Optimisation in R using Ucminf package - r

I am not able to apply ucminf function to minimise my cost function in R.
Here is my cost function:
costfunction <- function(X,y,theta){
m <- length(y);
J = 1/m * ((-t(y)%*%log(sigmoid(as.matrix(X)%*%as.matrix(theta)))) - ((1-t(y))%*%log(1-sigmoid(as.matrix(X)%*%as.matrix(theta)))))
}
Here is my sigmoid function:
sigmoid <- function(t){
g = 1./(1+exp(-t))
}
Here is my gradient function:
gradfunction <- function(X,y,theta){
grad = 1/ m * t(X) %*% (sigmoid(as.matrix(X) %*% as.matrix(theta) - y));
}
I am trying to do the following:
library("ucminf")
data <- read.csv("ex2data1.txt",header=FALSE)
X <<- data[,c(1,2)]
y <<- data[,3]
qplot(X[,1],X[,2],colour=factor(y))
m <- dim(X)[1]
n <- dim(X)[2]
X <- cbind(1,X)
initial_theta <<- matrix(0,nrow=n+1,ncol=1)
cost <- costfunction(X,y,initial_theta)
grad <- gradfunction(X,y,initial_theta)
This is where I want to call ucminf to find the minimum cost and values of theta. I am not sure how to do this.

Looks like you are trying to do the week2 problem of the machine learning course of Coursera.
No need to use ucminf packages here, you can simply use the R function optim it works
We will define the sigmoid and cost function first.
sigmoid <- function(z)
1 / (1 + exp(-z))
costFunction <- function(theta, X, y) {
m <- length(y)
J <- -(1 / m) * crossprod(c(y, 1 - y),
c(log(sigmoid(X %*% theta)), log(1 - sigmoid(X %*% theta))))
grad <- (1 / m) * crossprod(X, sigmoid(X %*% theta) - y)
list(J = J, grad = grad)
}
Let's load the data now, to make this code it reproductible, I put the data in my dropbox.
download.file("https://dl.dropboxusercontent.com/u/8750577/ex2data1.txt",
method = "curl", destfile = "/tmp/ex2data1.txt")
data <- matrix(scan('/tmp/ex2data1.txt', what = double(), sep = ","),
ncol = 3, byrow = TRUE)
X <- data[, 1:2]
y <- data[, 3, drop = FALSE]
m <- nrow(X)
n <- ncol(X)
X <- cbind(1, X)
initial_theta = matrix(0, nrow = n + 1)
We can then compute the result of the cost function at the initial theta like this
cost <- costFunction(initial_theta, X, y)
(grad <- cost$grad)
## [,1]
## [1,] -0.100
## [2,] -12.009
## [3,] -11.263
(cost <- cost$J)
## [,1]
## [1,] 0.69315
Finally we can use optim to ge the optimal theta
res <- optim(par = initial_theta,
fn = function(t) costFunction(t, X, y)$J,
gr = function(t) costFunction(t, X, y)$grad,
method = "BFGS", control = list(maxit = 400))
(theta <- res$par)
## [,1]
## [1,] -25.08949
## [2,] 0.20566
## [3,] 0.20089
(cost <- res$value)
## [1] 0.2035
If you have some problem with the function download.file, the data can be downloaded
here

As you did not provide a reproducible example it is hard to exactly give you the code you need, but the general idea is to hand the functions over to ucminf:
ucminf(start, costfunction, gradfunction, y = y, theta = initial_theta)
Note that start needs to be a vector of initial starting values which when handed over as X to the two functions need to produce a result. Usually you use random starting value (e.g., runif).

Related

How to predict and extract R Squared with .lm.fit?

