So, I have this function;
f <- function(a, b, q=1, f0=1000) {
#calculate R:
R <- (f0 - (a*b))*((q+1)/(a^(q+1)))
return(ifelse(a<=100,(R * a^q) + b, 0)) }
I am using it in another function, funk2
funk2 <- function(a,x,b,l,r) {
f(a-x,b) * exp(-(l/r)*(exp(-r*a)*(exp(r*x)-1))) }
funk2 is then used to evaluate another multi-variable function funk1 using integration;
funk1 <- function(x,b,l,r) {
sapply(x, function (s) {
integrate(funk2, lower = s, upper = s+56, x=s, b=b, l=l, r=r)$value }) }
when I try to evaluate funk1
funk1(10,100,1,1)
{or by putting any other values} I get an error saying that
Error in integrate(funk2, lower = s, upper = s + 100, x = s, b = b,
non-finite function value
I am not sure what am I doing wrong here? Please help!
Thanks in advance.
The problem is in f(a-x,b) inside funk2 when a is equal to x then f(0,b) will produce NaN. For example, if you modify your lower to lower = s*1.01 you get:
funk1(10,100,1,1)
[1] 4464.721
Related
I am working on code that uses the uniroot function to approximate the root of an equation. I am trying to plot the behaviour of the function being passed through uniroot as the value of a free variable changes:
library(Deriv)
f1 <- function(s) {
(1 - 2*s)^(-3/2)*exp((8*s)/(1-2*s))
}
f2 <- function(s) {
log(f1(s))
}
f3 <- Deriv(f2, 's')
f4 <- Deriv(f3, 's')
f5 <- Deriv(f4, 's')
upp_s <- 1/2 - 1e-20
f_est <- function(x) {
f3a <- function(s) {f3(s = s) - x}
s_ <- uniroot(f3a,
lower = -9,
upper = upp_s)$root
return(s_)
}
plot(f_est, from = 0, to=100, col="red", main="header")
The output of f_est works as expected. However, when passed through the plot function, uniroot seems to break:
> plot(f_est, from = 0, to=100, col="red", main="header")
Error in uniroot(f3a, lower = -9, upper = upp_s) :
f() values at end points not of opposite sign
In addition: Warning messages:
1: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
the condition has length > 1 and only the first element will be used
2: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
Error in uniroot(f3a, lower = -9, upper = upp_s) :
f() values at end points not of opposite sign
The function is set up such that the endpoints specified in uniroot are always of opposite sign, and that there is always exactly one real root. I have also checked to confirm that the endpoints are non-missing when f_est is run by itself. I've tried vectorising the functions involved to no avail.
Why is this happening?
I was able to get most of the way there with
upp_s <- 0.497
plot(Vectorize(f_est), from = 0.2, to = 100)
Not only is 1/2 - epsilon exactly equal to 1/2 for values of epsilon that are too small (due to floating point error), I found that f3() gives NaN for values >= 0.498. Setting upp_s to 0.497 worked OK.
plot() applied to a function calls curve(), which needs a function that can take a vector of x values.
The curve broke with "f() values at end points not of opposite sign" if I started the curve from 0.1; I didn't dig in further and try to diagnose what was going wrong.
PS. It is generally more numerically stable and efficient to do computations directly on the log scale where possible. In this case, that means using
f2 <- function(s) { (-3/2)*log(1-2*s) + (8*s)/(1-2*s) }
instead of
f1 <- function(s) {
(1 - 2*s)^(-3/2)*exp((8*s)/(1-2*s))
}
f2_orig <- function(s) {
log(f1(s))
}
## check
all.equal(f2(0.25), f2_orig(0.25)) ## TRUE
Doing this and setting the lower bound of uniroot() to -500 lets us get pretty close to the zero boundary (although it looks both analytically and computationally as though the function diverges to -∞ as x goes to 0).
f3 <- Deriv(f2, 's')
upp_s <- 1/2 - 1e-10
lwr_a <- -500
f_est <- function(x) {
f3a <- function(s) { f3(s = s) - x}
s_ <- uniroot(f3a,
lower = lwr_a,
upper = upp_s)$root
return(s_)
}
plot(Vectorize(f_est), from = 0.005, to = 100, log = "x")
You can also solve this analytically, or ask caracas (an R interface to sympy) to do it for you:
library(caracas)
x <- symbol("x"); s <- symbol("s")
## peek at f3() guts to find the expression for the derivative;
## could also do the whole thing in caracas/sympy
solve_sys((11 +16*(s/(1-s*2)))/(1-s*2), x, list(s))
sol <- function(x) { (2*x - sqrt(32*x + 9) -3)/(4*x) }
curve(sol, add = TRUE, col = 2)
I have the next function:
f <- function(x,y,z,w) {dpois(x, lambda = w*cos(y-z) ) }
I want to construct a function g(y,z,w) defined by the integral of f with respect to x. I can do it when I have two variables, the code is for example, something like this
margin <- function(y) { sapply(y, function(y) { integrate(function(x) f(x,y), llimx, ulimx)$value }) }
But I don't know how to do that for more than two variables.
