MATLAB金融计算试题0001.docx
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MATLAB金融计算试题0001.docx
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MATLAB金融计算试题0001
MATLAB金融计算试题(2014级研究生用)
(上机操作使用)
、利率期限结构(20分)
已知国债面值是100美元,各期收益率为
国债品种
票息
到期日
当期收益
3个月
17-Apr-2013
1.15
6个月
17-Jul-2013
1.18
2年
1.75
31-Dec-2014
1.68
5年
3.00
15-Nov-2017
2.97
10年
4.00
15-Nov-2022
4.01
30年
5.375
15-Feb-2041
4.92
试分析其利率期限结构。
MATLAB命令:
bonds=[datenum('04/17/2013')0100;
datenum('07/17/2013')0100;
datenum('12/31/2014')0.0175100;
datenum('11/15/2017')0.03100;
datenum('11/15/2022')0.04100;
datenum('02/15/2041')0.0537100];
yield=[0.01150.01180.01680.02970.04010.0492]';settle=datenum('01/17/2013');%结算日
[zerorates,curvedates]=zbtyield(bonds,yield,settle)datestr(curvedates)
plot(zerorates)运行结果:
zerorates=
0.055(.CCC1
0.05--
0.045--
0.04-/-
0.035・■
0.03・■
0.025・■
0.02「-
0.015--
0.01!
1111[c[J1
11.522.533.544.555.56
0.0115
0.0118
0.0168
0.0302
0.0418
0.0550
curvedates=
735341
735432
735964
737014
738840
745507
ans=17-Apr-2013
17-Jul-2013
31-Dec-2014
15-Nov-2017
15-Nov-2022
15-Feb-2041
二、期权定价(30分)
若股票现在价格为$50,期权执行价格为$52,无风险利率为0.1,股票波动标准差为0.4,期权的到期日为6个月,且若这一卖权在3.5月时有一次股息支付$2。
(1)使用Black-Scholes定价公式计算欧式卖权和买权的价值;
MATLAB命令:
price=50;
strike=52;
rate=0.1;
time=6/12;
volatility=0.4;
[callprice,putprice]=blsprice(price,strike,rate,time,volatility)
运行结果:
callprice=
5.8651
putprice=
5.3290
(2)利用二项式期权定价(二叉树(CRR)模型定价数值解)计算看涨看跌期权价格
MATLAB命令:
price=50;
strike=52;
rate=0.1;
time=6/12;
increment=1/12;
volatility=0.4;
flag=0;
dividentrate=0;
divident=2;
exdiv=3.5;
[price,option]=binprice(price,strike,rate,time,increment,volatility,f
lag,dividentrate,divident,exdiv)
运行结果:
得岀二叉树每个交点处的资产价格和期权价值
price=
50.0000
55.8985
62.5172
69.9441
76.2699
85.6054
96.0836
0
44.7755
50.0326
55.9315
60.5420
67.9524
76.2699
0
0
40.1226
44.8084
48.0575
53.9398
60.5420
0
0
0
35.9790
38.1474
42.8167
48.0575
0
0
0
0
30.2809
33.9873
38.1474
0
0
0
0
0
26.9787
30.2809
0
0
0
0
0
0
24.0366
option=
6.7016
3.9308
1.7652
0.4598
0
0
0
0
9.6686
6.2275
3.1393
0.9412
0
0
0
0
13.3762
9.5132
5.4560
1.9263
0
0
0
0
17.5811
13.8526
9.1833
3.9425
0
0
0
0
21.7191
18.0127
13.8526
0
0
0
0
0
25.0213
21.7191
0
0
0
0
0
0
27.9634
由结果可知,option第一行第一列就是看跌期权价格,该期权价格为6.7016元。
MATLAB命令:
price=50;
strike=52;
rate=0.1;
time=6/12;
increment=1/12;
volatility=0.4;
flag=1;
dividentrate=0;
divident=2;
exdiv=3.5;
[price,option]=binprice(price,strike,rate,time,increment,volatility,flag,dividentrate,divident,exdiv)运行结果:
得出二叉树每个交点处的资产价格和期权价值.
