I am new to both R and copulas. I'm trying to fit a Tawn Type 1 copula to data, but keep receiving the following error message:
Error in .local(u, copula, log, ...) : unused argument (checkPar = FALSE)
The data I'm using is Swedish Motor Insurance data, located here: http://instruction.bus.wisc.edu/jfrees/jfreesbooks/Regression%20Modeling/BookWebDec2010/data.html
The x is claims, and the y is payments.
Can anyone please help me discern what's wrong?
Thanks in advance!
# Estimate x (Claims) gamma distribution parameters and visually compare simulated vs observed data
x_mean <- mean(x)
x_var <- var(x)
x_rate <- x_mean / x_var
x_shape <- ( (x_mean)^2 ) / x_var
hist(x, breaks = 20, col = "green", density = 20)
hist(rgamma( nrow(mat), rate = x_rate, shape = x_shape), breaks = 20,col = "blue", add = T, density = 20, angle = -45)
# Estimate y (Payment) gamma distribution parameters and visually compare simulated vs observed data
y_mean <- mean(y)
y_var <- var(y)
y_rate <- y_mean / y_var
y_shape <- ( (y_mean)^2 ) / y_var
hist(y, breaks = 20, col = "green", density = 20)
hist(rgamma(nrow(mat), rate = y_rate, shape = y_shape), breaks = 20, col = "blue", add = T, density = 20, angle = -45)
# Looks good, so...
# Measure association using Kendall's Tau
cor(mat, method = "kendall")
##[,1] [,2]
##[1,] 1.0000000 0.8673572
##[2,] 0.8673572 1.0000000
# Now try Spearman's Rho
cor(mat, method = "spearman")
##[,1] [,2]
##[1,] 1.0000000 0.9624433
##[2,] 0.9624433 1.0000000
# Use Vine Copula pachage to select a copula
var_a <- pobs(mydata)[,1]
var_b <- pobs(mydata)[,2]
selectedCopula <- BiCopSelect(var_a, var_b, familyset = NA)
selectedCopula
## Bivariate copula: Tawn type 1 (par = 12.9, par2 = 0.96, tau = 0.89)
# Family
selectedCopula$family
##[1] 104
# Par
# par -> parameter of the copula
selectedCopula$par
##[1] 12.89932
# Estimate copula parameters
cop_model <- tawnT1Copula(param = c(2,0.5))
m <- pobs(as.matrix(mat))
fit <- fitCopula(cop_model, m, method = 'ml')
coef(fit)
Without a minimum working example, we cannot be sure. But, the error message is clear: a copula function is being fed a "checkPar" argument, which does not exist.
The code you provided does not have the word "checkPar", so it's probably in a dependency. Is your R up-to-date? Are your packages up-to-date?
Related
I have a training df with 2 columns like
a b
1 1000 20
2 1008 13
...
n ... ...
Now, as I am required to find a 95% CI for the estimate of 'b' based on a specific 'a' value, with a 'k' value of my choice and compare the CI result to other specific value of 'k's. My question is how can I perform bootstrap for this with 1000 bootstrap reps as I am required to use a fitted knn model for the training data with kernel = 'gaussian' and k can only be in range 1-20 ?
I have found that the best k for this model is k = 5, and had a go for bootstrap but it doesn't work
library(kknn)
library(boot)
boot.kn = function(formula, data, indices)
{
# Create a bootstrapped version
d = data[indices,]
# Fit a model for bs
fit.kn = fitted(train.kknn(formula,data, kernel= "gaussian", ks = 5))
# Do I even need this complicated block
target = as.character(fit.kn$terms[[2]])
rv = my.pred.stats(fit.kn, d[,target])
return(rv)
}
bs = boot(data=df, statistic=boot.kn, R=1000, formula=b ~ a)
boot.ci(bs,conf=0.95,type="bca")
Please inform me for more info if I'm not clear enough. Thank you.
Here is a way to regress b on a with the k-nearest neighbors algorithm.
