Amplification Effects: Negative Denominator

Question

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.

Answer 1

$\beta$ 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.