As the title suggest, I have seen some user mentioned that .lm.fit() functions has an advantage of more speed than a regular lm(), but when i look deeper at .lm.fit() in help, it is supposed to be a fitter functions, it returns a set of list instead of a model, which makes me to think is it still possible to extract components like R squared, Adj R Squared, and lastly do a predict() out of it?
Below is sample data and executions:
test_dat <- data.frame(y = rnorm(780, 20, 10))
for(b in 1:300){
name_var <- paste0("x",b)
test_dat[[name_var]] <- rnorm(780, 0.01 * b, 5)
}
tic()
obj_lm <- lm(y ~ ., data = test_dat)
print(class(obj_lm))
print(summary(obj_lm)$r.squared)
print(summary(obj_lm)$adj.r.squared)
predict(obj_lm)
toc() #approximately 0.4 seconds
tic()
datm <- as.matrix(test_dat)
obj_lm_fit <- .lm.fit(cbind(1,datm[,-1]), datm[,1])
print(class(obj_lm_fit))
toc() #approximately 0.2 seconds
Functions predict and resid are generic and since .lm.fit returns an object of class "list", all you have to do is to write methods implementing the definitions of what you want. Below are methods to compute fitted values, residuals and R^2.
set.seed(2023) # make the results reproducible
test_dat <- data.frame(y = rnorm(780, 20, 10))
for(b in 1:300){
name_var <- paste0("x",b)
test_dat[[name_var]] <- rnorm(780, 0.01 * b, 5)
}
obj_lm <- lm(y ~ ., data = test_dat)
datm <- as.matrix(test_dat)
obj_lm_fit <- .lm.fit(cbind(1,datm[,-1]), datm[,1])
#------------------------------------------------------------------------
# the methods for objects of class "list"
#
fitted.list <- function(object, X) {
X %*% coef(object)
}
resid.list <- residuals.list <- function(object, X, y) {
y_fitted <- fitted(object, X)
y - y_fitted
}
rsquared <- function(x, ...) UseMethod("rsquared")
rsquared.default <- function(x, ...) {
summary(x)$r.squared
}
rsquared.list <- function(object, X, y) {
e <- resid.list(object, X, y)
1 - sum(e^2)/sum( (y - mean(y))^2 )
}
rsquared(obj_lm_fit, cbind(1,datm[,-1]), datm[,1])
#> [1] 0.3948863
rsquared(obj_lm)
#> [1] 0.3948863
Created on 2023-01-03 with reprex v2.0.2
Edit 1
Added method to also calculate adj.R2
adj_rsquared_list <- function(object, X, y){
r2 <- rsquared.list(object, X, y)
k <- ncol(X) - 1
n <- nrow(X)
rate_of_error <- (1 - r2) * (n - 1) / (n - k - 1)
adj_r2 <- 1 - rate_of_error
return(adj_r2)
}
adj_rsquared_list(obj_lm_fit, cbind(1,datm[,-1]), datm[,1])
#> [1] 0.01590061
Edit 2
After the edit by Jovan, I have changed fitted.list above to use coef(), a function that extracts the first arguments list member "coefficients", if it exists, and rewrote the default and list methods of rsquared to accept a adj argument. The code to compute the adjusted R^2 is a copy&paste of Jovan's code.
rsquared <- function(x, ...) UseMethod("rsquared")
rsquared.default <- function(x, adj = FALSE, ...) {
if(adj) {
summary(x)$adj.r.squared
} else summary(x)$r.squared
}
rsquared.list <- function(object, X, y, adj = FALSE) {
e <- resid.list(object, X, y)
r2 <- 1 - sum(e^2)/sum( (y - mean(y))^2 )
if(adj) {
k <- ncol(X) - 1
n <- nrow(X)
rate_of_error <- (1 - r2) * (n - 1) / (n - k - 1)
adj_r2 <- 1 - rate_of_error
adj_r2
} else r2
}
# same as above
rsquared(obj_lm_fit, cbind(1,datm[,-1]), datm[,1])
#> [1] 0.3948863
rsquared(obj_lm)
#> [1] 0.3948863
# new, `adj = TRUE`
rsquared(obj_lm_fit, cbind(1,datm[,-1]), datm[,1], adj = TRUE)
#> [1] 0.01590061
rsquared(obj_lm, adj = TRUE)
#> [1] 0.01590061
Created on 2023-01-03 with reprex v2.0.2

