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Hi everyone im using R to try and simulate some economic models. We do this primarily through the use of the euler equation. I've figured out that applying shocks to values which are defined within the function (in this case it is k is pretty simple as seen in the code below, however I'm interested in applying a shock to parameters like delta, theta and rho.
For what its worth I'm using the R package deSolve. Any help is appreciated.
library('deSolve')
##############################################
#Computing the neoclassical growth model in R#
##############################################
#parameters and state space
A<-1
theta<- 0.1
alpha<-0.5
delta<-0.3
rho<-0.9
kinital <- c(k = 1)
times <- seq(from = 0, to = 100, by = 0.2)
#define euler equation
euler <- function(t, k, parms)
list((1/theta)*alpha*A*k^(alpha-1)-delta-rho)
#Compute
out <- ode(y = kinital, times = times, func = euler,
parms = NULL)
plot(out, main = "Euler equation", lwd = 2)
#########################
#Temporary Capital Shock#
########################
eventdat <- data.frame(var = c("k"),
time = c(30) ,
value = c(10),
method = c("add"))
eventdat1 <- data.frame(var = c("k"),
time = c(30) ,
value = c(-5),
method = c("add"))
out3<-ode(y=kinital,times=times,func=euler,events=list(data=eventdat))
out4<-ode(y=kinital,times=times,func=euler,events=list(data=eventdat1))
plot(out,out3,out4,main="Temporary Shock",lwd=3)
Not a great fix but the way to deal with this type of problem is by conditioning your values to take place over some interval. I do this for depreciation as follows:
##############################
#Temporary Depreciation Shock#
##############################
#New Vars
A<-1
theta<- 0.1
alpha<-0.5
delta<-0.3
rho<-0.9
kinital <- c(k = 17)
times <- seq(from = 0, to = 400, by = 0.2)
#Redefine Euler
euler2<-function(t,k,prams){
list((1/theta)*alpha*A*k^(alpha-1)-delta-rho)}
euler3<-function(t,k,prams){
list((1/theta)*alpha*A*k^(alpha-1)-(delta+0.05*(t>=30&t<=40))-rho)}
#Output
doutbase<-ode(y=kinital,times=times, func=euler2, parms=NULL)
doutchange<-ode(y=kinital,times=times, func=euler3, parms=NULL)
#plots
plot(doutbase,doutchange,main="Change in depreciation at t=30 until t=40",lwd=2)
A colleague off of stackexchange suggested a cleaner bit of code which is a bit cleaner. This is seen below:
A<-1
theta<- 0.1
alpha <- 0.5
rho<-0.9
init <- c(k = 17, delta = 0.3)
times <- seq(from = 0, to = 400, by = 0.2)
euler.function<-function(t,y, prams){
k <- y[1]
delta <- y[2]
dk <- (1/theta)*alpha*A*k^(alpha-1)-delta-rho
list(c(dk, 0))}
deventdat<- data.frame(var = c("delta", "delta"),
time = c(30, 51) ,
value = c(0.1, -0.1),
method = c("add"))
res<-ode(y=init,times=times, func=euler.function, parms=NULL, events=list(data=deventdat))
plot(res,lwd=2)
I am trying to solve this system of ODEs through deSolve, dX/dt = -X*a + (Y-X)b + c and dY/dt = -Ya + (X-Y)*b for time [0,200], a=0.30, b=0.2 but c is 1 for time [50,70] and 0 otherwise. The code I have been using is,
time <- seq(0, 200, by=1)
parameters <- c(a=0.33, b=0.2, c=1)
state <- c(X = 0, Y = 0)
two_comp <- function(time, state, parameters){
with(as.list(c(state, parameters)), {
dX = -X*a + (Y-X)*b + c
dY = -Y*a + (X-Y)*b
return(list(c(dX, dY)))
})
}
out <- ode(y = state, times = time, func = two_comp, parms = parameters)
out.df = as.data.frame(out)
I have left out the time varying part of the c parameter since I can't figure out a way to include it and run it smoothly. I tried including it in the function definitions, but to no avail.
The standard way is to use approxfun, i.e. create a time dependent signal, that we also call forcing variable:
library("deSolve")
time <- seq(0, 200, by=1)
parameters <- c(a=0.33, b=0.2, c=1)
state <- c(X = 0, Y = 0)
two_comp <- function(time, state, parameters, signal){
cc <- signal(time)
with(as.list(c(state, parameters)), {
dX <- -X * a + (Y - X) * b + cc
dY <- -Y * a + (X - Y) * b
return(list(c(dX, dY), c = cc))
})
}
signal <- approxfun(x = c(0, 50, 70, 200),
y = c(0, 1, 0, 0),
method = "constant", rule = 2)
out <- ode(y = state, times = time, func = two_comp,
parms = parameters, signal = signal)
plot(out)
Note also the deSolve specific plot function and that the time dependent variable cc is used as an additional output variable.
