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小勳
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小勳 發問時間: 教育與參考考試 · 8 年前

求救.....經濟學題目,贈20點

效用函數若為:U(X1,X2)=αlnX1+(1-α)lnX2。請計算出X1與X2之間的MRS、對X1和X2的需要函數,以及支出函數

已更新項目:

感謝這位大大....

在知識+確實很多只是高中生跟一般大學生

要回答這種題目也的確不容易

但是面對不會的題目

我也不知道要去問誰了

只好上來碰碰運氣....

你看,這不就讓我遇上你了

熱心又聰明的網友~~~~

我是想考在職經濟所

這是中山在職經濟所的考古題

不是本科系出身,只是有興趣而已

如果方便的話,還有一些題目在發問中......

感謝你的幫忙~~~~

1 個解答

評分
  • 8 年前
    最佳解答

    Ans:Your question is not difficult, but its answer is almost unlikely obtained in this website (Yahoo Knowledge +).

    In my experiences, 90% of users in this website are high school students or college students looking for the answer of their homeworks.

    How can you expect that they can give you the answer?

    To simplify the notations in your question, let α=a, X1=X, X2=Y.

    Hence, utility function is U=a*lnX+(1-a)*lnY1. MUx=dU/dX=a/X, MUy=dU/dY=(1-a)/Y,

    MRSxy=MUx/MUy = [a/(1-a)](Y/X)2. The first-order condition for equilibrium is MRSxy=Px/Py,

    [a/(1-a)](Y/X)=Px/Py ---> Y=[(1-a)/a](Px/Py)X -----(1)

    ---> Substitute (1) into budget constraint PxX+PyY=I

    ---> PxX+Py[(1-a)/a](Px/Py)X=I

    ---> X=aI/Px ----(2) (This is X's demand function)

    ---> Substitute (2) into (1), we can obtain Y=(1-a)I/Py (Y's demand function) 3. To find out expenditure function, we use the following model:

    min PxX+PyY=I s.t. U=a*lnX+(1-a)*lnY=ln[(X^a)(Y^(1-a))] (^表示次方)

    --> Again, we use the first-order condition for equilibrium.

    ---> Substitute (1) into utility constraint U=ln[(X^a)(Y^(1-a))]

    ---> U=ln{{[[(1-a)/a](Px/Py)]^(1-a)}X}

    ---> X={[(Py/Px)(a/(1-a))]^(1-a)}exp(U) ---(3)

    ---> Substitute (3) into (1), we can obtain Y={[(Px/Py)(a/(1-a))]^a}exp(U) ---(4)

    ---> Substitute (3) and (4) into PxX+PyY, we can obtain the expenditure function:

    E=Px*{[(Py/Px)(a/(1-a))]^(1-a)}exp(U)+Py*{[(Py/Px)(a/(1-a))]^(1-a)}exp(U)

    ---> It can be simplified as E={[(Px/a)^a]*[Py/(1-a)]^(1-a)}exp(U)

    .

    2012-01-28 00:55:50 補充:

    Sorry, the 3rd part shall be revised as follows:

    3. To find out expenditure function, .....

    min PxX+PyY ( not PxX+PyY=I)

    參考資料: myself
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