AS6-3 預測股票的投資報酬

主要議題：預測股票的投資報酬

學習重點：

• 先分群以後、再做預測性模型
• 集群分析的模型與預測方法

rm(list=ls(all=T))
Sys.setlocale("LC_ALL","C")

[1] "C"

options(digits=5, scipen=12)
library(dplyr)
library(caTools)
library(caret)
library(flexclust)

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1. 資料探索

1.1

Load StocksCluster.csv into a data frame called “stocks”.

A = read.csv('data/StocksCluster.csv')
nrow(A)

[1] 11580

How many observations are in the dataset?

+11580 +

1.2

mean(A$PositiveDec)

[1] 0.54611

table(A$PositiveDec == 1)

FALSE  TRUE 
 5256  6324 

6324/(5256+6324)

[1] 0.54611

What proportion of the observations have positive returns in December?

+0.54611 +

1.3

cor(A[1:11]) %>% sort %>% unique %>% tail %>% round(2) #unique()翠取資料中單一的

[1] 0.09 0.13 0.14 0.17 0.19 1.00

cor(A[1:11]) %>% sort %>% unique %>% tail %>% round(2)

[1] 0.09 0.13 0.14 0.17 0.19 1.00

What is the maximum correlation between any two return variables in the dataset? You should look at the pairwise correlations between ReturnJan, ReturnFeb, ReturnMar, ReturnApr, ReturnMay, ReturnJune, ReturnJuly, ReturnAug, ReturnSep, ReturnOct, and ReturnNov.

+0.09 0.13 0.14 0.17 0.19 1.00 +Nov. & Oct.

1.4

colMeans(A[,1:11]) %>% sort %>% barplot(las=2, cex.names=0.8, cex.axis=0.8)

colMeans(A[,1:11])

 ReturnJan  ReturnFeb  ReturnMar  ReturnApr  ReturnMay ReturnJune ReturnJuly  ReturnAug  ReturnSep 
 0.0126316 -0.0076048  0.0194023  0.0263081  0.0247366  0.0059379  0.0030509  0.0161983 -0.0147208 
 ReturnOct  ReturnNov 
 0.0056508  0.0113874 

Which month (from January through November) has the largest mean return across all observations in the dataset?

+ReturnApr +

Which month (from January through November) has the smallest mean return across all observations in the dataset?

+ReturnSep +

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2. 邏輯式回歸，單一模型

分割訓練、測試資料

Run the following commands to split the data into a training set and testing set, putting 70% of the data in the training set and 30% of the data in the testing set:

set.seed(144)

spl = sample.split(stocks$PositiveDec, SplitRatio = 0.7)

stocksTrain = subset(stocks, spl == TRUE)

stocksTest = subset(stocks, spl == FALSE)

library(caTools)
set.seed(144)
spl = sample.split(A$PositiveDec,0.7)
TR = subset(A, spl)
TS = subset(A, !spl)
sapply(list(A, TR, TS), function(x) mean(x$PositiveDec))

[1] 0.54611 0.54614 0.54606

library(caTools)
set.seed(144)
spl = sample.split(A$PositiveDec, SplitRatio = 0.7)
TR = subset(A, spl == TRUE) 
TS = subset(A, spl == FALSE)
sapply(list(A, TR, TS), function(x) mean(x$PositiveDec))

[1] 0.54611 0.54614 0.54606

2.1 單一模型：訓練準確率，\(\text{acc}_{train}\)

Then, use the stocksTrain data frame to train a logistic regression model (name it StocksModel) to predict PositiveDec using all the other variables as independent variables. Don’t forget to add the argument family=binomial to your glm command.

glm1 = glm(PositiveDec ~ .,  TR, family=binomial)
pred = predict(glm1, type='response')
table(TR$Pos, pred > 0.5) %>% {sum(diag(.))/sum(.)} 

[1] 0.57118

glm1 =glm(PositiveDec ~ ., TR , family = binomial)
pred = predict(glm1 ,type ='response')
table(TR$Pos, pred > 0.5) %>% {sum(diag(.))/sum(.)} 

[1] 0.57118

table(TR$Pos, pred > 0.5)

   
    FALSE TRUE
  0   990 2689
  1   787 3640

(990+3640)/(990+2689+787+3640)

[1] 0.57118

What is the overall accuracy on the training set, using a threshold of 0.5?

