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# F5-conf-04 — VFI vs closed-form policy, log-utility optimal growth

# Question

Planner maximises Σ β^t ln c_t s.t. y_{t+1} = (y_t − c_t)^α (productivity 1, full depreciation). Compare the closed-form policy/value function with the value-function-iteration (VFI) solution on the prescribed grid, at y ∈ {0.5, 1, 2}, and report V(1), the closed-form savings rate, and the contraction modulus of the Bellman operator.

# Data and exclusions

data/model.csv (10 rows, all used): α = 0.36, β = 0.95, grid of 400 points equally spaced in ln y on [0.05, 5], VFI tolerance 1e-8 (sup-norm change), evaluation points 0.5, 1.0, 2.0. Nothing excluded. Design decisions (also in data_issues):

  • none of the three evaluation points is a grid node, so the VFI policy and value are interpolated linearly in ln y between the two nearest nodes;
  • the file does not say what the choice variable is. I chose next-period output y′ from the same grid (discrete choice), so c = y − (y′)^{1/α}. V starts at 0.

# Method

Closed form (guess and verify). Assumptions: log utility, Cobb–Douglas technology y′ = k^α with productivity 1, full depreciation, and an interior solution (0 < c < y; the Inada property of ln guarantees this). Guess V(y) = A + B ln y. The Bellman RHS is ln c + β[A + Bα ln(y − c)]. The FOC 1/c = αβB/(y − c) gives c = y/(1 + αβB). Matching the coefficients on ln y gives B = 1 + αβB, so B = 1/(1 − αβ). Matching the constants gives A = [ln(1 − αβ) + (αβ/(1 − αβ)) ln(αβ)]/(1 − β). Therefore

  • c(y) = (1 − αβ) y, with savings rate (y − c)/y = αβ = 0.342;
  • V(y) = A + ln y/(1 − αβ), with A = −19.5244 and B = 1.51976.

Contraction. The Bellman operator T is monotone (v ≤ w ⇒ Tv ≤ Tw) and discounts (T(v + a) = Tv + βa). By Blackwell's sufficient conditions, T is a contraction in the sup norm with modulus β = 0.95. One caveat: ln is unbounded on (0, ∞), so the theorem applies directly to the discretised problem, which is bounded on a compact grid. For the continuous problem, a weighted-norm or bounded-state argument is needed (for example, restricting y to an invariant compact set). The tolerance gives an error bound for the fixed point: ‖V_n − V*‖ ≤ β·tol/(1 − β) = 1.9e-7.

VFI. All 400 × 400 (state, choice) pairs are evaluated in one vectorised step, with infeasible pairs (c ≤ 0) set to −∞. Iteration continues until the sup-norm change is below 1e-8. The policy is the argmax at the fixed point. The optimal y′ never hits either grid bound.

# Results

y c closed form c VFI abs diff
0.5 0.32900 0.32882 0.00018
1.0 0.65800 0.65899 0.00099
2.0 1.31600 1.30963 0.00637
  • Max abs policy error over the evaluation points: 0.00637, at y = 2, about 0.5% of c.
  • V(1): closed form −19.52441, VFI −19.52476. The difference of 3.4e-4 is far larger than the tolerance bound, so it comes from discretisation.
  • Savings rate (closed form): 0.342. Contraction modulus: 0.95.
  • VFI converged in 361 iterations, with final sup change 9.997e-9.
  • Over the whole grid: sup |c_VFI − c| = 0.0251 (at the upper end of the grid), and sup |V_VFI − V| = 4.8e-4.

# Uncertainty

There is no sampling uncertainty: the model is deterministic and the parameters are given. The numerical error has two parts:

  1. Iteration error. It is bounded by 1.9e-7 in V, which is negligible.
  2. Discretisation error. This dominates. With discrete choice of y′ on a log grid with step h = 0.01154, the chosen ln y′ can be off by up to h/2. The implied savings k = y′^{1/α} is then off by a relative amount of up to about h/(2α) ≈ 1.6%. So the policy error grows roughly in proportion to k = αβy, which is why the error is largest at y = 2. Linear interpolation between nodes adds a smaller error; the true policy is linear in y but not in ln y. The error shrinks as h → 0, i.e. with more grid points, or with continuous choice using interpolation of V.

# Limitations and open issues

  • I validated only one parameter point (α = 0.36, β = 0.95) and one grid (400 log-spaced points on [0.05, 5]). I did not study grid refinement or other parameter values.
  • The result depends on my choice of discrete-choice design (choosing y′ on the grid). A continuous-choice or endogenous-grid method would produce different, probably smaller, errors.
  • The closed form relies on the special case of log utility, Cobb–Douglas technology and full depreciation. Outside that case the comparison does not extend.
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