First, a data set. This is a subset of the iris data set, keeping the first two columns. One row is removed to later be the new data.
i <- which(iris$Sepal.Length == 5.3)
df1 <- iris[-i, 1:2]
newdata <- iris[i, 1:2]
names(df1) <- c("a", "b")
names(newdata) <- c("a", "b")
Now load the packages to be used and determine the optimal value for k with package kknn.
library(caret)
library(kknn)
library(boot)
fit <- kknn::train.kknn(
formula = b ~ a,
data = df1,
kmax = 15,
kernel = "gaussian",
distance = 1
)
k <- fit$best.parameters$k
k
#[1] 9
And bootstrap predictions for the new point a <- 5.3.
boot.kn <- function(data, indices, formula, newdata, k){
d <- data[indices, ]
fit <- knnreg(formula, data = d)
predict(fit, newdata = newdata)
}
set.seed(2021)
R <- 1e4
bs <- boot(df1, boot.kn, R = R, formula = b ~ a, newdata = newdata, k = k)
ci <- boot.ci(bs, level = 0.95, type = "bca")
ci
#BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
#Based on 10000 bootstrap replicates
#
#CALL :
#boot.ci(boot.out = bs, type = "bca", level = 0.95)
#
#Intervals :
#Level BCa
#95% ( 3.177, 3.740 )
#Calculations and Intervals on Original Scale
Plot the results.
old_par <- par(mfrow = c(2, 1),
oma = c(5, 4, 0, 0) + 0.1,
mar = c(1, 1, 1, 1) + 0.1)
hist(bs$t, main = "Histogram of bootstrap values")
abline(v = 3.7, col = "red")
abline(v = mean(bs$t), col = "blue")
abline(v = ci$bca[4:5], col = "blue", lty = "dashed")
plot(b ~ a, df1)
points(5.3, 3.7, col = "red", pch = 19)
points(5.3, mean(bs$t), col = "blue", pch = 19)
arrows(x0 = 5.3, y0 = ci$bca[4],
x1 = 5.3, y1 = ci$bca[5],
col = "blue", angle = 90, code = 3)
par(old_par)
I am trying to simulate PCA Data as follows:
q <- 5 # no. of PCs
p <- 20 # no. of variables
n <- 2000 # no. of individuals
eps <- 0.05 # error standard deviation
# Eigenvalues
Sig <- seq(3, 1, length.out = q)^2
Lambda <- diag(Sig)
# Matrix of Principal Components
H <- rmvnorm(n = n, mean = rep(0, q), sigma = Lambda)
# Add gaussian noise
E <- matrix(rnorm(n*p, sd = sqrt(eps)), ncol = p)
# Data matrix
Y <- H %*% t(Amat) + E
# Perform PCA
summary(m1 <- prcomp(Y, scale = T)) # and so on...
However, I have no idea how to create the matrix of Loadings Amat in a meaningful way.
Thanks for any help I receive from you and I appreciate it!
This is not using the same structure as the OP, but it simulates a PCA with 4 different groups (which could be species) which each have 3 "traits" (each of the trait have different means and sd based on some biological data found in the literature for example).
set.seed(123) # setting this so the random results will be repeatable
library(MASS)
# Simulating 3 traits for 4 different species
n = 200 # number of "individuals"
# Generate the groups
Amat1 = MASS::mvrnorm(n, mu = c(11.2,11.8,9.91), Sigma = diag(c(1.31,1.01,1.02)))
Amat2 = MASS::mvrnorm(n, mu = c(7.16,8.54,6.82), Sigma = diag(c(0.445,0.546,0.350)))
Amat3 = MASS::mvrnorm(n, mu = c(15.6,14.6,13.5), Sigma = diag(c(1.43,0.885,0.990)))
Amat4 = MASS::mvrnorm(n, mu = c(8.65,14.1,8.24), Sigma = diag(c(0.535,0.844,0.426)))
# Combine the data
Amat = rbind(Amat1,Amat2,Amat3,Amat4)
# Make group data
Amat.gr = cbind(Amat, gl(4,k=n,labels = c(1,2,3,4)))
# Calculate the covariance matrix for each group
by(Amat.gr[,1:3],INDICES = Amat.gr[,4],FUN = cov) # calculate covariance matrix for all groups
# Plot the result
summary(m1 <- prcomp(Amat, scale= T))
# biplot(m1, xlabs=rep(".", nrow(Amat)), cex = 2)
plot(vegan::scores(m1), asp = 1, pch = 19, col = gl(4,k=n,labels = c(1,2,3,4)))
plot(Amat[,1],Amat[,2], pch = 19, col = gl(4,k=n,labels = c(1,2,3,4)))
The plot on the left shows the PCA and on the right the raw data.
I added a toy example with data to show what is the algorithm to compute a PCA in R from Legendre and Legendre 2012.
