Let
dXt = -Yt dt +cos(Xt + Yt)/(sqrt(t+1))*dW1_t
dYt = Xt dt +sin(Xt + Yt)/(sqrt(t+1))*dW2_t
X0=1, Y0=1, T=1,
(W1_t, W2_t) is a Brownian Motion in dimension 2.
Could someone tell me how I could implement this system of SDEs in R?
I tried it with
library(Sim.DiffProc)
set.seed(1234)
fx <- expression(-y , x )
gx <- expression(cos(x+y)/sqrt(t+1),sin(x+y)/sqrt(t+1))
mod2d1 <-snssde2d(drift=fx,diffusion=gx,x0=c(x0=1,y0=1),M=1000)
the default solver in this package here is Euler Maruyama.What I need to calculate is Expected Value ((X_T)^2+(Y_T)^2) (with a different discretiziations and compare it with the analytic solution for this expected value. My question would be, how can I calculate this expected value of the sum of squares of the processes? summary(mod) gives me the mean, but how can I get the result that I want? do I have to square the expressions prior to input them to the solver?
Related
I am attempting to run a two state HMM using a lognormal distribution. I have read Michelot and Langrock (2019) regarding choosing starting parameters through inspecting the data in a histogram and then running iterations in parallel, which has worked for my gamma distribution. Identifying the starting parameters for the lognormal distribution is troubling me however. Do I plot the log of my step length distribution then attempt extracting starting parameters or use the same starting parameters as my gamma distribution and rely on stepDist="lnorm"?
My code for the lognormal attempt currently looks like this:
ncores <- detectCores() - 1
cl <- makeCluster(getOption("cl.cores", ncores))
clusterExport(cl, list("data", "fitHMM"))
niter <- 20
allPar0 <- lapply(as.list(1:niter), function(x) {
stepMean0 <- runif(2,
min = c(x,y),
max = c(y,z))
stepSD0 <- runif(2,
min = c(x,y),
max = c(y,z))
angleMean0 <- c(0, 0)
angleCon0 <- runif(2,
min = c(a,b),
max = c(a,b))
stepPar0 <- c(stepMean0, stepSD0)
anglePar0 <- c(angleMean0, angleCon0)
return(list(step = stepPar0, angle = anglePar0))
})
# Fit the niter models in parallel
logP <- parLapply(cl = cl, X = allPar0, fun = function(par0) {
m <- fitHMM(data = data, nbStates = 2, stepDist = "lnorm", stepPar0 = par0$step,
anglePar0 = par0$angle)
return(m)
})
# Extract likelihoods of fitted models
likelihoodL <- unlist(lapply(logP, function(m) m$mod$minimum))
likelihoodL
# Index of best fitting model (smallest negative log-likelihood)
whichbestpL <- which.min(likelihoodL)
bestL <- logP[[whichbestpL]]
bestL
If I use negative values from plotting the log of the step length of the data then I get the error:
Error in checkForRemoteErrors(val) :
7 nodes produced errors; first error: Check the step parameters bounds (the initial parameters should be strictly between the bounds of their parameter space).
If I use the same starting parameter values that I used for my gamma distribution then I get the error
Error in unserialize(node$con) :
embedded nul in string: 'X\n\0\0\0\003\0\004\002\0\0\003\005\0\0\0'
Please could someone shed some light on how I'm failing at this?
Thank you!
Unfortunately, I can't tell for sure what the problem is from the code you included. If you don't get an error when you run fitHMM outside of parLapply, then it suggests that the problem is in how you choose the values of x, y, and z in your code.
The first parameter of the log-normal distribution can be negative or positive, and it is actually the mean of the logarithm of the step length. So, to find good starting values for this, you should look at a histogram of the log step lengths (e.g., following the dedicated moveHMM vignette). The second parameter is the standard deviation of the log step lengths, and this should be strictly positive (but could also be chosen based on the spread of the histogram of log step lengths).
To summarise, you should choose all the initial values based on plots of the log step lengths (rather than the step lengths themselves), and you should not use the same ranges of values for stepMean0 and stepSD0 (because the former can be negative or positive, whereas the latter is positive). Hopefully, this should help you choose x, y, and z.
I am doing some projects related to statistics simulation using R based on "Introduction to Scientific Programming and Simulation Using R" and in the Students projects session (chapter 24) i am doing the "The pipe spiders of Brunswick" problem, but i am stuck on one part of an evolutionary algorithm, where you need to perform some data perturbation according to the sentence bellow:
"With probability 0.5 each element of the vector is perturbed, independently
of the others, by an amount normally distributed with mean 0 and standard
deviation 0.1"
What does being "perturbed" really mean here? I dont really know which operation I should be doing with my vector to make this perturbation happen and im not finding any answers to this problem.
Thanks in advance!
# using the most important features, we create a ML model:
m1 <- lm(PREDICTED_VALUE ~ PREDICTER_1 + PREDICTER_2 + PREDICTER_N )
#summary(m1)
#anova(m1)
# after creating the model, we perturb as follows:
#install.packages("perturb") #install the package
library(perturb)
set.seed(1234) # for same results each time you run the code
p1_new <- perturb(m1, pvars=c("PREDICTER_1","PREDICTER_N") , prange = c(1,1),niter=200) # your can change the number of iterations to any value n. Total number of iteration would come to be n+1
p1_new # check the values of p1
summary(p1_new)
Perturbing just means adding a small, noisy shift to a number. Your code might look something like this.
x = sample(10, 10)
ind = rbinom(length(x), 1, 0.5) == 1
x[ind] = x[ind] + rnorm(sum(ind), 0, 0.1)
rbinom gets the elements to be modified with probability 0.5 and rnorm adds the perturbation.
