Non-linear solver in R (log-linear distribution) - r

I need help in preparing something similar to Solver (from Excel) in R.
I try to develop a tool, which will take some points and create parameters of curve suitable for them. This curve will have a shape of log-linear distribution. I need 4 parameters, which could be useable in Excel formula:
y = b*loglindist(x*a, c, d), where b is a parameter using for the result, a is a parameter using for a value of distribution, c is a mean, and d is a standard deviation.
I have to minimize sse between actual points and points estimated with the curve.
My code is as follows:
input <- read.csv2("C:/Users/justyna.andrulewicz/Desktop/R estimator/data.csv", sep=",")
data <- as.matrix(input)
x <- nrow(data)
max_reach <- 90 ### max y
# solver
# constrains
a_min <- 0.000000001
b_min <- 0.5*max_reach
c_min <- 0.000000001
d_min <- 0.000000001
a_max <- 1000
b_max <- max_reach
c_max <- 1000
d_max <- 1000
constrains <- round(matrix(c(a_min,b_min,c_min,d_min,a_max,b_max,c_max,d_max), nrow=2, ncol=4, byrow=TRUE, dimnames=list(c("mins", "maxes"), c("a","b","c","d"))),1)
constrains
ui <- matrix(c(1,0,0,0, -1,0,0,0, 0,1,0,0, 0,-1,0,0, 0,0,1,0, 0,0,-1,0, 0,0,0,1, 0,0,0,-1), ncol=4, byrow=TRUE)
ci <- round(c(a_min, -a_max, b_min, -b_max, c_min, -c_max, d_min, -d_max), 1)
a <- 100
b <- 0.4*max_reach
c <- 1
d <- 1
par <-as.numeric(c(a,b,c,d))
par
spends <- as.numeric(data[,1])
estimated <- b*plnorm(a*spends, meanlog = c, sdlog = d, log = FALSE)
actual <- as.numeric(data[,2])
se <- estimated-actual
sse <- function(se) sum(se^2)
sse(se)
optimization <- constrOptim(par, sse, NULL, ui, ci, method="SANN")
results<-round(as.numeric(optimization$par,nrow=4,ncol=1),6)
results
but it doesn't work: the results make no sense, as you can see in the plot.
step <- 10^3
y <- 1:100
spends<-y*step
a_est<-optimization$par[1]
b_est<-optimization$par[2]
c_est<-optimization$par[3]
d_est<-optimization$par[4]
curve<-b_est*plnorm(a_est*spends, meanlog = c_est, sdlog = d_est, log = FALSE)
est <-plot(spends, curve, type="l", col="blue")
act <-plot(data, type="p", col="red")
Please help: maybe can I replace constOptim and use another function, which will better address my problem? Or maybe there is another way to solve my problem?

Related

Maximum likelihood estimation of a multivariate normal distribution of arbitrary dimesion in R - THE ULTIMATE GUIDE?

