<p>$\beta$ appears to be a <em>decision</em> variable, <strong>at firm level</strong>. Then its value should be determined optimally, under some criterion. So what is the criterion, what is the objective function which should be optimized over the value of $\beta$?</p>

<p>It appears that we should maximize the value of the option $V(\beta)$</p>

<p>$$V(\beta) = \frac{-c + q(\theta)(1-2\beta)J}{\rho - q(\theta)\beta}$$</p>

<p>The first derivative is (I will dispense with $\theta$)</p>

<p>$$\frac {\partial V(\beta)}{\partial \beta} = \frac{-2qJ\cdot[\rho - q\beta]+q\cdot[-c + q(1-2\beta)J]}{[\rho - q\beta]^2}$$</p>

<p>$$\implies \frac {\partial V(\beta)}{\partial \beta} = \frac{-2qJ\rho -cq + q^2J}{[\rho - q\beta]^2} = \frac{q}{[\rho - q\beta]^2}\cdot (-2J\rho -c + qJ)$$</p>

<p>$$\implies {\rm sign}\frac {\partial V(\beta)}{\partial \beta} = {\rm sign}\big\{(q-2\rho)J-c\big\} \tag {1}$$</p>

<p>So $V(\beta)$ is monotonic in $\beta$. Therefore, depending on the values taken by $J,\rho, c, q$, it will be optimal to set $\beta$ equal to either $0$ or $1$.  </p>

<p>Assume now that $\partial V(\beta)/\partial \beta &gt;0$. This requires that</p>

<p>$$(q-2\rho)J-c &gt; 0 \implies q &gt; \frac {c}{J} + 2\rho \implies q &gt; \rho \tag {2}$$</p>

<p>Then $\beta^* =1$ and</p>

<p>$$V^*|_{\beta =1} = \frac{-c - qJ}{\rho - q} = \frac{c + qJ}{q-\rho } &gt; 0$$</p>

<p>given $(2)$.  </p>

<p>In turn, assume that $\partial V(\beta)/\partial \beta &lt;0$. This requires that</p>

<p>$$(q-2\rho)J-c &lt; 0 \implies q &lt; \frac {c}{J} + 2\rho  \tag {3}$$</p>

<p>Then $\beta^* =0$ and</p>

<p>$$V^*|_{\beta =0} = \frac{-c + qJ}{\rho} $$</p>

<p>To have $V^*|_{\beta =0} &gt;0 $ we need</p>

<p>$$\frac {c}{J} &lt; q &lt; \frac {c}{J} + 2\rho \tag {4}$$</p>

<p>to respect also $(3)$. So here a necessary restriction on the parameters is imposed, in order to obtain a positive value for $V$.  </p>

<p>Combining, if we impose the restrictions</p>

<p>$$q &gt; \rho, \;\;\; q &gt; c/J$$</p>

<p>we allow for either $\beta =1$ or $\beta =0$ (it will depend on the exact parameter values), while at the same time guaranteeing that $V&gt;0$ in all cases.
This is not unusual in economic models -i.e. the need to restrict the parameters in such a way so as for the model to provide solutions that are economically meaningful. And we need $V$ to be positive, because otherwise the vacancy would not exist in the first place, as I understand the set up.</p>

CC BY-SA 3.0
2015-07-14T13:13:21.903Z
The second restriction has an intuitive meaning: Creation of vacancies has positive return (expected cost $c/q < J$ expected return). But my question is mainly, what the first restriction means: In my dynamic setup, that turns out not to be always true (as $q$ is a function that for a large enough $\theta$ can be arbitrarily small). In that case, the "amplification effect" (as in, decide to reject and come back as a new vacancy) is larger than the time preference. This is my question: What does it **mean** if that first condition does not hold (and in turn the denominator becomes negative).
FooBar
CC BY-SA 3.0
2015-07-14T18:46:56.050Z
@FooBar This appears to be a case of the model becoming unsuitable to usefully describe the real-world phenomenon under study. In the context of purposeful, optimizing behavior from the part of the firm, I cannot see any meaning in having a vacancy with a negative value.
Alecos Papadopoulos
CC BY-SA 3.0
2015-07-14T13:01:46.310Z
2015-07-14T13:01:46.310Z
Alecos Papadopoulos
<p>Let $0 &lt; \rho &lt; 1$ be the discount rate, $V$ some option value, and $F$ some fundamental.</p>

