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嘻嘻 · 2017年09月16日

能否详细解释下最后的correlation计算公式 问一道题:NO.PZ2016062402000020 [ FRM I ]

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源_品职助教 · 2017年09月17日


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NO.PZ2016062402000020 问题如下 Consir the following lineregression mol: Y=a+bX+e. Suppose a=0.05, b=1.2, SY) = 0.26, anSe) = 0.1. Whis the correlation between X anY? A.0.923 B.0.852 C.0.701 0.462 We cfinthe volatility of X from the variancomposition, Equation: V(y)=β2V(x)+V(e)V(y)=\beta^2V(x)+V(e)V(y)=β2V(x)+V(e). This gives V(x)=V(y)−V(e)β2=0.26∧2−0.10∧21.22=0.04V(x)=\frac{V(y)-V(e)}{\beta^2}=\frac{0.26^\wee2-0.10^\wee2}{1.2^2}=0.04V(x)=β2V(y)−V(e)​=1.220.26∧2−0.10∧2​=0.04. Then SX) = 0.2, anp=SX)∗bSY)=1.2×0.20.26=0.923p=\frac{S(X)^\ast b}}{S(Y)}}=\frac{1.2\times0.2}{0.26}=0.923p=SY)SX)∗b​=0.261.2×0.2​=0.923. 我试着自己推了一下,不知道是不是可以把这个当成一个结论。Y = a + + ε, 因此V(Y) = (bX)^2 + V(ε), 带入得 0.26^2 = 1.2^2 * V(X) +0.1^2, 得到V(X)=0.04。Cov(X,Y)= E[(X-E(X)]*E[(Y-E(Y))], 把= a + + ε 代入,得Cov (X,Y)= E[(X-E(X)] *E(a+bX+ε - a-b*E(X)) = E[(X-E(X)] * E(X-E(X)) = b*E(X-E(X))^2 = b*V(X)所以我们得到Cov(X,Y) = b*V(X) 这个等式,代入讲义中ρ的公式两边取平方,ρ^2 = b^2 * V(X)/V(Y) = 1.2^2 * 0.04 / 0.26^2, 因此ρ = 0.923

2023-07-07 17:40 1 · 回答

NO.PZ2016062402000020 问题如下 Consir the following lineregression mol: Y=a+bX+e. Suppose a=0.05, b=1.2, SY) = 0.26, anSe) = 0.1. Whis the correlation between X anY? A.0.923 B.0.852 C.0.701 0.462 We cfinthe volatility of X from the variancomposition, Equation: V(y)=β2V(x)+V(e)V(y)=\beta^2V(x)+V(e)V(y)=β2V(x)+V(e). This gives V(x)=V(y)−V(e)β2=0.26∧2−0.10∧21.22=0.04V(x)=\frac{V(y)-V(e)}{\beta^2}=\frac{0.26^\wee2-0.10^\wee2}{1.2^2}=0.04V(x)=β2V(y)−V(e)​=1.220.26∧2−0.10∧2​=0.04. Then SX) = 0.2, anp=SX)∗bSY)=1.2×0.20.26=0.923p=\frac{S(X)^\ast b}}{S(Y)}}=\frac{1.2\times0.2}{0.26}=0.923p=SY)SX)∗b​=0.261.2×0.2​=0.923.

2022-05-11 20:28 1 · 回答

NO.PZ2016062402000020 问题如下 Consir the following lineregression mol: Y=a+bX+e. Suppose a=0.05, b=1.2, SY) = 0.26, anSe) = 0.1. Whis the correlation between X anY? A.0.923 B.0.852 C.0.701 0.462 We cfinthe volatility of X from the variancomposition, Equation: V(y)=β2V(x)+V(e)V(y)=\beta^2V(x)+V(e)V(y)=β2V(x)+V(e). This gives V(x)=V(y)−V(e)β2=0.26∧2−0.10∧21.22=0.04V(x)=\frac{V(y)-V(e)}{\beta^2}=\frac{0.26^\wee2-0.10^\wee2}{1.2^2}=0.04V(x)=β2V(y)−V(e)​=1.220.26∧2−0.10∧2​=0.04. Then SX) = 0.2, anp=SX)∗bSY)=1.2×0.20.26=0.923p=\frac{S(X)^\ast b}}{S(Y)}}=\frac{1.2\times0.2}{0.26}=0.923p=SY)SX)∗b​=0.261.2×0.2​=0.923.

2022-05-10 19:32 1 · 回答

NO.PZ2016062402000020问题如下 Consir the following lineregression mol: Y=a+bX+e. Suppose a=0.05, b=1.2, SY) = 0.26, anSe) = 0.1. Whis the correlation between X anY?A.0.923B.0.852C.0.7010.462We cfinthe volatility of X from the variancomposition, Equation: V(y)=β2V(x)+V(e)V(y)=\beta^2V(x)+V(e)V(y)=β2V(x)+V(e). This gives V(x)=V(y)−V(e)β2=0.26∧2−0.10∧21.22=0.04V(x)=\frac{V(y)-V(e)}{\beta^2}=\frac{0.26^\wee2-0.10^\wee2}{1.2^2}=0.04V(x)=β2V(y)−V(e)​=1.220.26∧2−0.10∧2​=0.04. Then SX) = 0.2, anp=SX)∗bSY)=1.2×0.20.26=0.923p=\frac{S(X)^\ast b}}{S(Y)}}=\frac{1.2\times0.2}{0.26}=0.923p=SY)SX)∗b​=0.261.2×0.2​=0.923.请问最后p的公式是如何推导而出的?

2022-04-17 00:52 2 · 回答

NO.PZ2016062402000020问题如下Consir the following lineregression mol: Y=a+bX+e. Suppose a=0.05, b=1.2, SY) = 0.26, anSe) = 0.1. Whis the correlation between X anY?A.0.923B.0.852C.0.7010.462We cfinthe volatility of X from the variancomposition, Equation: V(y)=β2V(x)+V(e)V(y)=\beta^2V(x)+V(e)V(y)=β2V(x)+V(e). This gives V(x)=V(y)−V(e)β2=0.26∧2−0.10∧21.22=0.04V(x)=\frac{V(y)-V(e)}{\beta^2}=\frac{0.26^\wee2-0.10^\wee2}{1.2^2}=0.04V(x)=β2V(y)−V(e)​=1.220.26∧2−0.10∧2​=0.04. Then SX) = 0.2, anp=SX)∗bSY)=1.2×0.20.26=0.923p=\frac{S(X)^\ast b}}{S(Y)}}=\frac{1.2\times0.2}{0.26}=0.923p=SY)SX)∗b​=0.261.2×0.2​=0.923.先对原方程左右取方差求x方差;再把xy挪到等式一边取方差求corr,答案是一样的,过程这么想对么?

2022-04-11 13:42 1 · 回答