STAT647 Spatial Statistics

Transcription

1 STAT647 Spatial Statistics HW02 Date: 09/17/2012 Instructor: Dr. Mikyoung Jun Student: Chih-Hung Hsu (Aaron)

2 1. Variogram: Using the data set (data 1) from the course webpage, plot variograms (try variogram cloud, lines that give binned averages, boxplots of variograms, directional variograms, etc) Sol Variogram cloud library(fields) # # Read in data data1<-read.table("data1.txt",header=t) x<-data1[,1:2] y<-data1[,3] # # Calculate variogram myvgram<-vgram( x,y, N=13, lon.lat=true, dmax=13000) # # Plot variogram cloud plot( myvgram$d, sqrt(myvgram$vgram), xlab="distance(miles)", ylab="semivariogram") lines(myvgram$centers, myvgram$stats["mean",], col="white") Graphical result 1

3 1.2. Bloxplots of variograms library(fields) # # Read in data data1<-read.table("data1.txt",header=t) x<-data1[,1:2] y<-data1[,3] # # Calculate variogram myvgram<-vgram( x,y, N=13, lon.lat=true, dmax=13000) # Plot binned variogram with boxplot brk<- seq( 0, 13000,1000) bplot.xy( myvgram$d, sqrt(myvgram$vgram), breaks=brk,ylab="sqrt(vg)", xlab="distance(miles)",ylab="semivariogram") lines(myvgram$centers, myvgram$stats["mean",], col=4) Graphical result 2

4 1.3. Directional virograms library(geor) # # Calculate and plot direction virograms dirvar <- variog4(coords = data1[,1:2], data = data1[,3], max.dist=360) plot(dirvar) Graphical result 3

5 2. Optimization problem: Download the data file (data 2) from the course webpage Fit a function y = f (x) = a log (x) + bx 2 +ϵ(x) to the data when a and b are parameters Try least squares and assume ϵ(x) N(0, σ 2 ) for a constant σ Specifically use nlm function in R and get numerical estimates of a and b Try at least 5 different starting points (initial values) Report your convergence result (parameter estimates, number of iteration taken) for each starting points Now try the same with the constraints that a > 0, b < 0. Report the results in the same way. Sol Try lm function in R library(geor) # # Read in data data2<-read.table("data2.txt",header=t) x<-data2$x y<-data2$y logx<-log10(x) sqx<-x^2 # # Fit a linear model to data fit<-lm(y~logx+sqx-1) plot(x,y,xlab="x",ylab="y") # # Print fitting result (coefficients) in console window print(fit) Fitting result Call: lm(formula = y ~ logx + sqx - 1) Coefficients: logx sqx

6 2.2. Use nlm to find parameters library(geor) # # Read in data data2<-read.table("data2.txt",header=t) x<-data2$x y<-data2$y logx<-log10(x) sqx<-x^2 # # Define objective function and use ulm to find the parameters obj<-function(z) sum((y-z[1]*logx-z[2]*sqx)^2) # # Assign initial guess theta<-c(0,0) # # Find the optimum solution result<-nlm(obj,theta) print(result) Example result ( set (0,0) as starting point) $minimum [1] $estimate [1] $gradient [1] e e-05 $code [1] 1 $iterations [1] 5 Comparing results with respect to different starting points Starting point (a, b) Estimate (a, b) Objective value Result Code Iterations (0,0) ( ) (15,39) ( ) (96,19) ( ) (73,97) ( ) (18,52) ( ) Discussion The starting points are generated in a way that a and b are both randomly distributed between 1 and 100. The nlm results all converge within 9 iterations and to the same point with 10-3 accuracy. The nlm results are consist with the the result of lm function. 5

7 2.3. Considering constraints Given constraints that a>0, b<0 The parameter a is transformed to a 0, such that a=exp(a 0 ), or a 0 =ln(a); b=-exp(b 0 ), or b 0 =ln(-b). The function y = f(x) = a log (x) + bx 2 +ϵ(x) becomes y = f(x) = exp(a 0 ) log (x) - exp(b 0 )x 2 +ϵ(x) library(geor) # # Reading data data2<-read.table("data2.txt",header=t) x<-data2$x y<-data2$y logx<-log10(x) sqx<-x^2 # # With constraints # # Define objective function and use ulm to find the parameters obj<-function(z) sum((y-exp(z[1])*logx+exp(z[2])*sqx)^2) # # Assign initial guess theta<-c(0,0) # # Find the optimum solution result<-nlm(obj,theta) print(result) # # Display the value of a, b a<-exp(result$estimate[1]) b<--exp(result$estimate[2]) cat(a,b,"\n") Choose (a 0, b 0 )=(0,0) as starting point, that is (a, b)=(0,0), the nlm result is as following. $minimum [1] $estimate [1] $gradient [1] e e-03 $code [1] 1 $iterations [1] 37 6

8 The solution is a 0 = ; b 0 = , that is a=0, b= Comparing results with respect to different starting points Starting point Estimate Estimate Objective value Result Code Iterations (a 0, b 0) (a 0, b 0) (a, b) (0,0) ( , ) (0, ) (-90,-54) (-90,-54) (0, 0) (91,7) ( ,7) ( ) 15,508,285, (-51,79) (-51, ) (0, 0) (85,-76) ( ,-76) (0, 0) Discussion The starting points are generated in a way that a 0 and b 0 are both randomly distributed between -100 and 100. The best solution is when (a 0, b 0 )= ( , ), or (a,b)= (0, ) Some nlm results show that even the values of a 0 and b 0 changes, the values of a and b stay around a=0 and b=0, which is a local optimum in the (a,b) domain. Choosing starting points at (a 0, b 0 ) domain doesn t seem to give good results. An alternative way is selecting the starting points in (a, b) domain, then convert (a, b) to (a 0, b 0 ). Finally use the converted values as the initial guess in the optimization model to find the solutions. 7