Timeline for Euler equation through tangency conditions
Current License: CC BY-SA 4.0
10 events
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| Apr 2, 2019 at 13:58 | history | edited | user17900 | CC BY-SA 4.0 |
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| Apr 2, 2019 at 13:51 | history | edited | user17900 | CC BY-SA 4.0 |
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| Apr 24, 2018 at 10:32 | comment | added | user17900 | My point was that $\frac{0}{\beta f'(k+1)-1} = 0$, as another commenter has now pointed out, which is why your equation implies that $U'(c_t) = 0$ (which is not true at the optimum). | |
| Apr 23, 2018 at 21:19 | comment | added | ptr64 | I never implied $U'(c_t) = 0$. I just argued that at equilibrium the resource constraint should bind, meaning we should be able to use the gradient to get equilibrium conditions (one of which is the Euler equation). The gradient is given by the partials with respect to the controls right (ie $c_t, k_{t+1}$)?$\nabla U'(c_t) = \langle \beta^t u'(c_t), 0 \rangle$. The second should be zero since $c$ and $k$ are independent. My issue was that I kept getting the wrong subscripts using gradients. | |
| Apr 23, 2018 at 14:53 | comment | added | user17900 | @ptr. I am not completely clear on how you tried to solve the problem. However, the approach is not correct: the equation you get actually implies that $U'(C_t)=0$, which is incorrect. (Though in a very trivial way would actually imply the Euler equation!) Instead, I would suggest you approach the problem using either of the two methods I outlined. | |
| Apr 23, 2018 at 14:53 | comment | added | Maarten Punt | @afreelunch even if I might (or might not) agree with you I think your last remark about macroeconomics is uncalled for. | |
| Apr 23, 2018 at 14:42 | history | edited | user17900 | CC BY-SA 3.0 |
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| Apr 23, 2018 at 12:58 | vote | accept | ptr64 | ||
| Apr 23, 2018 at 12:57 | comment | added | ptr64 | Thank you, you answered my first question. But my second question was whether we can use the gradient of the utility function and the gradient of the resource constraint to use the tangency conditions. I just can't seem to find where my algebra is wrong. | |
| Apr 23, 2018 at 11:26 | history | answered | user17900 | CC BY-SA 3.0 |