More about this can be found:
in the ?forcings help page and
in a short tutorial on Github.
The interval limits where c is equal to 1 can be passed as parameters. Then, inside the differential function, use them to create a logical value
time >= lower & time <= upper
Since FALSE/TRUE are coded as the integers 0/1, every time this condition is false, c is multiplied by zero and the trick is done.
library(deSolve)
two_comp <- function(time, state, parameters){
with(as.list(c(state, parameters)), {
dX = -X*a + (Y-X)*b + c*(time >= lower & time <= upper)
dY = -Y*a + (X-Y)*b
return(list(c(dX, dY)))
})
}
time <- seq(0, 200, by=1)
parameters <- c(a=0.33, b=0.2, c=1, lower = 50, upper = 70)
state <- c(X = 0, Y = 0)
out <- ode(
y = state,
times = time,
func = two_comp,
parms = parameters
)
out.df <- as.data.frame(out)
head(out.df)
matplot(out.df$time, out.df[-1], type = "l", lty = "solid", ylim = c(0, 3))
legend("topright", legend = names(out.df)[-1], col = 1:2, lty = "solid")
I am trying to integrate the following function using a Monte Carlo Integration. The interval I want to integrate is x <- seq(0, 1, by = 0.01) and y <- seq(0, 1, by = 0.01).
my.f <- function(x, y){
result = x^2 + sin(x) + exp(cos(y))
return(result)
}
I calculated the integral using the cubature package.
library(cubature)
library(plotly)
# Rewriting the function, so it can be integrated
cub.function <- function(x){
result = x[1]^2 + sin(x[1]) + exp(cos(x[2]))
return(result)
}
cub.integral <- adaptIntegrate(f = cub.function, lowerLimit = c(0,0), upperLimit = c(1,1))
The result is 3.134606. But when I use my Monte Carlo Integration Code, see below, my result is about 1.396652. My code is wrong by more than a factor of 2!
What I did:
Since I need a volume to conduct a Monte Carlo Integration, I calculated the function values on the mentioned interval. This will give me an estimation of the maximum and minimum of the function.
# My data range
x <- seq(0, 1, by = 0.01)
y <- seq(0, 1, by = 0.01)
# The matrix, where I save the results
my.f.values <- matrix(0, nrow = length(x), ncol = length(y))
# Calculation of the function values
for(i in 1:length(x)){
for(j in 1:length(y)){
my.f.values[i,j] <- my.f(x = x[i], y = y[j])
}
}
# The maximum and minimum of the function values
max(my.f.values)
min(my.f.values)
# Plotting the surface, but this is not necessary
plot_ly(y = x, x = y, z = my.f.values) %>% add_surface()
So, the volume that we need is simply the maximum of the function values, since 1 * 1 * 4.559753 is simply 4.559753.
# Now, the Monte Carlo Integration
# I found the code online and modified it a bit.
monte = function(x){
tests = rep(0,x)
hits = 0
for(i in 1:x){
y = c(runif(2, min = 0, max = 1), # y[1] is y; y[2] is y
runif(1, min = 0, max = max(my.f.values))) # y[3] is z
if(y[3] < y[1]**2+sin(y[1])*exp(cos(y[2]))){
hits = hits + 1
}
prop = hits / i
est = prop * max(my.f.values)
tests[i] = est
}
return(tests)
}
size = 10000
res = monte(size)
plot(res, type = "l")
lines(x = 1:size, y = rep(cub.integral$integral, size), col = "red")
So, the result is completely wrong. But if I change the function a bit, suddenly is works.
monte = function(x){
tests = rep(0,x)
hits = 0
for(i in 1:x){
x = runif(1)
y = runif(1)
z = runif(1, min = 0, max = max(my.f.values))
if(z < my.f(x = x, y = y)){
hits = hits + 1
}
prop = hits / i
est = prop * max(my.f.values)
tests[i] = est
}
return(tests)
}
size = 10000
res = monte(size)
plot(res, type = "l")
lines(x = 1:size, y = rep(cub.integral$integral, size), col = "red")
Can somebody explain why the result suddenly changes? To me, both functions seem to do the exact same thing.
