Mean variance optimisation - r

I am doing a mean variance optimization to solve portfolios optimization problem. What I am trying to do is to minimize the variance with respect both constraints :
x1m1+x2m2+...+xnmn=m
x1+x2+...+xn=1
So this is the code I did:
################ Simulation for n=3 ################
################ Parameters ################
mu<-50 ## Mean of the portfolio
n<-3 ## Number of asset
m1<-30000 ## Size of the simulation
########### 3 Assets ############
x<- rnorm(m1,2,1)
y<- rnorm(m1,0.5,1.5)
z<- rnorm(m1,3.75,1)
d<-data.frame(x,y,z)
################ Solution Directe ################
Sol<-function(m1) {
A = matrix(nrow=n+2, ncol=n+2)
for (i in 1:n){
for (j in 1:n)
if(i==j) {
A[i,j] <- (2*var(d[,i]))
} else {
A[i,j] <- cov(d[,i],d[,j])
}
}
for (i in 1:n){
A[i,n+1] <- -mean(d[,i])
A[i,n+2] <- -1
}
for (j in 1:n){
A[n+1,j] <- mean(d[,j])
A[n+2,j] <- 1
}
for (i in 2:n+2){
for (j in 2:n+2)
if(i==j) {
A[i,j] <- 0
} else {
A[i,j] <- 0
}
}
A
Inv=solve(A)
Sol=Inv%*%c(0,0,0,m1,1)
result=list(x=Sol,A=A,Inv=Inv)
return(result)
}
Sol(mu)
Sol(mu)$x ## The solution
Sol(mu)$A
I known, I´m using a lot of bad things for R, but I could not figure out a better solution. So my question is it correct?
Any correction and suggestion to improve this process! please feel free to share your extant code in R.
Huge thanks!

One way is to minimize numerically by solnp() from the Rsolnp package. This also offers a way to add more restrictions (leverage constraints etc):
muVec <- colMeans(d) #mean-vector of assets
Sigma <- cov(d) #covariance-matrix
fmin <- function(x) as.numeric(t(x) %*% Sigma %*% x) #variance of portfolio to min.
eqn <- function(x) c(t(x) %*% muVec, sum(x)) #equality restriction
sol <- function(mu) Rsolnp::solnp(rep(0.5, 3), fun=fmin, eqfun=eqn, eqB=c(mu,1))
x <- sol(50)
after solving we can now print the parameters and portfolio variance:
> x$par
[1] -5.490106 -11.270906 17.761012
> x$vscale[1]
[1] 630.4916
In your simple case a closed solution exists and can be boiled down to:
S <- solve(Sigma)
A <- matrix( c(t(muVec) %*% S %*% muVec,
rep( t(muVec) %*% S %*% rep(1,3), 2),
t(rep(1,3)) %*% S %*% rep(1,3)), ncol=2
)
sol2 <- function(mu) S %*% cbind(muVec,1) %*% solve(A) %*% c(mu,1)
which "luckily" gives the same results:
> sol2(50)
[,1]
x -5.490106
y -11.270906
z 17.761012
> fmin(sol2(50))
[1] 630.4916

Related

Estimating risk difference using delta approach

I want to estimate the confidence interval using the delta approach; an image of an example model is attached output of delta approach. The intervals are very wide compared to the MLE or bootstrap estimates for bootstrap approach. Can anyone suggest a way to derive the accurate CI's using the delta approach? Thank you
Mod1 <- Mod0 <- model.matrix(object)[subset, ]
n <- nrow(Mod0)
Nvec <- matrix(rep(c(1/n,0,0,1/n),each=n), n*2, 2)
if (method == 'delta') {
if (scale == 'linear') {
df <- deriv( ~y/x, c('x', 'y')) # y depends on x
}
out <- sapply(parm, function(p) {
Mod0[, p] <- 0
Mod1[, p] <- 1
Mod <- rbind(Mod0, Mod1)
allpreds <- family(object)$linkinv(Mod %*% coef(object))
avgpreds <- t(Nvec) %*% allpreds
val <- f(avgpreds)
V <- sweep(chol(vcov(object)) %*% t(Mod), allpreds*(1-allpreds), '*', MARGIN = 2) %*% Nvec
V <- t(V) %*% V
dxdy <- matrix(attr(eval(df, list('x'=avgpreds[1], 'y'=avgpreds[2])), 'gradient'))
se <- sqrt(t(dxdy) %*% V %*% dxdy)
out <- c(val, se, z <- abs({val-null}/se), 2*pnorm(abs(val/se), lower.tail=FALSE), val + qnorm(cilevel[1])*se, val + qnorm(cilevel[2])*se)
names(out) <- c(name, 'Std. Error', 'Z-value', 'p-value', ciname)
out
})
out <- t(out)
rownames(out) <- names(cf)
return(out)
} ## end if delta

What is going on inside the varimax function in R?

