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在線雲佬~~SY
 
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只看樓主 倒序閱讀 使用道具 樓主   發表于: 2010-02-03
(25pts). Prove that if R^2 = 1, then the estimated slope of the regression of y on x
equals to the reciprocal of the estimated slope of the regression of x on y.

(20pts). Prove that the value of R^2 is unchanged if we change the units in both the
x and y–coordinates via x′i = c1 + c2xi and y′i = d1 + d2yi for i = 1, . . . , n.

我想問究竟應該要點樣做呀..唔係好明..
在線雲佬~~SY
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只看該作者 1  發表于: 2010-02-03
push!
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只看該作者 2  發表于: 2010-02-03
When I was studying Applied Maths in F6-7 and even some Statistics courses in university, I have also not seen the above topics.....sorry cannot help.

BTW, this topic is from which academic level? just curious.
離線PonG
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只看該作者 3  發表于: 2010-02-06
引用第2樓kylau2010-02-03 23:48發表的“”:
When I was studying Applied Maths in F6-7 and even some Statistics courses in university, I have also not seen the above topics.....sorry cannot help.
BTW, this topic is from which academic level? just curious.

大三
離線kylau
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只看該作者 4  發表于: 2010-02-06
引用第3樓pongpong1232010-02-06 12:07發表的“”:
大三 [表情]


I see. This should be for major in statistics if I am correct. Coz I haven't touched these in my Stat minor.

What I touched in universities are those concerning NPV, normal distributions, samplings, etc. Those things that should have been taught in F6-7 and finance courses.
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只看該作者 5  發表于: 2010-02-11
I think I now know that these questions asked for (after studing CFA level 2 in these few days...). May not be the correct proof but I think the idea will be as follows:

1. Prove that if R^2 = 1, then the estimated slope of the regression of y on x
equals to the reciprocal of the estimated slope of the regression of x on y.

If R^2 = 1, x and y are prefectly correlated,
i.e. for linear relationship y= a+bx+E , error term E = 0 for all x.
=> y = a+bx, slope = b
=> x = y/b - a/b, slope = 1/b, which is reciprocal of b.

2. Prove that the value of R^2 is unchanged if we change the units in both the
x and y–coordinates via x′i = c1 + c2xi and y′i = d1 + d2yi for i = 1, . . . , n.

The main idea is that changing x and y in a linear manner, the relationship between x and y are still linear. R^2 should be no effect.
e.g. y = ax+b, if let x′i = c1 + c2xi and y′i = d1 + d2yi => (x′i - c1) / c2 = xi , (y′i - d1) / d2 = yi,
substititon gets (y′i - d1) / d2 = a(x′i - c1) / c2 + b
=> y′i = ad2(x′i - c1) / c2 + bd2 + d1
=> y′i = ad2 / c2(x′i ) + (bd2 + d1 - ac1d2 / c2), also linear.
離線dream
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只看該作者 6  發表于: 2010-02-15
用戶被禁言,該主題自動屏蔽!
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只看該作者 7  發表于: 2010-02-15
引用第6樓dream2010-02-15 17:02發表的“”:
dun understand [表情]

你D 簽名檔有D 18 禁FEEL
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只看該作者 8  發表于: 2010-02-16
用戶被禁言,該主題自動屏蔽!
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只看該作者 9  發表于: 2010-02-16
引用第8樓dream2010-02-16 15:37發表的“”:
delete左D 簽名檔LA..O唔OK [表情]

OK