Maybe you can try the code below
f <- Vectorize(function(x,y,z,w) dpois(x, w*cos(y-z)),"x")
g <- function(y,z,w) integrate(f, lower = 0, upper =Inf, y, z, w)
I'm trying to write the density of a mixture Gaussian distribution to an arbitrary power, b, in R. Currently, I have two methods that works, but I prefer if I could avoid a for loop.
dnorm_mix_tempered_unnorm <- function(x, w, m, s, b) {
value <- 0
for (i in 1:length(w)) {value <- value + w[i]*dnorm(x, mean = m[i], sd = s[i])}
value <- value^(b)
return(value)
}
Alternatively, I can vectorise this to avoid the for loop:
dnorm_mix_tempered_unnorm <- function(x, w, m, s, b) {
return(sum(w*dnorm(x, mean = m, sd = s))^b)
}
Both of these give the same result, but the second is more efficient since it is vectorised. But I need to next normalise this so that the density integrates to 1, I do this by using:
dnorm_mix_tempered <- function(x, weights, means, sds, beta) {
norm_constant <- integrate(function(x) dnorm_mix_tempered_unnorm(x, w = weights,
m = means, s = sds, b = 1/beta), lower = -Inf,
upper = Inf)$value
value <- dnorm_mix_tempered_unnorm(x, w = weights, m = means, s = sds, b = 1/beta)
/ norm_constant
return(value)
}
If I define dnorm_mix_tempered_unnorm with for loops, this works with no problem, and I can use curve() to plot the density. But if I define dnorm_mix_tempered_unnorm by using vectorisation, then I get the following error:
Error in integrate(function(x) dnorm_mix_tempered_unnorm(x, w = weights, :
evaluation of function gave a result of wrong length
Does anyone know what is going on when I am vectorising instead and trying to integrate?
Thanks in advance,
R.
A possible option is
dnorm_mix_tempered_unnorm <- function(x, w, m, s, b) {
return(rowSums(mapply(dnorm, mean = m, sd = m, MoreArgs = list(x = x)))^b)
}
But I think it is quite similar to your first proposal.
So, I have these functions:
funk1 <- function(a,x,l,r) {
x^2*exp(-(l*(1-exp(-r*a))/r))}
funk2 <- function(x,l,r) {
sapply(x, function (s) {
integrate(funk1, lower = 0, upper = s, x=s, l=l, r=r)$value })}
which are used to explain the data y in,
z <- data.frame(ts = 1:100,
y = funk2(1:100, l = 1, r = 1) + rpois(100, 1:100))
I wish to use optim to maximise the likelihood, so I defined a likelihood function:
LL_funk <- function(l,r) {
n=nrow(z)
R = sum((funk2(ts,l,r) - y)^2)
logl = -((n/2)*log(R))
return(-logl)
}
and I tried to fit using optim
fit <- optim(par=c(0.5,0.5), fn= LL_funk, method="Nelder-Mead")
But I get an error:
Error in integrate(funk1, lower = 0, upper = s, x = s, l = l, r = r) :
a limit is missing
I am not sure why? I could run nls fitting funk2(x,l,r) to y
nls(y ~ funk2(ts,l,r), data = z, start = list(l = 0.5, r = 0.5))
That means funk2 is working. I guess its the problem with LL function that I have designed, which I cant figure out!! Please Help!
Yup! There were two problems with your function. This worked for me:
LL_funk <- function(params) {
n=nrow(z)
l = params[1]
r = params[2]
R = sum((funk2(z$ts,l,r) - z$y)^2)
logl = -((n/2)*log(R))
return(-logl)
}
Previous issues:
LL_funk only takes 1 argument, which is the vector of parameters.
In LHS of the assignment of R, ts and y were not actually referring to columns in your dataset.
I'm working on a problem where a parameter is estimated through minimizing the sum of squares. The equations needed are:
I used optim in the package stats:
# provide the values for a test dataset (the y estimated should be 1.41)
pvector <- c(0.0036,0.0156,0.0204,0.0325,0.1096,0.1446,0.1843,0.4518)
zobs <- c(0.0971,0.0914,0.1629,0.1623,0.3840,0.5155,0.3648,0.6639)
# make input of the C value
c <- function(y){
gamma(y)/((gamma(y*(1-pvector)))*(gamma(y*pvector)))
}
# make input of the gamma function
F1 <- function(y){
f1 <- function(x){
c*(1-x)^(y*(1-pvector)-1)*x^(y*pvector-1)
}
return (f1)
}
# integration over x
int <- function(y){
integrate (F1(y),lower =0.001, upper =1)
}
# write the function for minimization
f2 <- function(y) {
sum ((int-zobs)^2)
}
# minimization
optim(0.01,f2, method = "Brent", lower =0, upper = 1000, hessian=TRUE)
Which didn't work. I received the following error message:
Error in int - zobs : non-numeric argument to binary operator
I think there must be something fundamentally wrong with the way how the function was written.