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Richard ZHANG · 2021年08月13日

请问这个题为什么不能用correlation = COV(x,y) / SDxSDy 来求?还是不太明白

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=0.2620.1021.22=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.

请问这个题为什么不能用correlation = COV(x,y) / SDxSDy 来求?还是不太明白
3 个答案

品职答疑小助手雍 · 2022年01月04日

同学你好,回归里面最小二乘法结论的公式

品职答疑小助手雍 · 2021年09月20日

同学你好,讲义174页,最小二乘法定义。

这个公式很重要,需要记牢。

品职答疑小助手雍 · 2021年08月13日

嗨,爱思考的PZer你好:


用到了的。

β=cov除以σx方,而ρ=cov/(σx*σy)

那连起来就是β=ρ*(σx*σy)/σx方=ρ*σy/σx。

可以解出来ρ。

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就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

Jessie999 · 2021年09月19日

请问贝塔为什么等于cov除以希格玛x平方?

子瑶Eunice · 2022年01月04日

请问贝塔为什么等于cov除以希格玛x平方?

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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

2024-08-29 16:40 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. 第一步求Sx)V(y)=0.26^2 = 1.2^2 x Sx)^2 + 0.1^2Sx) = 0.2第二利用Beta公式求correlationBeta = correlation x Sy)/Sx)b = Beta = 1.21.2 = correlation x 0.26/0.2correlation = 0.923

2024-04-05 12:01 1 · 回答

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2024-04-05 11:54 1 · 回答

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2023-07-07 17:40 1 · 回答

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2022-05-11 20:28 1 · 回答