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Econ 7011C: Econometrics for Finance Fall 2022 Homework 3 Answer to question no 1 Answer to question no 1(a) Definition: Med: Producer Price Index by Industry: General Medical and Surgical...

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Econ 7011C: Econometrics for Finance
Fall 2022
Homework 3
Answer to question no 1
Answer to question no 1(a)
Definition:
Med: Producer Price Index by Industry: General Medical and Surgical Hospitals
Transp: Producer Price Index by Industry: Transportation and Warehousing Industries
Food: Consumer Price Index: Food and Beverages
Source: U.S. Bureau of Labor Statistics
Frequency: Monthly
Answer to question no 1(b)
Name Mean SD Max Min No. of obs
med XXXXXXXXXX XXXXXXXXXX
transp XXXXXXXXXX XXXXXXXXXX
food XXXXXXXXXX XXXXXXXXXX
Answer to question no 1(c)
(i) Regress med on transp
(ii) Regress food on transp

(iii) Regress food on med

(iv) Regress med and food on transp
Answer to question no 1(d)
Stationarity checking (med):
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first differenced). But
for second difference SD increased to 0.7302
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty high unlike the
first differenced variables and their lags
ADF test: Suggests I(1) stationary because │-9.94│>│-3.51│which means that we reject the null.
Overall: med is I(1) stationary.
Stationarity checking (transp):
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto 2.42 (from levels to first differenced). But
for second difference SD increased to 2.7635
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty high unlike the
first differenced variables and their lags
ADF test: Suggests I(1) stationary because │-6.61│>│-3.51│which means that we reject the null.
Overall transp is I(1) stationary.
Stationarity checking (food):
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first differenced). But
for second difference SD increased to 0.3942
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty high unlike the
first differenced variables and their lags.
ADF test: Suggests I(1) stationary because │-3.84│>│-3.51│which means that we reject the null.
Overall food is I(1) stationary.
Answer to question no 1(e)
Re- estimation regression
(i)Regress dmed on dtransp
(ii)Regress dfood on dtransp
(iii)Regress dfood on dmed
(iv) Regress dmed and dfood on dtransp
Answer to question1(f):
Comparison of regression med on transp: Out 1 regression between med and transp and out 5 regression between
dmed and dtransp tells us that both of the regression have same result that means there is no problem with the first
model before differentiation.
Comparison of regression of food on transp: Out 2 regressions between food and transp and out 6 regression
etween dfood and dtransp tells us that both of the regression have same result that means there is no problem with
the first model before differentiation.
Comparison of regression of food on med: Out 3 regressions of food on med and out 7 regression between dfood
and dmed tells us that first regression had impact of food on med. Because after the first differentiation regression
the relationship of food on med comes out non-significant.
Comparison of regression of med and food on transp: Out 4 regressions of med and food on transp and out 8
egression between dfood and dtransp tells us that both of the regression have same result that means there is no
problem with the first model before differentiation.
Overall we could say the all the first difference regression makes sense as it takes into account the first difference
variables and regression.
Answer to question 2
For my convenience I have converted the given variables names.
Stock name: Apple=i
SANDP= m
Inflation: inf
Industrial production=indus
USTB3m=rft
M1MONYSUPPLY=m1
CONSUMERCREDIT=credit
BAAAAASPREAD=spread
Answer to question no 2 (a)
Transforming the cpi into inflation I have estimated the regression model using “APPLE” stock as dependent
variable and all other variable on the right-hand side.

Plotting of the stock price:


Answer to question no 2(b)
From the above chart I can see that S&P 500, industrial production, risk free rate, m1 money supply,
consumer credit as the p value of those variables are less than 0.05.
Answer to question no 2(c)
Explanation: Using iterative elimination of variables having high p values and using AIC/BIC criteria
out3 is the best model because out 3 has the lowest AIC/BIC value.
Dropping inflation:

Dropping spread:


Choosing best model we get:

Answer to question no 2(d)
Variable (i)
Explanation:
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD increased to 1.9376
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at 10% level of significance because │-2.73│>│-3.45│which
means that we reject the null.
Overall: Variable (i) is I(1) stationary.


