这个公式，为啥，只乘以 weight的值？ 2*0.3*0.7？ 公式不还要乘以标准差么？
问题如下图：
选项：
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解释：
NO.PZ2015121801000042问题如下A portfolio manager creates the following portfolio:If the covarianof returns between the two securities is -0.0240, the expectestanrviation of the portfolio is closest to:A.2.4%.B.7.5%.C.9.2%. is correct.lσport=w12σ12+w22σ22+2w1w2Cov(R1R2)=(0.3)2(20%)2+(0.7)2(12%)2+2(0.3)(0.7)(−0.0240)=(0.3600%+0.7056%−1.008%)0.5=(0.0576%)0.5=2.4%{l}{\sigma _{port}} = \sqrt {w_1^2\sigma _1^2 + w_2^2\sigma _2^2 + 2{w_1}{w_2}Cov({R_1}{R_2})} \\ = \sqrt {{{(0.3)}^2}{{(20\% )}^2} + {{(0.7)}^2}{{(12\% )}^2} + 2(0.3)(0.7)( - 0.0240)} \\ = {(0.3600\% + 0.7056\% - 1.008\% )^{0.5}} = {(0.0576\% )^{0.5}} = 2.4\% lσport=w12σ12+w22σ22+2w1w2Cov(R1R2)=(0.3)2(20%)2+(0.7)2(12%)2+2(0.3)(0.7)(−0.0240)=(0.3600%+0.7056%−1.008%)0.5=(0.0576%)0.5=2.4%这道为什么百分比要写成0•12 而上一道12%直接代入12？
NO.PZ2015121801000042 问题如下 A portfolio manager creates the following portfolio:If the covarianof returns between the two securities is -0.0240, the expectestanrviation of the portfolio is closest to: A.2.4%. B.7.5%. C.9.2%. is correct.lσport=w12σ12+w22σ22+2w1w2Cov(R1R2)=(0.3)2(20%)2+(0.7)2(12%)2+2(0.3)(0.7)(−0.0240)=(0.3600%+0.7056%−1.008%)0.5=(0.0576%)0.5=2.4%{l}{\sigma _{port}} = \sqrt {w_1^2\sigma _1^2 + w_2^2\sigma _2^2 + 2{w_1}{w_2}Cov({R_1}{R_2})} \\ = \sqrt {{{(0.3)}^2}{{(20\% )}^2} + {{(0.7)}^2}{{(12\% )}^2} + 2(0.3)(0.7)( - 0.0240)} \\ = {(0.3600\% + 0.7056\% - 1.008\% )^{0.5}} = {(0.0576\% )^{0.5}} = 2.4\% lσport=w12σ12+w22σ22+2w1w2Cov(R1R2)=(0.3)2(20%)2+(0.7)2(12%)2+2(0.3)(0.7)(−0.0240)=(0.3600%+0.7056%−1.008%)0.5=(0.0576%)0.5=2.4% 能不能把其中的原理一下，谢谢！
这题2W1W2COV(R1R2)我直接带入得出0.0101.没有转换成百分比形式最后相加，这个为什么最后要转成百分比的形式1.008%？
老师，这个答案有问题吧。难道不是0.0036+0.0071-0.0002 再开平方么
请问求组合方差的公式，cov是可以直接替换corr的对吗？直接带入，不需要转换？
