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2023 PEA Q12

Household has endowment Lˉ\bar L, supplies LsL^s, enjoys Lˉ−Ls\bar L-L^s leisure. u=Cβ+(Lˉ−Ls)βu=C^\beta+(\bar L-L^s)^\beta, 0<β<10<\beta<1. Budget: PC=WLsPC=WL^s. Find LsL^s as a function of W/PW/P.

The available source text does not include a full option block for this item, so the question is marked needsReview.
Reveal answer and solution

Answer

A

Solution

  1. 1

    Substitute C=WPLsC=\frac{W}{P}L^s into uu and maximise:

  2. 2
    max⁡Ls  (WPLs)β+(Lˉ−Ls)β. \max_{L^s}\;\Bigl(\tfrac{W}{P}L^s\Bigr)^\beta+(\bar L-L^s)^\beta.
  3. 3

    FOC:

  4. 4
    β(WP)β(Ls)β−1−β(Lˉ−Ls)β−1=0, \beta\Bigl(\tfrac{W}{P}\Bigr)^\beta(L^s)^{\beta-1}-\beta(\bar L-L^s)^{\beta-1}=0,
  5. 5

    i.e.\

  6. 6
    (WP)β(Ls)β−1=(Lˉ−Ls)β−1  ⟹  (Lˉ−LsLs)β−1=(WP)β. \Bigl(\tfrac{W}{P}\Bigr)^\beta(L^s)^{\beta-1}=(\bar L-L^s)^{\beta-1} \;\Longrightarrow\; \Bigl(\tfrac{\bar L-L^s}{L^s}\Bigr)^{\beta-1}=\Bigl(\tfrac{W}{P}\Bigr)^\beta.
  7. 7

    Raise to power 1/(β−1)1/(\beta-1) (recall β−1<0\beta-1<0, so flips):

  8. 8
    Lˉ−LsLs=(WP)β/(β−1)=(WP)−β/(1−β). \tfrac{\bar L-L^s}{L^s}=\Bigl(\tfrac{W}{P}\Bigr)^{\beta/(\beta-1)}=\Bigl(\tfrac{W}{P}\Bigr)^{-\beta/(1-\beta)}.
  9. 9

    Therefore

  10. 10
    Lˉ=Ls[1+(WP)−β/(1−β)]  ⟹  Ls=Lˉ1+(W/P)−β/(1−β). \bar L=L^s\Bigl[1+\bigl(\tfrac{W}{P}\bigr)^{-\beta/(1-\beta)}\Bigr] \;\Longrightarrow\; L^s=\dfrac{\bar L}{1+\bigl(W/P\bigr)^{-\beta/(1-\beta)}}.

Answer structure / marking notes

Needs review: source TeX does not provide a full four-option MCQ block.

Content note

Imported from public/resources/isi/msqe/solutions/pea/2023/ISI_MSQE_PEA_2023_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.