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I have an R function defined as: f<-function(x){ return(dchisq(x,9,77)*((13.5/x)^5)*exp(-13.5/x)) } Numerically integrating this function, I observed a couple of things: A) Integrating the function f over the entire positive real line gives an error: > integrate(f,0,Inf) Error in integrate(f, 0, Inf) : the integral is probably divergent B) Increasing the interval of integration

Dear R useRs, i have the function f1(x, y, z) which i want to integrate for x and y. On the one hand i do this by first integrating for x and then for y, on the other hand i do this the other way round and i wondering why i doesn't get the same result each way? z <- c(80, 20, 40, 30) "f1" <- function(x, y, z) {dgamma(cumsum(z)[-length(z)], shape=x, rate=y)}

Dear R users, i try to integrate lgamma from 0 to Inf. But here i get the message "roundoff error is detected in the extrapolation table", if i use 1.0e120 instead of Inf the computation works, but this is against the suggestion of integrates help information to use Inf explicitly. Using stirlings approximation doesnt bring the solution too. ## Stirlings approximation lgammaApprox

The background: I'm trying to fit a Poisson-lognormal distrbutuion to some data. This is a way of modelling species abundances: N ~ Pois(lam) log(lam) ~ N(mu, sigma2) The number of individuals are Poisson distributed with an abundance drawn from a log-normal distrbution. To fit this to data, I need to integrate out lam. In principle, I can do it this way: PLN1 <- function(lam, Count,

...> > I ran the regression tests, but they fail when building against older > > OpenSSL versions due to the lack of GCM support. > > Please try this diff: > I applied the diff you supplied, along with the previous diff. The regression tests got further along, but now fail in integrity.sh: run test integrity.sh ... test integrity: hmac-sha1 @2300 RSA_public_decrypt failed: error:0407006A:rsa routines:RSA_padding_check_PKCS1_type_1:block type is not 01. key_verify failed for server_host_key. unexpected error mac hmac-sha1 at 2300 test integrity: hmac-sha1 @2301 RSA_public_decrypt...

In the help for ?integrate: >When integrating over infinite intervals do so explicitly, rather than just using a large number as the endpoint. This increases the chance of a correct answer ? any function whose integral over an infinite interval is finite must be near zero for most of that interval. I understand that and there are examples such as: ## a slowly-convergent integral integrand

I'm trying to provide different parameters to the integrate function for various probability functions. I'm using dnorm as the simplest example here. For instance integrate(dnorm, -1.96, 1.96) produces the correct answer for a normal distribution with mean 0 and standard deviation 1. I've tried two ways to use mean=2.0 and standard deviation 1, but with no luck. The examples follow.

Hi all, Is there any other packages to do numerical integration other than the default 'integrate'? Basically, I am integrating: integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value The integration is ok provided sigma is >0. However, when mu=-1.645074 and sigma=17535.26 It stopped working. On the other hand, Maple gives me a value of 0.5005299403. It is an

Dear All, Hello! I have some questoins in R programming as follows: Question 1- How to take the integral of this function with respect to y, such that x would appear in the output after taking integral. f(x,y)=(0.1766*exp(-exp(y+lnx))*-exp(y+lnx))/(1-exp(-exp(y+lnx))) y in (-6.907,-1.246) It is doable in maple but not in R. At least I could not find the way. p.s: result from maple is:

Full_Name: Bert De Boeck Version: R 2.2.0 OS: Windows Submission from: (NULL) (157.193.193.152) I think there is a bug when using integrate for integrating a function which is 0 in a whole sub-interval. For example: #define uniform function f<-function(x){ifelse(x<1,0,ifelse(x<3,1,0))} #this is the correct integral integrate(f,-10,10) #here there is a problem integrate(f,-50,50)

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

In 'Writing R Extensions' section 6.9 there is the paragraph There are interfaces (defined in header R_ext/Applic.h) for definite and for indefinite integrals. ?Indefinite? means that at least one of the integration boundaries is not finite. An indefinite integral usually means an antiderivative, not an integral over an infinite spread. Should that first sentence end with 'for

Dear List-Members, since 3 weeks I have been heavily working on reproducing the results of an economic paper. The method there uses the numerical solution of an integral within nonlinear least squares. Within the integrand there is also some parameter to estimate. Is that in the end possible to implement in R [Originally it was done in GAUSS]? I'm nearly into giving up. I constucted an

Hi guys im having a problem getting R to numerically integrate for some function, say f(bhat)*optimize(G(bhat)), over bhat. Where id like to integrate this over some finite range, so that here as we integrate over bhat optimize would return a different optimum. For instance consider this simple example for which I cannot get R to return the desired result: f <- function(bhat) exp(bhat) g

Hello, I need to integrate the absolute difference between two lines measured on different points. # For example : x <- seq(0, 1, 1/100) f_x <- runif(101) + x y <- seq(0, 1, 1/23) f_y <- runif(24) + (1 - y) plot(x, f_x, type="l") lines(y, f_y) Then I would like to compute Integral( | f_x - f_y | )dx. (This is not the same as | Integral(f_x)dx - Integral(f_y)dx |.)

Dear r-help, I need to draw some lattice plots and have stuck in strange problems. I have a data frame of 405 rows: > str(dframe) `data.frame': 405 obs. of 4 variables: $ year : num 1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ... $ month: num 1 2 3 4 5 6 7 8 9 10 ... $ V_A2 : num NA NA NA NA NA ... $ V_A3 : num NA NA NA NA NA ... Variable month changes from 1 to 27

I'm having a function of the form 1> f<-function(x){ 1+ 1+ return(x^p) 1+ 1+ } ,and I would like to integrate it with respect to x, where p should be any constant. One way would be to set a value for p globally and then call integrate function: p=2 integrate(f, lower = -1, upper = 1) However, I would like to use 'integrate' inside a function, so I could call it passing

Dear all, This is the first time I am posting to the r-devel list. On StackOverflow, they suggested that the strange behaviour of integrate() was more bug-like. I am providing a short version of the question (full one with plots: https://stackoverflow.com/q/55639401). Suppose one wants integrate a function that is just a product of two density functions (like gamma). The support of the

Hi R users, I have a couple of questions about some problems that I am facing with regard to numerical integration and optimization of likelihood functions. Let me provide a little background information: I am trying to do maximum likelihood estimation of an econometric model that I have developed recently. I estimate the parameters of the model using the monthly US unemployment rate series

The integral of any probability density from -Inf to Inf should equal 1, correct? I don't understand last result below. > integrate(function(x) dnorm(x, 0,1), -Inf, Inf) 1 with absolute error < 9.4e-05 > integrate(function(x) dnorm(x, 100,10), -Inf, Inf) 1 with absolute error < 0.00012 > integrate(function(x) dnorm(x, 500,50), -Inf, Inf) 8.410947e-11 with absolute error <