I have been trying to figure out the core part of the varimax function in R. I found a wiki link that writes out the algorithm. But why is B <- t(x) %*% (z^3 - z %*% diag(drop(rep(1, p) %*% z^2))/p) is computed? I also am not sure as to why SVD is computed of the matrix B. The iteration step is probably to maximize/minimize the variance, and the singular values would really be variances of Principal Components. But I am also unsure about that. I am pasting the whole code of varimax for convenience, but really the relevant part and therefore my question on what is actually happening under the hood, is within the for loop.
function (x, normalize = TRUE, eps = 1e-05)
{
nc <- ncol(x)
if (nc < 2)
return(x)
if (normalize) {
sc <- sqrt(drop(apply(x, 1L, function(x) sum(x^2))))
x <- x/sc
}
p <- nrow(x)
TT <- diag(nc)
d <- 0
for (i in 1L:1000L) {
z <- x %*% TT
B <- t(x) %*% (z^3 - z %*% diag(drop(rep(1, p) %*% z^2))/p)
sB <- La.svd(B)
TT <- sB$u %*% sB$vt
dpast <- d
d <- sum(sB$d)
if (d < dpast * (1 + eps))
break
}
z <- x %*% TT
if (normalize)
z <- z * sc
dimnames(z) <- dimnames(x)
class(z) <- "loadings"
list(loadings = z, rotmat = TT)
}
Edit: The algorithm is available in the book "Factor Analysis of Data Matrices" by Holt, Rinehart and Winston and the actual sources can be found therein. This book is also cited with the varimax function in R.

