(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..
雲佬~~SY
2010-02-03 19:23
push!
kylau
2010-02-03 23:48
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.
PonG
2010-02-06 12:07
引用第2樓kylau於2010-02-03 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
kylau
2010-02-06 14:15
引用第3樓pongpong123於2010-02-06 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.
kylau
2010-02-11 22:18
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.
dream
2010-02-15 17:02
用戶被禁言,該主題自動屏蔽!
PonG
2010-02-15 23:15
引用第6樓dream於2010-02-15 17:02發表的“”: dun understand [表情]