# Generate vectors (example from Legendre and Legendre 2012)
v1 = c(2,3,5,7,9)
v2 = c(1,4,0,6,2)
# If you want to play with sample size
# n = 100
# v1 = rnorm(n = n, mean = mean(v1), sd = sd(v1))
# v2 = rnorm(n = n, mean = mean(v2), sd = sd(v2))
# Get the y matrix
y = cbind(v1,v2)
# Centered y matrix
yc = apply(y, 2, FUN = function(x) x-mean(x))
# Dispersion matrix
s = 1/(nrow(y)-1)*t(yc) %*% yc
# Compute the single value decomposition to get the eigenvectors and
ev = svd(s)$v
# get the principal components
f = yc %*% ev
# This gives the identity matrix
round(t(svd(s)$v) %*% svd(s)$v,2)
# these are the eigen values
svd(s)$d
-svd(yc)$v #p. 104
plot(f, pch = 19); abline(h=0,v=0, lty = 3)
I am trying to use a Gaussian Process Regression (GPR) model to predict hourly streamflow discharges in a river. I've got good results applying the caret::kernlab train () function (thanks Kuhn!).
Since the uncertainty idea is one of the main inherent ones advantages of the GPR, I would like to know if anyone could help me to access the results related to the prediction inteval of the test dataset.
I'll put an extract of the code I've been working. Since my real data are huge (and sincerely, I don't know how to put it here), I'll example with the data(airquality). The main goal in this particular example is to predict airquality$Ozone, using as predictos the lag-variables of airquality$Temperature.
rm(list = ls())
data(airquality)
airquality = na.omit(as.data.frame(airquality)); str(airquality)
library(tidyverse)
library(magrittr)
airquality$Ozone %>% plot(type = 'l')
lines(airquality$Temp, col = 2)
legend("topleft", legend = c("Ozone", "Temperature"),
col=c(1, 2), lty = 1:1, cex = 0.7, text.font = 4, inset = 0.01,
box.lty=0, lwd = 1)
attach(airquality)
df_lags <- airquality %>%
mutate(Temp_lag1 = lag(n = 1L, Temp)) %>%
na.omit()
ESM_train = data.frame(df_lags[1:81, ]) # Training Observed 75% dataset
ESM_test = data.frame(df_lags[82:nrow(df_lags), ]) # Testing Observed 25% dataset
grid_gaussprRadial = expand.grid(.sigma = c(0.001, 0.01, 0.05, 0.1, 0.5, 1, 2)) # Sigma parameters searching for GPR
# TRAIN MODEL ############################
# Tuning set
library(caret)
set.seed(111)
cvCtrl <- trainControl(
method ="repeatedcv",
repeats = 1,
number = 20,
allowParallel = TRUE,
verboseIter = TRUE,
savePredictions = "final")
# Train (aprox. 4 seconds time-simulation)
attach(ESM_train)
set.seed(111)
system.time(Model_train <- caret::train(Ozone ~ Temp + Temp_lag1,
trControl = cvCtrl,
data = ESM_train,
metric = "MAE", # Using MAE since I intend minimum values are my focus
preProcess = c("center", "scale"),
method = "gaussprRadial", # Setting RBF kernel function
tuneGrid = grid_gaussprRadial,
maxit = 1000,
linout = 1)) # Regression type
plot(Model_train)
Model_train
ESM_results_train <- Model_train$resample %>% mutate(Model = "") # K-fold Training measures
# Select the interested TRAIN data and arrange them as dataframe
Ozone_Obs_Tr = Model_train$pred$obs
Ozone_sim = Model_train$pred$pred
Resid = Ozone_Obs_Tr - Ozone_sim
train_results = data.frame(Ozone_Obs_Tr,
Ozone_sim,
Resid)
# Plot Obs x Simulated train results
library(ggplot2)
ggplot(data = train_results, aes(x = Ozone_Obs_Tr, y = Ozone_sim)) +
geom_point() +
geom_abline(intercept = 0, slope = 1, color = "black")
# TEST MODEL ############################
# From "ESM_test" dataframe, we predict ESM Ozone time series, adding it in "ESM_forecasted" dataframe
ESM_forecasted = ESM_test %>%
mutate(Ozone_Pred = predict(Model_train, newdata = ESM_test, variance.model = TRUE))
str(ESM_forecasted)
# Select the interested TEST data and arrange them as a dataframe
Ozone_Obs = ESM_forecasted$Ozone
Ozone_Pred = ESM_forecasted$Ozone_Pred
# Plot Obs x Predicted TEST results
ggplot(data = ESM_forecasted, aes(x = Ozone_Obs, y = Ozone_Pred)) +
geom_point() +
geom_abline(intercept = 0, slope = 1, color = "black")
# Model performance #####
library(hydroGOF)
gof_TR = gof(Ozone_sim, Ozone_Obs_Tr)
gof_TEST = gof(Ozone_Pred,Ozone_Obs)
Performances = data.frame(
Train = gof_TR,
Test = gof_TEST
); Performances
# Plot the TEST prediction
attach(ESM_forecasted)
plot(Ozone_Obs, type = "l", xlab = "", ylab = "", ylim = range(Ozone_Obs, Ozone_Pred))
lines(Ozone_Pred , col = "coral2", lty = 2, lwd = 2)
legend("top", legend = c("Ozone Obs Test", "Ozone Pred Test"),
col=c(1, "coral2"), lty = 1:2, cex = 0.7, text.font = 4, inset = 0.01, box.lty=0, lwd = 2)
These last lines generate the following plot:
The next, and last, step would be to extract the prediction intervals, which is based on a gaussian distribution around each prediction point, to plot it together with this last plot.
