<p>When you take the integral of a $\max_{\{\cdot\}}$ operator, I think you have to split the integral into two separate integrals with different supports on them.</p>

<p>Even if your value function is complicated and there is no differentiability, you only need continuity for an existence of a solution to solve the optimization problem.</p>

CC BY-SA 3.0
2017-10-09T18:35:24.980Z
2017-10-09T18:35:24.980Z
Kitsune Cavalry
<p>Here is my attempt, where I assume an absolute upper limit on the support of $F$, $F(\overline{w})=1$, for simplicity.</p>

<p>Rewrite the first equation as
\begin{equation}
rV_1(w)=
w+\lambda_1\int_w^{\overline{w}}V_1(\tilde{x})dF(\tilde{x})
+\underbrace{\lambda_1\int_0^{w}V_1(w)dF(\tilde{x})}_{I}
-\lambda_1\int_0^{\overline{w}}V_1(w)dF(\tilde{x})
+\delta[V_0-V_1(w)] \ ,
\end{equation}
whereby
\begin{equation}
-\lambda_1\int_0^{\overline{w}}V_1(w)dF(\tilde{x})=
-\lambda_1\int_w^{\overline{w}}V_1(w)dF(\tilde{x})
-\underbrace{\lambda_1\int_0^{w}V_1(w)dF(\tilde{x})}_{II} \ .
\end{equation}</p>

<p>The terms $I$ and $II$ cancel out, so that arranging gives
\begin{equation}
(\delta +r)V_1(w)=
w+\lambda_1\int_w^{\overline{w}}[V_1(\tilde{x})-V_1(w)]dF(\tilde{x})
+\delta V_0 \ .
\end{equation}
If we apply Leibniz' rule know, we get
\begin{equation}
(\delta +r)V_1'(w)=
1-\lambda_1\int_w^{\overline{w}}V_1'(w)dF(\tilde{x})=1-\lambda_1V_1'(w)[1-F(w)] \ ,
\end{equation}
whereby the last equality follows from $F(\overline{w})=1$. Solving for $V_1'(w)$ gives the desired solution.</p>

CC BY-SA 3.0
2018-02-22T11:11:55.903Z
2018-02-22T11:11:55.903Z
daniel_JKU
<p>I'm currently making my way through Burdett and Mortensen's classic paper of on the job search. What should be an easy task of finding an expression for the reservation wage is made slightly more complicated by the presence of the max operator.  We are faced with the following Bellman equation for the value of a job paying a wage $w$. The bellman equations are standard. The value of a job paying $w$ consists of the wage $w$ plus the expected gain from searching and finding a better job discounted by the probability a job offer comes along $\lambda_1$ plus the loss of due to becoming unemployed when the job is destroyed at rate $\delta$. The value of unemployment $V_0$ consists of unemployment benefit $b$ plus the expected gain from becoming employed discounted by the probability an offer comes along $\lambda_0$. Note the probability an offer is made is different depending on whether someone is already employed or unemployed. The distribution of offers is given by $F$ \begin{equation} rV_1(w)=w+\lambda_1\bigg[\int \max\{V_1(w),V_1(\tilde{x})\}-V_1(w)\bigg]\;dF(\tilde{x})+\delta [V_0-V_1(w)]   \end{equation} \begin{equation}rV_0=b+\lambda_0 \bigg[\int \max\{V_0,V_1(\tilde{x})\}\;dF(\tilde{x})-V_0\bigg]\end{equation} Since $V_1(w)$ is increasing in $w$ and $V_0$ is independent of it we know a reservation wage exists such that if $w&gt;R\implies V_1(w)&gt;V_0$, $w&lt;R\implies V_1(w)&lt;V_0$ and $V_1(R)=V_0$. Standard arguments (integration by parts) shows that \begin{equation} R-b=(\lambda_0-\lambda_1)\int_R^\infty V_1'(\tilde{x})[1-F(\tilde{x})]\;d\tilde{x} \end{equation} from here I would like to take the derivative of the first equation and solve for $V_1'(w)$. However If I use <a href="https://en.wikipedia.org/wiki/Leibniz_integral_rule" rel="noreferrer">Leibniz integration rule</a> I need the integrand to be differentiable. The max of two continuous functions is usually not differentiable where they are equal so I have a problem. If I assume that I integrate over all $\tilde{x}\geq w$ then $V_1(\tilde{x})\geq V_1(w)$ (wage offers that will be induce a worker to switch jobs) and the result follows by Leibniz rule. But there are wages in the distribution that won't be accepted and this derivative wont hold. The derivative is \begin{equation} V'(\tilde{x})=\frac{1}{r+\delta+\lambda_1(1-F(\tilde{x}))} \end{equation} I imagine I'm missing something but I'm not sure what. If anyone could give me any advice I would really appreciate it. </p>

