Is it possible to plot in R the density function of a distribution?
For example suppose that I want to plot the density function of a Normal(0,5) or a Gamma(5,5).
f <- function(x) dnorm(x,0,5)
g <- function(x) dgamma(x,5,5)
par(mfrow=c(1,2)) # set up gaphics window for 2 plots
plot(f,xlim=c(-15,15),main="N(0,5)")
plot(g,xlim=c(0,3),main="Gamma(5,5)")
In R the distribution functions follow a pattern. For instance, for the normal distribution, the PDF is dnorm(...), the CDF is pnorm(...), the inverse CDF (quantile function) is qnorm(...), and the random number generator is rnorm(...).
One thing to watch out for is that R's convention for the arguments does not necessarily match what you find, for instance, on Wikipedia. For instance the arguments to dgamma(...) are x, shape, and rate, not x, k, and theta.
Related
I want to check the "probability integral transform" theorem using R.
Let's suppose X is an exponential random variable with lambda = 5.
I want to check that the random variable U = F_X = 1 - exp(-5*X) has a uniform (0,1) distribution.
How would you do it?
I would start in this way:
nsample <- 1000
lambda <- 5
x <- rexp(nsample, lambda) #1000 exponential observation
u <- 1- exp(-lambda*x) #CDF of x
Then I need to find the CDF of u and compare it with the CDF of a Uniform (0,1).
For the empirical CDF of u I could use the ECDF function:
ECDF_u <- ecdf(u) #empirical CDF of U
Now I should create the theoretical CDF of Uniform (0,1) and plot it on the same graph of the ECDF in order to compare the two graphs.
Can you help with the code?
You are almost there. You don't need to compute the ECDF yourself – qqplot will take care of this. All you need is your sample (u) and data from the distribution you want to check against. The lazy (and not quite correct) approach would be to check against a random sample drawn from a uniform distribution:
qqplot(runif(nsample), u)
But of course, it is better to plot against the theoretical quantiles:
# the actual plot
qqplot( qunif(ppoints(length(u))), u )
# add a line
qqline(u, distribution=qunif, col='red', lwd=2)
Looks pretty good to me.
I have a pdf $f(x)=4x^3$ of a random variable $X$ in which I need to simulate a draw from the distribution.
My solution consists of finding the cdf from the pdf (1st issue):
> pdf <- function(x){4*x^3}
> cdf <- integrate(pdf,lower=0,upper=x)
Error in integrate(pdf, lower = 0, upper = x) : object 'x' not found
Once I get the cdf $U$, I will set $X=F^-1(U)$. I notice that the pdf follows a Beta distribution with $\alpha=4$ and $\beta=1$.
Is it best to find the $F^-1$ via a inverse beta function? Is there a quick way to find the inverse of a beta function in R?
Since you have identified your pdf as beta, just use rbeta to sample.
s1 <- rbeta(5000,4,1)
In the case where the distribution is non-standard and you cannot solve analytically, you can use rejection sampling. Let's pretend we don't know your pdf is beta and we don't know how to integrate/inverse.
pdf <- function(x) 4*x^3 # on [0,1]
First we draw from our proposal distribution
p <- runif(50000)
Calculate the density values under our pdf
dp <- pdf(p)
And randomly accept/reject in proportion
s2 <- p[runif(50000) < dp/max(dp)]
You should find the distributions of s1 and s2 comparable, using histograms or, preferably, a qqplot.
This question already has answers here:
How do I best simulate an arbitrary univariate random variate using its probability function?
(4 answers)
Closed 8 years ago.
For a normalized probability density function defined on the real line, for example
p(x) = (2/pi) * (1/(exp(x)+exp(-x))
(this is just an example; the solution should apply for any continuous PDF we can define) is there a package in R to simulate from the distribution? I am aware of R's built-in simulators for many distributions.
I could numerically compute the inverse cumulative distribution function at a set of quantiles, store them in a table, and use the table to map from uniform variates to variates from the desired distribution. Is there already a package that does this?
Here is a way using the distr package, which is designed for this.
library(distr)
p <- function(x) (2/pi) * (1/(exp(x)+exp(-x))) # probability density function
dist <-AbscontDistribution(d=p) # signature for a dist with pdf ~ p
rdist <- r(dist) # function to create random variates from p
set.seed(1) # for reproduceable example
X <- rdist(1000) # sample from X ~ p
x <- seq(-10,10, .01)
hist(X, freq=F, breaks=50, xlim=c(-5,5))
lines(x,p(x),lty=2, col="red")
You can of course also do this is base R using the methodology described in any one of the links in the comments.
If this is the function that you're dealing with, you could just take the integral (or, if you're rusty on your integration rules like me, you could use a tool like Wolfram Alpha to do it for you).
In the case of the function provided, you can simulate with:
draw.val <- function(numdraw) log(tan(pi*runif(numdraw)/2))
A histogram confirms that we're sampling correctly:
hist(draw.val(10000), breaks=100, probability=T)
x <- seq(-10, 10, .001)
lines(x, (2/pi) * (1/(exp(x)+exp(-x))), col="red")
I want to turn a continuous random variable X with cdf F(x) into a continuous random variable Y with cdf F(y) and am wondering how to implement it in R.
For example, perform a probability transformation on data following normal distribution (X) to make it conform to a desirable Weibull distribution (Y).
(x=0 has CDF F(x=0)=0.5, CDF F(y)=0.5 corresponds to y=5, then x=0 corresponds to y=5 etc.)
There are many built in distribution functions, those starting with a 'p' will transform to a uniform and those starting with a 'q' will transform from a uniform. So the transform in your example can be done by:
y <- qweibull( pnorm( x ), 2, 6.0056 )
Then just change the functions and/or parameters for other cases.
The distr package may also be of interest for additional capabilities.
In general, you can transform an observation x on X to an observation y on Y by
getting the probability of X≤x, i.e. FX(x).
then determining what observation y has the same probability,
I.e. you want the probability Y≤y = FY(y) to be the same as FX(x).
This gives FY(y) = FX(x).
Therefore y = FY-1(FX(x))
where FY-1 is better known as the quantile function, QY. The overall transformation from X to Y is summarized as: Y = QY(FX(X)).
In your particular example, from the R help, the distribution functions for the normal distribution is pnorm and the quantile function for the Weibull distribution is qweibull, so you want to first of all call pnorm, then qweibull on the result.
I want to plot the fitted values versus the observed ones and want to put straight line showing the goodness of fit. However, I do not want to use abline() because I did not calculate the fitted values using lm command as my I used a model that R does not cover. I calculated the coefficients and used them to calculate the fitted values. So, what can I do to obtain such a plot in R or in winbugs?
Here is what I want
Still no data provided, but maybe this simple example using the curve function will inform the process:
x <- 1:10
y <- 2+ 3*(1:10) + rnorm(10)
plot(1:10, y)
curve( 2+3*x, 0, 10, add=TRUE)
Note to new R users. the expression y_i = 1 - xbeta + delta_i + e_i would fail in R in part because the x and beta are not separated by an operator. But if you do understand R's matrix syntax it might be a very compact expression even if "X" were multidimensional. All of htis depends on the specifics which we are so far lacking.