrt(2)+sin(pi*outer(i,u))
}
# Building the random functional process 'epsilon'
epsilon<-lapply(1:n, function(t)local({force(t);
function(u){
epsilon<-0
for (j in 1:10){
epsilon<-epsilon+2^(1-j)*Z[j,t]*zeta(u,j)
}
epsilon<-epsilon
}
}))
epsilon2<-lapply(1:n, function(t)local({force(t);
function(u){
crossprod(2^(1-(1:10))*Z[,t],zeta2(1:10)(u))
}
}))
epsilon3<-lapply(1:n, function(t)local({force(t);
function(u){
crossprod(2^(1-(1:10))*Z[,t],zeta3(u,1:10))
}
}))
###########################
# Building the function '...