I've asked a related question before which successfully received an answer. Now I want to sample values from an upside down bell curve but exclude a range of values that fall in the middle of it like shown on the picture below:
I have this code currently working:
min <- 1
max <- 20
q <- min + (max-min)*rbeta(10000, 0.5, 0.5)
How may I adapt it to achieve the desired output?
Say you want a sample of 10,000 from your distribution but don't want any numbers between 5 and 15 in your sample. Why not just do:
q <- min + (max-min)*rbeta(50000, 0.5, 0.5);
q <- q[!(q > 5 & q < 15)][1:10000]
Which gives you this:
hist(q)
But still has the correct size:
length(q)
#> [1] 10000
An "upside-down bell curve" compared to the normal distribution, with the exclusion of a certain interval, can be sampled using the following algorithm. I write it in pseudocode because I'm not familiar with R. I adapted it from another answer I just posted.
Notice that this sampler samples in a truncated interval (here, the interval [x0, x1], with the exclusion of [x2, x3]) because it's not possible for an upside-down bell curve extended to infinity to integrate to 1 (which is one of the requirements for a probability density).
In the pseudocode, RNDU01() is a uniform(0, 1) random number.
x0pdf = 1-exp(-(x0*x0))
x1pdf = 1-exp(-(x1*x1))
ymax = max(x0pdf, x1pdf)
while true
# Choose a random x-coordinate
x=RNDU01()*(x1-x0)+x0
# Choose a random y-coordinate
y=RNDU01()*ymax
# Return x if y falls within PDF
if (x<x2 or x>x3) and y < 1-exp(-(x*x)): return x
end
Related
I have 16068 datapoints with values that range between 150 and 54850 (mean = 3034.22). What would the R code be to generate a set of random numbers that grow in frequency exponentially between 54850 and 150?
I've tried using the rexp() function in R, but can't figure out how to set the range to between 150 and 54850. In my actual data population, the lambda value is 25.
set.seed(123)
myrange <- c(54850, 150)
rexp(16068, 1/25, myrange)
The call produces an error.
Error in rexp(16068, 1/25, myrange) : unused argument (myrange)
The hypothesized population should increase exponentially the closer the data values are to 150. I have 25 data points with a value of 150 and only one with a value of 54850. The simulated population should fall in this range.
This is really more of a question for math.stackexchange, but out of curiosity I provide this solution. Maybe it is sufficient for your needs.
First, ?rexp tells us that it has only two arguments, so we generate a random exponential distribution with the desired length.
set.seed(42) # for sake of reproducibility
n <- 16068
mr <- c(54850, 150) # your 'myrange' with less typing
y0 <- rexp(n, 1/25) # simulate exp. dist.
y <- y0[order(-y0)] # sort
Now we need a mathematical approach to rescale the distribution.
# f(x) = (b-a)(x - min(x))/(max(x)-min(x)) + a
y.scaled <- (mr[1] - mr[2]) * (y - min(y)) / (max(y) - min(y)) + mr[2]
Proof:
> range(y.scaled)
[1] 150.312 54850.312
That's not too bad.
Plot:
plot(y.scaled, type="l")
Note: There might be some mathematical issues, see therefore e.g. this answer.
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.
Forgive me if this has been asked before (I feel it must have, but could not find precisely what I am looking for).
Have can I draw one element of a vector of whole numbers (from 1 through, say, 10) using a probability function that specifies different chances of the elements. If I want equal propabilities I use runif() to get a number between 1 and 10:
ceiling(runif(1,1,10))
How do I similarly sample from e.g. the exponential distribution to get a number between 1 and 10 (such that 1 is much more likely than 10), or a logistic probability function (if I want a sigmoid increasing probability from 1 through 10).
The only "solution" I can come up with is first to draw e6 numbers from the say sigmoid distribution and then scale min and max to 1 and 10 - but this looks clumpsy.
UPDATE:
This awkward solution (and I dont feel it very "correct") would go like this
#Draw enough from a distribution, here exponential
x <- rexp(1e3)
#Scale probs to e.g. 1-10
scaler <- function(vector, min, max){
(((vector - min(vector)) * (max - min))/(max(vector) - min(vector))) + min
}
x_scale <- scaler(x,1,10)
#And sample once (and round it)
round(sample(x_scale,1))
Are there not better solutions around ?
I believe sample() is what you are looking for, as #HubertL mentioned in the comments. You can specify an increasing function (e.g. logit()) and pass the vector you want to sample from v as an input. You can then use the output of that function as a vector of probabilities p. See the code below.
logit <- function(x) {
return(exp(x)/(exp(x)+1))
}
v <- c(seq(1,10,1))
p <- logit(seq(1,10,1))
sample(v, 1, prob = p, replace = TRUE)
I want to draw a number of random variables from a series of distributions. However, the values returned have to be no higher than a certain threshold.
