session 6- assignment 6

Assignment 6

Assignment 1:

Create log(returndata). There are two ways to do
a>[Log St -Log S(t-1)]/Log S(t-1)
b>Log[(St-S(t-1))/S(t-1)]

Take the log returns and calculate the historical data

Commands:
> stockprice=read.csv(file.choose(), header=T)
> head(stockprice)
> closingprice<-stockprice[,5]
> closingprice<-ts(closingprice,frequency=252)
> closingprice.ts<-ts(closingprice,frequency=252)
> lagtable<-cbind(closingprice.ts,lag(closingprice.ts,k=-1),(closingprice.ts-lag(closingprice.ts,k=-1)))
> head(lagtable)
> returns<-(closingprice.ts-lag(closingprice.ts,k=-1))/lag(closingprice.ts,k=-1)
> returns
> LogReturn1<-log(closingprice.ts)-log(lag(closingprice.ts,k=-1))
> LogReturn<-LogReturn1/log(lag(closingprice.ts,k=-1))
> LogReturn
> T<-252^0.5
> historicalvolatility<-sd(returns)*T
>  historicalvolatility
> acf(LogReturn)
> adf.test(returns)

        Augmented Dickey-Fuller Test

data:  returns 
Dickey-Fuller = -5.6265, Lag order = 6, p-value = 0.01
alternative hypothesis: stationary 

Warning message:
In adf.test(returns) : p-value smaller than printed p-value

PIC 1:

Assignment 2:

Create Acf plot and interpret the output of above log returns data. Do ADF test and interpret.

Commands:

> acf(LogReturn)
> adf.test(LogReturn)

        Augmented Dickey-Fuller Test

data:  LogReturn
Dickey-Fuller = -5.6217, Lag order = 6, p-value = 0.01
alternative hypothesis: stationary

Warning message:
In adf.test(LogReturn) : p-value smaller than printed p-value

PLOT

Conclusion:
From ADF test, P-value< alpha i.e 0.01<0.05. We reject the null hypothesis and accept the alternate hypothesis. ACF plots the data within the confidence interval. The above test justifies the stationary property of  time series.