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I've been looking at the cake eating problem over a finite horizion and have been trying to figure out if we can derive a policy function for such a problem. My work is written below.



sequence form of such a problem is:

$$\max _{c_t,x_{t+1}}U(c_t)=\sum_{i=0}^T\beta^tu(c_t),\ \ \ \ \ \ 1>\beta>0$$ $$s.t. x_{t+1}=x_{t}-c_t$$

supposing that the instantaneous utility function for such preferences are $u(c_t)=\ln(c_t)$ it follows that the function equation for such a problem is:

$$v(x_{t+1})=\max_{x_{t+1}}\{\ln(x_t-x_{t+1})+\beta v(x_{t+1})\}$$

In this case im supposing $T=2$ (we have three periods starting from period zero) and suppose $x_0>0$. Solving the problem from backwards induction we know. $$v(x_2)=\ln(x_2)$$ it follows from this result that for the previous periods value function we have: $$v(x_1)=\max_{x_2}\{\ln(x_1-x_2)+\beta v(x_2)\}$$ $$v(x_1)=\max_{x_2}\{\ln(x_1-x_2)+\beta \ln(x_2)\}$$

after taking the first order condition for this result and solving for $x_2$ we get: $$x^* _2=\frac{\beta x_1}{1+\beta}$$

we know now our value function is properly defined as:

$$v(x_1)=\ln\left(x_1-\frac{\beta x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta} \right)$$

Simplifying: $$v(x_1)=\ln\left(\frac{ x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta}\right) $$
For our $v(x_0)$ we follow the same procedure before

$$v(x_0)=\max_{x_1}\{\ln(x_0-x_1)+\beta v(x_1)\}$$ $$v(x_0)=\max_{x_1} \left \{\ln(x_0-x_1)+\beta \left[\ln\left(\frac{ x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta}\right) \right]\right\}$$

upon solving such an optimization problem we find:

$$x_1^*=\frac{(\beta+\beta^2)x_0}{1+\beta+\beta^2}$$

From this work that I've done above we can see that the policy functions change each period.



Given the solving above do time invariant policy functions exist?

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  • $\begingroup$ Hi: That's neat what you did to solve it. It would be interesting to see what the pattern is if you add another period. your $x^{*}_1$ solution can obviously be writtten as $\frac{\beta\times(1+\beta)}{1+ \beta + \beta^2} x_{0}$ so maybe there is a pattern as you add more periods. $\endgroup$ – mark leeds Dec 25 '19 at 0:12
  • $\begingroup$ For example, maybe $x^{*}_2 = \frac{\beta^2 \times (1+\beta) x_{0}}{1+\beta + \beta^2 + \beta^3} $. Just a guess but you never know. $\endgroup$ – mark leeds Dec 25 '19 at 0:16
  • $\begingroup$ @markleeds you'd go about subbing $x_1^*$ into $x_2^*$ making it a function only of initial cake size and parameters. in this case $x_2^*=\frac{\beta^2 x_0}{1+\beta+\beta^2}$ it could be that such a pattern exists. You would have to use induction for that. $\endgroup$ – EconJohn Dec 25 '19 at 3:47
  • $\begingroup$ gotcha. thanks for interesting problem. $\endgroup$ – mark leeds Dec 25 '19 at 5:48
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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 found in a number of textbooks (Adda and Cooper, Sargent, as well as Stokey/Lucas).

In your example, the number of periods left determine the solution structure.

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  • $\begingroup$ Can you direct me to a specific quote in a textbook/page number? $\endgroup$ – EconJohn Dec 24 '19 at 23:37
  • $\begingroup$ The technical conditions can be found here (Blackwell's sufficient conditions etc.): web.stanford.edu/~pkurlat/teaching/… $\endgroup$ – ChinG Dec 24 '19 at 23:45
  • $\begingroup$ Essentially, what you are asking, is that what makes a problem stationary. $\endgroup$ – ChinG Dec 24 '19 at 23:46
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    $\begingroup$ The intuition is excellent. Thank you! $\endgroup$ – EconJohn Dec 25 '19 at 3:49

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