<labor-economics><search-and-matching>
Differentiation of the value function in Burdett Mortensen (1998)
mark
# Differentiation of the value function in Burdett Mortensen (1998)

(Asked by: username_0 on 2017-09-16)

I'm currently making my way through Burdett and Mortensen's classic paper of on the job search. What should be an easy task of finding an expression for the reservation wage is made slightly more complicated by the presence of the max operator. We are faced with the following Bellman equation for the value of a job paying a wage $w$. The bellman equations are standard. The value of a job paying $w$ consists of the wage $w$ plus the expected gain from searching and finding a better job discounted by the probability a job offer comes along $\lambda\_1$ plus the loss of due to becoming unemployed when the job is destroyed at rate $\delta$. The value of unemployment $V\_0$ consists of unemployment benefit $b$ plus the expected gain from becoming employed discounted by the probability an offer comes along $\lambda\_0$. Note the probability an offer is made is different depending on whether someone is already employed or unemployed. The distribution of offers is given by $F$ \begin{equation} rV\_1(w)=w+\lambda\_1\bigg[\int \max\{V\_1(w),V\_1(\tilde{x})\}-V\_1(w)\bigg]\;dF(\tilde{x})+\delta [V\_0-V\_1(w)] \end{equation} \begin{equation}rV\_0=b+\lambda\_0 \bigg[\int \max\{V\_0,V\_1(\tilde{x})\}\;dF(\tilde{x})-V\_0\bigg]\end{equation} Since $V\_1(w)$ is increasing in $w$ and $V\_0$ is independent of it we know a reservation wage exists such that if $w>R\implies V\_1(w)>V\_0$, $wLeibniz integration rule I need the integrand to be differentiable. The max of two continuous functions is usually not differentiable where they are equal so I have a problem. If I assume that I integrate over all $\tilde{x}\geq w$ then $V\_1(\tilde{x})\geq V\_1(w)$ (wage offers that will be induce a worker to switch jobs) and the result follows by Leibniz rule. But there are wages in the distribution that won't be accepted and this derivative wont hold. The derivative is \begin{equation} V'(\tilde{x})=\frac{1}{r+\delta+\lambda\_1(1-F(\tilde{x}))} \end{equation} I imagine I'm missing something but I'm not sure what. If anyone could give me any advice I would really appreciate it.

**Tags:** labor-economics, search-and-matching

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## 2 answers

### Answer by username_1 on 2017-10-09. Score: 2

When you take the integral of a $\max\_{\{\cdot\}}$ operator, I think you have to split the integral into two separate integrals with different supports on them.

Even if your value function is complicated and there is no differentiability, you only need continuity for an existence of a solution to solve the optimization problem.


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### Answer by username_2 on 2018-02-22. Score: 0

Here is my attempt, where I assume an absolute upper limit on the support of $F$, $F(\overline{w})=1$, for simplicity.

Rewrite the first equation as
\begin{equation}
rV\_1(w)=
w+\lambda\_1\int\_w^{\overline{w}}V\_1(\tilde{x})dF(\tilde{x})
+\underbrace{\lambda\_1\int\_0^{w}V\_1(w)dF(\tilde{x})}\_{I}
-\lambda\_1\int\_0^{\overline{w}}V\_1(w)dF(\tilde{x})
+\delta[V\_0-V\_1(w)] \ ,
\end{equation}
whereby
\begin{equation}
-\lambda\_1\int\_0^{\overline{w}}V\_1(w)dF(\tilde{x})=
-\lambda\_1\int\_w^{\overline{w}}V\_1(w)dF(\tilde{x})
-\underbrace{\lambda\_1\int\_0^{w}V\_1(w)dF(\tilde{x})}\_{II} \ .
\end{equation}

The terms $I$ and $II$ cancel out, so that arranging gives
\begin{equation}
(\delta +r)V\_1(w)=
w+\lambda\_1\int\_w^{\overline{w}}[V\_1(\tilde{x})-V\_1(w)]dF(\tilde{x})
+\delta V\_0 \ .
\end{equation}
If we apply Leibniz' rule know, we get
\begin{equation}
(\delta +r)V\_1'(w)=
1-\lambda\_1\int\_w^{\overline{w}}V\_1'(w)dF(\tilde{x})=1-\lambda\_1V\_1'(w)[1-F(w)] \ ,
\end{equation}
whereby the last equality follows from $F(\overline{w})=1$. Solving for $V\_1'(w)$ gives the desired solution.


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