ETF5952 Questions for Final Exam A
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SAMPLE-Exam A
ETF5952
Questions for Final Exam
Question 1 (25 points=5+5+5+10)
We consider a data set on weekly stock return. The variables in the data set are as follows:
Year: The year that the observation was recorded
Lag1: Percentage return for previous week
Lag2: Percentage return for 2 weeks previous
Lag3: Percentage return for 3 weeks previous
Lag4: Percentage return for 4 weeks previous
Lag5: Percentage return for 5 weeks previous
Volume: Volume of shares traded (average number of daily shares traded in billions)
Today: Percentage return for this week
Direction: A factor with levels Down and Up indicating whether the market had a positive or negative return on a given week
1. We obtain the sample statistics. Answer the average weekly return of this week’s variable.
Year Lag1 Lag2 Lag3 Lag4 Lag5
1996 : 53 Min . :-18 .1950 Min . :-18 .1950 Min . :-18 .1950 Min . :-18 .1950 Min . :-18 .1950 2007 : 53 1st Qu . : -1 .1540 1st Qu . : -1 .1540 1st Qu . : -1 .1580 1st Qu . : -1 .1580 1st Qu . : -1 .1660 1991 : 52 Median : 0 .2410 Median : 0 .2410 Median : 0 .2410 Median : 0 .2380 Median : 0 .2340 1992 : 52 Mean : 0 .1506 Mean : 0 .1511 Mean : 0 .1472 Mean : 0 .1458 Mean : 0 .1399 1993 : 52 3rd Qu . : 1 .4050 3rd Qu . : 1 .4090 3rd Qu . : 1 .4090 3rd Qu . : 1 .4090 3rd Qu . : 1 .4050 1994 : 52 Max . : 12 .0260 Max . : 12 .0260 Max . : 12 .0260 Max . : 12 .0260 Max . : 12 .0260 (Other):775
Volume Today Direction time
Min . :0 .08747 Min . :-18 .1950 Down:484 Min . : 1
1st Qu . :0 .33202 1st Qu . : -1 .1540 Up :605 1st Qu . : 273
Median :1 .00268 Median : 0 .2410 Median : 545
Mean :1 .57462 Mean : 0 .1499 Mean : 545
3rd Qu . :2 .05373 3rd Qu . : 1 .4050 3rd Qu . : 817
Max . :9 .32821 Max . : 12 .0260 Max . :1089
2. We estimate an autoregressitve with lag order of 1 or AR(1) model as reported below. Explain the effect of Lag1 on Today.
glm(formula = Today ~ Lag1, data = DATA)
Deviance Residuals:
Min 1Q Median 3Q Max
-19 .061 -1 .271 0 .113 1 .280 11 .235
Coefficients:
Estimate Std . Error t value Pr(>|t |)
(Intercept) 0 .16120 0 .07140 2 .258 0 .0242 *
Lag1 -0 .07503 0 .03024 -2 .481 0 .0133 *
---
Signif . codes: 0 *** 0 .001 ** 0 .01 * 0 .05 . 0 .1 1
3. We estimate an autoregressitve with lag order of 5 or AR(5) model as reported below. Explain the effect of Volume on Today (no more than 20 words).
glm(formula = Today ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = DATA)
Deviance Residuals:
Min -18 .5410
1Q
-1 .2622
Median
0 .0873
3Q
1 .2579
Max
11 .9316
Coefficients:
Estimate Std . Error t value Pr(>|t |)
(Intercept) 0 .25050 0 .09964 2 .514 0 .0121 *
Lag1 -0 .07143 0 .03046 -2 .346 0 .0192 *
Lag2 0 .04760 0 .03058 1 .556 0 .1199
Lag3 -0 .06857 0 .03045 -2 .252 0 .0245 *
Lag4 -0 .02215 0 .03047 -0 .727 0 .4674
Lag5 0 .01406 0 .03039 0 .463 0 .6437
Volume -0 .05441 0 .04274 -1 .273 0 .2033
---
Signif . codes: 0 *** 0 .001 ** 0 .01 * 0 .05 . 0 .1 1
4. We estimate the model with more variables and also obtain the AIC for all three models, as reported below (fit1, fit2 and fit3 are estimation results from Question1.2, Question1.3, Question1.4, respectively). In terms of R2 , the model here is better. Explain why this model is better or worse for predicting stock return. (no more than 30 words).
glm(formula = Today ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume +
Year, data = DATA)
Deviance Residuals:
Min -18 .0478
1Q
-1 .2246
Median
0 .0719
3Q
1 .2435
Max
12 .4662
Coefficients:
Estimate Std . Error t value Pr(>|t |)
(Intercept) -0 .05787 0 .34205 -0 .169 0 .86569
Lag1 -0 .09849 0 .03104 -3 .173 0 .00155 **
Lag2 0 .02080 0 .03149 0 .661 0 .50907
Lag3 -0 .09284 0 .03101 -2 .994 0 .00281 **
Lag4 -0 .04905 0 .03100 -1 .582 0 .11394
Lag5 -0 .01007 0 .03080 -0 .327 0 .74389
Volume -0 .06976 0 .15874 -0 .439 0 .66043
Year1991 0 .68642 0 .47174 1 .455 0 .14594
Year1992 0 .20143 0 .47144 0 .427 0 .66926
Year1993 0 .24403 0 .47139 0 .518 0 .60478
Year1994 0 .04760 0 .47122 0 .101 0 .91956
Year1995 0 .78668 0 .47503 1 .656 0 .09800 .
