similar to: bivariate vector numerical integration with infinite range

Dear R-users, I am facing problem with integrating in R a likelihood function which is a function of four parameters. It's giving me the result at the end but taking more than half an hour to run. I'm wondering is there any other efficient way deal with. The following is my code. I am ready to provide any other description of my function if you need to move forward.

Hi R-users. I'm having difficulty with an integration in R via the package "cubature". I'm putting it with a simple example here. I wish to integrate a function like: f(x1,x2)=2/3*(x1+x2) in the interval 0<x1<x2<7. To be sure I tried it by hand and got 114.33, but the following R code is giving me 102.6667.

Hi, I am using adaptIntegrate from Cubature to do numerical integration on a double integral with a 1 x 2 vector x. Say the function is something simple to start like f(x)=x1*x2 and I wish to integrate x1 over (0,365-x2) and x2 over (0,365) f <- function(x) {(x[2])*(x[1])} # "x" is vector int1<-adaptIntegrate(f, lowerLimit = c(0, 0), upperLimit = c(365-x[2], 365)) I recieve

Hello, I am trying to use adaptIntegrate function but I need to pass on a few additional parameters to the integrand. However, this function seems not to have the flexibility of passing on such additional parameters. Am I missing something or this is a known limitation. Is there a good alternative to such restrictions, if there at all are? Many thanks for your time. HC -- View this message in

Dear list, [cross-posting from Stack Overflow where this question has remained unanswered for two weeks] I'd like to perform a numerical integration in one dimension, I = int_a^b f(x) dx where the integrand f: x in IR -> f(x) in IR^p is vector-valued. integrate() only allows scalar integrands, thus I would need to call it many (p=200 typically) times, which sounds suboptimal. The

I want to measure the error in the estimation of a 2 dimensional density function that is calculated using an integral but run into problems trying to integrate with adaptIntegrate because the integrand also calls the function adaptIntegrate. In particular I want \int \hat{f}(x,y) - f(x,y) dx dy where \hat{f}(x,y) = \int K(a,b, x, y) da db and in this simulation study I know what the true value

Hello Dear List members, as you can see (and guess) from the code below adaptIntegrate(f,lowerLimit=c(-1,-1),upperLimit=c(.9999,.9999)) $integral [1] 9.997e-09 $error [1] 1.665168e-16 $functionEvaluations [1] 17 $returnCode [1] 0 > adaptIntegrate(f,lowerLimit=c(-1,-1),upperLimit=c(1,1)) the last command runs for 45 mins now. -this one takes only less than sec:

Dear all, I got an error message when running the following code. Can anyone give any suggestions on fixing this type of error? Thank you very much in advance. Hanna > integrand <- function(x, rho, a, b, z){ + x1 <- x[1] + x2 <- x[2] + Sigma <- matrix(c(1, rho, rho, 1), 2,2) + mu <- rep(0,2) + f <-

Hi there. Thanks for your time in advance. I am using R 2.2.0 and OS: Windows XP. My final goal is to calculate 1/2*integral of (f1(x)^1/2-f2(x)^(1/2))^2dx (Latex codes: $\frac{1}{2}\int^{{\infty}}_{\infty} (\sqrt{f_1(x)}-\sqrt{f_2(x)})^2dx $.) where f1(x) and f2(x) are two marginal densities. My problem: I have the following R codes using "adapt" package. Although "adapt"

Greetings, Sorry, the last message was sent by mistake! Here it is again: I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix

Hi folks, I am having a question about efficiently finding the integrals of a list of functions. To be specific, here is a simple example showing my question. Suppose we have a function f defined by f<-function(x,y,z) c(x,y^2,z^3) Thus, f is actually corresponding to three uni-dimensional functions f_1(x)=x, f_2(y)=y^2 and f_3(z)=z^3. What I am looking for are the integrals of these three

Hello to all, I am having trouble with intregrating a complicated uni-dimensional function of the following form Phi(x-a_1)*Phi(x-a_2)*...*Phi(x-a_{n-1})*phi(x-a_n). Here n is about 5000, Phi is the cumulative distribution function of standard normal, phi is the density function of standard normal, and x ranges over (-infty,infty). My idea is to to use quadrature to handle this integral. But

hello I work on the probabilities of bivariate normal distribution. I need integrate  the following function. f (x, y) = exp [- (x ^ 2 + y ^ 2 + x * y)] with - ∞ ≤ x ≤ 7.44 and - ∞ ≤ y ≤ 1.44   , either software R or  matlab Version R 2009a Thank you for helping me Regards Mezouara hicham PhD in Metrology Hicham_dess @ yahoo.fr [[alternative HTML version deleted]]

Hi all, it seems that there is problem with function "adaptIntegrate", when the integration limits is infinity. Please see the code below. The second integration does not seem to work. Can anyone familiar with this give some help? Thank you with much. Hanna library(mnormt) library(cubature) ff <- function(x, rho){ mu <- rep(0,3) Sigma

Hi All, I'm trying to use one of the (2D) numerical integration functions, which is not where the problem is. The function definition is as follows: adaptIntegrate(f, lowerLimit, upperLimit, ...) The problem is that I want to integrate a 3D function which has been parametrised such that it is a 2D function: eg. I want to integrate f(x_start+gradient_x*t1, y_start+gradient_y*t2) for t1 in

I need to calcuate the cumulative probability for the Natural Exponential Family - Hyperbolic secant distribution with a parameter theta between -pi/2 and pi/2. The integration should be between 0 and 1 as it is a probability. The function "integrate" works fine when the absolute value of theta is not too large. That is, the NEF-HS distribution is not too skewed. However, once the

Dear expeRts, I somehow don't see why the following does not work: integrand <- function(x, vec, mat, val) 1 # dummy return value A <- matrix(runif(16), ncol = 4) u <- c(0.4, 0.1, 0.2, 0.3) integrand(0.3, u, A, 4) integrate(integrand, lower = 0, upper = 1, vec = u, mat = A, val = 4) I would like to integrate a function ("integrand") which gets an "x" value (the

Hi All, Can any one help to explain why min and max function couldn't work in the integrate function directly. For example, if issue following into R: integrand <- function(x) {min(1-x, x^2)} integrate(integrand, lower = 0, upper = 1) it will return this: Error in integrate(integrand, lower = 0, upper = 1) : evaluation of function gave a result of wrong length However, as min(U,V) =

Hi, I've got an ugly but fairly simple function: mdevstdev <- function(a){ l <- dnorm(a)/(1-pnorm(a)) integrand <- function(z)(abs(z-l)*dnorm(z)) inted <- integrate(integrand, a, Inf) inted[[1]]/((1- pnorm(a))*sqrt((1 + a*l - l^2))) } I wanted to quickly produce a graph of this over the range [-3,3] so I used: plotit <-function(x=seq(-3,3,0.01),...){

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over