I have a equation y=9x+6. I want to extract 10 random points from this function. How should I proceed?
Generate 10 random x-values, in this example uniformly distributed (function runif), and then calculate the corresponding y-values following your equation.
You can control the x-range by setting different min and max parameters to function runif.
x <- runif(10, min = 0, max = 1)
y <- 9*x+6
plot(x,y)
Related
I would like to find out the x^2 density distribution by given x distribution. Currently my samples fall into triangular distribution(min = 1, max = 6, mode = 3). How to generate samples for x^2 distribution given x distribution is triangular in r ? Any help will be appreciate it.
library(ExtraDistr)
sample_size <- 100
x <- rtri(sample_size, 1, 6, 3) # ExtraDistr::triangular distribution (min, max, mode)
I can figure out by hand using inverting CDF. But it will give back two pdf functions. I do not know how to convert to sampling in R when there are two functions , which one x falls into (1,9), then another is in (9,36). Should I have half sample size for each pdf ?
I have the following challenge:
Let X be a V.A. discrete evenly distributed in the interval [1, 10]. Be Y
another V.A. each sample of Y is given by the median value of one
set of 10 X samples. Generate 10^6 Y samples.
I'm using R, with the following code:
x <- runif(10000000,1,10)
sample <- cut(x,breaks=10)
y = median(sample)
You could do:
mosaic::median(~x|sample)
or in base R:
tapply(x, sample, median)
I apply the sensitivity package in R. In particular, I want to use sobolroalhs as it uses a sampling procedure for inputs that allow for evaluations of models with a large number of parameters. The function samples uniformly [0,1] for all inputs. It is stated that desired distributions need to be obtained as follows
####################
# Test case: dealing with non-uniform distributions
x <- sobolroalhs(model = NULL, factors = 3, N = 1000, order =1, nboot=0)
# X1 follows a log-normal distribution:
x$X[,1] <- qlnorm(x$X[,1])
# X2 follows a standard normal distribution:
x$X[,2] <- qnorm(x$X[,2])
# X3 follows a gamma distribution:
x$X[,3] <- qgamma(x$X[,3],shape=0.5)
# toy example
toy <- function(x){rowSums(x)}
y <- toy(x$X)
tell(x, y)
print(x)
plot(x)
I have non-zero mean and standard deviations for some input parameter that I want to sample out of a normal distribution. For others, I want to uniformly sample between a defined range (e.g. [0.03,0.07] instead [0,1]). I tried using built in R functions such as
SA$X[,1] <- rnorm(1000, mean = 579, sd = 21)
but I am afraid this procedure messes up the sampling design of the package and resulted in odd results for the sensitivity indices. Hence, I think I need to adhere for the uniform draw of the sobolroalhs function in which and use the sampled value between [0, 1] when drawing out of the desired distribution (I think as density draw?). Does this make sense to anyone and/or does anyone know how I could sample out of the right distributions following the syntax from the package description?
You can specify mean and sd in qnorm. So modify lines like this:
x$X[,2] <- qnorm(x$X[,2])
to something like this:
x$X[,2] <- qnorm(x$X[,2], mean = 579, sd = 21)
Similarly, you could use the min and max parameters of qunif to get values in a given range.
Of course, it's also possible to transform standard normals or uniforms to the ones you want using things like X <- 579 + 21*Z or Y <- 0.03 + 0.04*U, where Z is a standard normal and U is standard uniform, but for some distributions those transformations aren't so simple and using the q* functions can be easier.
I have got n>2 independent continuous Random Variables(RV). For example say I have 4 Uniform RVs with different set of Upper and lowers.
W~U[-1,5], X~U[0,1], Y~[0,2], Z~[0.5,2]
I am trying to find out the approximate PDF for the sum of these RVs i.e. for T=W+X+Y+Z. As I don't need any closed form solution, I have sampled 1 million points for each of them to get 1 million samples for T. Is it possible in R to get the approximate PDF function or a way to get approximate probability of P(t<T)from this samples I have drawn. For example is there a easy way I can calculate P(0.5<T) in R. My priority here is to get probability first even if getting the density function is not possible.
Thanks
Consider the ecdf function:
set.seed(123)
W <- runif(1e6, -1, 5)
X <- runif(1e6, 0, 1)
Y <- runif(1e6, 0, 2)
Z <- runif(1e6, 0.5, 2)
T <- Reduce(`+`, list(W, X, Y, Z))
cdfT <- ecdf(T)
1 - cdfT(0.5) # Pr(T > 0.5)
# [1] 0.997589
See How to calculate cumulative distribution in R? for more details.
I want to generate some Weibull random numbers in a given interval. For example 20 random numbers from the Weibull distribution with shape 2 and scale 30 in the interval (0, 10).
rweibull function in R produce random numbers from a Weibull distribution with given shape and scale values. Can someone please suggest a method? Thank you in advance.
Use the distr package. It allows to do this kind of stuff very easily.
require(distr)
#we create the distribution
d<-Truncate(Weibull(shape=2,scale=30),lower=0,upper=10)
#The d object has four slots: d,r,p,q that correspond to the [drpq] prefix of standard R distributions
#This extracts 10 random numbers
d#r(10)
#get an histogram
hist(d#r(10000))
Using base R you can generate random numbers, filter which drop into target interval and generate some more if their quantity appears to be less than you need.
rweibull_interval <- function(n, shape, scale = 1, min = 0, max = 10) {
weib_rnd <- rweibull(10*n, shape, scale)
weib_rnd <- weib_rnd[weib_rnd > min & weib_rnd < max]
if (length(weib_rnd) < n)
return(c(weib_rnd, rweibull_interval(n - length(weib_rnd), shape, scale, min, max))) else
return(weib_rnd[1:n])
}
set.seed(1)
rweibull_interval(20, 2, 30, 0, 10)
[1] 9.308806 9.820195 7.156999 2.704469 7.795618 9.057581 6.013369 2.570710 8.430086 4.658973
[11] 2.715765 8.164236 3.676312 9.987181 9.969484 9.578524 7.220014 8.241863 5.951382 6.934886