[考試] 兩題計量經濟

看板Economics (經濟學)作者JimmyWr (獨立與自由)時間18年前 (2008/04/20 18:41)推噓4(4推 0噓 4→)

留言8則, 6人參與討論串1/1

來源: 經濟數學 Alpha C. Chian 科目: 經濟數學　Ch12 Optmization with equality Constrains 問題: 題目中說的12.22式 f(v)≧f(u) => { Σfj(u)(vj-uj) } { Σfj(v)(vj-uj) } ≧0 { } The function z=f(X1,X2)= X1X2 (X1,X2 ≧0） is quasiconcave . (半凹的嗎?) We shall now check it. Let u= (u1,u2) and v=(v1,v2) be any two points in the domain. Then f(u)=u1u2 and f(v)=V1V2 assume that: f(v)≧f(u) or v1v2≧u1u2 (v1 v2 u1 u2≧0) (12.27) since the partial derivitives of are f1=x2 and f2=x1 12.22式 amounts to the condition that (請問這邊f的偏導數根下面推出的方程式有何關係?) f1(u)(v1-u1)+f2(u)(v2-u2) = u2(v1-u1)-u1(v2-u2) ≧0 (為何透過導數可以判定需要都以 u去測度? 以及f1(u)和f2(u)為何等於 u2和u1 ?? 下面12.28的式子怎麼來的) or upon rearangement , u2(v1-u1)≧u1(u2-v2) (12.28) we need to consider four possiblities regarding the value of u1 and u2 First, if u1=u2=0 then 12.28 is trivially satisfied. Second , if u1=0 but u2>0 then 12.28 reduce to the condition u2v1≧0. which is again (違反的意思嗎?? 那為何可推出使u2 v1不為負成立) satisfied since u2 and v1 are both nonnegtive. Third , if u1>0 and u2=0 then (12.28) reduce to the condition 0≧-u1v2, (也同樣是違反0≧-u1v2嗎那又為何等是成立?) which is still satisfied. Last, suppose that u1 and u2 are both possitive so that u1 and u2 are both possitive , so that v1 and v2 are also positive. (這邊比較沒問題) Subtracting v2u1 from both sides of (12.27), we obtain v2(v1-u1)≧u1(u2-v2) (12.29) (這邊還看的懂只是不知道同時減掉 v2u1的用意??) Three subpossibilities now present themselves. (不懂這個單字?? 推測??) 1.If u2=v2 then v1≧u1 .In fact , we should have v1>u1 since (v1,v2) are distinct points. (不懂?? u2=v2不是會使等號右邊變成0? 那怎麼又推得 v1≧u1 以及distinct point 的 v1>u1? 不是也應該是 v1=u1?? The fact that u2=v2 and v1>u1 implies that condition(12.28 is satisfied) 2.If u2>v2 , then we must also have v1>u1 by(12.29) .Multiplying both sides of (12.29) by u2/v2 , we get u2(v1-u1)≧ u2/v2[u1(u2-v2)]>u1(u2-v2) [since u2/v2 >1 ] (12.30) Thus 12.28 is again satisfied (這邊我看的懂我比較沒意見) 3. The final subpossiblities is that u2<v2, implying that u2/v2 is a positive fraction. In this case . the first line of (12.30) still holds. The second line also holds , but now for a different reason: (以上我還能夠理解下面兩句就看不懂 ) a fraction (u2/v2) of a negative number (u2-v2) is greater than the latter number itself. 請問整個題目的中心主旨就是在映證原本透過12.27的假設而來的 12.28恆等式嗎 u2(v1-u1)≧u1(u2-v2) 題目太複雜了又有用到一些特別的英文煩請高手感激不盡 -- Anything, without blind insistence, is right. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.162.195.144