In your (first) code for monte, this line is in error:
y[3] < y[1]**2+sin(y[1])*exp(cos(y[2]))
Given your definition of my.f, it should surely be
y[3] < y[1]**2 + sin(y[1]) + exp(cos(y[2]))
Or..., given that you shouldn't be repeating yourself unnecessarily:
y[3] < my.f(y[1], y[2])
This is my first attempt at fitting a non-linear model in R, so please bear with me.
Problem
I am trying to understand why nls() is giving me this error:
Error in nlsModel(formula, mf, start, wts): singular gradient matrix at initial parameter estimates
Hypotheses
From what I've read from other questions here at SO it could either be because:
my model is discontinuous, or
my model is over-determined, or
bad choice of starting parameter values
So I am calling for help on how to overcome this error. Can I change the model and still use nls(), or do I need to use nls.lm from the minpack.lm package, as I have read elsewhere?
My approach
Here are some details about the model:
the model is a discontinuous function, a kind of staircase type of function (see plot below)
in general, the number of steps in the model can be variable yet they are fixed for a specific fitting event
MWE that shows the problem
Brief explanation of the MWE code
step_fn(x, min = 0, max = 1): function that returns 1 within the interval (min, max] and 0 otherwise; sorry about the name, I realize now it is not really a step function... interval_fn() would be more appropriate I guess.
staircase(x, dx, dy): a summation of step_fn() functions. dx is a vector of widths for the steps, i.e. max - min, and dy is the increment in y for each step.
staircase_formula(n = 1L): generates a formula object that represents the model modeled by the function staircase() (to be used with the nls() function).
please do note that I use the purrr and glue packages in the example below.
Code
step_fn <- function(x, min = 0, max = 1) {
y <- x
y[x > min & x <= max] <- 1
y[x <= min] <- 0
y[x > max] <- 0
return(y)
}
staircase <- function(x, dx, dy) {
max <- cumsum(dx)
min <- c(0, max[1:(length(dx)-1)])
step <- cumsum(dy)
purrr::reduce(purrr::pmap(list(min, max, step), ~ ..3 * step_fn(x, min = ..1, max = ..2)), `+`)
}
staircase_formula <- function(n = 1L) {
i <- seq_len(n)
dx <- sprintf("dx%d", i)
min <-
c('0', purrr::accumulate(dx[-n], .f = ~ paste(.x, .y, sep = " + ")))
max <- purrr::accumulate(dx, .f = ~ paste(.x, .y, sep = " + "))
lhs <- "y"
rhs <-
paste(glue::glue('dy{i} * step_fn(x, min = {min}, max = {max})'),
collapse = " + ")
sc_form <- as.formula(glue::glue("{lhs} ~ {rhs}"))
return(sc_form)
}
x <- seq(0, 10, by = 0.01)
y <- staircase(x, c(1,2,2,5), c(2,5,2,1)) + rnorm(length(x), mean = 0, sd = 0.2)
plot(x = x, y = y)
lines(x = x, y = staircase(x, dx = c(1,2,2,5), dy = c(2,5,2,1)), col="red")
my_data <- data.frame(x = x, y = y)
my_model <- staircase_formula(4)
params <- list(dx1 = 1, dx2 = 2, dx3 = 2, dx4 = 5,
dy1 = 2, dy2 = 5, dy3 = 2, dy4 = 1)
m <- nls(formula = my_model, start = params, data = my_data)
#> Error in nlsModel(formula, mf, start, wts): singular gradient matrix at initial parameter estimates
Any help is greatly appreciated.
I assume you are given a vector of observations of length len as the ones plotted in your example, and you wish to identify k jumps and k jump sizes. (Or maybe I misunderstood you; but you have not really said what you want to achieve.)
Below I will sketch a solution using Local Search. I start with your example data:
x <- seq(0, 10, by = 0.01)
y <- staircase(x,
c(1,2,2,5),
c(2,5,2,1)) + rnorm(length(x), mean = 0, sd = 0.2)
A solution is a list of positions and sizes of the jumps. Note that I use vectors to store these data, as it will become cumbersome to define variables when you have 20 jumps, say.
An example (random) solution:
k <- 5 ## number of jumps
len <- length(x)
sol <- list(position = sample(len, size = k),
size = runif(k))
## $position
## [1] 89 236 859 885 730
##
## $size
## [1] 0.2377453 0.2108495 0.3404345 0.4626004 0.6944078
We need an objective function to compute the quality of the solution. I also define a simple helper function stairs, which is used by the objective function.