The caret::kernlab train() appliance returned better prediction than, for instance, just kernlab::gaussprRadial(), or even tgp::bgp() packages. For both of them I could find the prediction interval.
For example, to pick up the prediction intervals via tgp::bgp(), it could be done typing:
Upper_Bound <- Ozone_Pred$ZZ.q2 #Ozone_Pred - 2 * sigma^2
Lower_Bound <- Ozone_Pred$ZZ.q1 #Ozone_Pred + 2 * sigma^2
Therefore, via caret::kernlab train(), I hope the required standard deviations could be found typing something as
Model_train$...
or maybe, with
Ozone_Pred$...
Moreover, at link: https://stats.stackexchange.com/questions/414079/can-mad-median-absolute-deviation-or-mae-mean-absolute-error-be-used-to-calc,
Stephan Kolassa author explained that we could estimate the prediction intervals through MAE, or even RMSE. But I didn't understand if this is my point, since the MAE I got is just the comparison between Obs x Predicted Ozone data, in this example.
Please, this solution is very important to me! I think I am near to obtain my main results, but I don't know anymore how to try.
Thanks a lot, friends!
I don't really know how the caret framework works, but getting a prediction interval for a GP regression with a Gaussian likelihood is easy enough to do manually.
First we just need a function for the squared exponential kernel, also called the radial basis function kernel, which is what you were using. sf here is the scale factor (unused in the kernlab implementation), and ell is the length scale, called sigma in the kernlab implementation:
covSEiso <- function(x1, x2 = x1, sf = 1.0, ell = 1.0) {
sf <- sf^2
ell <- -0.5 * (1 / (ell^2))
n <- nrow(x1)
m <- nrow(x2)
d <- ncol(x1)
result <- matrix(0, nrow = n, ncol = m)
for ( j in 1:m ) {
for ( i in 1:n ) {
result[i, j] <- sf * exp(ell * sum((x1[i, ] - x2[j, ])^2))
}
}
return(result)
}
I'm not sure what your code says about which length scale to use; below I will use a length scale of 25 and scale factor of 50 (obtained via GPML's hyperparameter optimization routines). Then we use the covSEiso() function above to get the relevant covariances, and the rest is application of basic Gaussian identities. I would refer you to Chapter 2 of Rasmussen and Williams (2006) (graciously provided for free online).