Portfolio Optimisation under weight constraints

With a lot of help from contributors to StackOverflow I have managed to put together a function to derive the weights of a 2-asset portfolio which maximises the Sharpe ratio. No short sales are allowed and the sum of weights add to 1. What I would like to do now is to constrain asset A to not being more or less than 10% from a user defined weight. As an example I would like to constrain the weight of asset A to be no less than 54% or more than 66% (i.e 60% +/- 10%). So on the below example I would end up with weights of (0.54,0.66) instead of the unsconstrained (0.243,0.7570) .I assume this can be done by tweaking bVect but I am not so sure how to go about it. Any help would be appreciated.
asset_A <- c(0.034320510,-0.001209628,0.031900161,0.023163947,-0.001872938,-0.010322489,0.006090395,-0.003270854,0.017778990,0.017204915)
asset_B <- c(0.047103261,0.055175057,0.021019816,0.020602347,0.007281368,-0.006547404,0.019155238,0.005494798,0.025429958,0.014929124)
require(quadprog)
HR_solve <- function(asset_A,asset_B) {
vol_A <- sd(asset_A)
vol_B <- sd(asset_B)
cor_AB <- cor(cbind(asset_A,asset_B),method="pearson")
ret_A_B <- as.matrix(c(mean(asset_A),mean(asset_B)))
vol_AB <- c(vol_A,vol_B)
covmat <- diag(as.vector(vol_AB))%*%cor_AB%*%diag(as.vector(vol_AB))
aMat <- cbind(rep(1,nrow(covmat)),diag(1,nrow(covmat)))
bVec <- c(1,0,0)
zeros <- array(0, dim = c(nrow(covmat),1))
minw <- solve.QP(covmat, zeros, aMat, bVec, meq = 1 ,factorized = FALSE)$solution
rp <- as.numeric(t(minw) %*% ret_A_B)
sp <- sqrt(t(minw) %*% covmat %*% minw)
wp <- t(matrix(minw))
sret <- sort(seq(t(minw) %*% ret_A_B,max(ret_A_B),length.out=100))
aMatt <- cbind(ret_A_B,aMat)
for (ri in sret[-1]){
bVect <- c(ri,bVec)
result <- tryCatch({solve.QP(covmat, zeros, aMatt, bVect, meq = 2,factorized = FALSE)},
warning = function(w){ return(NULL) } , error = function(w){ return(NULL)}, finally = {} )
if (!is.null(result)){
wp <- rbind(wp,as.vector(result$solution))
rp <-c(rp,t(as.vector(result$solution) %*% ret_A_B))
sp <- c(sp,sqrt(t(as.vector(result$solution)) %*% covmat %*% as.vector(result$solution))) }
}
HR_weights <- wp[which.max(rp/sp),]
as.matrix(HR_weights)
}
HR_solve(asset_A,asset_B)
[,1]
[1,] 0.2429662
[2,] 0.7570338
I think you should take a look at the link below.
http://economistatlarge.com/portfolio-theory/r-optimized-portfolio/r-code-graph-efficient-frontier
I think you'll learn a lot from that. I'll post the code here, in case the link gets shut down sometime in the future.
# Economist at Large
# Modern Portfolio Theory
# Use solve.QP to solve for efficient frontier
# Last Edited 5/3/13
# This file uses the solve.QP function in the quadprog package to solve for the
# efficient frontier.
# Since the efficient frontier is a parabolic function, we can find the solution
# that minimizes portfolio variance and then vary the risk premium to find
# points along the efficient frontier. Then simply find the portfolio with the
# largest Sharpe ratio (expected return / sd) to identify the most
# efficient portfolio
library(stockPortfolio) # Base package for retrieving returns
library(ggplot2) # Used to graph efficient frontier
library(reshape2) # Used to melt the data
library(quadprog) #Needed for solve.QP
# Create the portfolio using ETFs, incl. hypothetical non-efficient allocation
stocks <- c(
"VTSMX" = .0,
"SPY" = .20,
"EFA" = .10,
"IWM" = .10,
"VWO" = .30,
"LQD" = .20,
"HYG" = .10)
# Retrieve returns, from earliest start date possible (where all stocks have
# data) through most recent date
returns <- getReturns(names(stocks[-1]), freq="week") #Currently, drop index
#### Efficient Frontier function ####
eff.frontier <- function (returns, short="no", max.allocation=NULL,
risk.premium.up=.5, risk.increment=.005){
# return argument should be a m x n matrix with one column per security
# short argument is whether short-selling is allowed; default is no (short
# selling prohibited)max.allocation is the maximum % allowed for any one
# security (reduces concentration) risk.premium.up is the upper limit of the
# risk premium modeled (see for loop below) and risk.increment is the
# increment (by) value used in the for loop
covariance <- cov(returns)
print(covariance)
n <- ncol(covariance)
