1. discrete returns
讨论收益回报的离散性(例如, 每天每周等)
gross return: 总收益, 指的是 (不计成本的) 所有的收益
net return 通常称为净回报, 指的是扣除所有成本的真正收益
annualized gross return 年化总收益
annualized.NET return 年化净收益
为方便计算, 通常
事实上, 虽然是取极限等于, 其实在 n=10时约等号两边就以及很接近啦.
2.continuously compounded return (log return)
实际问题中可能直接将 continuously compounded return 称为 continuous return.
第二点的解释: 取的时间间隔很小时, log return 和 net return(净收益)几乎相等
NOTE: 第三点中虽然 k 时间段内的 log return 是期间所有的 log return 求和, 但未指明针对整段时间的情况下, 这个 continuous return 会是一个 list, 从 t0-t1 一直到 tn-1 - tn
3.Adjusted closed price (调整后的收盘价)
(约定俗成!!)通常使用 Adjusted closed price 来计算上述的各个 return
4.计算股票年收益时
mean and covariance matrix for the yearly returns of the stocks = mean and covariance matrix for the weekly returns of the stocks * 52
本年度最难知识点了..对不起小学老师, 可我真的忘记一年有52周了...一副扑克牌有多少张我也不知道 /(ㄒoㄒ)/~~
5.案例
- Consider the portfolios each having a different choice over the assets such that we cover all the following possibilities
- [0, 0, 1]
- [0, 0.1, 0.9]
- [0.1, 0.1, 0.8]
- .
- .
- [0.9, 0.1, 0]
- [1, 0, 0]
三只股票分别是 McDonald,Coca Cola 和 Microsoft, 时间为 1991 年1月1日到 2001 年1月1日
通过使用之前随笔中介绍的 hist_stock_data 获取.
- if ~exist('stocks', 'var')
- stocks=hist_stock_data('01011991','01012001','KO','MCD','MSFT','frequency','wk');
- CocaCola = stocks(1);
- McDonalds = stocks(2);
- Microsoft = stocks(3);
- end
- 5.1 For each of the portfolios calculate the yearly mean and yearly standard deviation.Create a similar graph of mean against standard deviation.
- % Caclualte log returns
- CCLR = log(CocaCola.AdjClose(2:end)./CocaCola.AdjClose(1:end-1) );
- McLR = log(McDonalds.AdjClose(2:end)./McDonalds.AdjClose(1:end-1) );
- MicLR = log(Microsoft.AdjClose(2:end)./Microsoft.AdjClose(1:end-1) );
- LogReturns = [ CCLR, McLR, MicLR ];
- yearlymean = 52 * mean(LogReturns)';
- yearlycov = 52 * cov(LogReturns)';
- S = yearlycov;
- % Now think about the portfolio
- % p is such that in total p_1 + p_2 + p_3 = 1 and we look at the following
- % combinations of the form (0.1, 0.1, 0.8), (0.1, 0.2, 0.7) and so forth
- x = zeros(67,0);
- portfolios = [];
- count = 1;
- for i = 0:10
- for j = 0:10-i
- k = 10 - i - j;
- % assume the portfolio weights as x
- x = [i; j; k] / 10;
- portfolios = [ portfolios; x'];
- port_mean(count) = x' * yearlymean;
- port_var(count) = x' * S * x;
- port_std(count) = sqrt(port_var(count));
- count = count+1;
- end
- end
- np = count -1;
- % Plot portfolios
- plot(port_std, port_mean, '.');
- title ('portfolio performance');
- xlabel('standard deviation');
- ylabel('mean');
得到的 Plot:
- 5.2 Which of the portfolios has the maximal mean
- % highest expected return
- max_r = find (port_mean == max(port_mean));
- portfolios(max_r, :)
- %pause;
- hold on;
- plot(port_std(max_r),port_mean(max_r), 'rx');
就是 mean 最高的点,(Y 轴最高)
- 5.3 Which of the portfolios has the lowest standard deviation
- % Lowest standatd deviation
- min_std = find (port_std == min(port_std));
- portfolios(min_std, :)
- %pause;
- hold on;
- plot(port_std(min_std),port_mean(min_std), 'rx');
类似上步, x 轴最小
- 5.4
- Which of the portfolios has the highest ratio of mean to stdev ( μx/σx )? why is this portfolio interesting?
- % Max slope (sharpe ratio)
- max_slope = find (port_mean ./ port_std == max(port_mean ./ port_std));
- %pause;
- hold on;
- plot(port_std(max_slope),port_mean(max_slope), 'rx');
- portfolios(max_slope, :)
- % Which portfolios dominate other portfolios
这个点对应的最佳投资, 为从 mean 轴对应 var 最小的点向曲线引出的切线与斜线的交点. 投资此处可以兼顾利润和风险
对应的最佳投资点:
来源: http://www.bubuko.com/infodetail-2874275.html