Homework 1 - 1940802 - LIU Ruiqi

学号：1940802 姓名：刘睿奇

首先导入基本的模块。

library(fBasics)
library(TTR)
library(xts)
library(zoo)
library(quantmod)

1、 1）试下载IBM公司股票每日的开盘价、收盘价、最高价、最低价；

getSymbols('IBM')

[1] "IBM"

head(IBM)

           IBM.Open IBM.High IBM.Low IBM.Close IBM.Volume IBM.Adjusted
2007-01-03    97.18    98.40   96.26     97.27    9196800     66.10009
2007-01-04    97.25    98.79   96.88     98.31   10524500     66.80686
2007-01-05    97.60    97.95   96.91     97.42    7221300     66.20204
2007-01-08    98.50    99.50   98.35     98.90   10340000     67.20779
2007-01-09    99.08   100.33   99.07    100.07   11108200     68.00288
2007-01-10    98.50    99.05   97.93     98.89    8744800     67.20101

chartSeries(IBM)

IBM_ohlc <- IBM[,1:4]
head(IBM_ohlc)

           IBM.Open IBM.High IBM.Low IBM.Close
2007-01-03    97.18    98.40   96.26     97.27
2007-01-04    97.25    98.79   96.88     98.31
2007-01-05    97.60    97.95   96.91     97.42
2007-01-08    98.50    99.50   98.35     98.90
2007-01-09    99.08   100.33   99.07    100.07
2007-01-10    98.50    99.05   97.93     98.89

2）提取其中的收盘价，并画出其时间序列图；

IBM_c <- IBM[,4]
head(IBM_c)

           IBM.Close
2007-01-03     97.27
2007-01-04     98.31
2007-01-05     97.42
2007-01-08     98.90
2007-01-09    100.07
2007-01-10     98.89

chartSeries(IBM_c)

3）计算收盘价的每日简单收益率、对数收益率；

IBM.srr <- 1 / (IBM_c$IBM.Close / diff(IBM_c$IBM.Close) - 1)
IBM.lrr <- diff(log(IBM_c$IBM.Close))

head(IBM.srr)

              IBM.Close
2007-01-03           NA
2007-01-04  0.010691899
2007-01-05 -0.009052996
2007-01-08  0.015191994
2007-01-09  0.011830111
2007-01-10 -0.011791756

head(IBM.lrr)

              IBM.Close
2007-01-03           NA
2007-01-04  0.010635145
2007-01-05 -0.009094223
2007-01-08  0.015077751
2007-01-09  0.011760682
2007-01-10 -0.011861830

4）画出收盘价对数收益率的时间序列图。

#chartSeries(IBM.srr)
chartSeries(IBM.lrr)

2. 考虑从2001年9月1日到2011年9月30日美国运通公司（AXP）、CRSP价值权重指数（VW）、CRSO的等权重指数（EW）以及S&P综合指数的日简单收益率。收益率中包含有支付的股息。数据来自d-axp3dx-0111.txt.

读取数据：

foo <- read.table(file = 'C:\\Users\\Ensom\\Documents\\Rscript\\d-axp3dx-0111.txt', header = T, row.names = 1)

head(foo)

1. 计算每个收益率序列的样本均值、标准差、偏度、超额峰度、最大值和最小值；

MEAN <- apply(foo, MARGIN = 2, FUN = mean)
SD <- apply(foo, MARGIN = 2, FUN = sd)
SKEW <- apply(foo, MARGIN = 2, FUN = skewness)
KURT <- apply(foo, MARGIN = 2, FUN = kurtosis)
EX_KURT <- KURT - 3
MAX <- apply(foo, MARGIN = 2, FUN = max)
MIN <- apply(foo, MARGIN = 2, FUN = min)
rbind(MEAN, SD, SKEW, EX_KURT, MAX, MIN)

                  axp            vw            ew            sp
MEAN     0.0005339487  0.0002240615  0.0006258233  0.0000942284
SD       0.0263684068  0.0136518353  0.0120803692  0.0137791170
SKEW     0.4597733639 -0.0983183709 -0.2474101701  0.0081519428
EX_KURT  6.5920528473  4.9821338456  5.1084277351  5.5326673290
MAX      0.2064850000  0.1148890000  0.1074220000  0.1158000000
MIN     -0.1759490000 -0.0897620000 -0.0782400000 -0.0903500000

