I have a couple of questions regarding to the piece of code shown below, the function called "Func1" will return a matrix as a result, the size of the matrix will be 50 rows and 15 columns, I called it "M", and "M2" is just the transpose of it. W0 is the initial value for the next part of the code, if I run the function called "Rowresult", then it also give me a 50*15 matrix.
My first question is: if I want to run the "Rowresult" function for different W0 values,such as W0 = 10,20,30. and I want to have 3 matrices in the size of 50*15 with different W0 values as results,how could I achieve it?
My second question is : if you tried my code in R, you will see a matrix called "wealth_result 2" as a result. once I got this big matrix, I would like to divide it (50*15 matrix) into three same size matrix, each matrix has a size of 50*5 (so they share the same rows but different columns, the first matrix takes the first 5 columns, the second takes 6-10 columns, third one takes 11-15 columns),and then I want to work out how many positive rows (rows with all numbers positive) among each of the 50 *5 matrix? How could I achieve this?
N=15
func1<-function(N){
alpha1 = 8.439e-02
beta1 = 8.352e-01
mu = 7.483e-03
omega = 1.343e-04
X_0 = -3.092031e-02
sigma_0 = 0.03573968
eps = rt (N,7.433e+00)
# loops
Xn= numeric (N)
sigma= numeric (N)
sigma[1] = sigma_0
Xn[1] = X_0
for (t in 2:N){
sigma[t] = sqrt (omega + alpha1 * (Xn[t-1])^2 + beta1* (sigma[t-1])^2)
Xn[t] = sigma[t] * eps[t]
}
Y = mu + Xn
}
# return matrix
M<-replicate(50,func1(N))
# returns matrix
M2<-t(M)
View(M2)
# wealth with initial wealth 10
W0=10
# 10,20,30,40
r= c(0.101309031, -0.035665516, -0.037377270, -0.005928941, 0.036612849,
0.062404039, 0.124240950, -0.034843633, 0.004770613, 0.005018101,
0.097685945, -0.090660099, 0.004863099, 0.029215984, 0.020835366)
Rowresult<- function(r){
const = exp(cumsum(r))
exp.cum = cumsum(1/const)
wealth=const*(W0 - exp.cum)
wealth
}
# wealth matrix
wealth_result <-apply(M2,1,Rowresult)
wealth_result2 <-t(wealth_result )
View(wealth_result2)
This delivers the desired counds of (all) "positive rows":
> sapply(1:3, function(m) sum( rowSums( wealth_result2[ , (1:5)+(m-1)*5 ] >0 )) )
[1] 250 230 2
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