MATLAB金融计算试题

1、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-20131.156 个月17-Jul-20131.182年1.7531-Dec-20141.685年3.0015-Nov-20172.9710年4.0015-Nov-20224.0130年5.37515-Feb-20414.92试分析其利率期限结构。MATLAB 命令： bonds=datenum('04/17/2013') 0 100; datenum('07/17/2013'

2、;) 0 100; datenum('12/31/2014') 0.0175 100; datenum('11/15/2017') 0.03 100; datenum('11/15/2022') 0.04 100; datenum('02/15/2041') 0.0537 100;yield=0.0115 0.0118 0.0168 0.0297 0.0401 0.0492' settle=datenum('01/17/2013'); % 结算日 zerorates,curvedates=zbtyield(

3、bonds,yield,settle) datestr(curvedates) plot(zerorates) 运行结果： zerorates =0.0550.050.0450.040.0350.030.0250.020.0150.011 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60.01150.01180.01680.03020.04180.0550 curvedates =735341735432735964737014738840745507ans = 17-Apr-201317-Jul-201331-Dec-201415-Nov-201715-Nov-202215-Fe

4、b-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 =5.8651pu

5、tprice =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,flag,dividentrate,divident,exdiv）运行结果： 得出二叉树每个交点处的资产价格和

6、期权价值 .price =50.000055.898562.517269.944176.269985.605496.0836044.775550.032655.931560.542067.952476.26990040.122644.808448.057553.939860.542000035.979038.147442.816748.0575000030.280933.987338978730.280900000024.0366option =6.70163.93081.76520.459800009.66866.22753.13930.9412000013.376

7、29.51325.45601.9263000017.581113.85269.18333.9425000021.719118.012713.85260000025.021321.719100000027.9634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,

8、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 =50.000055.898562.517269.944176.269985.605496.0836044.775550.032655.931560.542067.952476.26990040.122644.808448.057553.939860.542000035.979038.147442.816748.0575000030.280933.9873389

9、78730.280900000024.0366option =4.99967.879212.086417.944125.129434.036944.083602.11933.68096.259910.342716.3840 24.2699000.54731.08782.16224.29768.54200000000000000000000000000000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

10、=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*sigma2)*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 =5.50916.7865 三、搜集数

11、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=31.63 32.17 31.58 30.71 30.77 30.93 31.7931.58 32 33.91 33.12 34.98 35.3 35.5 34.65 35.46 35.9535.39 37.67 36.64 36.77 36.85 36.59 35

12、.81 35.18 35.7636.66 38.35 38.26 38.34 38.85 41.27 40.99 40.7 42.28' 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 =0.0171-0.0183-0.02750.0020率益收海虹控股 对数 收 益率直方图天数0.00520.0278-0.00660.01330.0597-0.023

13、30.05620.00910.0057-0.02390.02340.0138-0.01560.0644-0.02730.00350.0022-0.0071-0.0213-0.01760.0165 0.02520.0461-0.00230.00210.01330.0623-0.0068-0.00710.0388盛达矿业MATLAB 命令：TickSeries=13.07 12.88 13.19 12.98 12.78 12.49 12.7312.51 12.97 13.06 12.68 13.17 13.93 14.39 14.08 14.34 14.19 14.24 13.74 13.57 1

14、3.8 13.76 13.76 13.52 13.3 13.28 13.44 13.37 13.28 13.74 13.93 14.16 13.99 14.73 14.7'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 =-0.01450.02410.06-0.01590.05-0.01540.04-0.02270.01920.03-0.01730.

