Probability and Statistics

Spatial Distribution of IED

Improvised explosive devices (IED's) are planted in a city block of 25 square miles (exactly a 5-by-5-miles square). For every week the number of newly found IED's in this block is approximated by a Poisson distribution with $\lambda = 10$ , and the locations of these IED's are scattered uniformly over the city block.

week = 1:50
lambda = 10
X = NULL
Y = NULL
WK = NULL
n = 0
for(i in week){
  size = rpois(1,lambda)
  if(size > 0){
    X[(n+1):(n+size)] = runif(size,0,5)
    Y[(n+1):(n+size)] = runif(size,0,5)
    WK[(n+1):(n+size)] = i
    n = n + size
  }
}
WK = factor(WK,levels=week)

The location of every incident in 50 weeks is visualized in the entire 5-by-5-mile square. And the 2-by-2-mile square is highlighted at the south-west corner.

plot(c(0,5),c(0,5),type='n')
points(X,Y,pch=4,col='red')
text(X,Y,as.character(WK),pos=3)
lines(c(0,0,2,2,0),c(0,2,2,0,0),lty=2,col='blue')

The number of IED's in the entire 5-by-5-miles square follows a Poisson distribution with $\lambda = 10$ . Here is the comparison.

count = table(WK)
hist(count, freq=F, breaks=seq(-0.5,max(count)+0.5,by=1.0),col='green')
cat("\n count =", count)
cat("\n mean count =", mean(count))
cat("\n var of count =", var(count))
x = 0:max(count)
ppmf = dpois(x,lambda)
prob.mass(x,ppmf,col='red',density=10)
legend(x=0.7*max(count), y=max(ppmf)*1.2, legend=c('Data','Poisson'),
       col=c('green','red'),pch=c(15,22))

Suppose that you are responsible for sweeping and clearing the south-west corner of 2-by-2-mile square block every week. Then we can restrict the spetial data for the specified area.

SW = X <= 2 & Y <= 2
plot(c(0,2),c(0,2),type='n')
points(X[SW],Y[SW],pch=4,col='red')
text(X[SW],Y[SW],as.character(WK[SW]),pos=3)
lines(c(0,0,2,2,0),c(0,2,2,0,0),lty=2,col='blue')

To create the data for frequency at the specified area, we need to process the spetial data as follows:

count = table(WK[SW])
hist(count, freq=F, breaks=seq(-0.5,max(count)+0.5,by=1.0),col='green')
cat("\n count =", count)
cat("\n mean count =", mean(count))
cat("\n var of count =", var(count))

The average number of IED's in a week is $\lambda = 10$ . Since the IED's are scattered uniformly across the entire area, you expect $\lambda\times(4/25)$ in average at the specified area of 2-by-2-miles square. In fact, the number of IED's in this south-west area follows a Poisson distribution.

lambda2 = lambda * (4/25)
x = 0:max(count)
ppmf = dpois(x,lambda2)
prob.mass(x,ppmf,col='red',density=10)
legend(x=0.7*max(count), y=max(ppmf)*1.1, legend=c('Data','Poisson'),
       col=c('green','red'),pch=c(15,22))

Sample R code. You can download ied.R, and run it.