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

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

Thanks in advance!

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  • 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.

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