R optim with a hidden restriction

Background
I'm doing empirical likelihood maximization, and hit an error. The statistics is a bit complex and I describe the error like following.
Problem
Suppose I have multivariate function with domain:
,
And I want to minimize the function in the domain.
Here is the counter plot for my function, the optimum seems to be x = y = 0.5 and z = 0.
if(!require(plotly)){
install.packages('plotly')
require(plotly)
}else{
require(plotly)
}
myfun = function(x,y){
z <- 1-x-y
# want to minimiaze f
f = (x - 0.26)^2 + (y-0.51)^2 + 100*(z+0.6)^2
return(f) # return f, so that optim can minimize f
}
plot_ly(
x = seq(0,1,length.out = 101),
y = seq(0,1,length.out = 101),
z = outer(seq(0,1,length.out = 101), seq(0,1,length.out = 101),FUN = myfun),
type = "contour"
)
My fail attempt
since z = 1-x-y, I tried with 2 parameters (x,y) and their boundry [0,1].
Code:
myobj <- function(para){
x <- para[1]
y <- para[2]
z <- 1-x-y
# want to minimiaze f
f = (x - 0.26)^2 + (y-0.51)^2 + 100*(z+1.6)^2
return(f) # return f, so that optim can minimize f
}
# initialization x=y=0, lower bound (0,0), upper bound (1,1), without restriction on z
optim(par = c(0,0),fn = myobj,method = "L-BFGS-B",lower = c(0,0),upper = c(1,1))
Output:
$par
[1] 1 1
$value
[1] 36.7877
$counts
function gradient
2 2
$convergence
[1] 0
$message
[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"
The output shows that x=y=1 is the result, but when x=y=1, z = 1-1-1 = -1 does not in its domain.
I am wondering how can I put my restrition on z and get the right result,using function like optim?
Thanks a lot!
CVXR vs. constrOptim time comparation
library(microbenchmark)
library(CVXR)
method_const <- function(){
myobj <- function(para){
x <- para[1]
y <- para[2]
z <- 1 - x-y
# want to minimiaze f
f = (x - 0.26)^2 + (y-0.51)^2 + 100*(z+0.6)^2
return(f) # return f, so that optim can minimize f
}
res <- constrOptim(c(0.01,0.01), myobj, NULL,
ui = rbind(c(1,0),c(-1,0),c(0,1),c(0,-1),c(-1,-1)),
ci = c(0,-1,0,-1,-1))
return(round(c(res$par,res$value),4))
}
method_CVXR <-function(){
# declaration of variables x, y, and z
x <- Variable(1)
y <- Variable(1)
z <- Variable(1)
# setup constraints
cons <- list(sum(gg)==1,gg[1]>=0, gg[2]>=0, gg[3]>=0)
# formulate objective function
obj <- Minimize((gg[1] - 0.26)^2 + (gg[2]-0.51)^2 + 100*(gg[3]+0.6)^2)
problem <- Problem(obj,cons)
# solve the optimization problem
res <- solve(problem)
return(round(c(res$getValue(gg),res$value),4))
}
method_CVXR2 <-function(){
# declaration of variables x, y, and z
gg <- Variable(3)
# setup constraints
cons <- list(x+y+z==1,x>=0, y>=0, z>=0)
# formulate objective function
obj <- Minimize((x - 0.26)^2 + (y-0.51)^2 + 100*(z+0.6)^2)
problem <- Problem(obj,cons)
# solve the optimization problem
res <- solve(problem)