price=
50.0000
55.8985
62.5172
69.9441
76.2699
85.6054
96.0836
0
44.7755
50.0326
55.9315
60.5420
67.9524
76.2699
0
0
40.1226
44.8084
48.0575
53.9398
60.5420
0
0
0
35.9790
38.1474
42.8167
48.0575
0
0
0
0
30.2809
33.9873
38.1474
0
0
0
0
0
26.9787
30.2809
0
0
0
0
0
0
24.0366
option=
4.9996
7.8792
12.0864
17.9441
25.1294
34.0369
44.0836
0
2.1193
3.6809
6.2599
10.3427
16.384024.2699
0
0
0.5473
1.0878
2.1622
4.2976
8.5420
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
由结果可知,option第一行第一列就是看涨期权价格,该期权价格为4.9996元。
(3)假设股票价格服从几何布朗运动,试用蒙特卡洛模拟方法计算该期权价格。
MATLAB命令:
s0=50;
K=52;
r=0.1;
T=0.5;
sigma=0.4;Nu=1000;
randn('seed',0);%定义随机数发生器种子是0,
%这样保证每次模拟的结果相同
nuT=(r-0.5*sigmaA2)*T
sit=sigma*sqrt(T)
discpayoff=exp(-r*T)*max(0,s0*exp(nuT+sit*randn(Nu,1))-K);%期权到期时的
现金流
[eucall,varprice,ci]=normfit(discpayoff)
运行结果:
nuT=
0.0100
sit=
0.2828
eucall=
6.1478
varprice=
10.2924
ci=
5.5091
6.7865
三、搜集数据并计算画图(50分)
按照自己的研究生学号后两位数,在锐思金融数据库中搜集4种股票信息,包括最高价、
最低价、收盘价和开盘价,数据个数2个月左右,建立数据表格。
要求使用MATLAB编程
解决以下问题:
(1)将4种股票的收盘价格转化为收益率,并画出收益率直方图
海虹控股
MATLAB命令:
TickSeries=[31.6332.1731.5830.7130.7730.9331.79
31.583233.9133.1234.9835.335.534.6535.4635.95
35.3937.6736.6436.7736.8536.5935.8135.1835.76
36.6638.3538.2638.3438.8541.2740.9940.742.28]';
RetSeries=tick2ret(TickSeries)
bar(RetSeries)
xlabel('天数');ylabel('收益率');title('海虹控股对数收益率直方图');
运行结果:
RetSeries=
0.0171
-0.0183
-0.0275
0.0020
0.0052
0.0278
-0.0066
0.0133
0.0597
-0.0233
0.0562
0.0091
0.0057
-0.0239
0.0234
0.0138
-0.0156
0.0644
-0.0273
0.0035
0.0022
-0.0071
-0.0213
-0.0176
0.01650.0252
0.0461
-0.0023
0.0021
0.0133
0.0623
-0.0068
-0.0071
0.0388
盛达矿业
MATLAB命令:
TickSeries=[13.0712.8813.1912.9812.7812.4912.73
12.5112.9713.0612.6813.1713.9314.3914.0814.34
14.1914.2413.7413.5713.813.7613.7613.5213.313.2813.4413.3713.2813.7413.9314.1613.9914.7314.7]';
RetSeries=tick2ret(TickSeries)
bar(RetSeries)
xlabel('天数’);ylabel(
'收益率');
title('盛达矿业对数收益率直方图');
运行结果:
RetSeries=
-0.0145
0.0241
0.06
-0.0159
0.05
-0.0154
0.04
-0.0227
0.0192
0.03
-0.0173
0.0368
0.0069
率益收
-0.0291
0.0386
0.0577
0.0330
-0.0215
-0.02
-0.03
-0.04
5
-0.01
盛达矿业对数收益率直方图
10152025
天数
3035
02
O
0
O
0
0.0185
-0.0105
0.0035
-0.0351
-0.0124
0.0169
-0.0029
0
-0.0174
-0.0163
-0.0015
0.0120
-0.0052
-0.0067
0.0346
0.0138
0.0165
-0.0120
0.0529
-0.0020
恒逸石化
MATLAB命令:
TickSeries=[9.439.148.998.678.68.428.498.48.53
8.97
8.61
8.91
9.11
9.129.069.149.048.798.7
8.78
8.83
9.37
9.47
9.39.559.899.699.649.58
9.52
9.88
10.22
10.3
10.4510.84]';
RetSeries=tick2ret(TickSeries)
bar(RetSeries)
xlabel('天数’);ylabel('收益率');title('恒逸石化对数收益率直方图’);运行结果:
RetSeries=
-0.0308
0.08
-0.0164
-0.0356
-0.0081
0.06
0.04
-0.0209
0.0083
-0.0106
率益收
02
0
0.0155
-0.02
0.0516
-0.0401-0.04
0.0348
0.0224
0.0011
-0.06
0
510
15202530
天数
恒逸石化对数收益率直方图
35
-0.0066
0.0088
-0.0109
-0.0277
-0.0102
0.0092
0.0057
0.0612
0.0107
-0.0180
0.0269
0.0356
-0.0202
-0.0052
-0.0062
-0.0063
0.0378
0.0344
0.0078
0.0146
0.0373
金宇车城
MATLAB命令:
TickSeries=[10.911.1711.3211.3211.2211.0811.2711.1911.3111.5211.2511.7812.0712.1112.1512.29
12.4512.8712.7712.6312.5612.7112.7112.512.15
12.2312.1212.4812.612.8712.913.3313.513.513.42]';
RetSeries=tick2ret(TickSeries)bar(RetSeries)xlabel('天数’);ylabel('收益率');