The objective function abs_diff computes the average absolute difference between the fitted series (as defined by the solution) and y.
stairs <- function(len, position, size) {
ans <- numeric(len)
ans[position] <- size
cumsum(ans)
}
abs_diff <- function(sol, y, stairs, ...) {
yy <- stairs(length(y), sol$position, sol$size)
sum(abs(y - yy))/length(y)
}
Now comes the key component for a Local Search: the neighbourhood function that is used to evolve the solution. The neighbourhood function takes a solution and changes it slightly. Here, it will either pick a position or a size and modify it slightly.
neighbour <- function(sol, len, ...) {
p <- sol$position
s <- sol$size
if (runif(1) > 0.5) {
## either move one of the positions ...
i <- sample.int(length(p), size = 1)
p[i] <- p[i] + sample(-25:25, size = 1)
p[i] <- min(max(1, p[i]), len)
} else {
## ... or change a jump size
i <- sample.int(length(s), size = 1)
s[i] <- s[i] + runif(1, min = -s[i], max = 1)
}
list(position = p, size = s)
}
An example call: here the new solution has its first jump size changed.
## > sol
## $position
## [1] 89 236 859 885 730
##
## $size
## [1] 0.2377453 0.2108495 0.3404345 0.4626004 0.6944078
##
## > neighbour(sol, len)
## $position
## [1] 89 236 859 885 730
##
## $size
## [1] 0.2127044 0.2108495 0.3404345 0.4626004 0.6944078
I remains to run the Local Search.
library("NMOF")
sol.ls <- LSopt(abs_diff,
list(x0 = sol, nI = 50000, neighbour = neighbour),
stairs = stairs,
len = len,
y = y)
We can plot the solution: the fitted line is shown in blue.
plot(x, y)
lines(x, stairs(len, sol.ls$xbest$position, sol.ls$xbest$size),
col = "blue", type = "S")
Try DE instead:
library(NMOF)
yf= function(params,x){
dx1 = params[1]; dx2 = params[2]; dx3 = params[3]; dx4 = params[4];
dy1 = params[5]; dy2 = params[6]; dy3 = params[7]; dy4 = params[8]
dy1 * step_fn(x, min = 0, max = dx1) + dy2 * step_fn(x, min = dx1,
max = dx1 + dx2) + dy3 * step_fn(x, min = dx1 + dx2, max = dx1 +
dx2 + dx3) + dy4 * step_fn(x, min = dx1 + dx2 + dx3, max = dx1 +
dx2 + dx3 + dx4)
}
algo1 <- list(printBar = FALSE,
nP = 200L,
nG = 1000L,
F = 0.50,
CR = 0.99,
min = c(0,1,1,4,1,4,1,0),
max = c(2,3,3,6,3,6,3,2))
OF2 <- function(Param, data) { #Param=paramsj data=data2
x <- data$x
y <- data$y
ye <- data$model(Param,x)
aux <- y - ye; aux <- sum(aux^2)
if (is.na(aux)) aux <- 1e10
aux
}
data5 <- list(x = x, y = y, model = yf, ww = 1)
system.time(sol5 <- DEopt(OF = OF2, algo = algo1, data = data5))
sol5$xbest
OF2(sol5$xbest,data5)
plot(x,y)
lines(data5$x,data5$model(sol5$xbest, data5$x),col=7,lwd=2)
#> sol5$xbest
#[1] 1.106396 12.719182 -9.574088 18.017527 3.366852 8.721374 -19.879474 1.090023
#> OF2(sol5$xbest,data5)
#[1] 1000.424
I am wondering how I could make my function Bpp to accept a vector for its first argument t?
Bpp = function(t, n1, n2 = NULL){
N = ifelse(is.null(n2), n1, n1*n2/(n1+n2))
df = ifelse(is.null(n2), n1 - 1, n1 + n2 - 2)
H1 = integrate(function(delta)dcauchy(delta, 0, sqrt(2)/2)*dt(t, df, delta*sqrt(N)), -Inf, Inf)[[1]]
H0 = dt(t, df)
BF10 = H1/H0
p.value = 2*(1-pt(abs(t), df))
list(BF10 = BF10, p.value = p.value)
}
Bpp(t = -6:6, 20, 20) ## This will give error because `t` is now a vector?
Looks like I could give a quick answer without testing. Use the following in your Bpp:
# joint density
joint <- function(delta, t) dcauchy(delta, 0, sqrt(2)/2) * dt(t, df, delta*sqrt(N))
# marginal density of `t`
marginal.t <- function (t) integrate(joint, lower = -Inf, upper = Inf, t = t)[[1]]
H1 <- sapply(t, marginal.t)
So, here we also could use Vectorize how would that look like?
Use your original Bpp:
Bpp <- Vectorize(Bpp, vectorize.args = "t")
Bpp(-6:6, 20, 20)