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

What proportion of the observations have positive returns in December?

• 0.54611
• 利用該變數的平均值去計算得到正回報的比例

1.3

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.

• 0.19
• 先利用 cor() 去計算相關係數，再從中求出最高的相關係數
• 並記得將相關係數 = 1 的數值過濾掉 - 1 代表為變數自身的相關係數

1.4

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

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

• April
• 根據上圖結果可以得知
• 利用 colMeans() 去計算出每行平均值後以柱狀圖呈現

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

• September
• 根據上圖結果可以得知

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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

• 分割資料為 training / testing data
• 比例為 7:3
• 並觀察原始資料、training、testing 三個資料集的應變數平均值是否相仿

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

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

• 0.57118
• 計算 training set 的 ACC

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?

• 0.56707
• 將 testing data 丟進模型中並計算其 ACC

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

mean(TS$PositiveDec)

[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
• testing set 本身的 baseline model ACC

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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?

• Needing to know the dependent variable value to assign an observation to a cluster defeats the purpose of the methodology
• 不能讓應變數影響到分群，因為他是我們所要分析的目標

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:

mean(NTR$ReturnJan)

[1] 2.1006e-17

• 常態化 LTR 以及 LTS

mean(NTR$ReturnJan)

[1] 2.1006e-17

What is the mean of the ReturnJan variable in normTrain?

• 2.1006 e-17

mean(NTS$ReturnJan)

[1] -0.00041859

What is the mean of the ReturnJan variable in normTrain?

• -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
• 因為是已經常態化的數據，因此意味著兩個資料集的分佈是不同的

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)

• 計算 K-means 集群

table(km$cluster)

   1    2    3 
3157 4696  253 

Which cluster has the largest number of observations?

• 2
• 由上述結果可以得知

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)
CTR = predict(km.kcca)
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
• 可以依照先前我們針對 training data 所做的 k-means 集群分析
• 將 testing data 用同樣的集群方式分群

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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.

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

• stocksTrain1
• 觀察分群後的應變數平均值表現

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
• ReturnFeb
• ReturnMar
• ReturnJune
• ReturnAug
• ReturnOct
• 可以觀察出在不同的集群中
• 每個變數所帶來的影響強度有所差異、甚至完全相反

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
• 集群一之 ACC

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

• 0.55048
• 集群二之 ACC

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

• 0.64583
• 集群三之 ACC

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:

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
• 將全部的集群合併 ACC 計算出來

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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