Thursday, 7 February 2013

ASSIGNMENT1: 1.       Find returns of NSE data> 6months, having selected the 10th data pt as start and 95th point as end point. Then plot the returns as well.



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

> head(z)
       Date    Open    High     Low   Close Shares.Traded Turnover..Rs..Cr.
1 02-Jan-12 4640.20 4645.95 4588.05 4636.75     108460668           3590.96
2 03-Jan-12 4675.80 4773.10 4675.80 4765.30     146621115           5021.29
3 04-Jan-12 4774.95 4782.85 4728.85 4749.65     165938849           5661.16
4 05-Jan-12 4749.00 4779.80 4730.15 4749.95     177862936           5873.79
5 06-Jan-12 4724.15 4794.90 4686.85 4754.10     176057282           5234.69
6 07-Jan-12 4755.60 4759.40 4743.05 4746.90      18783880            414.88
> close<-z$Close[10:95]
> close.ts<-ts(close,deltat=1/252)
> close.ts
Time Series:
Start = c(1, 1)
End = c(1, 86)
Frequency = 252
 [1] 4831.25 4866.00 4873.90 4967.30 4955.80 5018.40 5048.60 5046.25 5127.35 5158.30 5204.70 5087.30 5199.25 5235.70 5269.90 5325.85 5361.65 5335.15 5368.15 5412.35
[21] 5381.60 5390.20 5416.05 5531.95 5521.95 5564.30 5607.15 5505.35 5483.30 5429.30 5281.20 5375.50 5385.20 5339.75 5359.35 5359.40 5280.35 5222.40 5220.45 5333.55
[41] 5359.55 5429.50 5463.90 5380.50 5317.90 5257.05 5274.85 5364.95 5228.45 5278.20 5184.25 5243.15 5194.75 5178.85 5295.55 5317.90 5358.50 5322.90 5234.40 5243.60
[61] 5226.85 5276.85 5207.45 5226.20 5289.70 5300.00 5332.40 5290.85 5200.60 5222.65 5202.00 5189.00 5190.60 5209.00 5248.15 5239.15 5188.40 5086.85 5114.15 4999.95
[81] 4974.80 4965.70 4928.90 4907.80 4942.80 4858.25
> summary(close.ts)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   4831    5185    5246    5238    5359    5607
> z.diff<-diff(close.ts)
> returns<-z.diff/lag(close.ts, k=-1)
> returns
Time Series:
Start = c(1, 2)
End = c(1, 86)
Frequency = 252
 [1]  7.192755e-03  1.623510e-03  1.916330e-02 -2.315141e-03  1.263166e-02  6.017854e-03 -4.654756e-04  1.607134e-02  6.036257e-03  8.995212e-03 -2.255654e-02
[12]  2.200578e-02  7.010627e-03  6.532078e-03  1.061690e-02  6.721932e-03 -4.942508e-03  6.185393e-03  8.233749e-03 -5.681451e-03  1.598038e-03  4.795740e-03
[23]  2.139936e-02 -1.807681e-03  7.669392e-03  7.700879e-03 -1.815539e-02 -4.005195e-03 -9.848084e-03 -2.727792e-02  1.785579e-02  1.804483e-03 -8.439798e-03
[34]  3.670584e-03  9.329490e-06 -1.474979e-02 -1.097465e-02 -3.733915e-04  2.166480e-02  4.874802e-03  1.305147e-02  6.335758e-03 -1.526382e-02 -1.163461e-02
[45] -1.144249e-02  3.385929e-03  1.708105e-02 -2.544292e-02  9.515248e-03 -1.779963e-02  1.136133e-02 -9.231092e-03 -3.060783e-03  2.253396e-02  4.220525e-03
[56]  7.634593e-03 -6.643650e-03 -1.662628e-02  1.757604e-03 -3.194370e-03  9.565991e-03 -1.315179e-02  3.600611e-03  1.215032e-02  1.947180e-03  6.113208e-03
[67] -7.791989e-03 -1.705775e-02  4.239895e-03 -3.953931e-03 -2.499039e-03  3.083446e-04  3.544870e-03  7.515838e-03 -1.714890e-03 -9.686686e-03 -1.957251e-02
[78]  5.366779e-03 -2.233020e-02 -5.030050e-03 -1.829219e-03 -7.410838e-03 -4.280874e-03  7.131505e-03 -1.710569e-02
> plot(returns)
>




ASSIGNMENT2:

> z<-read.csv(file.choose(),header=T)
> head(z)
  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
> data<- z[1:700,1:9]
> sapply(data,mean)
       age         ed     employ    address     income    debtinc   creddebt    othdebt    default 
34.8600000  1.7228571  8.3885714  8.2785714 45.6014286 10.2605714  1.5534571  3.0582286  0.2614286 
> data$ed<-factor(data$ed)
> logit.est<-glm(default~age+employ+address+income+debtinc+creddebt+othdebt,data=data,family="binomial")
> summary(logit.est)

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

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.3659  -0.6516  -0.2882   0.2625   2.9757  

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.376573   0.571560  -2.408   0.0160 *  
age          0.033712   0.017342   1.944   0.0519 .  
employ      -0.265086   0.031999  -8.284  < 2e-16 ***
address     -0.103960   0.023192  -4.483 7.38e-06 ***
income      -0.007566   0.008095  -0.935   0.3500    
debtinc      0.065099   0.030621   2.126   0.0335 *  
creddebt     0.628475   0.113759   5.525 3.30e-08 ***
othdebt      0.070761   0.077682   0.911   0.3623    
---
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: 552.20  on 692  degrees of freedom
AIC: 568.2

Number of Fisher Scoring iterations: 6

> confint.default(logit.est)
                    2.5 %      97.5 %
(Intercept) -2.4968094058 -0.25633710
age         -0.0002768342  0.06770153
employ      -0.3278025120 -0.20236959
address     -0.1494167558 -0.05850405
income      -0.0234310297  0.00829931
debtinc      0.0050836856  0.12511410
creddebt     0.4055112145  0.85143903
othdebt     -0.0814939858  0.22301505
> logit.eg2<-with(z[701:850,1:8],data.frame(age=mean(age),employ=mean(employ),address=mean(address),income=mean(income),debtinc=mean(debtinc),creddebt=mean(creddebt),othdebt=mean(othdebt),ed=factor(1:3)))
> logit.eg2$prob<-predict(logit.est,newdata=logit.eg2,type="response")
> head(logit.eg2)
    age   employ  address   income  debtinc creddebt  othdebt ed      prob
1 35.82 9.393333 8.806667 51.68667 9.756667   1.6852 3.174933  1 0.1143839
2 35.82 9.393333 8.806667 51.68667 9.756667   1.6852 3.174933  2 0.1143839
3 35.82 9.393333 8.806667 51.68667 9.756667   1.6852 3.174933  3 0.1143839




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