# Create initial Amat and bvec assuming only equality constraint
# (short-selling is allowed, no allocation constraints)
Amat <- matrix (1, nrow=n)
bvec <- 1
meq <- 1
# Then modify the Amat and bvec if short-selling is prohibited
if(short=="no"){
Amat <- cbind(1, diag(n))
bvec <- c(bvec, rep(0, n))
}
# And modify Amat and bvec if a max allocation (concentration) is specified
if(!is.null(max.allocation)){
if(max.allocation > 1 | max.allocation <0){
stop("max.allocation must be greater than 0 and less than 1")
}
if(max.allocation * n < 1){
stop("Need to set max.allocation higher; not enough assets to add to 1")
}
Amat <- cbind(Amat, -diag(n))
bvec <- c(bvec, rep(-max.allocation, n))
}
# Calculate the number of loops
loops <- risk.premium.up / risk.increment + 1
loop <- 1
# Initialize a matrix to contain allocation and statistics
# This is not necessary, but speeds up processing and uses less memory
eff <- matrix(nrow=loops, ncol=n+3)
# Now I need to give the matrix column names
colnames(eff) <- c(colnames(returns), "Std.Dev", "Exp.Return", "sharpe")
# Loop through the quadratic program solver
for (i in seq(from=0, to=risk.premium.up, by=risk.increment)){
dvec <- colMeans(returns) * i # This moves the solution along the EF
sol <- solve.QP(covariance, dvec=dvec, Amat=Amat, bvec=bvec, meq=meq)
eff[loop,"Std.Dev"] <- sqrt(sum(sol$solution*colSums((covariance*sol$solution))))
eff[loop,"Exp.Return"] <- as.numeric(sol$solution %*% colMeans(returns))
eff[loop,"sharpe"] <- eff[loop,"Exp.Return"] / eff[loop,"Std.Dev"]
eff[loop,1:n] <- sol$solution
loop <- loop+1
}
return(as.data.frame(eff))
}
# Run the eff.frontier function based on no short and 50% alloc. restrictions
eff <- eff.frontier(returns=returns$R, short="no", max.allocation=.50,
risk.premium.up=1, risk.increment=.001)
# Find the optimal portfolio
eff.optimal.point <- eff[eff$sharpe==max(eff$sharpe),]
# graph efficient frontier
# Start with color scheme
ealred <- "#7D110C"
ealtan <- "#CDC4B6"
eallighttan <- "#F7F6F0"
ealdark <- "#423C30"
ggplot(eff, aes(x=Std.Dev, y=Exp.Return)) + geom_point(alpha=.1, color=ealdark) +
geom_point(data=eff.optimal.point, aes(x=Std.Dev, y=Exp.Return, label=sharpe),
color=ealred, size=5) +
annotate(geom="text", x=eff.optimal.point$Std.Dev,
y=eff.optimal.point$Exp.Return,
label=paste("Risk: ",
round(eff.optimal.point$Std.Dev*100, digits=3),"\nReturn: ",
round(eff.optimal.point$Exp.Return*100, digits=4),"%\nSharpe: ",
round(eff.optimal.point$sharpe*100, digits=2), "%", sep=""),
hjust=0, vjust=1.2) +
ggtitle("Efficient Frontier\nand Optimal Portfolio") +
labs(x="Risk (standard deviation of portfolio)", y="Return") +
theme(panel.background=element_rect(fill=eallighttan),
text=element_text(color=ealdark),
plot.title=element_text(size=24, color=ealred))
ggsave("Efficient Frontier.png")
Ok I have found a way to do this... if you think there is a more elegant way please let me know...
require(quadprog)
HR_solve <- function(asset_A,asset_B,mean_A,range_A) {
vol_A <- sd(asset_A)
vol_B <- sd(asset_B)
cor_AB <- cor(cbind(asset_A,asset_B),method="pearson")
ret_A_B <- as.matrix(c(mean(asset_A),mean(asset_B)))
vol_AB <- c(vol_A,vol_B)
covmat <- diag(as.vector(vol_AB))%*%cor_AB%*%diag(as.vector(vol_AB))
bVec <- c(1,0,0)
aMat <- cbind(rep(1,nrow(covmat)),diag(1,nrow(covmat)))
zeros <- array(0, dim = c(nrow(covmat),1))
minw <- solve.QP(covmat, zeros, aMat, bVec, meq = 1 ,factorized = FALSE)$solution
rp <- as.numeric(t(minw) %*% ret_A_B)
sp <- sqrt(t(minw) %*% covmat %*% minw)
wp <- t(matrix(minw))
sret <- sort(seq(t(minw) %*% ret_A_B,max(ret_A_B),length.out=1000))
min_A <- mean_A * (1-range_A)
max_A <- mean_A * (1+range_A)
aMatt <- cbind(ret_A_B,aMat,-diag(2))
bVec <- c(1,min_A,0,-max_A,-1)
for (ri in sret[-1]){
bVect <- c(ri,bVec)
result <- tryCatch({solve.QP(covmat, zeros, aMatt, bVect, meq = 2,factorized = FALSE)},
warning = function(w){ return(NULL) } , error = function(w){ return(NULL)}, finally = {} )
if (!is.null(result)){
wp <- rbind(wp,as.vector(result$solution))
rp <-c(rp,t(as.vector(result$solution) %*% ret_A_B))
sp <- c(sp,sqrt(t(as.vector(result$solution)) %*% covmat %*% as.vector(result$solution))) }
}
HR_weights <- wp[which.max(rp/sp),]
as.matrix(HR_weights)
}
Just change aMat and bVec:
# sset A to be no less than 54% or more than 66%
aMat <- cbind(rep(1,nrow(covmat)),diag(1,nrow(covmat)),c(1,0),c(-1,0))
bVec <- c(1,0,0,.54,-.66)

