--------------------------------------
#      Einlesen von Daten und        # 
#    weitere Regressionsbeispiele    #
--------------------------------------


?read.table()

# Daten nach R importieren: den Datei-Pfad  "C:/..."  unten anpassen:
# ACHTUNG: Es muss so ein Schraegstrich / sein, nicht so \ einer, im 
# Windows-Dateiexplorer wird \ benutzt:

daxwithcomp = read.table("C:/Users/detlef/HSRM/Vorlesungen/WS201920/Oekonometrie/DAXwithComp.txt",header=TRUE,sep=";")

# alles mal anschauen:
daxwithcomp

# nur die ersten oder letzten 6 Zeilen:
head(daxwithcomp)
tail(daxwithcomp)
# 20 statt 6 Zeilen:
head(daxwithcomp,20)
tail(daxwithcomp,20)

# Goesse:
dim(daxwithcomp)

# kompakte Uebersicht der numerischen Daten:
summary(daxwithcomp)

# eventuelle Namen von Spalten oder Zeilen:
names(daxwithcomp)

# Informationen ueber die Objekt/Daten-Struktur: 
class(daxwithcomp)
str(daxwithcomp)
mode(daxwithcomp)     # ..mh, wuerde man eigentlich data.frame erwarten..


# let's take the last 4 years of data, 2011 - 2014:

daxwithcomp[1558,]
daxwithcomp[1559,]
daxwithcomp = daxwithcomp[-(1:1558),]
head(daxwithcomp)
tail(daxwithcomp)
dim(daxwithcomp)
# remove all rows with NA's:
daxwithcomp = na.omit(daxwithcomp)
dim(daxwithcomp)


# let's check some underlyings:

head(daxwithcomp)
dax = daxwithcomp[,1]  
alv = daxwithcomp[,3] 
bas = daxwithcomp[,4] 
plot(dax,type="l")
plot(alv,type="l")
plot(bas,type="l")

# normalize to 100 at start to compare all three:
ndax = dax/dax[1]*100
nalv = alv/alv[1]*100
nbas = bas/bas[1]*100

plot( ndax , type="l" , ylim = c( min(ndax,nalv,nbas) , max(ndax,nalv,nbas) ) )
lines(nalv,col="red")
lines(nbas,col="green")



# Now let's determine the DAX-Composition 
# through Regression:

uls = daxwithcomp[,2:31]          # 'uls' for 'underlyings' 
str(uls)                          # uls is dataframe

res = lm(dax ~ uls)               # doesn't work

# let's change uls from dataframe to matrix:
uls = as.matrix(uls)

res = lm(dax ~ uls)               # this works
summary(res)
plot(dax,type="l")
lines(res$fit,col=2)


# there is no constant term in DAX, 
# thus: Regression without intercept

res = lm(dax ~ -1 + uls)              
summary(res)
plot(dax,type="l")
lines(res$fit,col=2)




-----------------------------
#    S&P500 - Regression    #
-----------------------------

# Importieren der Daten nach R:

spx = read.table("C:/Users/detlef/HSRM/Vorlesungen/WS201920/Oekonometrie/SPX.txt",header=TRUE,sep=";")

spx                # quite large..

head(spx)
tail(spx)
names(spx)

# technical information:
class(spx)
str(spx)           # ist vom Daten-Typ "data.frame"
mode(spx)

# extract columns:
days = spx$Date
index = spx$Adj.Close

plot(index,type="l")
plot(log(index),type="l")

# versuchen wir, die tatsaechlichen Zeiten auf die 
# x-Achse zu bekommen:

plot(days,index)   # takes some time until we get something 
                   # which is not that what we want..

str(days)          # ist "Factor", muesste vielleicht sowas 
                   # wie "Date" sein..
class(days)
mode(days)
str(index)
class(index)
mode(index)



#-----------------------------------------------------------
# in the following we take a closer look to date-formatting,
# see also the 4 pdf-pages "Date Formatting in R" auf der 
# Vorlesungshomepage: 

days = as.Date(days)                         # doesn't work
days = as.Date(days,format="%d-%m-%y")

str(days)          # ok, ist jetzt "Date"
class(days)
mode(days)
head(days)
tail(days)

plot(days,index)           # technically we have Date format now,
                           # but 1950 has turned to 2050..

# we fix this by hand:
# dates can be added and subtracted, so let's try the following:

days[1]
startdate = as.Date("1950-01-03")
startdate
wrongstartdate = days[1]
wrongstartdate

days[1] - wrongstartdate + startdate
days[2] - wrongstartdate + startdate

# ok, that seem to work.  

# we have to correct only entries with year >= 2050:
# extract the 4-digit year from the date:

years = format(days,"%Y")
years
head(years)
tail(years)
length(years)

# let's try this:

days = ifelse( years>=2050, as.Date( days - wrongstartdate + startdate ), days)
head(days)

# internally, date-values are given by integers equal to the number of 
# days since January 1st 1970, with negative numbers for earlier dates. 
# days now is represented by these integers. Let's try to see the actual 
# dates again:

class(days)
days = as.Date(days)   # does not work
days2 = format(days, format="%d-%m-%Y")
days2                  # also does not work..

refDate = as.Date("1970-01-01")
refDate
class(refDate)

days3 = refDate + days

head(days3)
tail(days3)        # ok, finally we've made it...
                   # eventually there is a more quick solution..
days = days3

# End of we take a closer look to date-formating.
#-----------------------------------------------------------



# now we should get a nice plot:

plot(days,index)
plot(days,index,type="l")
plot(days,log(index))              # alright, looks all good now
plot(days,log(index),type="l")



---------------------------------
# Bestimmen der Wachstumsrate r #
---------------------------------

# in order to do the regression, we take the logarithm to obtain:
#
# log(SP500_t) = log(S_0) + r*(t-t_0)
#
# thus we can do a simple linear regression with 1 regressor being 
# the vector x = t_k - t_0 and y = log(SP500_{t_k}):

logindex = log(index)
times = (days-days[1])/365.25        # t-t_0 in year-fraction
head(times)
tail(times)                 # looks good, 65, almost 66 years.

# the fact that we calculate t-t_0 in year-fraction means that 
# r will have the meaning of a yearly growth rate:

# now the actual regression, just 1 line of code:

res = lm(logindex ~ times)
res

# beta0 = log(S_0) and beta1 = r:

res$coeff                   
r = res$coeff[2]
S0 = exp(res$coeff[1])
r                         # a growth rate of 6.9% per year
S0                        # comparable to spx[1,]
spx[1,]

# let's look at the fit:
plot(times,logindex)
points(times,res$fit,col="red")

plot(times,index)
points(times,exp(res$fit),col="red")