15、03680.0069-0.02910.03860.05770.0330-0.0215率益收-0.02-0.03-0.045-0.01盛达矿业对数收益率直方图10 15 20 25 天数30 3500.0185-0.01050.0035-0.0351-0.01240.0169-0.00290-0.0174-0.0163-0.00150.0120-0.0052-0.00670.03460.01380.0165-0.01200.0529-0.0020恒逸石化MATLAB 命令：TickSeries=9.43 9.14 8.99 8.67 8.6 8.42 8.49 8.4 8.538.978.618

16、.919.119.12 9.06 9.14 9.04 8.79 8.78.788.839.379.479.3 9.55 9.89 9.69 9.64 9.589.529.8810.2210.310.45 10.84'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 =-0.03080.08-0.0164-0.0356-0.00810.060.04-0.0

17、2090.0083-0.0106率益收0.0155-0.020.0516-0.0401 -0.040.03480.02240.0011-0.0605 1015 20 25 30 天数恒逸 石 化对 数 收 益率 直 方 图35-0.00660.0088-0.0109-0.0277-0.01020.00920.00570.06120.0107-0.01800.02690.0356-0.0202-0.0052-0.0062-0.00630.03780.03440.00780.01460.0373金宇车城MATLAB 命令：TickSeries=10.9 11.17 11.32 11.32 11.2

18、2 11.08 11.27 11.19 11.31 11.52 11.25 11.78 12.07 12.11 12.15 12.2912.45 12.87 12.77 12.63 12.56 12.71 12.71 12.5 12.1512.23 12.12 12.48 12.6 12.87 12.9 13.33 13.5 13.5 13.42'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

19、结果：RetSeries =0.02480.01340-0.0088-0.01250.0171-0.00710.01070.0186-0.02340.04710.02460.0033);0.0033 0.01150.01300.0337-0.0078-0.0110-0.00550.01190-0.0165-0.02800.0066-0.00900.02970.00960.02140.00230.03330.01280-0.0059（2）计算 4 种股票收盘价的协方差矩阵；MATLAB 命令：A=31.63 13.07 9.43 10.932.1712.889.1411.1731.5813.19

20、8.9911.3230.7112.988.6711.3230.7712.788.6 11.2230.9312.498.4211.0831.7912.738.4911.2731.5812.518.4 11.1932 12.97 8.53 11.3133.9113.068.9711.5233.1212.688.6111.2534.9813.178.9111.7835.313.939.1112.0735.514.399.1212.1134.6514.089.0612.1535.4614.349.1412.2935.9514.199.0412.4535.3914.248.7912.8737.6713.

21、748.7 12.7736.6413.578.7812.6336.7713.88.8312.5636.8513.769.3712.7136.5913.769.4712.7135.8113.529.3 12.535.1813.39.5512.1535.7613.289.8912.2336.6613.449.6912.1238.3513.379.6412.4838.2613.289.5812.638.3413.749.5212.8738.8513.939.8812.941.2714.1610.2213.3340.9913.9910.313.540.714.7310.4513.542.2814.71

22、0.8413.42cov(A)运行结果：ans =10.06081.47511.56872.30591.47510.37110.22700.38571.56870.22700.36820.33262.30590.38570.33260.58373）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= 10.0608 1.4751 1.5687 2.30591.4751 0.3711

23、0.2270 0.38571.5687 0.2270 0.3682 0.33262.3059 0.3857 0.3326 0.5837;PortWts=0.35 0.25 0.25 0.15;PortRisk,PortReturn=portstats(ExpReturn, ExpCovariance,PortWts)运行结果：PortRisk =1.4659PortReturn =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=0.3 0.25 0.2 0.15;ExpCovariance=10.0608 1.4751 1.5687 2

24、.30591.47510.3711 0.2270 0.38571.56872.30590.2270 0.3682 0.33260.3857 0.3326 0.5837;NumPorts=5;PortRink,PortReturn,PortWts=frontcon(ExpReturn,ExpCovariance,NumPor ts)运行结果：PortRink =0.54620.58201.17292.15853.1719PortReturn =0.22470.24360.26240.28120.3000PortWts =-0.00000.49490.50510.000000.87120.1288

25、-0.00000.24750.75250-0.00000.62370.37630-0.00001.000000.0000-0.00005）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=0.3 0.25 0.2 0.15;ExpCovariance=10.0608 1.4751 1.5687 2.30591.4751 0.3711 0.2270 0.38571.5687 0.2270 0.3682 0.33262.3059 0.3857 0.3326 0.5837;RisklessRate

