CFA错题笔记11Quantitative methodsWord格式文档下载.docx
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A.10.3%
B.2.1和1.449
2.2.TheBinomialDistribution
Equation
(1)
p
(
x
)
=
P
X
n
x
1
−
n
!
一个公司在过去10年中,年利润增长年数7年,下降3年,你决定将这些数据模型化,以二项式分布方法
A.在未来10年中有5年增长的概率是?
根据公式,10!
/(10-5)!
5!
*0.7^5*(1-0.7)^(10-5)=252*0.1681*0.0024=0.10
B.计算接下来10年的方差和标准差
根据公式,方差=10*0.7(1-0.7)=2.1,标准差=1.449
10.Statetheapproximateprobabilitythatanormalrandomvariablewillfallwithinthefollowingintervals:
A.Meanplusorminusonestandarddeviation.
B.Meanplusorminustwostandarddeviations.
C.Meanplusorminusthreestandarddeviations.
A.68%
B.95%
C.99%
3.2.TheNormalDistribution
Havingestablishedthatthenormaldistributionistheappropriatemodelforavariableofinterest,wecanuseittomakethefollowingprobabilitystatements:
∙Approximately50percentofallobservationsfallintheintervalμ±
(2/3)σ.
∙Approximately68percentofallobservationsfallintheintervalμ±
σ.
∙Approximately95percentofallobservationsfallintheintervalμ±
2σ.
∙Approximately99percentofallobservationsfallintheintervalμ±
3σ.
死记吧。
16.Thetotalnumberofparametersthatfullycharacterizesamultivariatenormaldistributionforthereturnsontwostocksis:
A.3.
B.4.
C.5.
C
Whenwehaveagroupofassets,wecanmodelthedistributionofreturnsoneachassetindividually,orthedistributionofreturnsontheassetsasagroup.“Asagroup”meansthatwetakeaccountofallthestatisticalinterrelationshipsamongthereturnseries.Onemodelthathasoftenbeenusedforsecurityreturnsisthemultivariatenormaldistribution.Amultivariatenormaldistributionforthereturnson
n
stocksiscompletelydefinedbythreelistsofparameters:
∙thelistofthemeanreturnsontheindividualsecurities(n
meansintotal);
∙thelistofthesecurities’variancesofreturn(n
variancesintotal);
and
∙thelistofallthedistinctpairwisereturncorrelations:
n(n
−1)/2distinctcorrelationsintotal.
在计算两只符合正态分布的股票参数时,一共要计算多少个?
根据考点可得,平均值和方差根据股票数来,有多少只股票就分别有多少个平均值和方差,相关系数即是2*1/2=1,一共是5个。
17.Aclienthasaportfolioofcommonstocksandfixed-incomeinstrumentswithacurrentvalueof£
1,350,000.Sheintendstoliquidate£
50,000fromtheportfolioattheendoftheyeartopurchaseapartnershipshareinabusiness.Furthermore,theclientwouldliketobeabletowithdrawthe£
50,000withoutreducingtheinitialcapitalof£
1,350,000.Thefollowingtableshowsfouralternativeassetallocations.
MeanandStandardDeviationforFourAllocations(inPercent)
A
D
Expectedannualreturn
16
12
10
9
Standarddeviationofreturn
24
17
11
Whatistheprobabilitythatthereturnonthesafety-firstoptimalportfoliowillbelessthantheshortfalllevel,
RL?
30%
3.3.ApplicationsoftheNormalDistribution
SFRatio=[E(RP)–
RL]/σP
1.Calculateeachportfolio’sSFRatio.
2.ChoosetheportfoliowiththehighestSFRatio.
一个投资人有个投资组合,目前价值1350000,他想取出50000,但他不想因为取出而降低组合价值,下面是可选择的资产分配方案,在安全优先原则下,求小于亏空水平的概率
RL=50000/1350000=3.7%
根据公式,求出安全第一比率:
A.0.5125B.0.488235C.0.525D.0.4848
选择安全第一比率最大的C,也就刚好处于亏空水平临界点
即求P(R<
3.7)=N(-0.525)=1-N(0.525)=1-N(0.523)=1-0.7019=0.2981
N(0.523)=0.7019,根据查表可得(Z分布)。
PleaserefertoExhibit1forQuestions18and19
Exhibit1. Z-TableValues,
P(Z
≤
z)=
N(z)for
z
≥0
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
0.5359
0.1
0.5398
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
0.2
0.5793
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
0.3
0.6179
0.6217
0.6255
0.6293
0.6331
0.6368
0.6406
0.6443
0.6480
0.6517
0.4
0.6554
0.6591
0.6628
0.6664
0.6700
0.6736
0.6772
0.6808
0.6844
0.6879
0.5
0.6915
0.6950
0.6985
0.7019
0.7054
0.7088
0.7123
0.7157
0.7190
0.7224
18.Aportfoliohasanexpectedmeanreturnof8percentandstandarddeviationof14percent.Theprobabilitythatitsreturnfallsbetween8and11percentis
closest
to:
A.8.3%
B.14.8%.
C.58.3%.
