1. **Problem Statement:** Derive the Euler equation and analyze the Ramsey–Cass–Koopmans model with discrete time, CRRA utility, and Cobb–Douglas production. 2. **Setup:** Maximize the lifetime utility $$\max_{\{c_t,k_{t+1}\}_{t=0}^\infty} \sum_{t=0}^\infty \beta^t \frac{c_t^{1-\sigma}-1}{1-\sigma}$$ subject to the capital accumulation equation $$k_{t+1} = A k_t^\alpha + (1-\delta) k_t - c_t$$ with initial capital $k_0 > 0$. 3. **Step 1: Derive the first-order conditions (FOCs)** Set up the Lagrangian: $$\mathcal{L} = \sum_{t=0}^\infty \beta^t \frac{c_t^{1-\sigma}-1}{1-\sigma} + \sum_{t=0}^\infty \lambda_t \left(A k_t^\alpha + (1-\delta) k_t - c_t - k_{t+1} \right)$$ FOCs w.r.t. $c_t$: $$\frac{\partial \mathcal{L}}{\partial c_t} = \beta^t c_t^{-\sigma} - \lambda_t = 0 \implies \lambda_t = \beta^t c_t^{-\sigma}$$ FOCs w.r.t. $k_{t+1}$: $$\frac{\partial \mathcal{L}}{\partial k_{t+1}} = -\lambda_t + \lambda_{t+1} \left( A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right) = 0$$ This implies $$\lambda_t = \lambda_{t+1} \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right)$$ Substitute the expression for $\lambda_t$: $$\beta^t c_t^{-\sigma} = \beta^{t+1} c_{t+1}^{-\sigma} \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right)$$ Divide both sides by $\beta^t$: $$c_t^{-\sigma} = \beta c_{t+1}^{-\sigma} \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right)$$ 4. **Euler Equation:** $$\boxed{c_t^{-\sigma} = \beta c_{t+1}^{-\sigma} \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right)}$$ This equation characterizes optimal consumption intertemporal choice. 5. **Step 2: Express as two-dimensional dynamic system** Define state variable $k_t$ and control variable $c_t$. From capital accumulation: $$k_{t+1} = A k_t^\alpha + (1-\delta) k_t - c_t$$ Rewrite the Euler equation: $$c_{t+1} = \left[ \beta \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right) \right]^{\frac{1}{\sigma}} c_t$$ Hence, system: $$\begin{cases} k_{t+1} = A k_t^\alpha + (1 - \delta) k_t - c_t \\ c_{t+1} = \left[ \beta \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right) \right]^{\frac{1}{\sigma}} c_t \end{cases}$$ 6. **Step 3: Steady State $(k^*, c^*)$** At steady state $k_{t+1}=k_t=k^*$ and $c_{t+1}=c_t=c^*$. From capital accumulation: $$k^* = A (k^*)^\alpha + (1-\delta) k^* - c^* \implies c^* = A (k^*)^\alpha - \delta k^*$$ From Euler equation: $$1 = \beta \left(A \alpha (k^*)^{\alpha-1} + 1 - \delta \right)$$ Rearranged: $$A \alpha (k^*)^{\alpha-1} = \frac{1}{\beta} - 1 + \delta$$ Solve for $k^*$: $$k^* = \left( \frac{\frac{1}{\beta} - 1 + \delta}{A \alpha} \right)^{\frac{1}{\alpha -1}}$$ Then $$c^* = A (k^*)^\alpha - \delta k^*$$ 7. **Step 4: Linearize system** Define functions: $$ F(k,c) = A k^\alpha + (1-\delta) k - c $$ $$ G(k', c) = \left[ \beta \left(A \alpha (k')^{\alpha -1} + 1 - \delta \right) \right]^{\frac{1}{\sigma}} c $$ Variables are $x_t = (k_t, c_t)$. Write linearization around $(k^*, c^*)$: Jacobian matrix $J$ with blocks $$ J = \begin{pmatrix} \frac{\partial k_{t+1}}{\partial k_t} & \frac{\partial k_{t+1}}{\partial c_t} \\ \frac{\partial c_{t+1}}{\partial k_t} & \frac{\partial c_{t+1}}{\partial c_t} \end{pmatrix} $$ Calculate derivatives: - $$\frac{\partial k_{t+1}}{\partial k_t} = A \alpha (k^*)^{\alpha -1} + (1-\delta)$$ - $$\frac{\partial k_{t+1}}{\partial c_t} = -1$$ Note $c_{t+1} = H(k_{t+1}) c_t$ where $$H(k_{t+1}) = \left[ \beta \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right) \right]^{\frac{1}{\sigma}}$$ Chain rule: - $$\frac{\partial c_{t+1}}{\partial k_t} = c_t H'(k_{t+1}) \frac{\partial k_{t+1}}{\partial k_t}$$ - $$\frac{\partial c_{t+1}}{\partial c_t} = H(k^*) + c_t H'(k^*) \frac{\partial k_{t+1}}{\partial c_t}$$ Where $$H'(k) = \frac{1}{\sigma} \left[ \beta \left(A \alpha k^{\alpha -1} + 1 - \delta \right) \right]^{\frac{1}{\sigma} - 1} \beta A \alpha (\alpha -1) k^{\alpha -2}$$ 8. **Step 5: Local stability analysis** Eigenvalues of $J$ determine stability. The steady state is a saddle path if one eigenvalue lies inside and one outside the unit circle. 9. **Step 6: Impulse Responses (Bonus)** Suppose $A_t = A (1 + \epsilon_t)$ with small shock $\epsilon_t$, temporary at $t=0$. Linearize system to compute how $k_t, c_t$ deviate from steady state after shock. Typically, consumption adjusts slowly due to smoothing ($\sigma$), while capital reacts with delay via accumulation. Impulse response functions (IRFs) show initial drop/increase in consumption, and gradual adjustment of capital. --- Final Euler equation: $$\boxed{c_t^{-\sigma} = \beta c_{t+1}^{-\sigma} \left(A \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right)}$$ Steady state: $$k^* = \left( \frac{\frac{1}{\beta} - 1 + \delta}{A \alpha} \right)^{\frac{1}{\alpha -1}}, \quad c^* = A (k^*)^\alpha - \delta k^*$$