Variable (m) :
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD increased to XXXXXXXXXX
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at because │-20.13│>│-3.45│which means that we reject the null.
Overall: Variable (m) is I(1) stationary.


Variable Indus:
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD increased to 1.1317
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at because │-5.29│>│-3.45│which means that we reject the null.
Overall: Variable (indus) is I(1) stationary.

Variable Inf:
Explanation:
Visual Inspection: suggests either I(0) stationary.
SD: Suggests it is I(0) stationary because SD increased from XXXXXXXXXXto XXXXXXXXXXand XXXXXXXXXXfrom levels
to first differenced and 2nd differenced).
ACFs: Suggests I(0) stationary because the autoco
elations among levels are reverting to mean.
ADF test: Suggests I(0) stationary at because │-12.41│>│-3.45│which means that we reject the null.
Overall: Variable (inf) is I(0) stationary.

Variable (rft):
Explanation:
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD increased to 0.0169.
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at because │-6.6│>│-3.45│which means that we reject the null.
Overall: Variable (rft) is I(1) stationary.

Variable M1:
Explanation:
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD decreased to XXXXXXXXXX.
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at because │-9.7│>│-3.45│which means that we reject the null.
Overall: Variable (m1) is I(1) stationary.

Variable credit:
Explanation:
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD increased to XXXXXXXXXX.
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at because │-10.43│>│-3.45│which means that we reject the null.
Overall: Variable (credit) is I(1) stationary.

Variable Spread:
Explanation:
Visual Inspection: suggests either I(1) or I(2) stationary
SD: Suggests it is I(1) stationary because SD declined from XXXXXXXXXXto XXXXXXXXXXfrom levels to first
differenced). But for second difference SD decreased to 0.2439
ACFs: Suggests I(1) stationary because the autoco
elations among levels and their lags are pretty
high unlike the first differenced variables and their lags.
ADF test: Suggests I(1) stationary at because │-14.32│>│-3.45│which means that we reject the null.
Overall: Variable (spread) is I(1) stationary.

Answer to question no 2(e)
For finding out the excess return of the stock apple and the market risk premium we have used the
following graph.
eri = ri – rf
erm = rm – rf
Plotting both on the graph:

Answer to question no 2(f):
Stationarity checking:
Ri:
Explanation:
Visual Inspection: suggests either I(0) stationary
SD: Suggests it is I(0) stationary because SD increased from XXXXXXXXXXto XXXXXXXXXXand XXXXXXXXXXfrom
levels to first differenced and second differenced).
ACFs: Suggests I(0) stationary because the autoco
elations among levels and their lags are towards
zero.
ADF test: Suggests I(1) stationary at because │-19.1│>│-3.45│which means that we reject the null.
Overall: Variable (ri) is I(0) stationary.

Variable rm:
Explanation:
Visual Inspection: suggests either I(0) stationary
SD: Suggests it is I(0) stationary because SD increased from XXXXXXXXXXto XXXXXXXXXXand XXXXXXXXXXfrom levels
to first differenced and second differenced).
ACFs: Suggests I(0) stationary because the autoco
elations among levels and their lags are towards
zero.
ADF test: Suggests I(1) stationary at because │-18.97│>│-3.45│which means that we reject the null.
Overall: Variable (rm) is I(0) stationary.