data(airquality)
library(tidyverse)
library(magrittr)
df_lags <- airquality %>%
mutate(Temp_lag1 = lag(n = 1L, Temp)) %>%
na.omit()
ESM_train <- data.frame(df_lags[1:81, ]) # Training Data 75% dataset
ESM_test <- data.frame(df_lags[82:nrow(df_lags), ]) # Testing Data 25% dataset
## For convenience I'll define separately the training and test inputs
X <- ESM_train[ , c("Temp", "Temp_lag1")]
Xstar <- ESM_test[ , c("Temp", "Temp_lag1")]
## Get the kernel manually
K <- covSEiso(X, ell = 25, sf = 50)
## We also need covariance between the test cases
Kstar <- covSEiso(Xstar, X, ell = 25, sf = 50)
Ktest <- covSEiso(Xstar, ell = 25, sf = 50)
## Now the 95% credible region for the posterior is
predictive_mean <- Kstar %*% solve(K + diag(nrow(K))) %*% ESM_train$Ozone
predictive_var <- Ktest - (Kstar %*% solve(K + diag(nrow(K))) %*% t(Kstar))
## Then for the prediction interval we only need to add the observation noise
z <- sqrt(diag(predictive_var)) + 25
interval_high <- predictive_mean + 2 * z
interval_low <- predictive_mean - 2 * z
Then we can check out the prediction intervals
This all is pretty easy to do via my gplmr package (available on GitHub) which can call GPML from R if you have Octave installed:
data(airquality)
library(tidyverse)
library(magrittr)
library(gpmlr)
df_lags <- airquality %>%
mutate(Temp_lag1 = lag(n = 1L, Temp)) %>%
na.omit()
ESM_train <- data.frame(df_lags[1:81, ]) # Training Data 75% dataset
ESM_test <- data.frame(df_lags[82:nrow(df_lags), ]) # Testing Data 25% dataset
X <- as.matrix(ESM_train[ , c("Temp", "Temp_lag1")])
y <- ESM_train$Ozone
Xs <- as.matrix(ESM_test[ , c("Temp", "Temp_lag1")])
ys <- ESM_test$Ozone
hyp0 <- list(mean = numeric(), cov = c(0, 0), lik = 0)
hyp <- set_hyperparameters(hyp0, "infExact", "meanZero", "covSEiso","likGauss",
X, y)
gp_res <- gp(hyp, "infExact", "meanZero", "covSEiso", "likGauss", X, y, Xs, ys)
predictive_mean <- gp_res$YMU
interval_high <- gp_res$YMU + 2 * sqrt(gp_res$YS2)
interval_low <- gp_res$YMU - 2 * sqrt(gp_res$YS2)
Then just plot the predictions, as above:
plot(NULL, xlab = "", ylab = "", xaxt = "n", yaxt = "n",
xlim = range(ESM_test$Temp), ylim = range(c(interval_high, interval_low)))
axis(1, tick = FALSE, line = -0.75)
axis(2, tick = FALSE, line = -0.75)
mtext("Temp", 1, 1.5)
mtext("Ozone", 2, 1.5)
idx <- order(ESM_test$Temp)
polygon(c(ESM_test$Temp[idx], rev(ESM_test$Temp[idx])),
c(interval_high[idx], rev(interval_low[idx])),
border = NA, col = "#80808080")
lines(ESM_test$Temp[idx], predictive_mean[idx])
points(ESM_test$Temp, ESM_test$Ozone, pch = 19)
plot(NULL, xlab = "", ylab = "", xaxt = "n", yaxt = "n",
xlim = range(ESM_test$Temp_lag1), ylim = range(c(interval_high, interval_low)))
axis(1, tick = FALSE, line = -0.75)
axis(2, tick = FALSE, line = -0.75)
mtext("Temp_lag1", 1, 1.5)
mtext("Ozone", 2, 1.5)
idx <- order(ESM_test$Temp_lag1)
polygon(c(ESM_test$Temp_lag1[idx], rev(ESM_test$Temp_lag1[idx])),
c(interval_high[idx], rev(interval_low[idx])),
border = NA, col = "#80808080")
lines(ESM_test$Temp_lag1[idx], predictive_mean[idx])
points(ESM_test$Temp_lag1, ESM_test$Ozone, pch = 19)
Thanks to a closed form formula (I work on risk neutral density, with this king of formula: RND formula, page 8), I have an incomplete distribution of this type:
My idea would be to fit this density with a student-t.
I already tried the MASS and fitdistrplus packages but just can't find how to perform my task. Everything I can do for now is to get the fitted parameters (m=1702.041, s=6.608536, df=15.18036), but from here I don't know how to get my fitted values for my distribution.
A sample of code:
temp = matrix(nrow=1000, ncol=3)
colnames(temp) = c("strikes", "first_density", "mulitply_first_density")
temp = as.data.frame(temp)
# we generate fake data
temp$strikes = seq(1000,2000,length=1000)
temp$first_density = runif(1000,max=0.006, min=1e-10)
# we multiply our first density to generate our sample
temp$mulitply_first_density = temp$first_density*1000000
# we generate our sample
vec = vector()
for (i in 1:nrow(temp))
{
vec = c(vec, rep(temp$strike[i], temp$mulitply_first_density[i]))
}
# we laod our library
library("MASS")
# we fir our parameters
fitted_parameters = fitdistr(vec, "t")
The formula for the t-density function using the location and scale parameters is given in the examples of the documentation as mydt.
