SESSION 6 - Business Application Lab

Question :  
(a) Create and plot the log of returns data for a 13 month period and interpret the stationarity.
(b) Calculate historical volatility of the same returns.
(c) Carry out adf test for returns and interpret the results.


> z<-read.csv(file.choose(),header=T)
> head(z)

         Date    Open    High     Low   Close Shares.Traded Turnover..Rs..Cr.
1 01-Dec-2011 1937.80 1973.40 1930.25 1945.50      91246016            791.90
2 02-Dec-2011 1947.90 1981.05 1936.55 1977.85      90348679            781.31
3 05-Dec-2011 1975.55 1986.40 1968.60 1975.85      88981133            691.75
4 07-Dec-2011 1978.10 2001.85 1973.50 1978.45      99171329            872.43
5 08-Dec-2011 1976.25 1976.25 1928.65 1934.50     104371626            820.01
6 09-Dec-2011 1920.80 1932.10 1901.90 1919.00      90902176            659.08

> closeprice<-z$Close
> closeprice.ts<-ts(closeprice, frequency=252)
> returns<-(closeprice.ts-lag(closeprice.ts,k=-1))/lag(closeprice.ts,k=-1)
> manipulate<-scale(returns)+10
> logreturns<-log(manipulate)
> logreturns
> acf(logreturns)

We can see from the graph that almost all the errors lie between the two dotted lines i.e. between 95% confidence interval. Thus, we can conclude that the time interval is stationary.


> T=(252)^0.5
> historicalvolatility<-sd(logreturns)*T

Warning message:
sd(<matrix>) is deprecated.
 Use apply(*, 2, sd) instead.

> historicalvolatility
[1] 1.620009

> adf.test(logreturns)

        Augmented Dickey-Fuller Test

data:  logreturns 

Dickey-Fuller = -5.2022, Lag order = 6, p-value = 0.01

alternative hypothesis: stationary 

Warning message:

In adf.test(logreturns) : p-value smaller than printed p-value

Since p value is less that 0.5, we cannot accept the null hypothesis and so can conclude with 95% confidence that the time series is stationary and further analysis can be done 

Session 5 - Business Application Lab

ASSIGNMENT 1 :

Converting the data in the time series format and then calculating the returns from it.
(Data taken - NSE MIDCAP 50 from July 31st to Dec 31st, 2012)

COMMANDS:

> z<-read.csv(file.choose(),header=T)
> Close<-z$Close
> Close

 [1] 1994.30 1993.30 2006.55 1990.00 2002.30 2033.70 2042.00 2046.85 2054.05
[10] 2057.85 2033.65 2063.55 2116.10 2155.80 2134.05 2191.65 2198.40 2203.40
[19] 2210.90 2216.90 2252.45 2269.65 2286.75 2298.00 2275.55 2255.90 2271.65
[28] 2238.95 2287.35 2286.05 2287.05 2254.00 2251.40 2281.30 2258.20 2258.80
[37] 2239.60 2228.80 2199.00 2188.10 2162.00 2174.40 2207.10 2226.45 2208.50
[46] 2214.35 2238.80 2242.30 2219.80 2229.75 2233.80 2233.70 2200.05 2178.80
[55] 2152.10 2168.00 2176.80 2176.10 2195.60 2226.20 2248.25 2288.45 2315.55
[64] 2332.05 2343.85 2369.60 2360.10 2377.95 2350.85 2361.85 2323.15 2347.85
[73] 2363.65 2388.25 2391.65 2379.35 2325.35 2327.45 2345.10 2334.00 2357.25
[82] 2369.50

> Close.ts<-ts(Close)
> Close.ts<-ts(Close,deltat=1/252)
> z1<-ts(data=Close.ts[10:95],frequency=1,deltat=1/252)
> z1.ts<-ts(z1)
> z1.ts

