We Moved @ statsravingmad.com/blog

It’s time,huh?

Well, i managed to take my personal webspace somewhere out there and of course the first thing to set up was my blog ;). WordPress, was great but there was limitations you know. At this point I’d like to present you my new blog address, statsravingmad.com/blog. I hope y’all follow me there. Please update you bookmarks…

CU!

In church…

Probability manipulations with Mathematica…

There comes a time that a statistician needs to do some ananytic calculations. There more than a bunch of tools to use but I usually prefer Mathematica or Maple. Today, I’m gonna use Mathematica to do a simple exhibition.

Let’s set this example upon the    distribution.

pfun = PDF[UniformDistribution[{2*Subscript[θ, 1] - Subscript[θ, 2],
2*Subscript[θ, 1] + Subscript[θ, 2]}], x]

One of the most intensive calculations is the characteristic function (eq. the moment generating function). This is straightforward to derive.

cfun=CharacteristicFunction[UniformDistribution[
{2*Subscript[θ, 1]-Subscript[θ, 2],2*Subscript[θ, 1]+Subscript[θ, 2]}],x]

.

The Table[] command calculates for us the raw moments for our distribution.

Table[Limit[D[cfun, {x, n}], x -> 0]/I^n, {n, 4}]

.

Calculate the sample statistics.

T=List[8.23,6.9,1.05,4.8,2.03,6.95];
{Mean[T],Variance[T]}

.

Now, we can use a simple moment matching technique to get estimates for the parameters.

Solve[{Mean[T]-2*Subscript[θ, 1]==0,-(2*Subscript[θ, 1])^2+
1/3 (12 Subscript[θ, 1]^2+\!\*SubsuperscriptBox[\(θ\), \(2\), \(2\)])-
Variance[T]==0},{Subscript[θ, 2],Subscript[θ, 1]}]

.

Check the true value for the .

Reduce[2 Subscript[θ, 1]-Subscript[θ, 2]<=2 Subscript[θ, 1]+Subscript[θ, 2],
Subscript[θ, 2]]

Then, .

A von Mises variate…

Inspired from a mail that came along the previous random generation post the following question rised :

How to draw random variates from the Von Mises distribution?

First of all let’s check the pdf of the probability rule, it is , for .

Ok, I admit that Bessels functions can be a bit frightening, but there is a work around we can do. The solution is a Metropolis algorithm simulation. It is not necessary to know the normalizing constant, because it will cancel in the computation of the ratio. The following code is adapted from James Gentle’s notes on Mathematical Statistics .

n <- 1000
x <- rep(NA,n)
a <-1
c <-3
yi <-3
j <-0

i<-2
while (i < n) {
	i<-i+1
	yip1 <- yi + 2*a*runif(1)- 1
	if (yip1 < pi & yip1 > - pi) {
		if (exp(c*(cos(yip1)-cos(yi))) > runif(1)) yi <- yip1
		else yi <- x[i-1]
		x[i] <- yip1
	}
}
hist(x,probability=TRUE,fg = gray(0.7), bty="7")
lines(density(x,na.rm=TRUE),col="red",lwd=2)

R 2.11.0 due date

This is the announcement as posted in the mailing list :

This is to announce that we plan to release R version 2.11.0 on Thursday,
April 22, 2010.

Those directly involved should review the generic schedule at
http://developer.r-project.org/release-checklist.html

The source tarballs will be made available daily (barring build
troubles) via
http://cran.r-project.org/src/base-prerelease/

For the R Core Team
Peter Dalgaard

The distribution of rho…

There was a post here about obtaining non-standard p-values for testing the correlation coefficient. The R-library

SuppDists

deals with this problem efficiently.

library(SuppDists)

plot(function(x)dPearson(x,N=23,rho=0.7),-1,1,ylim=c(0,10),ylab="density")
plot(function(x)dPearson(x,N=23,rho=0),-1,1,add=TRUE,col="steelblue")
plot(function(x)dPearson(x,N=23,rho=-.2),-1,1,add=TRUE,col="green")
plot(function(x)dPearson(x,N=23,rho=.9),-1,1,add=TRUE,col="red");grid()

legend("topleft", col=c("black","steelblue","red","green"),lty=1,
		legend=c("rho=0.7","rho=0","rho=-.2","rho=.9"))</pre>

This is how it looks like,


Now, let’s construct a table of critical values for some arbitrary or not significance levels.

q=c(.025,.05,.075,.1,.15,.2)
xtabs(qPearson(p=q, N=23, rho = 0, lower.tail = FALSE, log.p = FALSE) ~ q )
# q
#     0.025      0.05     0.075       0.1      0.15       0.2
# 0.4130710 0.3514298 0.3099236 0.2773518 0.2258566 0.1842217

We can calculate p-values as usual too…

1-pPearson(.41307,N=23,rho=0)
# [1] 0.0250003

Oh, mr counselor!

Xm…

There were two men trying to decide what to do for a living.  They went to see a counselor, and he decided that they had good problem solving skills.

He tried a test to narrow the area of specialty.  He put each man in a room with a stove, a table, and a pot of water on the table.  He said “Boil the water”.  Both men moved the pot from the table to the stove and turned on the burner to boil the water.  Next, he put them into a room with a stove, a table, and a pot of water on the floor.  Again, he said “Boil the water”.  The first man put the pot on the stove and turned on the burner.  The counselor told him to be an Engineer, because he could solve each problem individually.  The second man moved the pot from the floor to the table, and then moved the pot from the table to the stove and turned on the burner.  The counselor told him to be a mathematician because he reduced the problem to a previously solved problem.

– From The Mathematician, The Physicist and The Engineer (and Others)

In search of a random gamma variate…

One of the most common exersices given to Statistical Computing,Simulation or relevant classes is the generation of random numbers from a gamma distribution. At first this might seem straightforward in terms of the lifesaving relation that exponential and gamma random variables share. So, it’s easy to get a gamma random variate using the fact that

.

The code to do this is the following

rexp1 <- function(lambda, n) {
  u <- runif(n)
  x <- -log(u)/lambda
  }

rgamma1 <- function(k, lambda) {
  sum(rexp1(lambda, k))
}

This works unfortunately only for the case

\pi day!

It’s π-day today so we gonna have a little fun today with Buffon’s needle and of course R. A well known approximation to the value of is the experiment tha Buffon performed using a needle of length,. What I do in the next is only to copy from the following file the function estPi and to use an ergodic sample plot… Lame,huh?

estPi<- function(n, l=1, t=2) {
 m <- 0
 for (i in 1:n) {
 x <- runif(1)
 theta <- runif(1, min=0, max=pi/2)
 if (x < l/2 * sin(theta)) {
 m <- m +1
 }
 }
 return(2*l*n/(t*m))
}

So, an estimate would be…

A normal philosophy…

The following was sent by email to me. It originates to Youden.

                                 THE
                                NORMAL
                             LAW OF ERROR
                           STANDS OUT IN THE
                         EXPERIENCE OF MANKIND
                        AS ONE OF  THE BROADEST
                       GENERALIZATIONS OF NATURAL
                     PHILOSOPHY . IT SERVES AS THE
                   GUIDING INSTRUMENT IN RESEARCHES
                IN THE PHYSICAL AND SOCIAL SCIENCES AND
               IN MEDICINE AGRICULTURE AND ENGINEERING .
          IT IS AN INDISPENSABLE TOOL FOR THE ANALYSIS AND THE
INTERPRETATION OF THE BASIC DATA OBTAINED BY OBSERVATION AND EXPERIMENT

–W.J. Youden

Youden is one of the truly inspiring statisticians to me.