开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

claireteng · 2022年06月21日

B项展开是怎么样的?

NO.PZ2016062402000005

问题如下:

Given that x and y are random variables and a, b, c and d are constants, which one of the following definitions is wrong?

选项:

A.

E(ax+by+c)=aE(x)+bE(y)+cE{(ax+by+c)}=aE{(x)}+bE{(y)}+c

if x and y are correlated.

B.

V(ax+by+c)=V(ax+by)+cV{(ax+by+c)}=V{(ax+by)}+c,

if x and y are correlated.

C.

Cov(ax+by,cx+dy)=acV(x)+bdV(y)+(ad+bc)Cov(x,y)Cov{(ax+by,cx+dy)}=acV{(x)}+bdV{(y)}+{(ad+bc)}Cov{(x,y)},

if x and y are correlated.

D.

V(xy)=V(x+y)=V(x)+V(y)V{(x-y)}=V{(x+y)}=V{(x)}+V{(y)},

if x and y are uncorrelated.

解释:

Statement , as it is a linear operation. Statement C is correct, as in Equation:

V(Y)=σp2V(Y)=\sigma_p^2

=i=1nωi2σi2+i=1Nj=1,jiNωiωjσi,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+2i=1Nj<iNωiωjσi,j=\sum_{i=1}^N\omega_i^2\sigma_i^2+2\sum_{i=1}^N\sum_{j

Statement D is correct, as the covariance term is zero if the variables are uncorrelated. Statement B is false, as adding a constant c to a variable cannot change the variance. The constant drops 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

2 个答案

李坏_品职助教 · 2022年06月22日

嗨,爱思考的PZer你好:


看一下基础班讲义P49;

V(ax+by) = A^2 * V(x) + b^2 * V(Y) + 2ab * Cov(x,y)

----------------------------------------------
努力的时光都是限量版,加油!

李坏_品职助教 · 2022年06月21日

嗨,从没放弃的小努力你好:


计算方差的时候,常数c可以直接忽略,因为常数的方差是0,并且常数项和随机变量x、y之间都是不相关的。


所以B项的V(ax+by+c) = V(ax + by),不需要做什么复杂的变换。参考基础班讲义P20:

----------------------------------------------
就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

claireteng · 2022年06月22日

不会考对v(ax+by)的展开吗

  • 2

    回答
  • 0

    关注
  • 223

    浏览
相关问题

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=1n​ωi2​σi2​+∑i=1N​∑j=1,j​=iN​ωi​ωj​σi,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=1N​ωi2​σi2​+2∑i=1N​∑j iN​ωi​ωj​σi,j​Statement 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

2024-03-29 14:02 1 · 回答

NO.PZ2016062402000005 请问讲义第几页讲了相关知识

2021-12-25 15:23 1 · 回答

B的正确版本是不是Var(ax+by+c)=Var(ax+by)?

2020-03-17 14:54 1 · 回答

说A.B不相关,不想管不意味着一定独立,则A.B协方差不一定等于0,那么,是否错误

2020-02-20 19:08 1 · 回答