<p>$$ \rho V = \beta V + F$$</p>

<p>You have access to some option value $V$ that will always deliver you some fundamental flow value $F$, and you get access to another $\beta$ of the initial option value (some sort of amplification of the initial value/claim). Solving for the value of the option, we get</p>

<p>$$ V = \frac{F}{\rho - \beta}$$</p>

<p>Now that's maybe just me being confused talking, but typically here we only check whether $\rho - \beta$ is not exactly zero.</p>

<p>For $\rho &gt; \beta$, we have that $V$ is valued much more than the original $F$. How can I understand cases where $\rho &lt; \beta$? These are very possible, even for both $\beta$ and $\rho$ between zero and one. In that case, the denominator turns negative, and the value of the option $V$ is negative. What is happening?</p>

<h3>Extended Example</h3>

<p>Per popular request, here is a more general version of the model (still an abstraction, but hopefully this delivers enough context.</p>

<p>Think about $V$ as the value of a vacancy, in a search-and-matching context. Given market tightness $\theta$, you will find a match at rate $q(\theta)$. Vacancies are associated with flow costs $c$. </p>

<p>Now, once you've matched with an unemployed worker, you can decide to either accept that match ($\beta = 1)$, or to reject him ($\beta = 0$). In fact, the whole line $\beta \in [0, 1]$ is allowed, understanding it as a mixed strategy. In fact, the mixed strategy can be understood as population shares, given that the vacancy is representative.</p>

<p>So, given strategy $\beta$, the value of a vacancy is given by</p>

<p>$$ \rho V(\beta) = -c + q(\theta)[ \beta(V(\beta)-J) + (1-\beta)J]\\
\Leftrightarrow V(\beta) = \frac{-c + q(\theta)(1-2\beta)J}{\rho - q(\theta)\beta}
$$</p>

<p><em>Note that setting $F\equiv - c+ q(\theta)(1-2\beta)J$ almost recovers the initial equation.</em></p>

<p>Now, I tried to solve this (or rather, a more complicated setup) using a grid for $\beta$. But sometimes, the solution was given by a $\beta$ such that the denominator was negative (and small), and the numerator, too, was small. I tried wrapping my head around that, but I couldn't intuitively understand what that even means.</p>

<p><strong>Note that while the current answer goes a long way on solving the model for me, it does not actually answer my question - about the meaning of a negative denominator. Hence, at the current state, I will not award that bounty and invite more answers.</strong></p>

CC BY-SA 3.0
2015-07-12T16:11:47.383Z
@denesp It's originally from a Labor-Macro model, but the specifics are not relevant, so I tried to hold it as general as possible. $F$ is the value of a match, $V$ is the value of a vacancy (searching). After a match, you have the option of forgoing the match and keep searching, which is why $V$ appears on the right hand side. I deemed that irrelevant, but I could add it again if you think it matters.
FooBar
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2015-07-12T16:59:20.297Z
I'd be interested to see the paper it came from, for some context.
dismalscience
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2015-07-13T11:27:29.553Z
@FooBar You write in a comment "After a match, you have the option of forgoing the match and keep searching...". This appears to imply that you cannot have _both_ $\beta V$ _and_ $F$, that it is an "either the one or the other" situation. So I don't see how you can add them on the right hand side. More context?
Alecos Papadopoulos
CC BY-SA 3.0
2015-07-13T12:20:04.547Z
@AlecosPapadopoulos Let $\beta$ indicate your choice to forego it. Then you have $\rho V = \beta(V-F) + (1-\beta)F = \beta V + (1-2\beta)F$. In the question, I have removed the $(1-2\beta)$ term and rescaled $F$ for simplicity.
FooBar
CC BY-SA 3.0
2015-07-13T12:44:02.250Z
@FooBar. Now I am even more confused. The way you explain it in the comment points towards $\beta$ being an indicator function, a $0/1$ binary variable (choice to forego/not to forego). In the question $\beta$ is described differently.
Alecos Papadopoulos
CC BY-SA 3.0
2015-07-13T13:17:11.793Z
You may definitely add more context. If we go to labor matching function, V= (bJ-c)/(r+b) so it could be negative only because the prob of finding (b) is to low and V cannot degenerate to inf.
Yann
CC BY-SA 3.0
2015-07-13T13:47:15.900Z
@AlecosPapadopoulos The whole construct is much more complicated, which is why I did so much simplification. Trust me, you dont want want to go into the messy details. Also, I really thought that the underlying concept of an amplification effect (see also the related effect), was more general, as typically the feedback is in the denominator, versus the initial effect on the numerator. Now, if you (the commentators) agree on requiring more context, I will rewrite the question tonight or so.
FooBar
CC BY-SA 3.0
2015-07-13T14:12:06.103Z
@FooBar It appears that up to now this question, rather than answers, is getting comment-questions that radiate the uncertainty of the posters about how to approach the issue. So most likely a compromise towards a less abstract setting, will facilitate answers.
Alecos Papadopoulos
CC BY-SA 3.0
2015-07-14T10:44:42.053Z
Alright, all, I have added some context. @YM'fr to compare with your equation, I have $r-b$, rather than $r+b$ in the denominator - but I can't find any mistake in my equation.
FooBar
<labor-economics><search-and-matching>
Amplification Effects: Negative Denominator
FooBar
# Amplification Effects: Negative Denominator