return(round(c(res$getValue(x),res$getValue(y),res$value),4))
}
time_res = microbenchmark(method_const(),method_CVXR(),method_CVXR2())
print(time_res)
ggplot2::autoplot(time_res)
It will be more natural to solve this as a quadratic programming problem. Package quadprog provides such a QP solver, except that the problem formulation is a bit clumsy.
# objective function as quadratic problem
D <- diag(c(1, 1, 100))
d <- c(0.26, 0.51, -60)
# equality and bound constraints
A <- rbind(c(1,1,1), diag(3))
b <- c(1, 0,0,0)
meq <- 1
s <- solve.QP(D, d, t(A), b, meq)
s$solution
## [1] 0.375 0.625 0.000
Time comparisons on my computer are as follows:
Unit: milliseconds
expr mean median
method_quadprog() 0.035 0.032
method_solnl() 1.696 1.037
method_fmincon() 1.677 1.092
method_constroptim() 2.130 1.644
method_CVXR() 113.590 97.924
REMARK: Please note that fmincon is just a wrapper for solnl.
The solver behind CVXR is one of the fastest available in R, but building the model is taking some time. That is why CVXR is not efficient for very small problems such as this one, but can be hundreds of times faster than others for quite large problems.
Here is the performance comparison among several approaches
> time_res
Unit: milliseconds
expr min lq mean median uq max neval
method_constroptim() 1.8112 1.86370 3.140725 1.97750 2.07470 11.8188 20
method_fmincon() 1.1804 1.22620 1.633585 1.37365 1.45635 7.0064 20
method_solnl() 1.0980 1.17495 2.165110 1.27700 1.40575 9.3543 20
method_CVXR() 111.4424 121.00940 155.573570 129.92280 149.25700 414.5042 20
where the code for benchmark is given as below
library(microbenchmark)
library(CVXR)
library(pracma)
library(NlcOptim)
# objective function for minimization
f <- function(v) {
x <- v[1]
y <- v[2]
z <- 1- x - y
r <- (x - 0.26)^2 + (y-0.51)^2 + 100*(z +0.6)^2
}
# constrOptim()
method_constroptim <- function(){
v0 <- c(0.5,0.5)
ui <- rbind(c(1,0),c(-1,0),c(0,1),c(0,-1),c(-1,-1))
ci <- c(0,-1,0,-1,-1)
constrOptim(c(0.01,0.01), f, NULL, ui = ui,ci = ci)
}
# fmincon() from "pracma" package
method_fmincon <-function(){
v0 <- c(0.5,0.5,0)
Aeq <- t(matrix(c(1,1,1)))
beq <- 1
lb <- c(0,0,0)
ub <- c(1,1,1)
fmincon(v0,f,Aeq = Aeq, beq = beq, lb = lb, ub = ub)
}
# solnl() from "NlcOptim" package
method_solnl <- function() {
v0 <- c(0.5,0.5,0)
Aeq <- t(matrix(c(1,1,1)))
beq <- 1
lb <- c(0,0,0)
ub <- c(1,1,1)
solnl(v0,f,Aeq = Aeq, Beq = beq, lb = lb, ub = ub)
}
# solnl() from "CVXR" package
method_CVXR <-function(){
x <- Variable(1)
y <- Variable(1)
cons <- list(x>=0, y>=0, x+y<=1)
obj <- Minimize((x - 0.26)^2 + (y-0.51)^2 + 100*(1-x-y +0.6)^2)
problem <- Problem(obj,cons)
solve(problem)
}
time_res = microbenchmark(method_constroptim(),
method_fmincon(),
method_solnl(),
method_CVXR(),
times = 20)