I notice searching through stackoverflow for similar questions that this has been asked several times hasn't really been properly answered. Perhaps with help from other users this post can be a helpful guide to programming a numerical estimate of the parameters of a multivariate normal distribution.
I know, I know! The closed form solutions are available and trivial to implement. In my case I am interested in modifying the likelihood function for a specific purpose and I don't expect an exact analytic solution so this is a test case to check the procedure.
So here is my attempt. Please comment. Especially if I am missing opportunities for optimization. Note, I'm not a statistician so I'd appreciate any pointers.
ll_multN <- function(theta,X) {
# theta = c(mu, diag(Sigma), Sigma[upper.tri(Sigma)])
# X is an nxk dataset
# MLE: L = - (nk/2)*log(2*pi) - (n/2)*log(det(Sigma)) - (1/2)*sum_i(t(X_i-mu)^2 %*% Sigma^-1 %*% (X_i-mu)^2)
# summation over i is performed using a apply call for efficiency
n <- nrow(X)
k <- ncol(X)
# def mu
mu.vec <- theta[1:k]
# def Sigma
Sigma.diag <- theta[(k+1):(2*k)]
Sigma.offd <- theta[(2*k+1):length(theta)]
Sigma <- matrix(NA, k, k)
Sigma[upper.tri(Sigma)] <- Sigma.offd
Sigma <- t(Sigma)
Sigma[upper.tri(Sigma)] <- Sigma.offd
diag(Sigma) <- Sigma.diag
# compute summation
sum_i <- sum(apply(X, 1, function(x) (matrix(x,1,k)-mu.vec)%*%solve(Sigma)%*%t(matrix(x,1,k)-mu.vec)))
# compute log likelihood
logl <- -.5*n*k*log(2*pi) - .5*n*log(det(Sigma))
logl <- logl - .5*sum_i
return(-logl)
}
Simulated dataset generated using the rmvnorm() function in the package "mvtnorm". Random positive definite covariance matrix generated using the additional function Posdef() (taken from here: https://stat.ethz.ch/pipermail/r-help/2008-February/153708)
library(mvtnorm)
Posdef <- function (n, ev = runif(n, 0, 5)) {
# generates a random positive definite covariance matrix
Z <- matrix(ncol=n, rnorm(n^2))
decomp <- qr(Z)
Q <- qr.Q(decomp)
R <- qr.R(decomp)
d <- diag(R)
ph <- d / abs(d)
O <- Q %*% diag(ph)
Z <- t(O) %*% diag(ev) %*% O
return(Z)
}
set.seed(2)
n <- 1000 # number of data points
k <- 3 # number of variables
mu.tru <- sample(0:3, k, replace=T) # random mean vector
Sigma.tru <- Posdef(k) # random covariance matrix
eigen(Sigma.tru)$val # check positive def (all lambda > 0)
# Generate simulated dataset
X <- rmvnorm(n, mean=mu.tru, sigma=Sigma.tru)
# initial parameter values
pars.init <- c(mu=rep(0,k), sig_ii=rep(1,k), sig_ij=rep(0, k*(k-1)/2))
# limits for optimization algorithm
eps <- .Machine$double.eps # get a small value for bounding the paramter space to avoid things such as log(0).
lower.bound <- c(rep(-Inf,k), # bound on mu
rep(eps,k), # bound on sigma_ii
rep(-Inf,k)) # bound on sigma_ij i=/=j
upper.bound <- c(rep(Inf,k), # bound on mu
rep(100,k), # bound on sigma_ii
rep(100,k)) # bound on sigma_ij i=/=j
system.time(
o <- optim(pars.init,
ll_multN, X=X, method="L-BFGS-B",
lower = lower.bound,
upper = upper.bound)
)
plot(x=c(mu.tru,diag(Sigma.tru),Sigma.tru[upper.tri(Sigma.tru)]),
y=o$par,
xlab="Parameter",
ylab="Estimate",
pch=20)
abline(c(0,1), col="red", lty=2)
This currently runs on my laptop in
user system elapsed
47.852 24.014 24.611
and gives this graphical output:
Estimated mean and variance
In particular any advice on limit setting or algorithm choice would be much appreciated.
Thanks

r deSolve - plotting time evolution pde

suppose that we have a pde that describes the evolution of a variable y(t,x) over time t and space x, and I would like to plot its evolution on a three dimensional diagram (t,x,y). With deSolve I can solve the pde, but I have no idea about how to obtain this kind of diagram.
The example in the deSolve package instruction is the following, where y is aphids, t=0,...,200 and x=1,...,60:
library(deSolve)
Aphid <- function(t, APHIDS, parameters) {
deltax <- c (0.5, rep(1, numboxes - 1), 0.5)
Flux <- -D * diff(c(0, APHIDS, 0)) / deltax
dAPHIDS <- -diff(Flux) / delx + APHIDS * r
list(dAPHIDS )
}
D <- 0.3 # m2/day diffusion rate
r <- 0.01 # /day net growth rate
delx <- 1 # m thickness of boxes
numboxes <- 60
Distance <- seq(from = 0.5, by = delx, length.out = numboxes)
APHIDS <- rep(0, times = numboxes)
APHIDS[30:31] <- 1
state <- c(APHIDS = APHIDS) # initialise state variables
times <-seq(0, 200, by = 1)
out <- ode.1D(state, times, Aphid, parms = 0, nspec = 1, names = "Aphid")
"out" produces a matrix containing all the data that we need, t, y(x1), y(x2), ... y(x60). How can I produce a surface plot to show the evolution and variability of y in (t,x)?
The ways change a bit depending on using package. But you can do it with little cost because out[,-1] is an ideal matrix form to draw surface. I showed two examples using rgl and plot3D package.
out2 <- out[,-1]
AphID <- 1:ncol(out2)
library(rgl)
persp3d(times, AphID, out2, col="gray50", zlab="y")
# If you want to change color with value of Z-axis
# persp3d(times, AphID, out2, zlab="y", col=topo.colors(256)[cut(c(out2), 256)])
library(plot3D)
mat <- mesh(times, AphID)
surf3D(mat$x, mat$y, out2, bty="f", ticktype="detailed", xlab="times", ylab="AphID", zlab="y")

How can I improve the Integration and Parameterization of Convolved Distributions?