Year1996 0 .55581 0 .47238 1 .177 0 .23961
Year1997 0 .74433 0 .47768 1 .558 0 .11948
Year1998 0 .69017 0 .48118 1 .434 0 .15177
Year1999 0 .58028 0 .48512 1 .196 0 .23190
Year2000
Year2001
Year2002
-0 .04876 0 .49021
-0 .09390 0 .49905
-0 .39523 0 .50943
-0 .099 0 .92079
-0 .188 0 .85078
-0 .776 0 .43803
Year2003 0 .64860 0 .51383 1 .262 0 .20712
Year2004 0 .38829 0 .51445 0 .755 0 .45056
Year2005 0 .28258 0 .54832 0 .515 0 .60641
Year2006 0 .52796 0 .59237 0 .891 0 .37299
Year2007 0 .31644 0 .67343 0 .470 0 .63853
Year2008 -0 .48258 0 .88789 -0 .544 0 .58689
Year2009 0 .97455 0 .99170 0 .983 0 .32598
Year2010 0 .69020 0 .84774 0 .814 0 .41573
---
Signif . codes: 0 *** 0 .001 ** 0 .01 * 0 .05 . 0 .1 1
> c(AIC(fit1), AIC(fit2), AIC(fit3) )
[1] 4956 .627 4956 .663 4969 .627
Question 2 (25 points=5+5+5+10)
We consider a data set on credit card default. The variables in the data set are
● default: A factor with levels No and Yes indicating whether the customer defaulted on their debt
● student: A factor with levels No and Yes indicating whether the customer is a student
● balance: The average balance that the customer has remaining on their credit card after making their monthly payment
● income: Income of customer
1. We obtain summary statistics and a frequency table as below. Obtain the probability of default = YES and Student = NO and the probability of default = YES condition on student = NO (2 decimal places).
> summary(DATA)
default student balance income
No :9667 No :7056 Min . : 0 .0 Min . : 772
Yes: 333 Yes:2944 1st Qu . : 481 .7 1st Qu . :21340
Median : 823 .6 Median :34553
Mean : 835 .4 Mean :33517
3rd Qu . :1166 .3 3rd Qu . :43808
Max . :2654 .3 Max . :73554
> table(DATA$default, DATA$student)
No Yes
No 6850 2817
Yes 206 127
2. We obtain an estimate result as below. Interpret the estimated coefficient of income.
glm(formula = default ~ balance + income + student, family = "binomial", data = DATA)
Deviance Residuals:
Min 1Q Median 3Q Max
-2 .4691 -0 .1418 -0 .0557 -0 .0203 3 .7383
Coefficients:
Estimate Std . Error z value Pr(>|z|)
(Intercept) -1 .087e+01 4 .923e-01 -22 .080 < 2e-16 ***
balance 5 .737e-03 2 .319e-04 24 .738 < 2e-16 ***
income 3 .033e-06 8 .203e-06 0 .370 0 .71152
studentYes -6 .468e-01 2 .363e-01 -2 .738 0 .00619 **
---
Signif . codes: 0 *** 0 .001 ** 0 .01 * 0 .05 . 0 .1 1
3. Using the previous estimation result, interpret the estimated coefficient of student.
4. We obtain another estimate result as below. Interpret the effect of balance in the result (no more than 30 words, 4 decimal places).
glm(formula = default ~ balance * student + income, family = "binomial", data = DATA)
Deviance Residuals:
Min 1Q Median 3Q Max
-2 .4949 -0 .1417 -0 .0555 -0 .0202 3 .7576
Coefficients:
Estimate Std . Error z value
(Intercept)
balance
studentYes
income
balance:studentYes -2 .184e-04 4 .781e-04 -0 .457
---
Signif . codes: 0 *** 0 .001 ** 0 .01 * 0 .05 . 0 .1
Pr(>|z|)
<2e-16 ***
<2e-16 ***
0 .729
0 .714
0 .648
1
Question 3 (25 points=5+5+5+10)
We consider a data set on cars. The data set includes the following variables:
mpg: miles per gallon
cylinders: Number of cylinders between 4 and 8
displacement: Engine displacement (cu. inches)
horsepower: Engine horsepower
weight: Vehicle weight (lbs.)
acceleration: Time to accelerate from 0 to 60 mph (sec.)
