<p>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).</p>

<p>In your example, the number of periods left determine the solution structure.</p>

CC BY-SA 4.0
2019-12-24T23:37:38.303Z
Can you direct me to a specific quote in a textbook/page number?
EconJohn
CC BY-SA 4.0
2019-12-24T23:45:39.603Z
The technical conditions can be found here (Blackwell's sufficient conditions etc.): https://web.stanford.edu/~pkurlat/teaching/5%20-%20The%20Bellman%20Equation.pdf
ChinG
CC BY-SA 4.0
2019-12-24T23:46:24.917Z
Essentially, what you are asking, is that what makes a problem stationary.
ChinG
CC BY-SA 4.0
2019-12-25T03:49:13.883Z
The intuition is excellent. Thank you!
EconJohn
CC BY-SA 4.0
2019-12-24T23:11:54.130Z
2019-12-24T23:11:54.130Z
ChinG
<p>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.</p>

<hr>

<hr>

<p>sequence form of such a problem is:</p>

<p><span class="math-container">$$\max _{c_t,x_{t+1}}U(c_t)=\sum_{i=0}^T\beta^tu(c_t),\ \ \ \ \ \ 1&gt;\beta&gt;0$$</span>
<span class="math-container">$$s.t. x_{t+1}=x_{t}-c_t$$</span></p>

<p>supposing that the instantaneous utility function for such preferences are <span class="math-container">$u(c_t)=\ln(c_t)$</span> it follows that the function equation for such a problem is:</p>

<p><span class="math-container">$$v(x_{t+1})=\max_{x_{t+1}}\{\ln(x_t-x_{t+1})+\beta v(x_{t+1})\}$$</span></p>

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

<p>after taking the first order condition for this result and solving for <span class="math-container">$x_2$</span> we get:
<span class="math-container">$$x^* _2=\frac{\beta x_1}{1+\beta}$$</span></p>

<p>we know now our value function is properly defined as:</p>

<p><span class="math-container">$$v(x_1)=\ln\left(x_1-\frac{\beta x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta} \right)$$</span></p>

<p>Simplifying:
<span class="math-container">$$v(x_1)=\ln\left(\frac{ x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta}\right) $$</span>
<br>
For our <span class="math-container">$v(x_0)$</span> we follow the same procedure before</p>

<p><span class="math-container">$$v(x_0)=\max_{x_1}\{\ln(x_0-x_1)+\beta v(x_1)\}$$</span>
<span class="math-container">$$v(x_0)=\max_{x_1} \left \{\ln(x_0-x_1)+\beta \left[\ln\left(\frac{ x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta}\right) \right]\right\}$$</span></p>

<p>upon solving such an optimization problem we find:</p>

<p><span class="math-container">$$x_1^*=\frac{(\beta+\beta^2)x_0}{1+\beta+\beta^2}$$</span></p>

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

<hr>

<hr>

<p>Given the solving above do time invariant policy functions exist? </p>

CC BY-SA 4.0
2019-12-25T00:12:58.547Z
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
CC BY-SA 4.0
2019-12-25T00:16:09.627Z
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
CC BY-SA 4.0
2019-12-25T03:47:57.753Z
@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
CC BY-SA 4.0
2019-12-25T05:48:42.493Z
gotcha. thanks for interesting problem.
mark leeds
<macroeconomics><bellman-equations><recursive-macroeconomics>
Do policy functions exist for Finite Horizon Dynamic programming problems?
EconJohn
# Do policy functions exist for Finite Horizon Dynamic programming problems?

(Asked by: username_0 on 2019-12-24)

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?

**Tags:** macroeconomics, bellman-equations, recursive-macroeconomics

> **Comments:**
> **username_1**: 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.  
> **username_1**: 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.  
> **username_0**: @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.  
> **username_1**: gotcha. thanks for interesting problem.  

---

## 1 answers

### Answer by username_2 on 2019-12-24. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Can you direct me to a specific quote in a textbook/page number?  
> **username_2**: The technical conditions can be found here (Blackwell's sufficient conditions etc.): https://web.stanford.edu/~pkurlat/teaching/5%20-%20The%20Bellman%20Equation.pdf  
> **username_2**: Essentially, what you are asking, is that what makes a problem stationary.  
> **username_0**: The intuition is excellent. Thank you!  

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