# 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.