I have the following question regarding the mc2d package for Monte Carlo simulations.
Given a mc node, i.e. a mc object. How can we get the uncertainty for the values of the distribution?
For instance, as an input distribution I am using an uniform distribution, where the min is e.g. equal to 2, and the max equal to 8. Given this, we produce a mc object, apply it to mc.
The summary function produces values such as the median, mean, 97.5% etc. etc.
But as I said, how can be get an estimate of uncertainty for a given value?
Thanks in advance!
Well, you'd have to collect second momentum
Then
v = <x^2> - <x>^2
u = sqrt(v)/sqrt(N-1)
a = <x> +-u
To make things more clear, you sample events
x = 2 + (8-2)*U(0,1)
somewhere in the summary function you compute sum of events
m = m + x
so after running N events you report mean=m/N
You have to add code to collect second momentum, something like
m2 = m2 + x*x
So after run you could compute
v = m2/N - mean*mean
u = sqrt(v)/sqrt(N-1)
and report mean with uncertainty as mean +-u
i'm comparing different measures of distance and similarity for vector profiles (Subtest results) in R, most of them are easy to compute and/or exist in dist().
Unfortunately, one that might be interesting and is to difficult for me to calculate myself is Cattel's Rp. I can not find it in R.
Does anybody know if this exists already?
Or can you help me to write a function?
The formula (Cattell 1994) of Rp is this:
(2k-d^2)/(2k + d^2)
where:
k is the median for chi square on a sample of size n;
d is the sum of the (weighted=m) difference between the two profiles,
sth like: sum(m(x(i)-y(i)));
one thing i don't know is, how to get the chi square median in there
Thank you
What i get without defining the k is:
Rp.Cattell <- function(x,y){z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);return(z)}
Vector examples are:
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
They are measures by the same device, but related to different bodyparts. They don't need to be standartised or weighted, i would say.
This page gives a general formula for k, and then gives a more thorough method using SAS/IML which pretty much gives the same results. So I used the general formula, added calculation of degrees of freedom, which leads to this:
Rp.Cattell <- function(x,y) {
dof <- (2-1) * (length(y)-1)
k <- (1-2/(9*dof))^3
z <- (2*k-sum(sum(x-y))^2)/(2*k+sum(sum(x-y))^2)
return(z)
}
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
Rp.Cattell(x, y)
# [1] -0.9012083
Does this figure appear to make sense?
Trying to verify the function, I found out now that the median of chisquare is the chisquare value for 50% probability - relating to random. So the function should be:
Rp.Cattell <- function(x,y){
dof <- (2-1) * (length(y)-1)
k <- qchisq(.50, df=dof)
z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);
return(z)}
It is necessary though to standardize the Values before, so the results are distributed correctly.
So:
library ("stringr")
# they are centered already
x <- as.vector(scale(c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758),center=F, scale=T))
y <- as.vector(scale(c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925),center=F, scale=T))
Rp.Cattell(x, y) -0.584423
This sounds reasonable now - or not?
I consider calculation of z is incorrect.
You need to calculate the sum of the squared differences. Not the square of the sum of differences. Besides product operator is missing in 2k.
It should be
z <- (2*k-sum((x-y)^2))/(2*k+sum((x-y)^2))
Do you agree?
I need your helps to explain how I can obtain the same result as this function does:
gini(x, weights=rep(1,length=length(x)))
http://cran.r-project.org/web/packages/reldist/reldist.pdf --> page 2. Gini
Let's say, we need to measure the inocme of the population N. To do that, we can divide the population N into K subgroups. And in each subgroup kth, we will take nk individual and ask for their income. As the result, we will get the "individual's income" and each individual will have particular "sample weight" to represent for their contribution to the population N. Here is example that I simply get from previous link and the dataset is from NLS
rm(list=ls())
cat("\014")
library(reldist)
data(nls);data
help(nls)
# Convert the wage growth from (log. dollar) to (dollar)
y <- exp(recent$chpermwage);y
# Compute the unweighted estimate
gini_y <- gini(y)
# Compute the weighted estimate
gini_yw <- gini(y,w=recent$wgt)
> --- Here is the result----
> gini_y = 0.3418394
> gini_yw = 0.3483615
I know how to compute the Gini without WEIGHTS by my own code. Therefore, I would like to keep the command gini(y) in my code, without any doubts. The only thing I concerned is that the way gini(y,w) operate to obtain the result 0.3483615. I tried to do another calculation as follow to see whether I can come up with the same result as gini_yw. Here is another code that I based on CDF, Section 9.5, from this book: ‘‘Relative
Distribution Methods in the Social Sciences’’ by Mark S. Handcock,
#-------------------------
# test how gini computes with the sample weights
z <- exp(recent$chpermwage) * recent$wgt
gini_z <- gini(z)
# Result gini_z = 0.3924161
As you see, my calculation gini_z is different from command gini(y, weights). If someone of you know how to build correct computation to obtain exactly
gini_yw = 0.3483615, please give me your advices.
Thanks a lot friends.
function (x, weights = rep(1, length = length(x)))
{
ox <- order(x)
x <- x[ox]
weights <- weights[ox]/sum(weights)
p <- cumsum(weights)
nu <- cumsum(weights * x)
n <- length(nu)
nu <- nu/nu[n]
sum(nu[-1] * p[-n]) - sum(nu[-n] * p[-1])
}
This is the source code for the function gini which can be seen by entering gini into the console. No parentheses or anything else.
EDIT:
This can be done for any function or object really.
This is bit late, but one may be interested in concentration/diversity measures contained in the [SciencesPo][1] package.