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You should be careful about the way you wrote your two equations because the $\max$ operators are different. In the first equation, you maximize an objective function on a set, meaning you look for the best value of $y$ given that $y$ belongs to a known set. You did not define this set. In the second case, you choose the maximum among two values. Contrary to ...


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The original problem was probably of the form $$\max_{\{W_t\}_{t=1}^\infty}\sum_{t=0}^\infty \beta^t u(W_t-W_{t+1}),$$ $$\mbox{s.t. } W_{t+1}\in[0,W_t] \ \forall \ t, ~~ W_0 \mbox{ given}$$ When formulated that way there is no need to solve for any function $V$, but the infinite elements of the sequence ${\{W_t\}}_{t=1}^\infty$ are all choice variables. ...


3

Starting with your original equation: $max_{c_t, m_t, b_t} E_0\sum_{t=0}^\infty U(c_t, m_t)$ s.t. (1) $y+\frac{m_{t-1}}{1+\pi_t}+\frac{1+i_{t-1}}{1+\pi_t}b_{t-1}=c_t+m_t+b_t+\tau_t$ Here: $R_{t-1} =1+i_{t-1}$ and $1+\pi_t=\frac{P_t}{P_{t-1}}$ Note that in this problem, you have have two state variables, $m_{t-1}$ and $b_{t-1}$, and your main issue have ...


3

The "second" constraint appears redundant and it confuses matters. Re-arrange the first one to obtain $$\tilde{a}_{t+1} = \big[\tilde a_t+(1-\delta )Y_t-\tilde{c}_t\big]R_t$$ This tells us that wealth in the beginning of next period is fully determined by current-period decisions and known states, without any uncertainty whatsoever: we start with the ...


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I recommend the manuscript here by Lawrence Evans. The related example is described as 'Rocket Railroad Car'. The first instance is on pages 9-12 where a geometric solution is provided. The choice set is not constrained to a discrete set in principle but the first part shows optimality when only the highest and lowest actions are chosen. On pages 35-36, ...


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The value iteration operator is a contraction with respect to the supremum norm. Your example probably provides a counterexample for the statement that it's a contraction with respect to the Manhattan norm.


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No, they cannot by definition. To derive a time-invariant policy function, you need to have an infinite horizon problem. This is because the structure of the solution remains the same, no matter when you look at it. Intuitively, this is because any sequence of periods from any time period to infinity looks the same. The technical conditions for this can be ...


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Thanks to @Boaten I was able to find the solution. For those interested here are the steps for deriving the Fisher relation: Combining the FOC $[b_{t}]$ and the envelope for $[b_{t-1}]$ we get $U_{c_{t}} = \beta E_{t}\frac{U_{c_{t+1}}(1+i_{t})}{1+\pi _{t+1}}$ Assuming utility is log in both arguments this can be simplified as $\frac{c_{t+1}}{c_{t}} = \...


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In my experience, it's mainly just for cleanliness for results. Consider an infinite horizon repeated game, with discounted payoff representation (where I use $\delta = (1-\lambda)$ in your notation) $$ (1-\delta)\sum_{t=0}^{\infty}\delta^t R_t $$ where $0 < \delta < 1$. Suppose I play a strategy that gives me the same payoff, say $a$, for each ...


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Are you referring to repeated games, in which player $i$ obtains a payoff of the form $$(1-\delta)\sum_{t=0}^{\infty}\delta^{t}u_{i}(x_{i}^{t})$$ from the sequence of payoffs $\left\{x_{i}^{t}\right\}$? The $(1-\delta)$ is appended to the front to give the average per period payoff.


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What you want to maximize is the discounted stream of utility over time. To that end, you can set control variables over time (subject to some constraints, like technology, budget...). We are ultimately interested in the optimal setting of these variables (called the optimal policy). Bellman's main insight is the principle of optimality (reminiscent of ...


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I am by no means an expert on this, but maybe this helps. Here is a simple example for a bellman equation $V(y) = \max_x u(x,y) + \beta V(y')$ $s.t. \, y' = f(x,y)$ This is a functional equation in an unknown function V. A solution to this problem is a function V that satisfies the equation above. If you look at the equation, it's pretty clear that the ...


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You iterate towards a fixed point, so you want to reach a situation where plugging in your current iterated value produces itself. Now using your notation, we are told that we should calculate $$V_{n+1}(a) = V_n(a) + \Delta$$ where $$\Delta = \ u(c(a^*)) + \dfrac{\partial V_n(a)}{\partial a}da_t(a^*) - \rho V_n(a)$$ Insert the second into the first to ...


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