I am trying to solve for the parameters of a gamma distribution that is convolved with both normal and lognormal distributions. I can experimentally derive parameters for both the normal and lognormal components, hence, I just want to solve for the gamma params.
I have attempted 3 approaches to this problem:
1) generating convolved random datasets (i.e. rnorm()+rlnorm()+rgamma()) and using least-squares regression on the linear- or log-binned histograms of the data (not shown, but was very biased by RNG and didn't optimize well at all.)
2) "brute-force" numerical integration of the convolving functions (example code #1)
3) numerical integration approaches w/ the distr package. (example code #2)
I have had limited success with all three approaches. Importantly, these approaches seem to work well for "nominal" values for the gamma parameters, but they all begin to fail when k(shape) is low and theta(scale) is high—which is where my experimental data resides. please find the examples below.
Straight-up numerical Integration
# make the functions
f.N <- function(n) dnorm(n, N[1], N[2])
f.L <- function(l) dlnorm(l, L[1], L[2])
f.G <- function(g) dgamma(g, G[1], scale=G[2])
# make convolved functions
f.Z <- function(z) integrate(function(x,z) f.L(z-x)*f.N(x), -Inf, Inf, z)$value # L+N
f.Z <- Vectorize(f.Z)
f.Z1 <- function(z) integrate(function(x,z) f.G(z-x)*f.Z(x), -Inf, Inf, z)$value # G+(L+N)
f.Z1 <- Vectorize(f.Z1)
# params of Norm, Lnorm, and Gamma
N <- c(0,5)
L <- c(2.5,.5)
G <- c(2,7) # this distribution is the one we ultimately want to solve for.
# G <- c(.5,10) # 0<k<1
# G <- c(.25,5e4) # ballpark params of experimental data
# generate some data
set.seed(1)
rN <- rnorm(1e4, N[1], N[2])
rL <- rlnorm(1e4, L[1], L[2])
rG <- rgamma(1e4, G[1], scale=G[2])
Z <- rN + rL
Z1 <- rN + rL + rG
# check the fit
hist(Z,freq=F,breaks=100, xlim=c(-10,50), col=rgb(0,0,1,.25))
hist(Z1,freq=F,breaks=100, xlim=c(-10,50), col=rgb(1,0,0,.25), add=T)
z <- seq(-10,50,1)
lines(z,f.Z(z),lty=2,col="blue", lwd=2) # looks great... convolution performs as expected.
lines(z,f.Z1(z),lty=2,col="red", lwd=2) # this works perfectly so long as k(shape)>=1
# I'm guessing the failure to compute when shape 0 < k < 1 is due to
# numerical integration problems, but I don't know how to fix it.
integrate(dgamma, -Inf, Inf, shape=1, scale=1) # ==1
integrate(dgamma, 0, Inf, shape=1, scale=1) # ==1
integrate(dgamma, -Inf, Inf, shape=.5, scale=1) # !=1
integrate(dgamma, 0, Inf, shape=.5, scale=1) # != 1
# Let's try to estimate gamma anyway, supposing k>=1
optimFUN <- function(par, N, L) {
print(par)
-sum(log(f.Z1(Z1[1:4e2])))
}
f.G <- function(g) dgamma(g, par[1], scale=par[2])
fitresult <- optim(c(1.6,5), optimFUN, N=N, L=L)
par <- fitresult$par
lines(z,f.Z1(z),lty=2,col="green3", lwd=2) # not so great... likely better w/ more data,
# but it is SUPER slow and I observe large step sizes.
Attempting convolving via distr package
# params of Norm, Lnorm, and Gamma
N <- c(0,5)
L <- c(2.5,.5)
G <- c(2,7) # this distribution is the one we ultimately want to solve for.
# G <- c(.5,10) # 0<k<1
# G <- c(.25,5e4) # ballpark params of experimental data
# make the distributions and "convolvings'
dN <- Norm(N[1], N[2])
