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徐威廉 · 2024年04月05日

这个步骤对不对？并且为什么b = beta = 1.2???

  1. FRM I
  2. Quantitative

NO.PZ2016062402000020

问题如下：

Consider the following linear regression model: Y=a+bX+e. Suppose a=0.05, b=1.2, SD(Y) = 0.26, and SD(e) = 0.1. What is the correlation between X and Y?

选项：

A.

0.923

B.

0.852

C.

0.701

D.

0.462

解释：

We can find the volatility of X from the variance decomposition， Equation: V(y)=β2V(x)+V(e)V(y)=\beta^2V(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^\wedge2-0.10^\wedge2}{1.2^2}=0.04. Then SD(X) = 0.2, and p=SD(X)∗bSD(Y)=1.2×0.20.26=0.923p=\frac{SD{(X)^\ast b}}{SD{(Y)}}=\frac{1.2\times0.2}{0.26}=0.923.

第一步：求SD(x)

V（y）=0.26^2 = 1.2^2 x SD(x)^2 + 0.1^2

SD(x) = 0.2

第二：利用Beta公式求correlation

Beta = correlation x SD(y)/SD(x)

b = Beta = 1.2

1.2 = correlation x 0.26/0.2

correlation = 0.923

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品职答疑小助手雍 · 2024年04月05日

同学你好，你写的这个步骤没有错。

b=1.2这个是题目里给的条件。

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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.0.923B.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.有点奇怪啊，看了答案也没在讲义找到，相关例题，我这个是刚学完Quant Section2 筛选题库的题看到的

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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. beta 为什么等于1.2？

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