title('金宇车城对数收益率直方图运行结果:
RetSeries=
0.0248
0.0134
0
-0.0088
-0.0125
0.0171
-0.0071
0.0107
0.0186
-0.0234
0.0471
0.0246
0.0033
);
0.00330.0115
0.0130
0.0337
-0.0078
-0.0110
-0.0055
0.01190-0.0165-0.0280
0.0066-0.0090
0.0297
0.0096
0.0214
0.0023
0.0333
0.0128
0
-0.0059
(2)计算4种股票收盘价的协方差矩阵;
MATLAB命令:
A=[31.6313.079.4310.9
32.17
12.88
9.14
11.17
31.58
13.19
8.99
11.32
30.71
12.98
8.67
11.32
30.77
12.78
8.611.22
30.93
12.49
8.42
11.08
31.79
12.73
8.49
11.27
31.58
12.51
8.411.19
3212.978.5311.31
33.91
13.06
8.97
11.52
33.12
12.68
8.61
11.25
34.98
13.17
8.91
11.78
35.3
13.93
9.11
12.07
35.5
14.39
9.12
12.11
34.65
14.08
9.06
12.15
35.46
14.34
9.14
12.29
35.95
14.19
9.04
12.45
35.39
14.24
8.79
12.87
37.67
13.74
8.712.77
36.64
13.57
8.78
12.63
36.77
13.8
8.83
12.56
36.85
13.76
9.37
12.71
36.59
13.76
9.47
12.71
35.81
13.52
9.312.5
35.18
13.3
9.55
12.15
35.76
13.28
9.89
12.23
36.66
13.44
9.69
12.12
38.35
13.37
9.64
12.48
38.26
13.28
9.58
12.6
38.34
13.74
9.52
12.87
38.85
13.93
9.88
12.9
41.27
14.16
10.22
13.33
40.99
13.99
10.3
13.5
40.7
14.73
10.45
13.5
42.28
14.7
10.84
13.42]
cov(A)
运行结果:
ans=
10.0608
1.4751
1.5687
2.3059
1.4751
0.3711
0.2270
0.3857
1.5687
0.2270
0.3682
0.3326
2.3059
0.3857
0.3326
0.5837
(3)若给出这4种股票预期收益率分别为0.3、0.25、0.2和0.15,且购买权重分别0.35、0.25、0.25和0.15,求总资产的标准差和期望收益;
MATLAB命令:
ExpReturn=[0.3,0.25,0.2,0.15];
ExpCovariance=[10.06081.47511.56872.3059
1.47510.37110.22700.3857
1.56870.22700.36820.3326
2.30590.38570.33260.5837];
PortWts=[0.350.250.250.15];
[PortRisk,PortReturn]=portstats(ExpReturn,ExpCovariance,PortWts)
运行结果:
PortRisk=
1.4659
PortReturn=
0.2400
(4)求该资产组合有效前沿(有效前沿的个数选为5);
MATLAB命令:
ExpReturn=[0.30.250.20.15];
ExpCovariance=[10.06081.47511.56872.3059
1.4751
0.37110.22700.3857
1.5687
2.3059
0.22700.36820.3326
0.38570.33260.5837];
NumPorts=5;
[PortRink,PortReturn,PortWts]=frontcon(ExpReturn,ExpCovariance,NumPor
ts)
运行结果:
PortRink=
0.5462
0.5820
1.1729
2.1585
3.1719
PortReturn=
0.2247
0.2436
0.2624
0.2812
0.3000
PortWts=
-0.0000
0.4949
0.5051
0.0000
0
0.8712
0.1288
-0.0000
0.2475
0.7525
0
-0.0000
0.6237
0.3763
0
-0.0000
1.0000
0
0.0000
-0.0000
(5)无风险利率为0.35,借贷利率为0.5,投资者风险厌恶系数为3,求考虑无风险资
产及借贷情况下的最优资产配置。
MATLAB命令:
ExpReturn=[0.30.250.20.15];
ExpCovariance=[10.06081.47511.56872.3059
1.47510.37110.22700.3857
1.56870.22700.36820.3326
2.30590.38570.33260.5837];
RisklessRate=0.035;
BorrowRate=0.5;
RiskAversion=3;
[PortRisk,PortReturn,PortWts]=portopt(ExpReturn,ExpCovarianee)[RiskyRink,RiskyReturn,RiskyWts,RiskyFraction,0verallRick,OverallReturn]=portalloc(PortRisk,...
PortReturn,PortWts,RisklessRate,BorrowRate,RiskAversion)运行结果:
PortRisk=
0.5462
0.5534
0.5747
0.6084
0.9663
1.3864
1.8246
2.2704
2.7200
3.1719
PortReturn=
0.2247
0.2331
0.2415
0.2498
0.2582
0.2666
0.2749
0.2833
0.2916
0.3000
PortWts=
10000
0.4949
0.5051
0.0000
0
0.6621
0.3379
-0.0000
0.0000
0.8294
0.1706
-0.0000
0
0.9966
0.0034
-0.0000
0.1638
0.8362
0
-0.0000
0.3311
0.6689
0
0.0000
0.4983
0.5017
0
-0.0000
0.6655
0.3345
0
-0.0000
0.8328
0.1672
0
0.0000
1.0000
0
0.0000
-0.0000
RiskyRink=
0.5427
RiskyReturn=
0.2302
RiskyWts=
-0.00000.60400.3960-0.00
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