#simulated data
set.seed(42)
x <- rt(1e4, 7, 10)
plot(density(x))
library(MASS)
fitted_parameters = fitdistr(x, "t", start = list(df = 10, m = 10, s = 5))
# df m s
# 3.81901649 10.56816146 2.66905346
#( 0.15295551) ( 0.03448627) ( 0.03361758)
mydt <- function(x, m, s, df) dt((x-m)/s, df)/s
curve(do.call(mydt, c(list(x), as.list(fitted_parameters$estimate))), add = TRUE, col = "red")
legend("topright", legend = c("kernel density estimate", "fitted t distribution"),
col = c("black", "red"), lty = 1)
I am trying to fit a non-linear regression model where the mean-function is the bivariate normal distribution. The parameter to specify is the correlation rho.
The problem: "gradient of first iteration step is singular". Why?
I have here a little example with simulated data.
# given values for independent variables
x1 <- c(rep(0.1,5), rep(0.2,5), rep(0.3,5), rep(0.4,5), rep(0.5,5))
x2 <- c(rep(c(0.1,0.2,0.3,0.4,0.5),5))
## 1 generate values for dependent variable (incl. error term)
# from bivariate normal distribution with assumed correlation rho=0.5
fun <- function(b) pmnorm(x = c(qnorm(x1[b]), qnorm(x2[b])),
mean = c(0, 0),
varcov = matrix(c(1, 0.5, 0.5, 1), nrow = 2))
set.seed(123)
y <- sapply(1:25, function(b) fun(b)) + runif(25)/1000
# put it in data frame
dat <- data.frame(y=y, x1=x1, x2=x2 )
# 2 : calculate non-linear regression from the generated data
# use rho=0.51 as starting value
fun <- function(x1, x2,rho) pmnorm(x = c(qnorm(x1), qnorm(x2)),
mean = c(0, 0),
varcov = matrix(c(1, rho, rho, 1), nrow = 2))
nls(formula= y ~ fun(x1, x2, rho), data= dat, start=list(rho=0.51),
lower=0, upper=1, trace=TRUE)
This yields an error message:
Error in nls(formula = y ~ fun(x1, x2, rho), data = dat, start = list(rho = 0.51), :
singulärer Gradient
In addition: Warning message:
In nls(formula = y ~ fun(x1, x2, rho), data = dat, start = list(rho = 0.51), :
Obere oder untere Grenzen ignoriert, wenn nicht algorithm= "port"
What I don't understand is
I have only one variable (rho), so there is only one gradient which must be =0 if the matrix of gradients is supposed to be singular. So why should the gradient be =0?
The start value cannot be the problem as I know the true rho=0.5. So the start value =0.51 should be fine, shouldn't it?
The data cannot be completely linear dependent as I added an error term to y.
I would appreciate help very much. Thanks already.
Perhaps "optim" does a better job than "nls":
library(mnormt)
# given values for independent variables
x1 <- c(rep(0.1,5), rep(0.2,5), rep(0.3,5), rep(0.4,5), rep(0.5,5))
x2 <- c(rep(c(0.1,0.2,0.3,0.4,0.5),5))
## 1 generate values for dependent variable (incl. error term)
# from bivariate normal distribution with assumed correlation rho=0.5
fun <- function(b) pmnorm(x = c(qnorm(x1[b]), qnorm(x2[b])),
mean = c(0, 0),
varcov = matrix(c(1, 0.5, 0.5, 1), nrow = 2))
set.seed(123)
y <- sapply(1:25, function(b) fun(b)) + runif(25)/1000
# put it in data frame
dat <- data.frame(y=y, x1=x1, x2=x2 )
# 2 : calculate non-linear regression from the generated data
# use rho=0.51 as starting value
fun <- function(x1, x2,rho) pmnorm(x = c(qnorm(x1), qnorm(x2)),
mean = c(0, 0),
varcov = matrix(c(1, rho, rho, 1), nrow = 2))
f <- function(rho) { sum( sapply( 1:nrow(dat),
function(i){
(fun(dat[i,2],dat[i,3],rho) - dat[i,1])^2
} ) ) }
optim(0.51, f, method="BFGS")
The result is not that bad:
> optim(0.51, f, method="BFGS")
$par
[1] 0.5043406
$value
[1] 3.479377e-06
$counts
function gradient
14 4
$convergence
[1] 0
$message
NULL
Maybe even a little bit better than 0.5:
> f(0.5043406)
[1] 3.479377e-06
> f(0.5)
[1] 1.103484e-05
>
Let's check another start value:
> optim(0.8, f, method="BFGS")
$par
[1] 0.5043407
$value
[1] 3.479377e-06
$counts
function gradient
28 6
$convergence
[1] 0
$message
NULL