#---------------------------#
#    Loesungen UeBlatt13    #
#---------------------------#


---------------
#  Aufgabe 1  #
---------------

bevdata = read.table("C:/Users/detlef/OneDrive/hochschule/Vorlesungen/WS2122/Oekonometrie/Weltbevoelkerung.csv",header=TRUE,sep=";")
bevdata

bev = bevdata[,4]
bev
zeiten = bevdata[,3]
zeiten
plot(zeiten,bev)

zeiten[14]
zeiten2 = zeiten[1:14]
bev2 = bev[1:14]
zeiten2 = zeiten2 -1950
zeiten = zeiten -1950
logbev2 = log(bev2)

# Geben wir zunaechst noch einmal die 
# Regressionskoeffizienten an:

# Modell (1):
res1 = lm(bev2 ~ zeiten2)
a0 = res1$coef[1]
a1 = res1$coef[2]
a0
a1

# Modell (2):
X = cbind(zeiten2,zeiten2^2)
res2 = lm(bev2 ~ X )
b0 = res2$coef[1]
b1 = res2$coef[2]
b2 = res2$coef[3]
b0
b1
b2

# Modell (3):
res3 = lm(logbev2 ~ zeiten2)
beta0 = res3$coef[1]
beta1 = res3$coef[2]
B0 = exp(beta0)
r = beta1
B0
r


# Die Vertrauensintervalle sind gegeben durch
#
#   hat_beta plus/minus stddev * q_alpha
#
# wobei die Standardabweichung stddev aus den Re-
# gressionsresultaten abgelesen werden kann und 
# bei Vertauenslevel alpha=90%
#
#   q90 = qt(0.95,df=n-p)
#
# wenn n der Stichprobenumfang und p die Anzahl 
# der Regressoren ist (mit intercept oder konstantem 
# Term).

n = 14

# Modell (1):

summary(res1)               # also:
stddev_a0 = 58278
stddev_a1 = 1524

p = 2
q90 = qt(0.95,df=n-p)

# damit:
a0up = a0 + stddev_a0*q90
a0down = a0 - stddev_a0*q90
a1up = a1 + stddev_a1*q90
a1down = a1 - stddev_a1*q90

confint_a0 = c(a0down,a0up)
confint_a1 = c(a1down,a1up)
confint_a0
confint_a1


# Modell (2):

summary(res2)               # also:
stddev_b0 = 3.910*10^4
stddev_b1 = 2.793*10^3
stddev_b2 = 4.143*10^1

p = 3
q90 = qt(0.95,df=n-p)

# damit:
b0up = b0 + stddev_b0*q90
b0down = b0 - stddev_b0*q90
b1up = b1 + stddev_b1*q90
b1down = b1 - stddev_b1*q90
b2up = b2 + stddev_b2*q90
b2down = b2 - stddev_b2*q90

confint_b0 = c(b0down,b0up)
confint_b1 = c(b1down,b1up)
confint_b2 = c(b2down,b2up)
confint_b0
confint_b1
confint_b2


# Modell (3):

summary(res3)               # also:
stddev_beta0 = 0.0156
stddev_beta1 = 0.0004079
p = 2
q90 = qt(0.95,df=n-p)
q90

# damit:
beta0up = beta0 + stddev_beta0*q90
beta0down = beta0 - stddev_beta0*q90
beta1up = beta1 + stddev_beta1*q90
beta1down = beta1 - stddev_beta1*q90

confint_B0 = c(exp(beta0down),exp(beta0up))
confint_r = c(beta1down,beta1up)
confint_B0
confint_r




---------------
#  Aufgabe 2  #
---------------

# 2a) store the Excel-data as csv-file and make sure that numbers in Excel are 
#     formatted properly (ohne Hochkomma oder aehnliches..)

bip = read.table("C:/Users/detlef/OneDrive/hochschule/Vorlesungen/WS2122/Oekonometrie/BIP-Deutschland.csv",header=FALSE,sep=";")
bip

y = bip[,2]
y
x = bip[,1]
x
x = x - 1991
x

plot(x,y)


# 2b)

res = lm(y ~ x)

betas = res$coeff
beta0hat = betas[1]
beta1hat = betas[2]
beta0hat             # = B_{t_0}
beta1hat             # = r

plot(x,y)
lines(x,res$fit,col=2)


# 2c) Vertrauensintervalle:
#
# beta0hat +/- q_alpha stddev_beta0
# beta1hat +/- q_alpha stddev_beta1

n = length(x)
n
p = 2
q90 = qt(95/100,df=n-p)
q90
q95 = qt(97.5/100,df=n-p)
q95
q99 = qt(99.5/100,df=n-p)
q99

summary(res)         # also:
stddev_beta1 = 1.346

beta1up90 = beta1hat + q90*stddev_beta1
beta1up95 = beta1hat + q95*stddev_beta1
beta1up99 = beta1hat + q99*stddev_beta1
beta1down90 = beta1hat - q90*stddev_beta1
beta1down95 = beta1hat - q95*stddev_beta1
beta1down99 = beta1hat - q99*stddev_beta1

confint1_90 = c(beta1down90,beta1up90)
confint1_95 = c(beta1down95,beta1up95)
confint1_99 = c(beta1down99,beta1up99)

confint1_90
confint1_95
confint1_99