## 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!

Categories: Uncategorized

## In church…

Categories: Uncategorized

## 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  $U(2 \theta _1-\theta _2\leq x\leq 2 \theta _1+\theta _2)$  distribution.

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

$\begin{cases} \frac{1}{2 \theta _2} & 2 \theta _1-\theta _2\leq x\leq 2 \theta _1+\theta _2 \\ 0 & \text{True} \end{cases}$

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]

$-\frac{i \left(-e^{i x \left(2 \theta _1-\theta _2\right)}+e^{i x \left(2 \theta _1+\theta _2\right)}\right)}{2 x \theta _2}$.

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

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

$\left\{2 \theta _1,\frac{1}{3} \left(12 \theta _1^2+\theta _2^2\right),2 \theta _1 \left(4 \theta _1^2+\theta _2^2\right),16 \theta _1^4+8 \theta _1^2 \theta _2^2+\frac{\theta _2^4}{5}\right\}$.

Calculate the sample statistics.

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

$\{4.99333,8.46171\}$.

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]}]

$\left\{\left\{\theta _1\to 2.49667,\theta _2\to -5.03836\right\},\left\{\theta _1\to 2.49667,\theta _2\to 5.03836\right\}\right\}$.

Check the true value for the $\theta _2$.

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

$\theta _1\in \text{Reals}\&\&\theta _2\geq 0 .$

Then, $\left\{\left\{\theta _1\to 2.49667,\theta _2\to 5.03836\right\}\right\}$.

Categories: probability, statistics

## 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 $f(x):=\frac{e^{b \text{Cos}[y-a]}}{2 \pi \text{BesselI}[0,b]}$, for $-\pi \leq x\leq \pi$.

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)

Categories: probability, statistics

## 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

Categories: infos Tags: , , ,

## 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")

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
Categories: statistics

## 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.

Categories: infos Tags: , , , , ,
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