<p>Suppose <span class="math-container">$x'&gt;x$</span>. We'll show that <span class="math-container">$g(x')\geq g(x)$</span>. Suppose that is not the case and we have <span class="math-container">$x'&gt;x$</span> but <span class="math-container">$g(x')&lt;g(x)$</span>. Since <span class="math-container">$0\leq g(x') &lt; g(x) \leq f(x)&lt; f(x')$</span>, <span class="math-container">$g(x')$</span> is feasible in case of <span class="math-container">$x$</span>, and <span class="math-container">$g(x)$</span> is feasible in case of <span class="math-container">$x'$</span>, consequently both of the following inequalities are true:
<span class="math-container">\begin{eqnarray*}V(x')=u(f(x')-g(x'))+\beta V(g(x')) &amp;\geq &amp; u(f(x')-g(x))+\beta V(g(x)) \\
V(x)=u(f(x)-g(x))+\beta V(g(x))&amp;\geq &amp; u(f(x)-g(x'))+\beta V(g(x'))\end{eqnarray*}</span></p>
<p>Adding them we get the following inequality:</p>
<p><span class="math-container">$u(f(x')-g(x'))+\beta V(g(x'))+u(f(x)-g(x))+\beta V(g(x))\geq u(f(x')-g(x))+\beta V(g(x))+u(f(x)-g(x'))+\beta V(g(x'))$</span></p>
<p>which yields
<span class="math-container">$u(f(x')-g(x'))-u(f(x)-g(x'))\geq u(f(x')-g(x))-u(f(x)-g(x))$</span></p>
<p>which, by strict concavity of <span class="math-container">$u$</span>, implies that
<span class="math-container">$g(x')\geq g(x)$</span> contradicting that <span class="math-container">$g(x') &lt; g(x)$</span>.</p>

CC BY-SA 4.0
2023-11-17T02:07:26.507Z
2023-11-17T03:00:15.387Z
2023-11-17T03:00:15.387Z
Amit
<p>I'm going to assume that everything is smooth and that the optimal solution <span class="math-container">$g(x)$</span> is interior (in <span class="math-container">$]0, f(x)[$</span>)</p>
<p>First you can show that the function <span class="math-container">$V$</span> is concave as the Bellman operator maps concave functions to concave functions.</p>
<p>Now assume that the solution <span class="math-container">$g(x)$</span> is interior. Then the first order condition (with respect to <span class="math-container">$y$</span>) gives:
<span class="math-container">$$
-u'(f(x) - g(x)) + \beta V'(g(x)) = 0.
$$</span>
Taking the derivative with respect to <span class="math-container">$x$</span> gives:
<span class="math-container">$$
-u''(f(x) - g(x))f'(x) + u''(f(x) - g(x))g'(x) + \beta V''(g(x)) g'(x) = 0,
$$</span>
So:
<span class="math-container">$$
g'(x) = \frac{u''(f(x) - g(x))f'(x)}{u''(f(x) - g(x)) + \beta V''(g(x))}.
$$</span>
Both numerator and denominator on the right side are negative, so <span class="math-container">$g'(x) &gt; 0$</span>.</p>

CC BY-SA 4.0
2023-11-16T07:10:26.473Z
Since $f$ is not assumed to be concave, it is not clear to me why the value function should be concave.
Michael Greinecker
CC BY-SA 4.0
2023-11-16T08:04:02.587Z
@Michael Greinecker yes indeed. One could probably change the state variable from $x$ to $f(x)$, then show that $V$ is concave in $f(x)$ and therefore $g$ is monotone in $f(x)$. As $f$ is monotone, $g$ will then also be increasing in $x$.
tdm
CC BY-SA 4.0
2023-11-16T08:29:45.687Z
Good point, that might work.
Michael Greinecker
CC BY-SA 4.0
2023-11-16T12:56:27.433Z
what if $u$ and $f$ are not assumed to be ddifferentiable?
Ludwig Gershwin
CC BY-SA 4.0
2023-11-16T06:25:57.420Z
2023-11-16T06:25:57.420Z
tdm
<p>Consider the following functional equation:
<span class="math-container">$$V(x)=\max_{y\in [0,f(x)]}[u(f(x)-y)+\beta V(y)]$$</span>
where <span class="math-container">$u$</span> is continuous, strictly increasing, and strictly concave; the function <span class="math-container">$f$</span> is continuous and strictly increasing, and <span class="math-container">$\beta\in (0,1)$</span>. Let <span class="math-container">$g(x)$</span> be the decision rule associated with this functional equation:
<span class="math-container">$$V(x)=u[f(x)-g(x)]+\beta V[g(x)]$$</span>
How can I show that <span class="math-container">$g$</span> is non-decreasing (if <span class="math-container">$x'&gt;x$</span>, then <span class="math-container">$g(x')\geq g(x)$</span>)?</p>
<p><strong>Assume that <span class="math-container">$f$</span> and <span class="math-container">$u$</span> are not necessarily differentiable.</strong></p>

