I’m struggling with a problem regarding the beta delta agent.

enter image description here

Can someone give me a hint how I can come up with the levels of utility?

Thanks in advance!

enter image description here

  • 2
    $\begingroup$ Can you show us your attempts so far? I (and many others) are happy to help, but we don't want to do the work for you. $\endgroup$ – user11305 Jun 22 '18 at 1:54
  • $\begingroup$ So sorry, of course! I uploaded a picture of my attempts up top, I hope that gives a better overview. I think that it is wrong though, because for the "partying case" I combined the cost for writing (-8) and the additional value (+3) in one period. But for the "non partying case" I split the cost and the additional value of +10. If I would've combined both of them, I would come up with a result of 2*delta > 0, which doesn't make any sense to me. Thank you so much for you help! $\endgroup$ – Christian Schmitt Jun 22 '18 at 10:28

One caveat, I've written this up while writing the world cup so there may be plenty of mistakes. In addition, I confess that the problem is not terribly transparent. However, here is how I'd interpret it.

First, when viewed in period $0$, the discounted weights for utilities accrued are $1, \beta \delta, \beta \delta^2$. I make the following assumption: that the given utilities for the various actions are realized immediately--this, I think, is what the writer of this problem wants, but it is not very clear (and indeed, I think it would actually be more reasonable from a modeling perspective if the essay benefit were realized at the end).

Write $P$ for "Attend Party", write $E$ for "Write Essay" and $0$ for "Do Nothing". Write a Plan as a triple $\cdot, \cdot, \cdot$.

I suppose that the agent cannot go to the party and write the essay on the same night. Moreover, I suppose that there is party on Sunday night $t=2$ (which may or may not be realistic). I also assume that if he parties on $t=0$, he is so hung over that he gets only $3$ addition value for writing the essay on $t=2$. If not, and he recovered fully, the problem would be trivial: he'd party on $t=0$ and write the essay on Sunday. To see this note that his optimal choice could be narrowed down to $P, 0, E$, which yields $10 + 2\beta\delta^2$ or $E, P, 0$, which yields $2 + 10\beta \delta$. Clearly $P, 0, E \succeq E,P,0$ iff $10 + 2\beta\delta^2 \geq 2 + 10\beta \delta$ iff $\delta \leq 1$, which is always true (with strict inequality if $\delta \neq 1$). Furthermore, note that both of these plans would be followed by the agent at $t=1$ and $t=2$; no commitment device is needed.

The agent has the following plans.

$$\begin{split} &0, E, 0\\ &0, 0, E\\ &0, P, E\\ &P, 0, E\\ &P, E, 0\\ &E, P, 0\\ &E, 0, 0\\ &E, 0, P\\ &0, E, P \end{split}$$

These yield to him the following utilities (when viewed from period $0$, should he be able to commit to this plan):

$$\begin{split} &2\beta\delta\\ &2\beta\delta^2\\ &10\beta\delta - 5\beta\delta^2\\ &10 - 5\beta\delta^2\\ &10 - 5\beta\delta\\ &2 + 10\beta\delta\\ &2\\ &2 + 10\beta\delta^2\\ &2\beta\delta + 10\beta\delta^2 \end{split}$$

The possible commitment solutions are:

$$\begin{split} &P, 0, E\\ &E, P, 0\\ \end{split}$$

These are also clearly optimal for the $t=1$ and $t=2$ player as well. Finally, we have $P, 0, E \succeq E, P, 0$ iff $10 - 5\beta\delta^2 \geq 2 + 10\beta\delta$, which holds iff $$\delta \leq \frac{\left(\sqrt{5}\cdot\sqrt{\frac{\left(5\beta+8\right)}{\beta}}-5\right)}{5}$$

Again, to whomever: please feel free to make any edits or corrections necessary.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.