2.2 單一模型：測試準確率，\(\text{acc}_{test}\)

pred = predict(glm1, TS, type='response')
table(TS$Pos, pred > 0.5) %>% {sum(diag(.))/sum(.)} 

[1] 0.56707

Now obtain test set predictions from StocksModel. What is the overall accuracy of the model on the test, again using a threshold of 0.5?

2.3 單一模型：底線準確率，\(\text{acc}_{baseline}\)

mean(TS$PositiveDec)

[1] 0.54606

table(TS$PositiveDec)

   0    1 
1577 1897 

1897/(1577+1897)

[1] 0.54606

What is the accuracy on the test set of a baseline model that always predicts the most common outcome (PositiveDec = 1)?

+0.54606 +

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3. 集群分析

3.1 移除目標變數

Now, let’s cluster the stocks. The first step in this process is to remove the dependent variable using the following commands:

LTR = TR[,1:11]
LTS = TS[,1:11]

Why do we need to remove the dependent variable in the clustering phase of the cluster-then-predict methodology?

+需要知道因變量值來為群集分配觀察 +在cluster-then-predict下，我們想要知道利用前面11個月我們能不能夠成功預測第12個月，因此12月是應變變數，所以這裡我們先將它拿掉；如果我們把她留著，那麼預測出來的值是沒有意義的

3.2 區隔變數常態化

In the market segmentation assignment in this week’s homework, you were introduced to the preProcess command from the caret package, which normalizes variables by subtracting by the mean and dividing by the standard deviation.

In cases where we have a training and testing set, we’ll want to normalize by the mean and standard deviation of the variables in the training set. We can do this by passing just the training set to the preProcess function:

library(caret)
preproc = preProcess(LTR)
NTR = predict(preproc, LTR)
NTS = predict(preproc, LTS)

mean(NTR$ReturnJan)

[1] 2.1006e-17

What is the mean of the ReturnJan variable in normTrain?

+2.1006e-17 +

mean(NTS$ReturnJan)

[1] -0.00041859

What is the mean of the ReturnJan variable in normTest?

• -0.00041859

3.3 測試資料的常態化結果

Why is the mean ReturnJan variable much closer to 0 in normTrain than in normTest?

• The distribution of the ReturnJan variable is different in the training and testing set
• From mean(stocksTrain\(ReturnJan) and mean(stocksTest\)ReturnJan), we see that the average return in January is slightly higher in the training set than in the testing set. Since normTest was constructed by subtracting by the mean ReturnJan value from the training set, this explains why the mean value of ReturnJan is slightly negative in normTest.

3.4 K-Means集群

Set the random seed to 144 (it is important to do this again, even though we did it earlier). Run k-means clustering with 3 clusters on normTrain, storing the result in an object called km.

set.seed(144)
km <- kmeans(NTR, 3)

table(km$cluster)

   1    2    3 
3157 4696  253 

Which cluster has the largest number of observations?

+cluster 2 +

3.5

Recall from the recitation that we can use the flexclust package to obtain training set and testing set cluster assignments for our observations (note that the call to as.kcca may take a while to complete):

library(flexclust)
km.kcca = as.kcca(km, NTR)

Found more than one class "kcca" in cache; using the first, from namespace 'kernlab'
Also defined by 'flexclust'
Found more than one class "kcca" in cache; using the first, from namespace 'kernlab'
Also defined by 'flexclust'

CTR = predict(km.kcca)

Found more than one class "kcca" in cache; using the first, from namespace 'kernlab'
Also defined by 'flexclust'

CTS = predict(km.kcca, newdata=NTS)

table(CTS)

CTS
   1    2    3 
1298 2080   96 

How many test-set observations were assigned to Cluster 2?