Iterative optimization of alternative glm family

I'm setting up an alternative response function to the commonly used exponential function in poisson glms, which is called softplus and defined as $\frac{1}{c} \log(1+\exp(c \eta))$, where $\eta$ corresponds to the linear predictor $X\beta$
I already managed optimization by setting parameter $c$ to arbitrary fixed values and only searching for $\hat{\beta}$.
BUT now for the next step I have to optimize this parameter $c$ as well (iteratively changing between updated $\beta$ and current $c$).
I tried to write a log-lik function, score function and then setting up a Newton Raphson optimization (using a while loop)
but I don't know how to seperate the updating of c in an outer step and updating \beta in an inner step..
Are there any suggestions?
# Response function:
sp <- function(eta, c = 1 ) {
return(log(1 + exp(abs(c * eta)))/ c)
}
# Log Likelihood
l.lpois <- function(par, y, X){
beta <- par[1:(length(par)-1)]
c <- par[length(par)]
l <- rep(NA, times = length(y))
for (i in 1:length(l)){
l[i] <- y[i] * log(sp(X[i,]%*%beta, c)) - sp(X[i,]%*%beta, c)
}
l <- sum(l)
return(l)
}
# Score function
score <- function(y, X, par){
beta <- par[1:(length(par)-1)]
c <- par[length(par)]
s <- matrix(rep(NA, times = length(y)*length(par)), ncol = length(y))
for (i in 1:length(y)){
s[,i] <- c(X[i,], 1) * (y[i] * plogis(c * X[i,]%*%beta) / sp(X[i,]%*%beta, c) - plogis(c * X[i,]%*%beta))
}
score <- rep(NA, times = nrow(s))
for (j in 1:length(score)){
score[j] <- sum(s[j,])
}
return(score)
}
# Optimization function
opt <- function(y, X, b.start, eps=0.0001, maxiter = 1e5){
beta <- b.start[1:(length(b.start)-1)]
c <- b.start[length(b.start)]
b.old <- b.start
i <- 0
conv <- FALSE
while(conv == FALSE){
eta <- X%*%b.old[1:(length(b.old)-1)]
s <- score(y, X, b.old)
h <- numDeriv::hessian(l.lpois,b.old,y=y,X=X)
invh <- solve(h)
# update
b.new <- b.old + invh %*% s
i <- i + 1
# Test
if(any(is.nan(b.new))){
b.new <- b.old
warning("convergence failed")
break
}
# convergence reached?
if(sqrt(sum((b.new - b.old)^2))/sqrt(sum(b.old^2)) < eps | i >= maxiter){
conv <- TRUE
}
b.old <- b.new
}
eta <- X%*%b.new[1:(length(b.new)-1)]
# covariance
invh <- solve(numDeriv::hessian(l.lpois,b.new,y=y,X=X))
fitted <- sp(eta, b.new[length(b.new)])
result <- list("coefficients" = c(beta = b.new),
"fitted.values" = fitted,
"covariance" = invh)
}
# Running fails ..
n <- 100
x <- runif(n, 0, 1)
Xdes <- cbind(1, x)
eta <- 1 + 2 * x
y <- rpois(n, sp(eta, c = 1))
opt(y,Xdes,c(0,1,1))
You have 2 bugs:
line 25:
(y[i] * plogis(c * X[i,]%*%beta) / sp(X[i,]%*%beta, c) - plogis(c * X[i,]%*%beta))
this returns matrix so you must convert to numeric:
as.numeric(y[i] * plogis(c * X[i,]%*%beta) / sp(X[i,]%*%beta, c) - plogis(c * X[i,]%*%beta))
line 23:
) is missing:
you have:
s <- matrix(rep(NA, times = length(y)*length(par), ncol = length(y))
while it should be:
s <- matrix(rep(NA, times = length(y)*length(par)), ncol = length(y))

