Timeline for Extreme Value Theorem in Economics
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 7, 2021 at 10:59 | history | edited | Michael Greinecker | CC BY-SA 4.0 |
Added detail
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| Dec 12, 2020 at 18:54 | comment | added | develarist | the talk about optimal plans might suggest looking into optimal mass transportation theory using the Wasserstein distance via linear programming, which I have not seen applied to EVT problems yet | |
| Nov 3, 2020 at 20:54 | comment | added | Michael Greinecker | @AndrewCoyle What ultimately matters is the possible consumption paths and in defining them, you need to take into account which sector one invests in influences the resulting paths. But the set of feasible consumption paths will still be compact. | |
| Nov 3, 2020 at 20:43 | comment | added | EB3112 | Thanks Michael. With two sectors, would C and K still remain the only concerns in the feasible set? | |
| Nov 3, 2020 at 19:44 | comment | added | Michael Greinecker | @AndrewCoyle None of that should cause a problem. Things are actually easier with a finite time horizon and you would not need the assumption that $u$ is bounded. | |
| Nov 3, 2020 at 18:54 | comment | added | EB3112 | Can I ask what difference would be made to this problem was made a two-sector problem (with a final goods sector, and say, some intermediate sector), or the fact that the problem is in finite-horizon? I see for example a lot of reference to infinity above. | |
| Nov 2, 2020 at 23:15 | comment | added | Michael Greinecker | @Michael Exactly. If $u$ was unbounded, one would need to make some assumption that guarantees that utility cannot grow too fast along a feasible path. | |
| Nov 2, 2020 at 22:52 | comment | added | Michael | @MichaelGreinecker Right, my bad. Compact sets can be of the form $\Pi_i [-a_i, a_i]$ where $a_i \rightarrow \infty$. (The example then seems a bit funny economically. The production function gives "free lunch" but since $u$ is bounded the problem is well-posed.) | |
| Nov 2, 2020 at 0:28 | comment | added | Michael Greinecker | @Michael Compactness in the product topology is very different than compactness in the norm topology on $\ell^\infty$. The product topology is the topology of pointwise convergence in contrast to the norm topology on $\ell^\infty$, which is the topology of uniform convergence. The production function in your example causes no problems. | |
| Nov 1, 2020 at 22:05 | comment | added | Michael | Seems that additional assumption is needed to ensure that $f(k) \leq k$, which then implies $F$ is bounded in $l^{\infty}$. E.g. take $f(k) = \sqrt{k} + y$ where $y > 0$ is a constant, which is concave. Take $c_t = 0$ and $k_{t+1} = f(k_t)$, then the feasible consumption/production plan $\{ (0, k_t) \}_{t\geq 0}$ would not lie in a compact set of the Tychonoff type. (That $f(0) = y >0$ is admittedly uneconomical, and the resulting problem is not well-posed. Requiring $f(0) = 0$ would address the both issues.) | |
| Nov 1, 2020 at 14:16 | comment | added | Michael Greinecker | @AndrewCoyle It is not the production function that is compact, it is the set of feasible production ad consumption paths. | |
| Oct 31, 2020 at 13:48 | comment | added | EB3112 | Hi Michael, could I just confirm with you that I am understanding the point above? Namely, the utility function is well-defined and continuous; and the production function is compact (by Tychonoff's theorem). Therefore the EVT is confirmed/an optimal plan exists. | |
| Oct 31, 2020 at 1:11 | comment | added | Michael Greinecker | @Michael The set $\prod_{t=1}^\infty [0,f^{t-1}(k)]^2$ is compact by Tychonoff's theorem and the continuity of $f$ implies that $F$ is a closed subset. | |
| Oct 31, 2020 at 0:41 | comment | added | Michael | Why is $F$ compact? | |
| Oct 30, 2020 at 20:16 | comment | added | EB3112 | Thank you Michael, very informative! Alongside that, thank you for the book recomendation! | |
| Oct 30, 2020 at 15:49 | history | answered | Michael Greinecker | CC BY-SA 4.0 |