similar to: Integration of two dimension function

Dear all, Y, X are bivariate normal with all the parameters known. I would like to generate numbers from the distribution Y | X > c where c is a constant. Does there exist an R function generating random numbers from such a distribution? Thank you very much. hannah [[alternative HTML version deleted]]

Sorry. I meant in the previous email that the function h() is a monotone decreasing function. Thanks very much. 2018-02-06 13:32 GMT-05:00 li li <hannah.hlx at gmail.com>: > Hi all, > The function h below is a function of c and it should be a monotone > increasing function since the integrand is nonnegative and integral is > taken from c to infinity. However, as we can see

Hi Hanna, your function is essentially zero outside a short interval around 9. And the help page states: "If the function is approximately constant (in particular, zero) over nearly all its range it is possible that the result and error estimate may be seriously wrong." You could try to integrate over a finite interval, say (7, 12). G?ran Brostr?m On 2018-02-06 19:40, li li wrote:

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

Hi all, I have some some problem with regard to finding the integral of a function containing an indicator function. please see the code below: func1 <- function(x, mu){ (mu^2)*dnorm(x, mean = mu, sd = 1)*dgamma(x, shape=2)} m1star <- function(x){ integrate(func1, lower = 0, upper = Inf,x)$val} T <- function(x){ 0.3*dnorm(x)/(0.3*dnorm(x)+0.7*m1star(x))} func2 <-

Dear all, I need to generate numbers from multivariate normal with large dimensions (5,000,000). Below is my code and the error I got from R. Sigma in the code is the covariance matrix. Can anyone give some idea on how to take care of this error. Thank you. Hannah > m <- 5000000 > m1 <- 0.5*m > rho <- 0.5 > Sigma <- rho* matrix(1, m, m)+diag(1-rho, m)

Hi all, The function h below is a function of c and it should be a monotone increasing function since the integrand is nonnegative and integral is taken from c to infinity. However, as we can see from the plot, it is not shown to be monotone. Something wrong with the usage of integrate function? Thanks so much for your help. Hanna h <- function(c){ g <- function(x){pnorm(x-8.8,

Hi all, I have some some problem with regard to finding the integral of a function containing an indicator function. please see the code below: func1 <- function(x, mu){ (mu^2)*dnorm(x, mean = mu, sd = 1)*dgamma(x, shape=2)} m1star <- function(x){ integrate(func1, lower = 0, upper = Inf,x)$val} T <- function(x){ 0.3*dnorm(x)/(0.3*dnorm(x)+0.7*m1star(x))} func2 <-

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

Also, it is possible to change "ylim" also? 2012/7/1 li li <hannah.hlx@gmail.com> > Dear all, > I have a few questions regarding the boxplot output from the > "geom_boxplot" function. > Attached is the output I get. Below are my questions: > > 1. How can I define the xlab and ylab myself? > Also I would like to remove

Have you tried P2["20", "10", "0"] ? Jean On Thu, Jun 1, 2017 at 3:10 PM, li li <hannah.hlx at gmail.com> wrote: > Hi all, > I have a three dimensional array with the corresponding dimension names. > I would like to subset the array according to the dimension names. For > example, > suppose I want to extract the values corresponding to A=20,

First I wanted to thank both Marc Schwartz Greg Snow and for their reply. Then I needed to add a level of complexity to the problem. I would be able to create the biggest possible matrix. In other way does it exist a method to ask smthing like the following : max number of rows for a matrix if column=x? Thank you ------------------------------------------------------ Passa a Infostrada.

Dear all, I need to do some calculation where the code used are below. I get error message when I choose k to be large, say greater than 25. The error message is "Error in integrate(temp, lower = 0, upper = 1, k, x, rho, m) : the integral is probably divergent". Can anyone give some help on resolving this. Thanks. Hannah m <- 100 alpha <- 0.05 rho <- 0.1 F0

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, For two sets of random variables, say, x <- rnorm(1000, 10, 10) and y <- rnorm(1000. 3, 20). Is there any way to overlay the histograms (and density curves) of x and y on the plot of y vs. x? The histogram of x is on the x axis and that of y is on the y axis. The density curve here is to approximate the shape of the distribution and does not have to have area 1. Thank you

Hi all, I am trying to download the Rcolorbrewer package from Cran http://cran.r-project.org/web/packages/RColorBrewer//index.html It seems the files have been removed. Does anyone know where can I download the package? Thanks. Hannah [[alternative HTML version deleted]]

Hi all, I want to add 6 plots in the format of 2 columns and 3 rows as one figure in latex. The plots are in .eps file. I know how to add 2 plots side by side, but could not figure out how to do multiple rows. I know this may not be the right place to ask such a question. But I do not know who to ask, so just try my luck here. Thank you in advance.

For a data frame, we can add an additional row or column easily. For example, we can add an additional row of zero and an additional row of column as follows. Is there an easy and similar way to add zeros in each dimension? For example, for array(1:12, dim=c(2,2,3))? Thanks for your help!! Hanna > x <- as.data.frame(matrix(1:20,4,5))> x[5,] <- 0> x[,6] <- 0> x V1 V2

Hi all, I have a function rho.f which gives a list of estimators. I have the following problems. rho.f(0.3) gives me the right answer. However, if I use rho.f(corr[4]) give me a different answer, even though corr[4]==0.3. This prevents me from using a for loop. Can someone give me some help? Thank you very much in advance. Hannah >