@@ -107,7 +107,7 @@ First let's consider the household side.
107107
108108### Consumer's problem
109109
110- Suppose that utility for individuals born at time $t$ take the form
110+ Suppose that utility for individuals born at time $t$ takes the form
111111
112112``` {math}
113113:label: eq_crra
@@ -263,7 +263,7 @@ them to zero:
263263
264264### Demand
265265
266- Using our assumption $\ell_1 = 1$ allows us to write
266+ Using our assumption $\ell_t = 1$ allows us to write
267267
268268``` {math}
269269:label: wage_one
@@ -444,7 +444,14 @@ In particular, since $w_t = (1-\alpha)k_t^\alpha$, we have
444444If we iterate on this equation, we get a sequence for capital stock.
445445
446446
447- Let's plot the 45 degree diagram.
447+ Let's plot the 45 degree diagram of these dynamics, which we write as
448+
449+ $$
450+ k_{t+1} = g(k_t)
451+ \quad \text{where }
452+ g(k) := \frac{\beta}{1+\beta} (1-\alpha)(k)^{\alpha}
453+ $$
454+
448455
449456``` {code-cell} ipython3
450457def k_update(k, α, β):
@@ -481,14 +488,15 @@ plt.show()
481488The diagram shows that the model has a unique positive steady state, which we
482489denote by $k^* $.
483490
484- We can solve for $k^* $ by setting $k _ {t+1} = k_t = k^* $ in [ ] ( law_of_motion_capital ) , which yields
491+ We can solve for $k^* $ by setting $k^* = g(k^ * )$, or
485492
486493``` {math}
487494:label: steady_state_1
488495 k^* = \frac{\beta (1-\alpha) (k^*)^{\alpha}}{(1+\beta)}
489496```
490497
491- We can solve this equation to obtain
498+ Solving this equation yields
499+
492500``` {math}
493501:label: steady_state_2
494502 k^* = \left (\frac{\beta (1-\alpha)}{1+\beta} \right )^{1/(1-\alpha)}
@@ -707,7 +715,7 @@ If $k_t$ is given then $f$ is a function of unknown $k_{t+1}$.
707715
708716Then we can use ` scipy.optimize.newton ` to solve $f(k_ {t+1}, k_t)=0$ for $k_ {t+1}$.
709717
710- First let define $f$.
718+ First let's define $f$.
711719
712720``` {code-cell} ipython3
713721def f(k_prime, k, model):
774782Unlike the log preference case, the CRRA utility steady state $k^*$
775783cannot be obtained analytically.
776784
777- Instead, solve for $k^*$ using newton 's method.
785+ Instead, we solve for $k^*$ using Newton 's method.
778786
779787```
780788
@@ -783,12 +791,12 @@ Instead, solve for $k^*$ using newton's method.
783791:class: dropdown
784792```
785793
786- We introduce a function $g $ such that
787- positive steady state is the root of $g $.
794+ We introduce a function $h $ such that
795+ positive steady state is the root of $h $.
788796
789797``` {math}
790798:label: crra_newton_2
791- g (k^*) = k^*
799+ h (k^*) = k^*
792800 \left [
793801 1 + \beta^{-1/\gamma} (\alpha (k^*)^{\alpha-1})^{(\gamma-1)/\gamma}
794802 \right ] - (1-\alpha)(k^*)^{\alpha}
@@ -797,7 +805,7 @@ positive steady state is the root of $g$.
797805Here it is in Python
798806
799807``` {code-cell} ipython3
800- def g (k_star, model):
808+ def h (k_star, model):
801809 α, β, γ = model.α, model.β, model.γ
802810 z = (1 - α) * k_star**α
803811 R1 = α ** (1-1/γ)
@@ -809,7 +817,7 @@ def g(k_star, model):
809817Let's apply Newton's method to find the root:
810818
811819``` {code-cell} ipython3
812- k_star = optimize.newton(g , 0.2, args=(model,))
820+ k_star = optimize.newton(h , 0.2, args=(model,))
813821print(f"k_star = {k_star}")
814822```
815823
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