CC BY-SA 4.0
2023-11-15T22:52:34.830Z
Why do you think it is?
Michael Greinecker
CC BY-SA 4.0
2023-11-16T00:57:40.847Z
@MichaelGreinecker I honestly have no idea. Could you give me a hint?
Ludwig Gershwin
<dynamic-programming>
How can I show that the policy function is non-decreasing?
Ludwig Gershwin
# How can I show that the policy function is non-decreasing?

(Asked by: username_0 on 2023-11-15)

Consider the following functional equation:
$$V(x)=\max\_{y\in [0,f(x)]}[u(f(x)-y)+\beta V(y)]$$
where $u$ is continuous, strictly increasing, and strictly concave; the function $f$ is continuous and strictly increasing, and $\beta\in (0,1)$. Let $g(x)$ be the decision rule associated with this functional equation:
$$V(x)=u[f(x)-g(x)]+\beta V[g(x)]$$
How can I show that $g$ is non-decreasing (if $x'>x$, then $g(x')\geq g(x)$)?

**Assume that $f$ and $u$ are not necessarily differentiable.**

**Tags:** dynamic-programming

> **Comments:**
> **username_1**: Why do you think it is?  
> **username_0**: @MichaelGreinecker I honestly have no idea. Could you give me a hint?  

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## 2 answers

### Answer by username_2 on 2023-11-17. Score: 2 | Accepted answer

Suppose $x'>x$. We'll show that $g(x')\geq g(x)$. Suppose that is not the case and we have $x'>x$ but $g(x'). Since $0\leq g(x') < g(x) \leq f(x)< f(x')$, $g(x')$ is feasible in case of $x$, and $g(x)$ is feasible in case of $x'$, consequently both of the following inequalities are true:
\begin{eqnarray\*}V(x')=u(f(x')-g(x'))+\beta V(g(x')) &\geq & u(f(x')-g(x))+\beta V(g(x)) \\
V(x)=u(f(x)-g(x))+\beta V(g(x))&\geq & u(f(x)-g(x'))+\beta V(g(x'))\end{eqnarray\*}

Adding them we get the following inequality:

$u(f(x')-g(x'))+\beta V(g(x'))+u(f(x)-g(x))+\beta V(g(x))\geq u(f(x')-g(x))+\beta V(g(x))+u(f(x)-g(x'))+\beta V(g(x'))$

which yields
$u(f(x')-g(x'))-u(f(x)-g(x'))\geq u(f(x')-g(x))-u(f(x)-g(x))$

which, by strict concavity of $u$, implies that
$g(x')\geq g(x)$ contradicting that $g(x') < g(x)$.


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### Answer by username_3 on 2023-11-16. Score: 1

I'm going to assume that everything is smooth and that the optimal solution $g(x)$ is interior (in $]0, f(x)[$)

First you can show that the function $V$ is concave as the Bellman operator maps concave functions to concave functions.

Now assume that the solution $g(x)$ is interior. Then the first order condition (with respect to $y$) gives:
$$
-u'(f(x) - g(x)) + \beta V'(g(x)) = 0.
$$
Taking the derivative with respect to $x$ gives:
$$
-u''(f(x) - g(x))f'(x) + u''(f(x) - g(x))g'(x) + \beta V''(g(x)) g'(x) = 0,
$$
So:
$$
g'(x) = \frac{u''(f(x) - g(x))f'(x)}{u''(f(x) - g(x)) + \beta V''(g(x))}.
$$
Both numerator and denominator on the right side are negative, so $g'(x) > 0$.

> **Comments:**
> **username_1**: Since $f$ is not assumed to be concave, it is not clear to me why the value function should be concave.  
> **username_3**: @username_1 yes indeed. One could probably change the state variable from $x$ to $f(x)$, then show that $V$ is concave in $f(x)$ and therefore $g$ is monotone in $f(x)$. As $f$ is monotone, $g$ will then also be increasing in $x$.  
> **username_1**: Good point, that might work.  
> **username_0**: what if $u$ and $f$ are not assumed to be ddifferentiable?  

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