HW1
Kunal Dolas
2/23/2021
This question should be answered using the Weekly data set, which is part of the ISLR package. This data is similar in nature to the Smarket data discussed in class, except that it contains 1,089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)
library(faraway)
data(Weekly)
head(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Downsummary(Weekly)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## dim(Weekly)## [1] 1089 9cor(Weekly[,-9])## Year Lag1 Lag2 Lag3 Lag4
## Year 1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1 -0.03228927 1.000000000 -0.07485305 0.05863568 -0.071273876
## Lag2 -0.03339001 -0.074853051 1.00000000 -0.07572091 0.058381535
## Lag3 -0.03000649 0.058635682 -0.07572091 1.00000000 -0.075395865
## Lag4 -0.03112792 -0.071273876 0.05838153 -0.07539587 1.000000000
## Lag5 -0.03051910 -0.008183096 -0.07249948 0.06065717 -0.075675027
## Volume 0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today -0.03245989 -0.075031842 0.05916672 -0.07124364 -0.007825873
## Lag5 Volume Today
## Year -0.030519101 0.84194162 -0.032459894
## Lag1 -0.008183096 -0.06495131 -0.075031842
## Lag2 -0.072499482 -0.08551314 0.059166717
## Lag3 0.060657175 -0.06928771 -0.071243639
## Lag4 -0.075675027 -0.06107462 -0.007825873
## Lag5 1.000000000 -0.05851741 0.011012698
## Volume -0.058517414 1.00000000 -0.033077783
## Today 0.011012698 -0.03307778 1.000000000attach(Weekly)
plot(Volume)
plot(Volume,Year)
#There is a trend of increase in the volume over the period of 1989 to 2010.
- Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
gmod<-glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(gmod)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4##From the summary, Lag2 appears to be statistically significant.
- Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
glm.probs<-predict(gmod, type="response")
glm.probs[1:10]## 1 2 3 4 5 6 7 8
## 0.6086249 0.6010314 0.5875699 0.4816416 0.6169013 0.5684190 0.5786097 0.5151972
## 9 10
## 0.5715200 0.5554287contrasts(Direction)## Up
## Down 0
## Up 1glm.pred=rep("Down", 1089)
glm.pred[glm.probs > 0.5]="Up"
table(glm.pred, Direction)## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557(557+54)/1089## [1] 0.5610652mean(glm.pred==Direction)## [1] 0.5610652Confusion matrix is telling the ratio of correct prediction by the regression model. here the value is 0.5610652, suggesting not a good performance by the model.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train=(Year <2008)
length(train)## [1] 1089test=Weekly [! train, ]
dim(test)## [1] 156 9Direction.test= Direction [! train]
gmod2 <- glm(Direction~Lag2, data=Weekly, family=binomial, subset=train)
glm.probs<-predict(gmod2, test, type="response")
glm.pred=rep("Down", 156)
glm.pred[glm.probs > 0.5]="Up"
table(glm.pred, Direction.test)## Direction.test
## glm.pred Down Up
## Down 7 5
## Up 65 79(7+79)/156## [1] 0.5512821mean(glm.pred==Direction.test)## [1] 0.5512821from the above result, the prediction is little more poor than the previous model.