(Asked by: username_0 on 2015-07-10)

Let $0 < \rho < 1$ be the discount rate, $V$ some option value, and $F$ some fundamental.

$$ \rho V = \beta V + F$$

You have access to some option value $V$ that will always deliver you some fundamental flow value $F$, and you get access to another $\beta$ of the initial option value (some sort of amplification of the initial value/claim). Solving for the value of the option, we get

$$ V = \frac{F}{\rho - \beta}$$

Now that's maybe just me being confused talking, but typically here we only check whether $\rho - \beta$ is not exactly zero.

For $\rho > \beta$, we have that $V$ is valued much more than the original $F$. How can I understand cases where $\rho < \beta$? These are very possible, even for both $\beta$ and $\rho$ between zero and one. In that case, the denominator turns negative, and the value of the option $V$ is negative. What is happening?

### Extended Example

Per popular request, here is a more general version of the model (still an abstraction, but hopefully this delivers enough context.

Think about $V$ as the value of a vacancy, in a search-and-matching context. Given market tightness $\theta$, you will find a match at rate $q(\theta)$. Vacancies are associated with flow costs $c$.

Now, once you've matched with an unemployed worker, you can decide to either accept that match ($\beta = 1)$, or to reject him ($\beta = 0$). In fact, the whole line $\beta \in [0, 1]$ is allowed, understanding it as a mixed strategy. In fact, the mixed strategy can be understood as population shares, given that the vacancy is representative.

So, given strategy $\beta$, the value of a vacancy is given by

$$ \rho V(\beta) = -c + q(\theta)[ \beta(V(\beta)-J) + (1-\beta)J]\\
\Leftrightarrow V(\beta) = \frac{-c + q(\theta)(1-2\beta)J}{\rho - q(\theta)\beta}
$$

*Note that setting $F\equiv - c+ q(\theta)(1-2\beta)J$ almost recovers the initial equation.*

Now, I tried to solve this (or rather, a more complicated setup) using a grid for $\beta$. But sometimes, the solution was given by a $\beta$ such that the denominator was negative (and small), and the numerator, too, was small. I tried wrapping my head around that, but I couldn't intuitively understand what that even means.

**Note that while the current answer goes a long way on solving the model for me, it does not actually answer my question - about the meaning of a negative denominator. Hence, at the current state, I will not award that bounty and invite more answers.**

**Tags:** labor-economics, search-and-matching
> **Comments:**
> **username_0**: @denesp It's originally from a Labor-Macro model, but the specifics are not relevant, so I tried to hold it as general as possible. $F$ is the value of a match, $V$ is the value of a vacancy (searching). After a match, you have the option of forgoing the match and keep searching, which is why $V$ appears on the right hand side. I deemed that irrelevant, but I could add it again if you think it matters.  
> **username_1**: I'd be interested to see the paper it came from, for some context.  
> **username_2**: @FooBar You write in a comment "After a match, you have the option of forgoing the match and keep searching...". This appears to imply that you cannot have _both_ $\beta V$ _and_ $F$, that it is an "either the one or the other" situation. So I don't see how you can add them on the right hand side. More context?  
> **username_0**: @AlecosPapadopoulos Let $\beta$ indicate your choice to forego it. Then you have $\rho V = \beta(V-F) + (1-\beta)F = \beta V + (1-2\beta)F$. In the question, I have removed the $(1-2\beta)$ term and rescaled $F$ for simplicity.  
> **username_2**: @FooBar. Now I am even more confused. The way you explain it in the comment points towards $\beta$ being an indicator function, a $0/1$ binary variable (choice to forego/not to forego). In the question $\beta$ is described differently.  
> **username_3**: You may definitely add more context. If we go to labor matching function, V= (bJ-c)/(r+b) so it could be negative only because the prob of finding (b) is to low and V cannot degenerate to inf.  
> **username_0**: @AlecosPapadopoulos The whole construct is much more complicated, which is why I did so much simplification. Trust me, you dont want want to go into the messy details. Also, I really thought that the underlying concept of an amplification effect (see also the related effect), was more general, as typically the feedback is in the denominator, versus the initial effect on the numerator. Now, if you (the commentators) agree on requiring more context, I will rewrite the question tonight or so.  
> **username_2**: @FooBar It appears that up to now this question, rather than answers, is getting comment-questions that radiate the uncertainty of the posters about how to approach the issue. So most likely a compromise towards a less abstract setting, will facilitate answers.  
> **username_0**: Alright, all, I have added some context. @YM'fr to compare with your equation, I have $r-b$, rather than $r+b$ in the denominator - but I can't find any mistake in my equation.  

