# Do policy functions exist for Finite Horizon Dynamic programming problems?

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?

• 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. – mark leeds Dec 25 '19 at 0:12
• 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. – mark leeds Dec 25 '19 at 0:16
• @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. – EconJohn Dec 25 '19 at 3:47
• gotcha. thanks for interesting problem. – mark leeds Dec 25 '19 at 5:48