#--------------------------#
#   Loesungen UeBlatt 8    #
#--------------------------#


#  Aufgabe 1  #


# 1a)

f = function(x,y)
{
  w = sqrt(x^2+y^2)
  if(w==0)
  { res = 1 }                     # lim_{x->0} sin(x)/x = 1
  else
  { res = sin(w)/(w)  }
  return(res)
}

# 1b)

x = seq(-20,20,by=0.2)
y = x
nx = length(x)
ny = length(y)
z = matrix( 0 , nrow=nx , ncol=ny )

for(i in 1:nx)
{
  for(j in 1:ny)
  {
    z[i,j] = f(x[i],y[j])  
  }
}

persp(x,y,z)

require(plot3D)

persp3D(x,y,z)

# 1c) 

contour(x,y,z)

contour(x,y,z,nlevels=100)





#  Aufgabe 2  #


# we can use the code from week8.txt:

# log-Likelihood-Funktion fuer das ARCH(1)-Modell:

logL = function( bsvol , w0 )
{
  w1 = 1 - w0
  vol = sqrt(w0*bsvol^2+w1*ret^2)     # man kann den Vektor ret benutzen, obwohl 
                                      # der nicht als Argument uebergeben wurde
  # shift vol by 1 element:
  n = length(ret)
  vol = vol[-n]                       # remove last one
  vol = c(bsvol,vol)                  # add a new first one
  terms = log(vol) + 1/2*ret^2/vol^2
  result = -sum(terms)
  return(result)
}


# load SPX.txt data and generate return vector ret:

spx = read.table("C:/Users/detlef/OneDrive/hochschule/Vorlesungen/SS2025/Datenanalyse-mit-R/SPX.txt",header=TRUE,sep=";")
head(spx)
summary(spx)     # no NA's
S = spx[,2]
n = length(S)
ret = rep(0,n)       
for(i in 2:n)
{
  ret[i] = (S[i]-S[i-1])/S[i-1]
}
plot(ret,type="l")               # returns look ok


# compute spx-logL and make setup for contour-plot:

bsvol = seq( from=0.1/100 , to=4/100 , by=0.1/100 )
w0    = seq( from=0.05    , to=0.95  , by=0.05 )
nbs = length(bsvol)
nw  = length(w0)
z   = matrix( 0 , nrow=nbs , ncol=nw )
for(i in 1:nbs)
{
  for(j in 1:nw )
  {
    z[i,j] = logL( bsvol[i] , w0[j] )         
  }
}

# let's look at the result:

contour( bsvol , w0 , z )
contour( bsvol , w0 , z , nlevels = 50 )
contour( bsvol , w0 , z , nlevels = 50 , zlim=c(60000,80000) )
contour( bsvol , w0 , z , nlevels = 100 , zlim=c(60000,80000) )

# we take a more refined look at the region w0 = 0.5 to 0.9
# and bsvol = 0.5% to 1.5%:

bsvol = seq( from=0.5/100 , to=1.5/100 , by=0.02/100 )
w0    = seq( from=0.5 , to=0.9 , by=0.01 )
nbs = length(bsvol)
nw  = length(w0)
z   = matrix( 0 , nrow=nbs , ncol=nw )
for(i in 1:nbs)
{
  for(j in 1:nw )
  {
    z[i,j] = logL( bsvol[i] , w0[j] )         
  }
}

contour( bsvol , w0 , z , nlevels = 100 , zlim=c(60000,80000) )
contour( bsvol , w0 , z , nlevels = 100 , zlim=c(69000,70000) )

max(z)
logL(sd(ret),0.69)
sd(ret)

# thus: Maximum approximately at bsvol = sd(ret) = 0.97% and w0 = 0.69





# same analysis for GE.txt data:

# load GE.txt data and generate return vector ret:

ge = read.table("C:/Users/detlef/OneDrive/hochschule/Vorlesungen/SS2025/Datenanalyse-mit-R/GE.txt",header=TRUE,sep=";")

head(ge)
summary(ge)     # no NA's
S = ge[,2]
n = length(S)
ret = rep(0,n)       
for(i in 2:n)
{
  ret[i] = (S[i]-S[i-1])/S[i-1]
}
plot(ret,type="l")              # returns look ok


# compute GE-logL and make setup for contour-plot:

bsvol = seq( from=0.1/100 , to=4/100 , by=0.1/100 )
w0    = seq( from=0.05 , to=0.95 , by=0.05 )
nbs = length(bsvol)
nw  = length(w0)
z = matrix( 0 , nrow=nbs , ncol=nw )
for(i in 1:nbs)
{
  for(j in 1:nw )
  {
    z[i,j] = logL( bsvol[i] , w0[j] )         
  }
}

# let's look at the result:

contour( bsvol , w0 , z )
contour( bsvol , w0 , z , nlevels = 50 )
contour( bsvol , w0 , z , nlevels = 50 , zlim=c(40000,50000) )

# we take a more refined look at the region w0 = 0.5 to 0.9
# and bsvol = 1.0% to 2.0%:

bsvol = seq( from=1.0/100 , to=2.0/100 , by=0.02/100 )
w0    = seq( from=0.5 , to=0.9 , by=0.01 )
nbs = length(bsvol)
nw  = length(w0)
z = matrix( 0 , nrow=nbs , ncol=nw )
for(i in 1:nbs)
{
  for(j in 1:nw )
  {
    z[i,j] = logL( bsvol[i] , w0[j] )         
  }
}

contour( bsvol , w0 , z , nlevels = 50 , zlim=c(40000,50000) )
contour( bsvol , w0 , z , nlevels = 100 , zlim=c(49500,50000) )

max(z)
logL(sd(ret),0.68)
sd(ret)

# thus: Maximum approximately at bsvol = sd(ret) = 0.97% and w0 = 0.69