## Chapter 5
## November 17, 2007
## Alan T. Arnholt
#
# Example 5.2
library(adapt)    # Only for R
# part a.
fxy <- function(v){1}
fxy
adapt(ndim=2,lower=c(0,0),upper=c(.6,.8),functn=fxy)$value
# part b.
adapt(ndim=2,lower=c(0.1,0.25),upper=c(0.9,0.75),functn=fxy)$value
#
# Example 5.8
# part f. ... similar to Figure 5.2
function.draw <- function(f, low=-1, hi=1, n=30){
r <- seq(low, hi, length=n)
z <- outer (r,r,f)
persp(r,r,z,xlab="X",ylab="Y",zlab="Z",theta=300,phi=20,ticktype="detailed")}
f3 <- function(x,y) {ifelse(x >= y, 8*x*y, 0)}
function.draw(f3,0,1,25)
#
# Example 5.12
X1 <- c(58,72,72,86,86,100,100,114,114,128)
Y1 <- c(80,80,90,90,100,100,110,110,120,120)
covar <- function(x,y,f){sum((x-mean(x))*(y-mean(y))*f)}
covar(X1,Y1,1/10)
# Example 5.13
cor(X1,Y1)
#
# Bivariate Normal Distribution
# Ideas for Figure 5.4
function1.draw <- function(f, low = -1, hi = 1, n = 50){
r <- seq(low, hi, length = n)
z <- outer(r, r, f)
persp(r, r, z, axes=FALSE, box=TRUE)}
par(mfrow=c(1,3), pty="s")
f1 <- function(x,y){
exp( (x^2-2*0.5*x*y+y^2) / (-2*(1-0.5^2)) )/
(2*pi*sqrt(1-0.5^2))}
x <- seq(-3,3,length=100)
y <- x
function1.draw(f1,-3,3,20)
contour(x,y,outer(x,y,f1),nlevels=10)
image(x,y,outer(x,y,f1),zlim=range(outer(x,y,f1)),add = FALSE)
par(mfrow=c(1,1))
#
# Example 5.18
# part a.
pnorm(1.8,2.4,0.6)
# part b.
pnorm(1.8,1.77,0.48)
# part c.
1 - pnorm(3, 2.4, 0.6)
# part d.
1 - pnorm(3, 1.77, 0.48)
################################################################################