Newton Raphson for logistic regression

I did code for Newton Raphson for logistic regression. Unfortunately I tried many data there is no convergence. there is a mistake I do not know where is it. Can anyone help to figure out what is the problem.
First the data is as following; y indicate the response (0,1) , Z is 115*30 matrix which is the exploratory variables. I need to estimate the 30 parameters.
y = c(rep(0,60),rep(1,55))
X = sample(c(0,1),size=3450,replace=T)
Z = t(matrix(X,ncol=115))
#The code is ;
B = matrix(rep(0,30*10),ncol=10)
B[,1] = matrix(rep(0,30),ncol=1)
for(i in 2 : 10){
print(i)
p <- exp(Z %*%as.matrix(B[,i])) / (1 + exp(Z %*% as.matrix(B[,i])))
v.2 <- diag(as.vector(1 * p*(1-p)))
score.2 <- t(Z) %*% (y - p) # score function
increm <- solve(t(Z) %*% v.2 %*% Z)
B[,i] = as.matrix(B[,i-1])+increm%*%score.2
if(B[,i]-B[i-1]==matrix(rep(0.0001,30),ncol=1)){
return(B)
}
}
Found it! You're updating p based on B[,i], you should be using B[,i-1] ...
While I was finding the answer, I cleaned up your code and incorporated the results in a function. R's built-in glm seems to work (see below). One note is that this approach is likely to be unstable: fitting a binary model with 30 predictors and only 115 binary responses, and without any penalization or shrinkage, is extremely optimistic ...
set.seed(101)
n.obs <- 115
n.zero <- 60
n.pred <- 30
y <- c(rep(0,n.zero),rep(1,n.obs-n.zero))
X <- sample(c(0,1),size=n.pred*n.obs,replace=TRUE)
Z <- t(matrix(X,ncol=n.obs))
R's built-in glm fitter does work (it uses iteratively reweighted least squares, not N-R):
g1 <- glm(y~.-1,data.frame(y,Z),family="binomial")
(If you want to view the results, library("arm"); coefplot(g1).)
## B_{m+1} = B_m + (X^T V_m X)^{-1} X^T (Y-P_m)
NRfit function:
NRfit <- function(y,X,start,n.iter=100,tol=1e-4,verbose=TRUE) {
## used X rather than Z just because it's more standard notation
n.pred <- ncol(X)
B <- matrix(NA,ncol=n.iter,
nrow=n.pred)
B[,1] <- start
for (i in 2:n.iter) {
if (verbose) cat(i,"\n")
p <- plogis(X %*% B[,i-1])
v.2 <- diag(c(p*(1-p)))
score.2 <- t(X) %*% (y - p) # score function
increm <- solve(t(X) %*% v.2 %*% X)
B[,i] <- B[,i-1]+increm%*%score.2
if (all(abs(B[,i]-B[,i-1]) < tol)) return(B)
}
B
}
matplot(res1 <- t(NRfit(y,Z,start=coef(g1))))
matplot(res2 <- t(NRfit(y,Z,start=rep(0,ncol(Z)))))
all.equal(res2[6,],unname(coef(g1))) ## TRUE

customized ridge regression function

I am working on Ridge regression, I want to make my own function. It tried the following. It work for individual value of k but not for array for sequence of values.
dt<-longley
attach(dt)
library(MASS)
X<-cbind(X1,X2,X3,X4,X5,X6)
X<-as.matrix(X)
Y<-as.matrix(Y)
sx<-scale(X)/sqrt(nrow(X)-1)
sy<-scale(Y)/sqrt(nrow(Y)-1)
rxx<-cor(sx)
rxy<-cor(sx,sy)
for (k in 0:1){
res<-solve(rxx+k*diag(rxx))%*%rxy
k=k+0.01
}
Need help for optimized code too.
poly.kernel <- function(v1, v2=v1, p=1) {
((as.matrix(v1) %*% t(v2))+1)^p
}
KernelRidgeReg <- function(TrainObjects,TrainLabels,TestObjects,lambda){
X <- TrainObjects
y <- TrainLabels
kernel <- poly.kernel(X)
design.mat <- cbind(1, kernel)
I <- rbind(0, cbind(0, kernel))
M <- crossprod(design.mat) + lambda*I
#crossprod is just x times traspose of x, just looks neater in my openion
M.inv <- solve(M)
#inverse of M
k <- as.matrix(diag(poly.kernel(cbind(TrainObjects,TrainLabels))))
#Removing diag still gives the same MSE, but will output a vector of prediction.
Labels <- rbind(0,as.matrix(TrainLabels))
y.hat <- t(Labels) %*% M.inv %*% rbind(0,k)
y.true <- Y.test
MSE <-mean((y.hat - y.true)^2)
return(list(MSE=MSE,y.hat=y.hat))
}
Kernel with p=1, will give you ridge regression.
Solve built-in R function sometimes return singular matrix. You may want to write your own function to avoid that.

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