Eri: Variable:
Explanation:
Visual Inspection: suggests either I(0) stationary
SD: Suggests it is I(0) stationary because SD increased from 12
Answered Same Day Oct 24, 2022

Solution

Mohd answered on Oct 24 2022
52 Votes
-
-
-
2022-10-24
li
ary(readr)
Fred_data <- read_csv("Fred_data.csv", col_types = cols(DATE = col_date(format = "%m/%d/%Y")))
View(Fred_data)
model_1<-lm(Unemployment_Rate~CPI,data=Fred_data)
model_2<-lm(Unemployment_Rate~PCE,data=Fred_data)
model_3<-lm(PCE~CPI,data=Fred_data)
model_4<-lm(PCE~CPI+Unemployment_Rate,data=Fred_data)
summary(model_1)
##
## Call:
## lm(formula = Unemployment_Rate ~ CPI, data = Fred_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.6000 -1.1571 -0.1901 0.3581 9.3738
##
## Coefficients:
## Estimate Std. E
or t value Pr(>|t|)
## (Intercept) 7.0239 0.5154 13.629 < 2e-16 ***
## CPI -0.7601 0.1905 -3.991 0.000124 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard e
or: 1.702 on 103 degrees of freedom
## Multiple R-squared: 0.1339, Adjusted R-squared: 0.1255
## F-statistic: 15.92 on 1 and 103 DF, p-value: 0.0001236
summary(model_2)
##
## Call:
## lm(formula = Unemployment_Rate ~ PCE, data = Fred_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.4976 -1.0211 -0.4509 0.4673 9.3972
##
## Coefficients:
## Estimate Std. E
or t value Pr(>|t|)
## (Intercept) 6.998e+00 1.187e+00 5.897 4.75e-08 ***
## PCE -1.403e-04 8.572e-05 -1.636 0.105
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard e
or: 1.806 on 103 degrees of freedom
## Multiple R-squared: 0.02534, Adjusted R-squared: 0.01588
## F-statistic: 2.678 on 1 and 103 DF, p-value: 0.1048
summary(model_3)
##
## Call:
## lm(formula = PCE ~ CPI, data = Fred_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15370.3 -1171.2 110.4 926.8 2584.9
##
## Coefficients:
## Estimate Std. E
or t value Pr(>|t|)
## (Intercept) 12568.4 617.5 20.36 <2e-16 ***
## CPI 438.2 228.2 1.92 0.0576 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard e
or: 2039 on 103 degrees of freedom
## Multiple R-squared: 0.03456, Adjusted R-squared: 0.02519
## F-statistic: 3.687 on 1 and 103 DF, p-value: 0.05759
summary(model_4)
##
## Call:
## lm(formula = PCE ~ CPI + Unemployment_Rate, data = Fred_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15210.8 -1183.0 -3.4 1126.7 2574.8
##
## Coefficients:
## Estimate Std. E
or t value Pr(>|t|)
## (Intercept) 13407.4 1033.7 12.970 <2e-16 ***
## CPI 347.4 245.2 1.417 0.160
## Unemployment_Rate -119.4 118.0 -1.012 0.314
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard e
or: 2039 on 102 degrees of freedom
## Multiple R-squared: 0.04416, Adjusted R-squared: 0.02541
## F-statistic: 2.356 on 2 and 102 DF, p-value: 0.09994
li
ary(tseries)
## Registered S3 method overwritten by 'quantmod':
## method from
## as.zoo.data.frame zoo
adf.test(Fred_data$CPI)
## Warning in adf.test(Fred_data$CPI): p-value greater than printed p-value
##
## Augmented Dickey-Fuller Test
##
## data: Fred_data$CPI
## Dickey-Fuller = 0.85597, Lag order = 4, p-value = 0.99
## alternative hypothesis: stationary
acf(Fred_data$CPI)
adf.test(Fred_data$Unemployment_Rate)
##
## Augmented Dickey-Fuller Test
##
## data: Fred_data$Unemployment_Rate
## Dickey-Fuller = -2.7089, Lag order = 4, p-value = 0.2827
## alternative hypothesis: stationary
acf(Fred_data$Unemployment_Rate)
adf.test(Fred_data$PCE)
##
## Augmented Dickey-Fuller Test
##
## data: Fred_data$PCE
## Dickey-Fuller = -2.7877, Lag order = 4, p-value = 0.25
## alternative hypothesis: stationary
acf(Fred_data$PCE)
skimr::skim(Fred_data)
Data summary
    Name
    Fred_data
    Number of rows
    105
    Number of...
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