问题如下图：
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轻轻问相关系数=b*X的标准差/Y的标准差，是哪个公式？
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}=0.83σP2β2σM2=0.9772×0.16720.1562=0.83 决策系数怎么就是拟合优度了？
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}=0.83σP2β2σM2=0.9772×0.16720.1562=0.83 请问这个公式源自哪里
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}=0.83σP2β2σM2=0.9772×0.16720.1562=0.83 老师，R-square的英文表述是coefficient of termination,那r(xy)呢？
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}=0.83σP2β2σM2=0.9772×0.16720.1562=0.83 p是coefficient 那coefficient termination就代表p的平方吗
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}=0.83σP2β2σM2=0.9772×0.16720.1562=0.83 请问这个题目用的公式在讲义那页？