Stochastic gradient descent from gradient descent implementation in R

I have a working implementation of multivariable linear regression using gradient descent in R. I'd like to see if I can use what I have to run a stochastic gradient descent. I'm not sure if this is really inefficient or not. For example, for each value of α I want to perform 500 SGD iterations and be able to specify the number of randomly picked samples in each iteration. It would be nice to do this so I could see how the number of samples influences the results. I'm having trouble through with the mini-batching and I want to be able to easily plot the results.
This is what I have so far:
# Read and process the datasets
# download the files from GitHub
download.file("https://raw.githubusercontent.com/dbouquin/IS_605/master/sgd_ex_data/ex3x.dat", "ex3x.dat", method="curl")
x <- read.table('ex3x.dat')
# we can standardize the x vaules using scale()
x <- scale(x)
download.file("https://raw.githubusercontent.com/dbouquin/IS_605/master/sgd_ex_data/ex3y.dat", "ex3y.dat", method="curl")
y <- read.table('ex3y.dat')
# combine the datasets
data3 <- cbind(x,y)
colnames(data3) <- c("area_sqft", "bedrooms","price")
str(data3)
head(data3)
################ Regular Gradient Descent
# http://www.r-bloggers.com/linear-regression-by-gradient-descent/
# vector populated with 1s for the intercept coefficient
x1 <- rep(1, length(data3$area_sqft))
# appends to dfs
# create x-matrix of independent variables
x <- as.matrix(cbind(x1,x))
# create y-matrix of dependent variables
y <- as.matrix(y)
L <- length(y)
# cost gradient function: independent variables and values of thetas
cost <- function(x,y,theta){
gradient <- (1/L)* (t(x) %*% ((x%*%t(theta)) - y))
return(t(gradient))
}
# GD simultaneous update algorithm
# https://www.coursera.org/learn/machine-learning/lecture/8SpIM/gradient-descent
GD <- function(x, alpha){
theta <- matrix(c(0,0,0), nrow=1)
for (i in 1:500) {
theta <- theta - alpha*cost(x,y,theta)
theta_r <- rbind(theta_r,theta)
}
return(theta_r)
}
# gradient descent α = (0.001, 0.01, 0.1, 1.0) - defined for 500 iterations
alphas <- c(0.001,0.01,0.1,1.0)
# Plot price, area in square feet, and the number of bedrooms
# create empty vector theta_r
theta_r<-c()
for(i in 1:length(alphas)) {
result <- GD(x, alphas[i])
# red = price
# blue = sq ft
# green = bedrooms
plot(result[,1],ylim=c(min(result),max(result)),col="#CC6666",ylab="Value",lwd=0.35,
xlab=paste("alpha=", alphas[i]),xaxt="n") #suppress auto x-axis title
lines(result[,2],type="b",col="#0072B2",lwd=0.35)
lines(result[,3],type="b",col="#66CC99",lwd=0.35)
}
Is it more practical to find a way to use sgd()? I can't seem to figure out how to have the level of control I'm looking for with the sgd package