Equation(4)
Z
=(X
–μ)/σ
一个投资组合平均值是8%,标准差是14%,求组合收益率在8%~11%的概率:
第一步,求出收益率在11%的概率
当收益率是11%,Z=(11-8)/14=0.2143,查表得知N(0.21)=0.583,即11%的概率
第二步,求出收益率在8%的概率
因为8%是平均值,所以概率是0.5
第三步,求出区间概率
0.583-0.5=0.083
24.Thecumulativedistributionfunctionforadiscreterandomvariableisshowninthefollowingtable.
X=x
CumulativeDistributionFunction
F(x)=
P(X
x)
0.15
2
0.25
3
0.50
4
0.60
5
0.95
6
1.00
Theprobabilitythat
X
willtakeonavalueofeither2or4is
A.0.20.
B.0.35.
C.0.85.
R102.
1.TheDiscreteUniformDistribution
离散随机变量的累积分布函数如下表所示,求X为2或4的概率:
这道题我会做错的原因是没有审题题目问的是2或4,我看成2~4,求了区间概率
P
(2)=0.25-0.15=0.1
P(4)=0.6-0.5=0.1
两者相加即可,即0.2
27.Astockispricedat$100.00andfollowsaone-periodbinomialprocesswithanupmovethatequals1.05andadownmovethatequals0.97.If1millionBernoullitrialsareconducted,andtheaverageterminalstockpriceis$102.00,theprobabilityofanupmove(p)is
A.0.375.
B.0.500.
C.0.625.
2.2.TheBinomialDistribution
股票价格目前是100,符合二叉树分布原理,上涨是1.05倍,下降是0.97倍,如果实行了1亿的伯努利试验后,股票平均价格是102,那么上涨的概率是多少?
设上涨的概率是p
102=p*100*1.05+(1-p)*100*0.97
求得p=0.625
R11samplingandestimation
1.PeterBiggswantstoknowhowgrowthmanagersperformedlastyear.Biggsassumesthatthepopulationcross-sectionalstandarddeviationofgrowthmanagerreturnsis6percentandthatthereturnsareindependentacrossmanagers.
HowlargearandomsampledoesBiggsneedifhewantsthestandarddeviationofthesamplemeanstobe1percent?
36
R11
4.3.SelectionofSampleSize
彼得·
逼格想知道去年的经理业绩,逼格猜想总体标准差是6%,如果标准误是1%,那么需要多少样本量?
根据公式标准误=标准差/根号n,得,1%=6%/根号n,求得n=36.
2.PetraMunziwantstoknowhowvaluemanagersperformedlastyear.Munziestimatesthatthepopulationcross-sectionalstandarddeviationofvaluemanagerreturnsis4percentandassumesthatthereturnsareindependentacrossmanagers.
Munziwantstobuilda95percentconfidenceintervalforthemeanreturn.HowlargearandomsampledoesMunzineedifshewantsthe95percentconfidenceintervaltohaveatotalwidthof1percent?
246
4.2.ConfidenceIntervalsforthePopulationMean
∙90percentconfidenceintervals:
Use
z0.05
=1.65
∙95percentconfidenceintervals:
z0.025
=1.96
∙99percentconfidenceintervals:
z0.005
=2.58
标准差=4%,置信区间95%,置信区间宽度1%,求样本数量
2*1.96*4%/根号n=1%,求得n=246(注意,计算器调成保留5位小数:
按2nd再按format)
3.Assumethattheequityriskpremiumisnormallydistributedwithapopulationmeanof6percentandapopulationstandarddeviationof18percent.Overthelastfouryears,equityreturns(relativetotherisk-freerate)haveaveraged−2.0percent.Youhavealargeclientwhoisveryupsetandclaimsthatresultsthispoorshould
neveroccur.Evaluateyourclient’sconcerns.
Whatistheprobabilityofa−2.0percentorloweraveragereturnoverafour-yearperiod?
18.67%
4.2.ConfidenceIntervalsforthePopulationMean
z
¯
μ
/
σ
√
假设股票风险溢价符合正太分布,总体均值是6%,总体标准差是18%过去的四年,股票的收益(相对于无风险收益来说)平均值是-2%,你有一个大客户很烦恼这个结果,并认为是不可能发生的,你分析下。
收益小于-2%的概率是多少?
因为符合正太分布,所以可以使用Z-分布方法求概率,根据公式求出Z,注意这里的Z和R103.2正太分布的Z公式区别(18题,分母没有根号n,是因为只有一个n=1),这里的分母多一个根号n,是因为n>
1.
7.Tenanalystshavegiventhefollowingfiscalyearearningsforecastsforastock:
Forecast(Xi)
NumberofAnalysts(ni)
1.40
1.43
1.44
1.45
1.47
1.48
1.50
Becausethesampleisasmallfractionofthenumberofanalystswhofollowthisstock,
assumethatwecanignorethefinitepopulationcorrectionfactor.Assumethattheanalystforecastsarenormallydistributed.
Providea95percentconfidenceintervalforthepopulationmeanoftheforecasts.
1.43~1.47
10个分析师给出股票收益预测:
由于样本较小,假设我们暂时忽略总体相关系数的因素,并且样本分析符合正态分布,构建一个预测总体均值的95%置信区间
我做错的原因是我用了Z-分布,没有理解清楚题目的意思,由题目可知,该题目是不知道总体方差的,因此使用t-分布,查表得知t0.025=2.262
19.Asamplemeaniscomputedfromapopulationwithavari