Time Series:
Start = 1
End = 86
Frequency = 1
 [1] 2057.85 2033.65 2063.55 2116.10 2155.80 2134.05 2191.65 2198.40 2203.40
[10] 2210.90 2216.90 2252.45 2269.65 2286.75 2298.00 2275.55 2255.90 2271.65
[19] 2238.95 2287.35 2286.05 2287.05 2254.00 2251.40 2281.30 2258.20 2258.80
[28] 2239.60 2228.80 2199.00 2188.10 2162.00 2174.40 2207.10 2226.45 2208.50
[37] 2214.35 2238.80 2242.30 2219.80 2229.75 2233.80 2233.70 2200.05 2178.80
[46] 2152.10 2168.00 2176.80 2176.10 2195.60 2226.20 2248.25 2288.45 2315.55
[55] 2332.05 2343.85 2369.60 2360.10 2377.95 2350.85 2361.85 2323.15 2347.85
[64] 2363.65 2388.25 2391.65 2379.35 2325.35 2327.45 2345.10 2334.00 2357.25
[73] 2369.50      NA      NA      NA      NA      NA      NA      NA      NA
[82]      NA      NA      NA      NA      NA

> z1.diff<-diff(z1)
> z2<-lag(z1.ts,K=-1)
> Returns<-z1.diff/z2
> plot(Returns,main="10th to 95th day returns")
> z3<-cbind(z1.ts,z1.diff,Returns)
> plot(z3,main="Data from 10th to 95th day, Difference, Returns")

ASSIGNMENT 2 :

Do logit analysis for 700 data points and then predict for 150 data points.


COMMANDS:

> z<-read.csv(file.choose(),header=T)
> z1<-z[1:700,1:9]
> head(z1)

  age ed employ address income debtinc creddebt othdebt default
1  41  3     17      12    176     9.3    11.36    5.01       1
2  27  1     10       6     31    17.3     1.36    4.00       0
3  40  1     15      14     55     5.5     0.86    2.17       0
4  41  1     15      14    120     2.9     2.66    0.82       0
5  24  2      2       0     28    17.3     1.79    3.06       1
6  41  2      5       5     25    10.2     0.39    2.16       0

> z1$ed<-factor(z1$ed)

> z1.est<-glm(default ~ age + ed + employ + address + income + debtinc + creddebt + othdebt, data=z1, family = "binomial")
> summary(z1.est)

Call:
glm(formula = default ~ age + ed + employ + address + income +
    debtinc + creddebt + othdebt, family = "binomial", data = z1)

Deviance Residuals:
    Min       1Q   Median       3Q      Max
-2.4322  -0.6463  -0.2899   0.2807   3.0255

Coefficients:
             Estimate Std. Error z value Pr(>|z|)  
(Intercept) -1.589302   0.605324  -2.626  0.00865 **
age          0.035514   0.017588   2.019  0.04346 *
ed2          0.307623   0.251629   1.223  0.22151  
ed3          0.352448   0.339937   1.037  0.29983  
ed4         -0.085359   0.472938  -0.180  0.85677  
ed5          0.874942   1.293734   0.676  0.49886  
employ      -0.260737   0.033410  -7.804 5.99e-15 ***
address     -0.105426   0.023264  -4.532 5.85e-06 ***
income      -0.007855   0.007782  -1.009  0.31282  
debtinc      0.070551   0.030598   2.306  0.02113 *
creddebt     0.625177   0.112940   5.535 3.10e-08 ***
othdebt      0.053470   0.078464   0.681  0.49558  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 804.36  on 699  degrees of freedom
Residual deviance: 549.56  on 688  degrees of freedom
AIC: 573.56

Number of Fisher Scoring iterations: 6

> forecast<-z[701:850,1:8]
> forecast$ed<-factor(forecast$ed)
> forecast$probability<-predict(z1.est, newdata=forecast, type="response")
> head(forecast)

    age ed employ address income debtinc creddebt othdebt probability
701  36  1     16      13     32    10.9     0.54    2.94  0.00783975
702  50  1      6      27     21    12.9     1.32    1.39  0.07044926
703  40  1      9       9     33    17.0     4.88    0.73  0.63780431
704  31  1      5       7     23     2.0     0.05    0.41  0.07471587
705  29  1      4       0     24     7.8     0.87    1.01  0.34464735
706  25  2      1       3     14     9.9     0.23    1.15  0.45584645