Sticking with what you have now
## all of this is the same
download.file("https://raw.githubusercontent.com/dbouquin/IS_605/master/sgd_ex_data/ex3x.dat", "ex3x.dat", method="curl")
x <- read.table('ex3x.dat')
x <- scale(x)
download.file("https://raw.githubusercontent.com/dbouquin/IS_605/master/sgd_ex_data/ex3y.dat", "ex3y.dat", method="curl")
y <- read.table('ex3y.dat')
data3 <- cbind(x,y)
colnames(data3) <- c("area_sqft", "bedrooms","price")
x1 <- rep(1, length(data3$area_sqft))
x <- as.matrix(cbind(x1,x))
y <- as.matrix(y)
L <- length(y)
cost <- function(x,y,theta){
gradient <- (1/L)* (t(x) %*% ((x%*%t(theta)) - y))
return(t(gradient))
}
I added y to your GD function and created a wrapper function, myGoD, to call yours but first subsetting the data
GD <- function(x, y, alpha){
theta <- matrix(c(0,0,0), nrow=1)
theta_r <- NULL
for (i in 1:500) {
theta <- theta - alpha*cost(x,y,theta)
theta_r <- rbind(theta_r,theta)
}
return(theta_r)
}
myGoD <- function(x, y, alpha, n = nrow(x)) {
idx <- sample(nrow(x), n)
y <- y[idx, , drop = FALSE]
x <- x[idx, , drop = FALSE]
GD(x, y, alpha)
}
Check to make sure it works and try with different Ns
all.equal(GD(x, y, 0.001), myGoD(x, y, 0.001))
# [1] TRUE
set.seed(1)
head(myGoD(x, y, 0.001, n = 20), 2)
# x1 V1 V2
# V1 147.5978 82.54083 29.26000
# V1 295.1282 165.00924 58.48424
set.seed(1)
head(myGoD(x, y, 0.001, n = 40), 2)
# x1 V1 V2
# V1 290.6041 95.30257 59.66994
# V1 580.9537 190.49142 119.23446
Here is how you can use it
alphas <- c(0.001,0.01,0.1,1.0)
ns <- c(47, 40, 30, 20, 10)
par(mfrow = n2mfrow(length(alphas)))
for(i in 1:length(alphas)) {
# result <- myGoD(x, y, alphas[i]) ## original
result <- myGoD(x, y, alphas[i], ns[i])
# red = price
# blue = sq ft
# green = bedrooms
plot(result[,1],ylim=c(min(result),max(result)),col="#CC6666",ylab="Value",lwd=0.35,
xlab=paste("alpha=", alphas[i]),xaxt="n") #suppress auto x-axis title
lines(result[,2],type="b",col="#0072B2",lwd=0.35)
lines(result[,3],type="b",col="#66CC99",lwd=0.35)
}
You don't need the wrapper function--you can just change your GD slightly. It is always good practice to explicitly pass arguments to your functions rather than relying on scoping. Before you were assuming that y would be pulled from your global environment; here y must be given or you will get an error. This will avoid many headaches and mistakes down the road.
GD <- function(x, y, alpha, n = nrow(x)){
idx <- sample(nrow(x), n)
y <- y[idx, , drop = FALSE]
x <- x[idx, , drop = FALSE]
theta <- matrix(c(0,0,0), nrow=1)
theta_r <- NULL
for (i in 1:500) {
theta <- theta - alpha*cost(x,y,theta)
theta_r <- rbind(theta_r,theta)
}
return(theta_r)
}