• 2080
• 分成3類 cluster2 有2080個

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4. 邏輯式回歸，分群模型

4.1 依集群分析的結果切割資料

Using the subset function, build data frames stocksTrain1, stocksTrain2, and stocksTrain3, containing the elements in the stocksTrain data frame assigned to clusters 1, 2, and 3, respectively (be careful to take subsets of stocksTrain, not of normTrain). Similarly build stocksTest1, stocksTest2, and stocksTest3 from the stocksTest data frame.

mean(TR1$PositiveDec)

[1] 0.60247

mean(TR2$PositiveDec)

[1] 0.51405

mean(TR3$PositiveDec)

[1] 0.43874

Which training set data frame has the highest average value of the dependent variable?

+TR1 +

4.2 分群模型，模型係數

Build logistic regression models StocksModel1, StocksModel2, and StocksModel3, which predict PositiveDec using all the other variables as independent variables. StocksModel1 should be trained on stocksTrain1, StocksModel2 should be trained on stocksTrain2, and StocksModel3 should be trained on stocksTrain3.

M = lapply(split(TR, CTR), function(x) 
  glm(PositiveDec~., data=x, family=binomial) )
sapply(M, function(x) coef(summary(x))[,1])

                    1        2          3
(Intercept)  0.172240  0.10293 -0.1818958
ReturnJan    0.024984  0.88451 -0.0097893
ReturnFeb   -0.372074  0.31762 -0.0468833
ReturnMar    0.595550 -0.37978  0.6741795
ReturnApr    1.190478  0.49291  1.2814662
ReturnMay    0.304209  0.89655  0.7625116
ReturnJune  -0.011654  1.50088  0.3294339
ReturnJuly   0.197692  0.78315  0.7741644
ReturnAug    0.512729 -0.24486  0.9826054
ReturnSep    0.588327  0.73685  0.3638068
ReturnOct   -1.022535 -0.27756  0.7822421
ReturnNov   -0.748472 -0.78747 -0.8737521

Which variables have a positive sign for the coefficient in at least one model and a negative sign for the coefficient in at least one model? Select all that apply.

• ReturnJan、ReturnFFeb、ReturnMar、ReturnJune、ReturnAug、ReturnOct

4.3 分群模型：分群測試準確率，\(\text{acc}_{test}^{1,2,3}\)

Using StocksModel1, make test-set predictions called PredictTest1 on the data frame stocksTest1. Using StocksModel2, make test-set predictions called PredictTest2 on the data frame stocksTest2. Using StocksModel3, make test-set predictions called PredictTest3 on the data frame stocksTest3.

Pred = lapply(1:3, function(i) 
  predict(M[[i]], TS[CTS==i,], type='response') )
sapply(1:3, function(i) 
  table(TS$Pos[CTS==i], Pred[[i]] > 0.5) %>% {sum(diag(.))/sum(.)}  )

[1] 0.61941 0.55048 0.64583

What is the overall accuracy of StocksModel1 on the test set stocksTest1, using a threshold of 0.5?

+0.61941 +

What is the overall accuracy of StocksModel2 on the test set stocksTest3, using a threshold of 0.5?

+0.55048 +

What is the overall accuracy of StocksModel3 on the test set stocksTest3, using a threshold of 0.5?

+0.64583 +

4.4 分群模型：整體測試準確率，\(\text{acc}_{test}^{1+2+3}\)

To compute the overall test-set accuracy of the cluster-then-predict approach, we can combine all the test-set predictions into a single vector and all the true outcomes into a single vector:

rr table( do.call(c, split(TS$Pos,CTS)), do.call(c, Pred) > 0.5 ) %>% {sum(diag(.))/sum(.)}

[1] 0.57887

What is the overall test-set accuracy of the cluster-then-predict approach, again using a threshold of 0.5?

+0.57887 +

We see a modest improvement over the original logistic regression model. Since predicting stock returns is a notoriously hard problem, this is a good increase in accuracy. By investing in stocks for which we are more confident that they will have positive returns (by selecting the ones with higher predicted probabilities), this cluster-then-predict model can give us an edge over the original logistic regression model.

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