2. 把简单收益率转换成对数收益率，计算每个对数收益率的样本均值、标准差、偏度、超额峰度、最大值和最小值；

foo_l <- log(foo + 1)  #将简单收益率转换为对数收益率，并存入foo_l

MEAN <- apply(foo_l, MARGIN = 2, FUN = mean)
SD <- apply(foo_l, MARGIN = 2, FUN = sd)
SKEW <- apply(foo_l, MARGIN = 2, FUN = skewness)
KURT <- apply(foo_l, MARGIN = 2, FUN = kurtosis)  #峰度
EX_KURT <- KURT - 3                             #超额峰度（峰度-3）
MAX <- apply(foo_l, MARGIN = 2, FUN = max)
MIN <- apply(foo_l, MARGIN = 2, FUN = min)

rbind(MEAN, SD, SKEW, EX_KURT, MAX, MIN)

                  axp            vw            ew            sp
MEAN     0.0001880014  0.0001307504  0.0005525758 -7.486380e-07
SD       0.0262943876  0.0136703590  0.0120995505  1.379041e-02
SKEW     0.0209917876 -0.3003522828 -0.4273150537 -2.063569e-01
EX_KURT  6.0204992951  4.8800820200  5.0177118109  5.322826e+00
MAX      0.1877111733  0.1087548484  0.1020347916  1.095716e-01
MIN     -0.1935228578 -0.0940491752 -0.0814703930 -9.469537e-02

3、几何（X）和代数（Y）考试分数如下表。 X 79 75 77 73 78 81 76 72 70 Y 80 82 76 77 84 81 72 70 75

先在Excel中将原表的行列转置如下： X Y 79 80 75 82 77 76 73 77 78 84 81 81 76 72 72 70 70 75

读取数据：

da = read.table("clipboard",header = T)  #需要先复制上面的表格
print(da)

（1）请建立由X估计Y的回归方程；

plot(da$X, da$Y)

z <- lm(da$Y~da$X)
print(z)

Call:
lm(formula = da$Y ~ da$X)

Coefficients:
(Intercept)         da$X  
    18.1722       0.7833  

故，回归方程为 Y == 18.1722 + 0.7833 * X

（2）检验回归方程的显著性；

summary(z)

Call:
lm(formula = da$Y ~ da$X)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.7056 -2.4889 -0.0556  1.9944  5.0778 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  18.1722    30.6805   0.592   0.5723  
da$X          0.7833     0.4051   1.934   0.0944 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.051 on 7 degrees of freedom
Multiple R-squared:  0.3482,    Adjusted R-squared:  0.2551 
F-statistic:  3.74 on 1 and 7 DF,  p-value: 0.09439

X后有“.”标记，可知在0.95的置信度下是显著的

（3）几何得分为71分的学生，其代数分数为几何？

x1 = 71
y1 = 18.1722 + 0.7833 * x1
print(y1)

[1] 73.7865

故，估计其代数分数为74分

4、 1）试画出正态分布N(5,100)的密度函数图和分布函数图；

curve(dnorm(x, 5, sqrt(100)), xlim = c(5-30, 5+30), col = 'darkblue', lwd = 2)  #密度函数图

curve(pnorm(x, 5, sqrt(100)), xlim = c(5-40, 5+40), col = 'darkred', lwd = 2)  #分布函数图

2）试求出服从正态分布N(5,100)的随机数1000个，并画出直方图，上面套上正态分布拟合图和非参数核函数拟合图（用不同颜色）；

N <- 1000  #随机数容量
rand_set <- rnorm(N, 5, 10)  #产生随机数

hist(rand_set, xlim = c(min(rand_set)-5, max(rand_set))+5, probability = T, nclass = max(rand_set) - min(rand_set) + 1, col = 'lightblue', main = 'N(5,100)')  #直方图

se <- seq(-40,40,length.out = 1000)
ye <- dnorm(se, 5, 10)
lines(se, ye, col="darkblue", lwd = 2)  #正态分布拟合图

lines(density(rand_set), col = 'red', lwd = 2)  #核函数拟合图

3）对随机数进行正态性检验，并给出结论。

normalTest(rand_set)

Title:
 Shapiro - Wilk Normality Test

Test Results:
  STATISTIC:
    W: 0.9984
  P VALUE:
    0.517 

Description:
 Thu Jun 18 23:46:56 2020 by user: Ensom

W = 0.9984 接近1，P VALUE=0.517 > 0.05，不能拒绝随机数服从正态分布的假设。

此外还可以使用QQ图检验：

qqnorm(rand_set, main="Normality Check via QQ Plot")   #QQ图
qqline(rand_set, col='red')     

可见，随机数点基本都在qq线附近，比较符合正态分布。