C选项为什么用这个求资产组合的方差公式？
问题如下图：
选项：
A. 
B. 
C. 
D. 
解释：
NO.PZ2016062402000005 问题如下 Given thx any are ranm variables anc anare constants, whione of the following finitions is wrong? A.E(ax+by+c)=aE(x)+bE(y)+cE{(ax+by+c)}=aE{(x)}+bE{(y)}+cE(ax+by+c)=aE(x)+bE(y)+c ，if x any are correlate B.V(ax+by+c)=V(ax+by)+cV{(ax+by+c)}=V{(ax+by)}+cV(ax+by+c)=V(ax+by)+c,if x any are correlate C.Cov(ax+by,cx+)=acV(x)+b(y)+(abc)Cov(x,y)Cov{(ax+by,cx+)}=acV{(x)}+b{(y)}+{(abc)}Cov{(x,y)}Cov(ax+by,cx+)=acV(x)+b(y)+(abc)Cov(x,y),if x any are correlate V(x−y)=V(x+y)=V(x)+V(y)V{(x-y)}=V{(x+y)}=V{(x)}+V{(y)}V(x−y)=V(x+y)=V(x)+V(y), if x any are uncorrelate Statement , it is a lineoperation. Statement C is correct, in Equation: V(Y)=σp2V(Y)=\sigma_p^2V(Y)=σp2=∑i=1nωi2σi2+∑i=1N∑j=1,j≠iNωiωjσi,j=\sum_{i=1}^n\omega_i^2\sigma_i^2+\sum_{i=1}^N\sum_{j=1,j\neq i}^N\omega_i\omega_j\sigma_{i,j}=∑i=1nωi2σi2+∑i=1N∑j=1,j=iNωiωjσi,j=∑i=1Nωi2σi2+2∑i=1N∑j iNωiωjσi,j=\sum_{i=1}^N\omega_i^2\sigma_i^2+2\sum_{i=1}^N\sum_{j i}^N\omega_i\omega_j\sigma_{i,j}=∑i=1Nωi2σi2+2∑i=1N∑j iNωiωjσi,jStatement is correct, the covarianterm is zero if the variables are uncorrelate Statement B is false, aing a constant c to a variable cannot change the variance. The constant ops out because it is also in the expectation. Statement B is false, aing a constant c to a variable cannot change the variance.
NO.PZ2016062402000005问题如下Given thx any are ranm variables anc anare constants, whione of the following finitions is wrong?A.E(ax+by+c)=aE(x)+bE(y)+cE{(ax+by+c)}=aE{(x)}+bE{(y)}+cE(ax+by+c)=aE(x)+bE(y)+c ，if x any are correlateB.V(ax+by+c)=V(ax+by)+cV{(ax+by+c)}=V{(ax+by)}+cV(ax+by+c)=V(ax+by)+c,if x any are correlateC.Cov(ax+by,cx+)=acV(x)+b(y)+(abc)Cov(x,y)Cov{(ax+by,cx+)}=acV{(x)}+b{(y)}+{(abc)}Cov{(x,y)}Cov(ax+by,cx+)=acV(x)+b(y)+(abc)Cov(x,y),if x any are correlateV(x−y)=V(x+y)=V(x)+V(y)V{(x-y)}=V{(x+y)}=V{(x)}+V{(y)}V(x−y)=V(x+y)=V(x)+V(y), if x any are uncorrelateStatement , it is a lineoperation. Statement C is correct, in Equation: V(Y)=σp2V(Y)=\sigma_p^2V(Y)=σp2=∑i=1nωi2σi2+∑i=1N∑j=1,j≠iNωiωjσi,j=\sum_{i=1}^n\omega_i^2\sigma_i^2+\sum_{i=1}^N\sum_{j=1,j\neq i}^N\omega_i\omega_j\sigma_{i,j}=∑i=1nωi2σi2+∑i=1N∑j=1,j=iNωiωjσi,j=∑i=1Nωi2σi2+2∑i=1N∑j iNωiωjσi,j=\sum_{i=1}^N\omega_i^2\sigma_i^2+2\sum_{i=1}^N\sum_{j i}^N\omega_i\omega_j\sigma_{i,j}=∑i=1Nωi2σi2+2∑i=1N∑j iNωiωjσi,jStatement is correct, the covarianterm is zero if the variables are uncorrelate Statement B is false, aing a constant c to a variable cannot change the variance. The constant ops out because it is also in the expectation.老师好，请问c是怎么推导出来的
NO.PZ2016062402000005 问题如下 Given thx any are ranm variables anc anare constants, whione of the following finitions is wrong? A.E(ax+by+c)=aE(x)+bE(y)+cE{(ax+by+c)}=aE{(x)}+bE{(y)}+cE(ax+by+c)=aE(x)+bE(y)+c ，if x any are correlate B.V(ax+by+c)=V(ax+by)+cV{(ax+by+c)}=V{(ax+by)}+cV(ax+by+c)=V(ax+by)+c,if x any are correlate C.Cov(ax+by,cx+)=acV(x)+b(y)+(abc)Cov(x,y)Cov{(ax+by,cx+)}=acV{(x)}+b{(y)}+{(abc)}Cov{(x,y)}Cov(ax+by,cx+)=acV(x)+b(y)+(abc)Cov(x,y),if x any are correlate V(x−y)=V(x+y)=V(x)+V(y)V{(x-y)}=V{(x+y)}=V{(x)}+V{(y)}V(x−y)=V(x+y)=V(x)+V(y), if x any are uncorrelate Statement , it is a lineoperation. Statement C is correct, in Equation: V(Y)=σp2V(Y)=\sigma_p^2V(Y)=σp2=∑i=1nωi2σi2+∑i=1N∑j=1,j≠iNωiωjσi,j=\sum_{i=1}^n\omega_i^2\sigma_i^2+\sum_{i=1}^N\sum_{j=1,j\neq i}^N\omega_i\omega_j\sigma_{i,j}=∑i=1nωi2σi2+∑i=1N∑j=1,j=iNωiωjσi,j=∑i=1Nωi2σi2+2∑i=1N∑j iNωiωjσi,j=\sum_{i=1}^N\omega_i^2\sigma_i^2+2\sum_{i=1}^N\sum_{j i}^N\omega_i\omega_j\sigma_{i,j}=∑i=1Nωi2σi2+2∑i=1N∑j iNωiωjσi,jStatement is correct, the covarianterm is zero if the variables are uncorrelate Statement B is false, aing a constant c to a variable cannot change the variance. The constant ops out because it is also in the expectation. B项展开的公式是什么？这部分讲义讲的比较简单，何老师没有展开讲，做题时感觉都不会
NO.PZ2016062402000005 请问讲义第几页讲了相关知识
B的正确版本是不是Var（ax+by+c）=Var（ax+by）？
