金融時間序列分析 第一次作業(yè)

1、1.1a.用百分數(shù)表示簡單月收益率（略）表1.三支股票的簡單日收益率（百分比）的描述性統(tǒng)計AmericanExpressCaterpillarStarbucksMean0.09718%0.08473%0.145%StardDeviation2.250181%2.153127%3.049927%Skewness0.0883880670.1237930240.039442047ExcessKurtosis2.3916842.2356725.

2、319044MinimumValue13.596%12.162%28.286%MaximumValue12.770%10.845%14.706%b.簡單收益率換成對數(shù)收益率（略）c.把對數(shù)收益率用百分比表示出來（略）表2.三支股票的對數(shù)日收益率（百分比）的描述性統(tǒng)計AmericanExpressCaterpillarStarbucksMean0.07188%0.06157%0.09839%StardDeviation2.248449%2

3、.150728%3.055389%Skewness0.0616688930.0153422160.354252488ExcessKurtosis2.5135872.3905807.884276MinimumValue14.61362%12.96760%33.24842%MaximumValue12.01802%10.29626%13.72021%d.對數(shù)收益率零均值檢驗（）=0.051對于AmericanExpressdata:dail

4、ylogre$V2t=1.6044df=2518pvalue=0.1087alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:0.015969930.15972365sampleestimates:meanofx0.07187686故接受原假設(shè)2對于Caterpillardata:dailylogre$V3t=1.4369df=2518pvalue

5、=0.1509alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:0.022455580.14560208sampleestimates:meanofx0.06157325故接受原假設(shè)3對于Starbucks2對于VWdata:dailylogre$V3t=4.3846df=347pvalue=1.542e05alternativehypothes

6、is:truemeanisnotequalto095percentconfidenceinterval:0.59553591.5644506sampleestimates:meanofx1.079993故拒絕原假設(shè)3對于EWdata:dailylogre$V4t=4.6899df=347pvalue=3.937e06alternativehypothesis:truemeanisnotequalto095percentconfidenc

7、einterval:0.82505212.0169110sampleestimates:meanofx1.420982故拒絕原假設(shè)4對于SPdata:dailylogre$V5t=3.3417df=347pvalue=0.0009235alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:0.3293961.271850sampleestimates