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

A小鼹鼠 · 2018年08月06日

问一道题:NO.PZ2016062402000022 [ FRM I ]

问题如下图:

选项:

A.

B.

C.

D.

解释:

请问这个计算公式怎么来的?
1 个答案

orange品职答疑助手 · 2018年08月07日

同学你好,在一元线性回归中有这样的结论:R^2=XY相关系数的平方,而相关系数=b*X的标准差/Y的标准差。把相关系数平方一下,就得到R方了

  • 1

    回答
  • 0

    关注
  • 636

    浏览
相关问题

NO.PZ2016062402000022问题如下 A portfolio manager is interestein the systematic risk of a stoportfolio, so he estimates the lineregression: RPt−RF=αP+βP[RMt−RF]+εPtR_{Pt}-R_F=\alpha_P+\beta_P{\lbraR_{Mt}-R_F\rbrack}+\varepsilon_{Pt}RPt​−RF​=αP​+βP​[RMt​−RF​]+εPt​,where RPtR_{Pt}RPt​ is the return of the portfolio time t, RMtR_{Mt}RMt​ is the return of the market portfolio time t, anRFR_FRF​ is the risk-free rate, whiis constant over time. Suppose thα = 0.008, β = 0.977, σ(RP)\sigma{(R_P)}σ(RP​) = 0.167, anσ(RM)\sigma{(R_M)}σ(RM​) = 0.156.Whis the approximate coefficient of termination in this regression? 0.913 0.834 0.977 0.955 the R-squareis given β2σM2σP2=0.9772×0.15620.1672=0.83\frac{\beta^2\sigma_M^2}{\sigma_P^2}=0.977^2\times\frac{0.156^2}{0.167^2}=0.83σP2​β2σM2​​=0.9772×0.16720.1562​=0.83 决策系数怎么就是拟合优度了?

2023-03-21 11:10 1 · 回答

NO.PZ2016062402000022问题如下 A portfolio manager is interestein the systematic risk of a stoportfolio, so he estimates the lineregression: RPt−RF=αP+βP[RMt−RF]+εPtR_{Pt}-R_F=\alpha_P+\beta_P{\lbraR_{Mt}-R_F\rbrack}+\varepsilon_{Pt}RPt​−RF​=αP​+βP​[RMt​−RF​]+εPt​,where RPtR_{Pt}RPt​ is the return of the portfolio time t, RMtR_{Mt}RMt​ is the return of the market portfolio time t, anRFR_FRF​ is the risk-free rate, whiis constant over time. Suppose thα = 0.008, β = 0.977, σ(RP)\sigma{(R_P)}σ(RP​) = 0.167, anσ(RM)\sigma{(R_M)}σ(RM​) = 0.156.Whis the approximate coefficient of termination in this regression? 0.913 0.834 0.977 0.955 the R-squareis given β2σM2σP2=0.9772×0.15620.1672=0.83\frac{\beta^2\sigma_M^2}{\sigma_P^2}=0.977^2\times\frac{0.156^2}{0.167^2}=0.83σP2​β2σM2​​=0.9772×0.16720.1562​=0.83 请问这个公式源自哪里

2022-11-17 15:57 1 · 回答

NO.PZ2016062402000022 0.834 0.977 0.955 the R-squareis given β2σM2σP2=0.9772×0.15620.1672=0.83\frac{\beta^2\sigma_M^2}{\sigma_P^2}=0.977^2\times\frac{0.156^2}{0.167^2}=0.83σP2​β2σM2​​=0.9772×0.16720.1562​=0.83 老师,R-square的英文表述是coefficient  of termination,那r(xy)呢?

2022-03-29 07:58 1 · 回答

NO.PZ2016062402000022 0.834 0.977 0.955 the R-squareis given β2σM2σP2=0.9772×0.15620.1672=0.83\frac{\beta^2\sigma_M^2}{\sigma_P^2}=0.977^2\times\frac{0.156^2}{0.167^2}=0.83σP2​β2σM2​​=0.9772×0.16720.1562​=0.83 p是coefficient 那coefficient termination就代表p的平方吗

2022-01-16 13:42 1 · 回答

NO.PZ2016062402000022 0.834 0.977 0.955 the R-squareis given β2σM2σP2=0.9772×0.15620.1672=0.83\frac{\beta^2\sigma_M^2}{\sigma_P^2}=0.977^2\times\frac{0.156^2}{0.167^2}=0.83σP2​β2σM2​​=0.9772×0.16720.1562​=0.83 请问这个题目用的公式在讲义那页?

2021-08-22 14:21 1 · 回答