dL <- Lnorm(L[1], L[2])
dG <- Gammad(G[1], G[2])
d.NL <- d(convpow(dN+dL,1))
d.NLG <- d(convpow(dN+dL+dG,1)) # for large values of theta, no matter how I change
# getdistrOption("DefaultNrFFTGridPointsExponent"), grid size is always wrong.
# Generate some data
set.seed(1)
rN <- r(dN)(1e4)
rL <- r(dL)(1e4)
rG <- r(dG)(1e4)
r.NL <- rN + rL
r.NLG <- rN + rL + rG
# check the fit
hist(r.NL, freq=F, breaks=100, xlim=c(-10,50), col=rgb(0,0,1,.25))
hist(r.NLG, freq=F, breaks=100, xlim=c(-10,50), col=rgb(1,0,0,.25), add=T)
z <- seq(-10,50,1)
lines(z,d.NL(z), lty=2, col="blue", lwd=2) # looks great... convolution performs as expected.
lines(z,d.NLG(z), lty=2, col="red", lwd=2) # this appears to work perfectly
# for most values of K and low values of theta
# this is looking a lot more promising... how about estimating gamma params?
optimFUN <- function(par, dN, dL) {
tG <- Gammad(par[1],par[2])
d.NLG <- d(convpow(dN+dL+tG,1))
p <- d.NLG(r.NLG)
p[p==0] <- 1e-15 # because sometimes very low probabilities evaluate to 0...
# ...and logs don't like that.
-sum(log(p))
}
fitresult <- optim(c(1,1e4), optimFUN, dN=dN, dL=dL)
fdG <- Gammad(fitresult$par[1], fitresult$par[2])
fd.NLG <- d(convpow(dN+dL+fdG,1))
lines(z,fd.NLG(z), lty=2, col="green3", lwd=2) ## this works perfectly when ~k>1 & ~theta<100... but throws
## "Error in validityMethod(object) : shape has to be positive" when k decreases and/or theta increases
## (boundary subject to RNG).
Can i speed up the integration in example 1? can I increase the grid size in example 2 (distr package)? how can I address the k<1 problem? can I rescale the data in a way that will better facilitate evaluation at high theta values?
Is there a better way all-together?
Help!
Well, convolution of function with gaussian kernel calls for use of Gauss–Hermite quadrature. In R it is implemented in special package: https://cran.r-project.org/web/packages/gaussquad/gaussquad.pdf
UPDATE
For convolution with Gamma distribution this package might be useful as well via Gauss-Laguerre quadrature
UPDATE II
Here is quick code to convolute gaussian with lognormal,
hopefully not a lot of bugs and and prints some reasonable looking graph
library(gaussquad)
n.quad <- 170 # integration order
# get the particular weights/abscissas as data frame with 2 observables and n.quad observations
rule <- ghermite.h.quadrature.rules(n.quad, mu = 0.0)[[n.quad]]
# test function - integrate 1 over exp(-x^2) from -Inf to Inf
# should get sqrt(pi) as an answer
f <- function(x) {
1.0
}
q <- ghermite.h.quadrature(f, rule)
print(q - sqrt(pi))
# convolution of lognormal with gaussian
# because of the G-H rules, we have to make our own function
# for simplicity, sigmas are one and mus are zero
sqrt2 <- sqrt(2.0)
c.LG <- function(z) {
#print(z)
f.LG <- function(x) {
t <- (z - x*sqrt2)
q <- 0.0
if (t > 0.0) {
l <- log(t)
q <- exp( - 0.5*l*l ) / t
}
q
}
ghermite.h.quadrature(Vectorize(f.LG), rule) / (pi*sqrt2)
}
library(ggplot2)
p <- ggplot(data = data.frame(x = 0), mapping = aes(x = x))
p <- p + stat_function(fun = Vectorize(c.LG))
p <- p + xlim(-1.0, 5.0)
print(p)