year: Model year (modulo 100)
origin: Origin of car (1. American, 2. European, 3. Japanese)
name: Vehicle name
The summary statistics is as follows:
mpg
Min . : 9 .00 1st Qu . :17 .00 Median :22 .75 Mean :23 .45 3rd Qu . :29 .00 Max . :46 .60
origin Min . :1 .000 1st Qu . :1 .000 Median :1 .000 Mean :1 .577 3rd Qu . :2 .000 Max . :3 .000
cylinders displacement Min . :3 .000 Min . : 68 .0 1st Qu . :4 .000 1st Qu . :105 .0 Median :4 .000 Median :151 .0 Mean :5 .472 Mean :194 .4 3rd Qu . :8 .000 3rd Qu . :275 .8 Max . :8 .000 Max . :455 .0
name
amc matador : 5
ford pinto : 5
toyota corolla : 5
amc gremlin : 4
amc hornet : 4
chevrolet chevette: 4
(Other) :365
horsepower weight acceleration year Min . : 46 .0 Min . :1613 Min . : 8 .00 73 : 40 1st Qu . : 75 .0 1st Qu . :2225 1st Qu . :13 .78 78 : 36 Median : 93 .5 Median :2804 Median :15 .50 76 : 34 Mean :104 .5 Mean :2978 Mean :15 .54 75 : 30 3rd Qu . :126 .0 3rd Qu . :3615 3rd Qu . :17 .02 82 : 30 Max . :230 .0 Max . :5140 Max . :24 .80 70 : 29
(Other):193
1. We obtain a regression result as below. Explain the effect of weight on miles per gallon (no more than 30 words.)
glm(formula = log(mpg) ~ log(weight) + cylinders + displacement +
horsepower + acceleration + year, data = DATA)
Deviance Residuals:
Min -0 .34100
1Q
-0 .05913
Median
0 .00180
3Q
0 .05995
Max
0 .37563
Coefficients:
Estimate Std . Error t value Pr(>|t |)
(Intercept) log(weight) cylinders displacement horsepower |
1 .133e+01 -1 .056e+00 -2 .045e-02 1 .166e-04 -1 .656e-03 |
acceleration -3 .262e-03
year71
year72
year73
year74
year75
1 .097e+00 1 .560e-01 1 .062e-02 2 .351e-04 4 .706e-04 3 .236e-03 3 .095e-02 2 .996e-02 2 .721e-02 3 .222e-02 3 .131e-02
10 .334 -6 .772 -1 .924 0 .496 -3 .520 -1 .008 0 .050 -0 .950 -1 .964 1 .061 1 .227
< 2e-16 ***
4 .96e-11 ***
0 .055082 .
0 .620209
0 .000485 ***
0 .314127
0 .960290
0 .342860
0 .050319 .
0 .289294
0 .220567
year76 6 .273e-02 3 .010e-02 2 .084 |
0 .037830 * |
year77 1 .163e-01 3 .067e-02 3 .791 |
0 .000175 *** |
year78 1 .316e-01 2 .903e-02 4 .533 |
7 .84e-06 *** |
year79 2 .196e-01 3 .072e-02 7 .148 |
4 .67e-12 *** |
year80 3 .328e-01 3 .238e-02 10 .280 |
< 2e-16 *** |
year81 2 .518e-01 3 .186e-02 7 .904 |
3 .07e-14 *** |
year82 2 .926e-01 3 .105e-02 9 .423 --- |
< 2e-16 *** |
Signif . codes: 0 *** 0 .001 ** 0 .01 * 0 .05 |
. 0 .1 1 |
2. We apply the lasso procedure, where the dependent variables is log of mpg and the rest of variables in the data are used as regressors. Explain why we should not simply run standard regression.
> x = sparse .model .matrix(log(mpg) ~ . , data=DATA)[,-1]
> y = log(DATA$mpg)
> dim(x)
[1] 392 321
> x = sparse .model .matrix(log(mpg) ~ . , data=DATA)[,-1]
> x = sparse .model .matrix(log(mpg) ~ . , data=DATA)[,-1]
> y = log(DATA$mpg)
> dim(x)
[1] 392 321
> sclasso = gamlr(x, y, family = "gaussian", nfold=10)
> sum(coef(sclasso) !=0)
[1] 58
3. In outputs from the lasso above, explain whether the lasso works properly (no more than 20 words).
4. We obtain a plot after we implement the lasso procedure above. This plot shows trajectories of estimated coefficients. Explain why lines spread widely on the left side and does not on the right side (no more than
40 words).
Question 4 (25 points=5+5+5+10)
We have a data set collected from 506 geographical areas and the data set contains a list of variables:
price: median housing price in US$1,000,
crime: crimes committed per capita
nox: nitric oxide pollution per 100m
rooms: average number of rooms
dist: weighted distance to city center
radial: access index to radial highways (radial),
stratio: average student-teacher ratio
We estimate a regression tree by using price as the dependent variable and the other variables as regressors. The estimation result is reported in Figure 1.
Figure 1: Regression Tree Result
1. The far-right node has 45 and 6%. Explain those numbers (no more than 20 words.)
2. Interpret prediction from the above estimation result for an area with rooms = 4, stratio = 15, dist = 2, crime = 8 and nox = 3 (no more than 20 words).
3. Find the node with 18 and 8% and explain characters of observations in the node.
4. Explain what possibly causes low housing prices at the far-left terminal node of the regression tree result.
2022-06-16