26、=0.035;BorrowRate=0.5;RiskAversion=3; PortRisk,PortReturn,PortWts=portopt(ExpReturn,ExpCovariance) RiskyRink, RiskyReturn, RiskyWts, RiskyFraction,OverallRick, OverallReturn=portalloc(PortRisk, .PortReturn, PortWts, RisklessRate,BorrowRate, RiskAversion) 运行结果：PortRisk =0.54620.55340.57470.60840.9663

27、1.38641.82462.27042.72003.1719PortReturn =0.22470.23310.24150.24980.25820.26660.27490.28330.29160.3000PortWts =.00000.49490.50510.000000.66210.3379-0.00000.00000.82940.1706-0.000000.99660.0034-0.00000.16380.83620-0.00000.33110.668900.00000.49830.50170-0.00000.66550.33450-0.00000.83280.167200.00001.0

28、00000.0000-0.0000RiskyRink =0.5427RiskyReturn =0.2302RiskyWts =-0.0000 0.6040 0.3960 -0.0000 RiskyFraction =0.2209OverallRick =0.1199OverallReturn =0.0781（6）绘制这 4 种股票的最高价、最低价、收盘价和开盘价的烛型图。 海虹控股MATLAB命令：a=31.9 30.5431.6331.2932.3530.7132.1731.332.631.5631.5832.0832 30.7 30.71 31.8231.5930.4530.7730.55

29、31.230 30.93 30.632.230.9431.7930.9432.1831.5231.5831.7832.1131.4332 31.633.9131.9133.9132.135.5732.533.1232.8335.433.2634.9833.2635.5734.2835.334.9636.4935.135.535.335.834.4534.6535.335.634.635.4634.7536.3535.3335.9535.6636.2634.935.3935.737.935 37.67 35.0137.6636.4236.6437.338.2936.336.7736.6437.5

30、836 36.85 37.337.3236.3536.5937.2536.89 35.8 35.81 36.5936.4134.7535.1835.936.0335.1835.7635.437.2235.536.6635.738.5536.7238.3537.0538.6538.0838.2638.4439.1738.0338.3438.2539.238.0738.8538.541.4939.0841.2739.0842.3940.6840.9941.240.9940.0140.740.7542.640.5942.2840.69;candle(a(:,1),a(:,2),a(:,3),a(:,

31、4) candle(a(:,1),a(:,2),a(:,3),a(:,4), 'r' ) title( ' 海虹控股 ' );运行结果：盛达矿业MATLAB 命令：a=13.11 12.08 13.07 12.212.9912.6512.8812.913.4912.8813.1912.9913.2712.8812.9813.113.112.7712.7812.9512.7712.4212.4912.712.812.4212.7312.4212.7712.3212.5112.7213.2212.512.9712.5113.3812.9513.0613.1513.2

32、712.4612.6812.9813.22 12.65 13.17 12.6814.1213.2213.9313.2914.6513.714.3913.9514.8414 14.08 14.3514.3513.8814.341414.3413.9214.1914.1514.314 14.24 14.114.213.713.7414.1413.7613.513.5713.513.913.413.813.514 13.71 13.76 13.7914.0113.7213.7613.813.9913.513.5213.9113.4713.213.313.313.4413.0313.2813.2713

33、.4513.1813.4413.2513.5413.313.3713.4513.4113.0913.2813.3313.7713.2913.7413.3414.0513.7213.9313.814.2313.8114.1613.9414.213.813.9914.0814.813.9314.7313.9314.9514.5614.714.78candle(a(:,1),a(:,2),a(:,3),a(:,4)candle(a(:,1),a(:,2),a(:,3),a(:,4), 'r' ) title( ' 盛达矿业 ' );运行结果：恒逸石化MATLAB 命令

34、：a=9.6 8.75 9.43 9.049.64 9.1 9.14 9.359.16 8.86 8.99 9.08 9 8.65 8.67 8.998.858.58 8.6 8.588.858.38 8.42 8.638.548.3 8.49 8.438.578.34 8.4 8.510.35;'r' )9.318.538.978.539.388.418.618.78.978.638.918.639.218.859.118.919.258.979.129.119.3 9.01 9.06 9.139.158.8 9.14 9.059.278.959.049.19.098.748.799.048.868.618.7 8.78.858.638.788.728.938.258.838.49.668.829.378.8210.29.469.479