Is it possible to sample from a conditional density in R given some conditional data?

In R, using the np package, I have created the bandwidths for a conditional density. What I would like to do is, given some new conditional vector, sample from the resulting distribution.
Current code:
library('np')
# Generate some test data.
somedata = data.frame(replicate(10,runif(100, 0, 1)))
# Conditional variables.
X <- data.frame(somedata[, c('X1', 'X2', 'X3')])
# Dependent variables.
Y <- data.frame(somedata[, c('X4', 'X5', 'X6')])
# Warning, this can be slow (but shouldn't be too bad).
bwsome = npcdensbw(xdat=X, ydat=Y)
# TODO: Given some vector t of conditional data, how can I sample from the resulting distribution?
I am quite new to R, so while I did read the package documentation, I haven't been able to figure out if what I vision makes sense or is possible. If necessary, I would happily use a different package.
Here is the Example 2.49 from: https://cran.r-project.org/web/packages/np/vignettes/np_faq.pdf , it gives the following
solution for for 2 variables:
###
library(np)
data(faithful)
n <- nrow(faithful)
x1 <- faithful$eruptions
x2 <- faithful$waiting
## First compute the bandwidth vector
bw <- npudensbw(~x1 + x2, ckertype = "gaussian")
plot(bw, view = "fixed", ylim = c(0, 3))
## Next generate draws from the kernel density (Gaussian)
n.boot <- 1000
i.boot <- sample(1:n, n.boot, replace = TRUE)
x1.boot <- rnorm(n.boot,x1[i.boot],bw$bw[1])
x2.boot <- rnorm(n.boot,x2[i.boot],bw$bw[2])
## Plot the density for the bootstrap sample using the original
## bandwidths
plot(npudens(~x1.boot+x2.boot,bws=bw$bw), view = "fixed")
Following this hint from #coffeejunky, the following is a possible
solution to your problem with 6 variables:
## Generate some test data.
somedata = data.frame(replicate(10, runif(100, 0, 1)))
## Conditional variables.
X <- data.frame(somedata[, c('X1', 'X2', 'X3')])
## Dependent variables.
Y <- data.frame(somedata[, c('X4', 'X5', 'X6')])
## First compute the bandwidth vector
n <- nrow(somedata)
bw <- npudensbw(~X$X1 + X$X2 + X$X3 + Y$X4 + Y$X5 + Y$X6, ckertype = "gaussian")
plot(bw, view = "fixed", ylim = c(0, 3))
## Next generate draws from the kernel density (Gaussian)
n.boot <- 1000
i.boot <- sample(1:n, n.boot, replace=TRUE)
x1.boot <- rnorm(n.boot, X$X1[i.boot], bw$bw[1])
x2.boot <- rnorm(n.boot, X$X2[i.boot], bw$bw[2])
x3.boot <- rnorm(n.boot, X$X3[i.boot], bw$bw[3])
x4.boot <- rnorm(n.boot, Y$X4[i.boot], bw$bw[4])
x5.boot <- rnorm(n.boot, Y$X5[i.boot], bw$bw[5])
x6.boot <- rnorm(n.boot, Y$X6[i.boot], bw$bw[6])
## Plot the density for the bootstrap sample using the original
## bandwidths
ob1 <- npudens(~x1.boot + x2.boot + x3.boot + x4.boot + x5.boot + x6.boot, bws = bw$bw)
plot(ob1, view = "fixed", ylim = c(0, 3))

How to plot the probabilistic density function of a function?

Assume A follows Exponential distribution; B follows Gamma distribution
How to plot the PDF of 0.5*(A+B)
This is fairly straight forward using the "distr" package:
library(distr)
A <- Exp(rate=3)
B <- Gammad(shape=2, scale=3)
conv <- 0.5*(A+B)
plot(conv)
plot(conv, to.draw.arg=1)
Edit by JD Long
Resulting plot looks like this:
If you're just looking for fast graph I usually do the quick and dirty simulation approach. I do some draws, slam a Gaussian density on the draws and plot that bad boy:
numDraws <- 1e6
gammaDraws <- rgamma(numDraws, 2)
expDraws <- rexp(numDraws)
combined <- .5 * (gammaDraws + expDraws)
plot(density(combined))
output should look a little like this:
Here is an attempt at doing the convolution (which #Jim Lewis refers to) in R. Note that there are probably much more efficient ways of doing this.
lower <- 0
upper <- 20
t <- seq(lower,upper,0.01)
fA <- dexp(t, rate = 0.4)
fB <- dgamma(t,shape = 8, rate = 2)
## C has the same distribution as (A + B)/2
dC <- function(x, lower, upper, exp.rate, gamma.rate, gamma.shape){
integrand <- function(Y, X, exp.rate, gamma.rate, gamma.shape){
dexp(Y, rate = exp.rate)*dgamma(2*X-Y, rate = gamma.rate, shape = gamma.shape)*2
}
out <- NULL
for(ix in seq_along(x)){
out[ix] <-
integrate(integrand, lower = lower, upper = upper,
X = x[ix], exp.rate = exp.rate,
gamma.rate = gamma.rate, gamma.shape = gamma.shape)$value
}
return(out)
}
fC <- dC(t, lower=lower, upper=upper, exp.rate=0.4, gamma.rate=2, gamma.shape=8)
## plot the resulting distribution
plot(t,fA,
ylim = range(fA,fB,na.rm=TRUE,finite = TRUE),
xlab = 'x',ylab = 'f(x)',type = 'l')
lines(t,fB,lty = 2)
lines(t,fC,lty = 3)
legend('topright', c('A ~ exp(0.4)','B ~ gamma(8,2)', 'C ~ (A+B)/2'),lty = 1:3)
I'm not an R programmer, but it might be helpful to know that for independent random variables with PDFs f1(x) and f2(x), the PDF
of the sum of the two variables is given by the convolution f1 * f2 (x) of the two input PDFs.

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