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Suggested Solutions to Homework # ECONS 500, Washington State University, Fall 2023
- Dynamic Utility with Geometric Discounting
(a) Because u′(ct) = c−t σ> 0 for all ct > 0 with 0 < σ < +∞, so u (ct) is strictly increasing. Since u′′(ct) = −σc− t σ−^1 < 0 for every ct > 0 we should get u (ct) is strictly concave. Finally, lim ct→ 0 u′^ (ct) = lim ct→ 0 c−t σ= lim ct→ 0
1 cσt^ = +∞^ with 0^ < σ <^ +∞. (b) If σ < 0, since u′(ct) = c−t σ> 0 always hold when ct > 0, so u (ct) is strictly increasing. Nevertheless, u′′(ct) = −σc− t σ−^1 > 0 so u (ct) is not a strictly concave function. Meanwhile, lim ct→ 0 u′^ (ct) = lim ct→ 0 c−t σ= 0, which is not +∞ anymore. If σ = 0, u′(ct) = 1 > 0, so u (ct) is strictly increasing. Nevertheless, u′′(ct) = 0, so u (ct) is not a strictly concave function. Meanwhile, lim ct→ 0 u′^ (ct) = 1, so Inada condition doesn’t hold here. (c) When σ → 1, both numerator and denominator in u (ct) converges to 0. By L’Hospital Rule of the 00 Type,
lim σ→ 1 u (ct) = lim σ→ 1
∂(c^1 t −σ− (^1) ) ∂σ ∂(1−σ) ∂σ
= lim σ→ 1 c^1 t −σlog (ct) = log (ct)
(d) The coefficient of relative risk aversion of u(c) is rR = −cu
′′(c) u′(c) =^ −
c·(−σc−σ−^1 ) c−σ^ =^ σ, which is a constant. (e) Let
g(U ) = (1 − σ)U +
X^ T
t=
βt
! (^1) −^1 σ .
It is easy to prove that g(U ) is a strictly increasing function of U. And we also have
g (U (c 0 , c 1 , ..., cT )) =
α 0 c^10 − σ+ α 1 c^11 − σ+ ... + αT c^1 T−σ
(^1) − (^1) σ where αt = βt.
So U (c 0 , c 1 , ..., cT ) =
PT
t=0 β
t c^1 t− σ−^1 1 −σ belongs to the CES utility family. (f) When σ = 1, U (c 0 , c 1 , ..., cT ) =
PT
t=0 β
t (^) log (ct)(plug in the solution in (c)). Let g(U ) = exp U , Then g (U (c 0 , c 1 , ..., cT )) = cα 0 0 cα 1 1 ...cα TT ,where αt = βt. (g) Using the result in (e), When σ ̸= 1, we have
g (U (αc 0 , αc 1 , ..., αcT )) = [α 0 · (αc 0 )^1 −σ^ + α 1 · (αc 1 )^1 −σ^ + ... + αT · (αcT )^1 −σ] 1 −^1 σ
= α
α 0 c^10 − σ+ α 1 c^11 − σ+ ... + αT c^1 T−σ
(^1) −^1 σ
= αg (U (c 0 , c 1 , ..., cT ))
which proves U is homothetic. When σ = 1, U (c 0 , c 1 , ..., cT ) =
PT
t=0 β
t (^) log (ct). Define g (U ) = exp{U/ PT t=0 β
t}, and it’s easy to check that g (U ) is homogeneous of degree one.
- Neoclassical Production Function
(a) It follows from the following steps.
F (tk, tn) = A[α(tk)^1 −ρ^ + (1 − α) (tn)^1 −ρ]
1 1 −ρ
= A[αt^1 −ρk^1 −ρ^ + (1 − α) t^1 −ρn^1 −ρ] 1 −^1 ρ
= At(αt^1 −ρk^1 −ρ^ + (1 − α) n^1 −ρ) 1 −^1 ρ = tF (k, n)
(b) It follows from the following steps.
lim ρ−→ 1 log F (k, n) = log A + lim ρ−→ 1
log (αk^1 −ρ^ + (1 − α) n^1 −ρ) 1 − ρ
= log A + lim ρ−→ 1
∂ log(αk^1 −ρ+(1−α)n^1 −ρ) ∂ρ ∂(1−ρ) ∂ρ
= log A + lim ρ−→ 1
α (log k) k^1 −ρ^ + (1 − α)(log n)n^1 −ρ αk^1 −ρ^ + (1 − α) n^1 −ρ
= log A + lim ρ−→ 1
α (log k) + (1 − α) log n(n/k)^1 −ρ α + (1 − α) (n/k)^1 −ρ = log A + α log k + (1 − α) log n lim ρ−→ 1 F (k, n) = Akαn^1 −α
(c) Fk (k, n) = αA (n/k)^1 −α^ > 0. lim k−→ 0 Fk (k, n) = ∞ Fn (k, n) = (1 − α) A (k/n)α^ > 0. lim k−→∞ Fk (k, n) = 0
(d) Fk (k, n) k = αA (n/k)^1 −α^ k = αAkαn^1 −α^ = αF (k, n). Fn (k, n) n = (1 − α) A (k/n)α^ n = (1 − α) Akαn^1 −α^ = (1 − α) F (k, n). Divide both sides by F (k, n) and we can get the result.
- Saving Rate and Dynamics in Growth Models
(a) If saving rate is constant, kt+1 = g(kt) = sktα + (1 − δ) kt, g′(kt) = αsk tα −^1 + 1 − δ, g′′(kt) = α (α − 1) skα t −^2 < 0, lim kt−→ 0
g′(kt) = +∞, lim kt−→+∞
g′(kt) = 1 − δ < 1. When α = 0. 33 , δ = 0. 96 , s = 0.5, the graph is given below.
Figure 3: Q3(c)
once and from below. Given this, any point away from k∗^ diverges over time. The dynamics is similar to Part (c), except for the fact that the curve is strictly convex. I skip the details here.
(e) g (k) crosses 45 degree line once and from above.