How do I find the maximum likelihood of a specific multivariate normal log likelihood in R?

I'm having trouble optimizing a multivariate normal log-likelihood in R. If anyone has a good solution for that, please let me know. Specifically, I cannot seem to keep the variance-covariance matrix positive-definite and the parameters in a reasonable range.
Let me introduce the problem more completely. I am essentially trying to simultaneously solve these two regression equations using MLE:
$$
y_1 = \beta_1 + \beta_2 x_1 + \beta_3 x_2 \\
y_2 = \beta_4 + \beta_3 x_1 + \beta_5 x_2
$$
The fact that $\beta_3$ is in both equations is not a mistake. I try to solve this using MLE by maximizing the likelihood of the multivariate normal distribution for $Y = (y_1, y_2)^\top$ where the mean is parameterized as above in the regression equations.
I've attached the log-likelihood function as I believe it should be, where I constrain the variance covariance matrix to be positive-definite by recreating it from necessarily positive eigenvalues and a cholesky decomposition.
mvrestricted_ll <- function(par, Y, X) {
# Indices
n <- nrow(X)
nbetas <- (2 + 3 * (ncol(Y) - 1))
# Extract parameters
beta <- par[1:nbetas]
eigvals <- exp(par[(nbetas + 1):(nbetas + ncol(Y))]) # constrain to be positive
chole <- par[(nbetas + ncol(Y) + 1):(nbetas + ncol(Y) + ncol(Y)*(ncol(Y)+1)/2)]
# Build Sigma from positive eigenvalues and cholesky (should be pos def)
L <- diag(ncol(Y))
L[lower.tri(L, diag=T)] <- chole
Sigma <- diag(eigvals) + tcrossprod(L)
# Linear predictor
# Hard coded for 2x2 example for now
mu <- cbind(beta[1] + beta[2]*X[,1] + beta[3]*X[,2],
beta[4] + beta[3]*X[,1] + beta[5]*X[,2])
yminmu <- Y - mu
nlogs <- n * log(det(Sigma))
invSigma <- solve(Sigma)
meat <- yminmu %*% tcrossprod(invSigma, yminmu)
return(- nlogs - sum(diag(meat)))
}
# Create fake data
n <- 1000
p <- 2
set.seed(20160201)
X <- matrix(rnorm(n*p), nrow = n)
set.seed(20160201)
Y <- matrix(rnorm(n*p), nrow = n)
# Initialize parameters
initpars <- c(rep(0, (2 + 3 * (ncol(Y) - 1)) + ncol(Y) + ncol(Y)*(ncol(Y)+1)/2))
# Optimize fails with BFGS
optim(par = initpars, fn = mvrestricted_ll, X=X, Y=Y, method = "BFGS")
# Optim does not converge with Nelder-mead, if you up the maxits it also fails
optim(par = initpars, fn = mvrestricted_ll, X=X, Y=Y)
Any help would be greatly appreciated.
EDIT: I should note that just letting Sigma be a vector in the parameters and then returning a very large value whenever it is not positive definite does not work either.
I have no idea if the code/answer is correct, but
invSigma <- try(solve(Sigma))
if (inherits(invSigma, "try-error")) return(NA)
and running
optim(par = initpars, fn = mvrestricted_ll, X=X, Y=Y,
control = list(maxit = 1e5))
gets me a little farther to a convergence code of 10 (degenerate Nelder-Mead simplex).
$par
[1] 1.361612e+01 4.674349e+01 -3.050170e+01 3.305013e+01 6.731194e+01
[6] -3.117192e+01 -5.408598e+00 -6.326897e-07 -1.987449e+01 -1.795924e+01
$value
[1] -1.529013e+19
$counts
function gradient
1219 NA
$convergence
[1] 10
I suspect that a real solution will involve looking more carefully at the code to see if it's really doing what you think it's doing (sorry); understanding why solve() errors occur might be a good first step. You can work on troubleshooting this by putting a cat(par, "\n") as the first line of the function and running it without the try/NA-return code. That will allow you to isolate an example data set that throws the error — then you can work your way through your code a line at a time (with debug() or by hand) to see what's happening.
You can consider using the following approach :
library(DEoptim)
fn <- function(par, mat_X, mat_Y)
{
X <- mat_X
Y <- mat_Y
n <- nrow(X)
nbetas <- (2 + 3 * (ncol(Y) - 1))
beta <- par[1 : nbetas]
eigvals <- exp(par[(nbetas + 1) : (nbetas + ncol(Y))])
chole <- par[(nbetas + ncol(Y) + 1) : (nbetas + ncol(Y) + ncol(Y) * (ncol(Y) + 1) / 2)]
L <- diag(ncol(Y))
L[lower.tri(L, diag = TRUE)] <- chole
Sigma <- tryCatch(diag(eigvals) + tcrossprod(L), error = function(e) NA)
if(is.null(dim(Sigma)))
{
return(10 ^ 30)
}else
{
mu <- cbind(beta[1] + beta[2] * X[,1] + beta[3] * X[,2],
beta[4] + beta[3] * X[,1] + beta[5] * X[,2])
yminmu <- Y - mu
nlogs <- n * log(det(Sigma))
invSigma <- tryCatch(solve(Sigma), error = function(e) NA)
if(is.null(dim(invSigma)))
{
return(10 ^ 30)
}else
{
meat <- yminmu %*% tcrossprod(invSigma, yminmu)
log_Lik <- - nlogs - sum(diag(meat))
if(is.na(log_Lik) | is.nan(log_Lik) | is.infinite(log_Lik))
{
return(10 ^ 30)
}else
{
return(-log_Lik)
}
}
}
}
n <- 1000
p <- 2
set.seed(20160201)
mat_X <- matrix(rnorm(n * p), nrow = n)
set.seed(2436537)
mat_Y <- matrix(rnorm(n * p), nrow = n)
lower <- rep(-10, 10)
upper <- rep(10, 10)
DEoptim(fn = fn, lower = lower, upper = upper,
control = list(itermax = 10000, parallelType = 1), mat_X = mat_X, mat_Y = mat_Y)

Resources