---

## 1 answers

### Answer by username_2 on 2015-07-14. Score: 2

$\beta$ appears to be a *decision* variable, **at firm level**. Then its value should be determined optimally, under some criterion. So what is the criterion, what is the objective function which should be optimized over the value of $\beta$?

It appears that we should maximize the value of the option $V(\beta)$

$$V(\beta) = \frac{-c + q(\theta)(1-2\beta)J}{\rho - q(\theta)\beta}$$

The first derivative is (I will dispense with $\theta$)

$$\frac {\partial V(\beta)}{\partial \beta} = \frac{-2qJ\cdot[\rho - q\beta]+q\cdot[-c + q(1-2\beta)J]}{[\rho - q\beta]^2}$$

$$\implies \frac {\partial V(\beta)}{\partial \beta} = \frac{-2qJ\rho -cq + q^2J}{[\rho - q\beta]^2} = \frac{q}{[\rho - q\beta]^2}\cdot (-2J\rho -c + qJ)$$

$$\implies {\rm sign}\frac {\partial V(\beta)}{\partial \beta} = {\rm sign}\big\{(q-2\rho)J-c\big\} \tag {1}$$

So $V(\beta)$ is monotonic in $\beta$. Therefore, depending on the values taken by $J,\rho, c, q$, it will be optimal to set $\beta$ equal to either $0$ or $1$.

Assume now that $\partial V(\beta)/\partial \beta >0$. This requires that

$$(q-2\rho)J-c > 0 \implies q > \frac {c}{J} + 2\rho \implies q > \rho \tag {2}$$

Then $\beta^\* =1$ and

$$V^\*|\_{\beta =1} = \frac{-c - qJ}{\rho - q} = \frac{c + qJ}{q-\rho } > 0$$

given $(2)$.

In turn, assume that $\partial V(\beta)/\partial \beta <0$. This requires that

$$(q-2\rho)J-c < 0 \implies q < \frac {c}{J} + 2\rho \tag {3}$$

Then $\beta^\* =0$ and

$$V^\*|\_{\beta =0} = \frac{-c + qJ}{\rho} $$

To have $V^\*|\_{\beta =0} >0 $ we need

$$\frac {c}{J} < q < \frac {c}{J} + 2\rho \tag {4}$$

to respect also $(3)$. So here a necessary restriction on the parameters is imposed, in order to obtain a positive value for $V$.

Combining, if we impose the restrictions

$$q > \rho, \;\;\; q > c/J$$

we allow for either $\beta =1$ or $\beta =0$ (it will depend on the exact parameter values), while at the same time guaranteeing that $V>0$ in all cases.
This is not unusual in economic models -i.e. the need to restrict the parameters in such a way so as for the model to provide solutions that are economically meaningful. And we need $V$ to be positive, because otherwise the vacancy would not exist in the first place, as I understand the set up.

> **Comments:**
> **username_0**: The second restriction has an intuitive meaning: Creation of vacancies has positive return (expected cost $c/q < J$ expected return). But my question is mainly, what the first restriction means: In my dynamic setup, that turns out not to be always true (as $q$ is a function that for a large enough $\theta$ can be arbitrarily small). In that case, the "amplification effect" (as in, decide to reject and come back as a new vacancy) is larger than the time preference. This is my question: What does it **mean** if that first condition does not hold (and in turn the denominator becomes negative).  
> **username_2**: @FooBar This appears to be a case of the model becoming unsuitable to usefully describe the real-world phenomenon under study. In the context of purposeful, optimizing behavior from the part of the firm, I cannot see any meaning in having a vacancy with a negative value.  

---