Let’s say I want to use the gamma distribution and the threshold is 10 and I need n=100 random numbers. I now want 100 random number between 0 and 10. (Say scale and shape are 1.)
Getting 100 random variables is obviously easy...
rgamma(100, shape = 1, rate = 1)
But how can I accomplish that these values range from 0 to 100?
EDIT
To make my question clearer. The 100 values drawn should be scaled beween 0 and 10. So that the highest drawn value is 10 and the lowest 0. Sorry if this was not clear...
EDIT No2
To add some context to the random numbers I need: I want to draw "system repair times" that follow certain distributions. However, within the system simulation there is a binomial probability of repairs beeing "simple" (i.e. short repair time) and "complicated" (i.e. long repair time). I now need a function that provides "short repair times" and one that provides "long repair times". The threshold would be the differentiation between short and long repair times. Again, I hope this makes my question a little clearer.
This is not possible with a gamma distribution.
The support of a distribution determine the range of sample data drawn from it.
As the support of the gamma distribution is (0,inf) this is not possible.(see https://en.wikipedia.org/wiki/Gamma_distribution).
If you really want to have a gamma distribution take a rejection sampling approach as Alex Reynolds suggests.
Otherwise look for a distribution with a bounded/finite support (see https://en.wikipedia.org/wiki/List_of_probability_distributions)
e.g. uniform or binomial
Well, fill vector with rejection, untested code
v <- rep(-1.0, 100)
k <- 1
while (TRUE) {
q <- rgamma(1, shape=1, rate=1)
if (q > 0.0 && q < 100) {
v[k] <- q
k<-k+1
if (k>100)
break
}
}
I'm not sure you can keep the properties of the original distribution, imposing additional conditions... But something like this will do the job:
Filter(function(x) x < 10, rgamma(1000,1,1))[1:100]
For the scaling - beware, the outcome will not follow the original distribution (but there's no way to do it, as the other answers pointed out):
# rescale numeric vector into (0, 1) interval
# clip everything outside the range
rescale <- function(vec, lims=range(vec), clip=c(0, 1)) {
# find the coeficients of transforming linear equation
# that maps the lims range to (0, 1)
slope <- (1 - 0) / (lims[2] - lims[1])
intercept <- - slope * lims[1]
xformed <- slope * vec + intercept
# do the clipping
xformed[xformed < 0] <- clip[1]
xformed[xformed > 1] <- clip[2]
xformed
}
# this is the requested data
10 * rescale(rgamma(100,1,1))
Use truncdist package. It truncates any distribution between upper and lower bounds.
Hope that helped.
Let me begin by saying this is a class assignment for an intro to R course.
First, in VGAM why are there dparetoI, ParetoI, pparetoI, qparetoI & rparetoI?
Are they not the same things?
My problem:
I would like to generate 50 random numbers in a pareto distribution.
I would like the range to be 1 – 60 but I also need to have 30% of the values <= 4.
Using VGAM I have tried a variety of functions and combinations of pareto from what I could find in documentation as well as a few things online.
I experimented with fit, quantiles and forcing a sequence from examples I found but I'm new and didn't make much sense of it.
I’ve been using this:
alpha <- 1 # location
k <- 2 # shape
mySteps <- rpareto(50,alpha,k)
range(mySteps)
str(mySteps[mySteps <= 4])
After enough iterations, the range will be acceptable but entries <= 4 are never close.
So my questions are:
Am I using the right pareto function?
If not, can you point me in the right direction?
If so, do I just keep running it until the “right” data comes up?
Thanks for the guidance.
So reading the Wikipedia entry for Pareto Distribution, you can see that the CDF of the Pareto distribution is given by:
FX(x) = 1 - (xm/x)α
The CDF gives the probability that X (your random variable) < x (a given value). You want Pareto distributions where
Prob(X < 4) ≡ FX(4) = 0.3
or
0.3 = 1 - (xm/4)α
This defines a relation between xm and α
xm = 4 * (0.7)1/α
In R code:
library(VGAM)
set.seed(1)
alpha <- 1
k <- 4 * (0.7)^(1/alpha)
X <- rpareto(50,k,alpha)
quantile(X,0.3) # confirm that 30% are < 4
# 30%
# 3.891941
Plot the histogram and the distribution
hist(X, breaks=c(1:60,Inf),xlim=c(1,60))
x <- 1:60
lines(x,dpareto(x,k,alpha), col="red")
If you repeat this process for different alpha, you will get different distribution functions, but in all cases ~30% of the sample will be < 4. The reason it is only approximately 30% is that you have a finite sample size (50).