similar to: Inversion function of dnorm ?

Hello all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context:

Dear all, Suppose I plot two normal distributions (A and B) side by side and add vertical line which hipotheticaly represent alpha value; e.g.: x <- seq(-3.5,5, length=1000) y <- dnorm(x) # Plot distribution A plot(y~x, type='l',axes=F,xlab="",ylab="",lwd=2) # Plot distribution B y2 <- dnorm(x-1.5) lines(y2~x,lwd=2) # Plot vertical line for alpha value

Hi, Having a weird problem with the integrate function. I have a function which calculates a loss density: I'd like to integrate it to get the distribution. The loss density function is: lossdensity<-function(p,Beta,R=0.4){ # the second derivative of the PDF # p is the default probability of the pool at which we are evaluating the lossdensity # Beta is the correlation with the market

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

I'm wondering what's the best way to shift a huge matrix left or right. My current implementation is the following: shiftMatrixL <- function(X, shift, padding=0) { cbind(X[, -1:-shift], matrix(padding, dim(X)[1], shift)) } X <- shiftMatrixL(X, 1)*3 + shiftMatrixL(X,2)*5... However, it's still slow due to heavy use of this function. The resulting matrix will only be read once

Hello, could You please help: I am looking for a way to formulate test accuracy measures such as test sensitivity, specificity, predictive values, and correct classification rate using p/d/qnorm. The tests' primary values follow a bimodal distribution, which is modelled by a mixture of two normal distributions: p * dnorm ((x - u1) / s1) / s1 + (1 - p) * dnorm ((x - u2) / s2) / s2)

Hi, I wonder if anyone knows how to control color and darkness of the background grid generated by ' type = c("g", "p") ' in a lattice plot (e.g., in xyplot). The documentation does not seem to offer a way to change them. Thanks a lot! Wen

Hi, This problems has bothered me for the lase couple of hours. > 1e-100==0 [1] FALSE > (1-1e-100)==1 [1] TRUE How can I tell R that 1-1e-100 does not equal to 1, actually, I found out that > (1-1e-16)==1 [1] FALSE > (1-1e-17)==1 [1] TRUE The reason I care about this is that I was try to use qnorm() in my code, for example, > qnorm(1e-100) [1] -21.27345 and if I want to

Hello, I wanted to fit a linear mixed model to a data that is similar in terms of design to the 'Machines' data in 'nlme' package except that each worker (with triplicates) only operates one machine. I created a subset of observations from 'Machines' data such that it looks the same as the data I wanted to fit the model with (see code below). I fitted a model in

Dear All I am trying to find a uniroot of a function within another function (see example) but I am getting an error message (f()values at end points not of opposite sign). I was wondering if you would be able to advise how redefine my function so that I can find the solution. In short my first function calculates the intergrale which is function of "t" , I need to find the uniroot of

Is there a function to compute the derivative of the probit (qnorm) function in R, or in any of the packages? Thanks, -Andrew [[alternative HTML version deleted]]

Howdee, One of my student spotted something I can't explain: a probability >1 vs a normal probability density function. > dnorm(x=1, mean=1, sd=0.4) [1] 0.9973557 > dnorm(x=1, mean=1, sd=0.39) [1] 1.022929 > dnorm(x=1, mean=1, sd=0.3) [1] 1.329808 > dnorm(x=1, mean=1, sd=0.1) [1] 3.989423 > dnorm(x=1, mean=1, sd=0.01) [1] 39.89423 > dnorm(x=1, mean=1, sd=0.001) [1]

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.

Dear R-users Idea: Plot a dnorm line using specific mean/sd to complete a histogram (skewed). xs:range of y-values, ys: dnorm function Problem: I expected to multiply the ys function with the sample size (n=250-300). I was wondering about a factor between 12'000 and 30'000 to match the size of the dnorm line with the specific histogram. Thanks Sibylle hist(Biotree[Ld,]$Height2008,

Dear R community, I was trying to estimate density at point zero of a multivariate distribution (9 dimensions) and for this I was using a multinormal approximation and the function dmvnorm , gtools package. To have a sense of the error I tried to look the mismatch between a unidimensional version of my distribution and estimate density at point zero with function density, dmvnorm and dnorm. At

Hi All, Why would calls to dnorm and dmvnorm return values that are above 1? For example, > dnorm(0.00003,mean=0, sd=0.1) [1] 3.989423 This is happening on two different installations of R that I have. Thank you. Hailu [[alternative HTML version deleted]]

I am attempting to wrap the histogram function in my own custom function, so that I can quickly generate some standard plots. A part of what I want to do is to draw a normal curve over the histogram: > x <- rnorm(1000) > hist(x, freq=F) > curve(dnorm(x), lty=3, add=T) (for normal use, x would be a vector of empirical values, but the rnorm() function works for testing) That

Hi, Apropos of a recent Inf question, I've previously wondered if dnorm "does the right thing" with dnorm(0, 0, -Inf) which gives zero. Should that be zero or NaN (or NA)? The help says "'sd < 0' is an error and returns 'NaN'" and since -Inf < 0 is TRUE, then... is this a bug? Thank you, Stephen Rochester, MN USA

Here's another example of my plotmath whipping boy, the Normal distribution. A colleague asks for a Normal plotted above a series of axes that represent various other distributions (T, etc). I want to use vectors of equations in plotmath to do this, but have run into trouble. Now I've isolated the problem down to a relatively small piece of working example code (below). If you would

Full_Name: Marco Vezzoli Version: 2.2.0 2.1.0 2.0.0 OS: Solaris, Windows, Linux Submission from: (NULL) (57.78.11.38) The dnorm functions yield a wrong value when the standard deviation is near to 1e-1 e.g. > dnorm(0,mean=0.04,sd=0.3) [1] 1.318039 this error is consistent in various version and os.