I'm trying to do something that might be very simple, but because of my lack of knowledge about working with dynamic systems, I'm unsure that my methods are sound.
Consider the following value functions, where $I_{F,t}$ is the value of a match to the firm (by this I mean the value of having a person employed by that firm), and $I_{V,t}$ is the value of an open vacancy to a firm: \begin{align*} I_{F,t} &= y-w + \frac{1}{1+r}\Big[(1-\delta)I_{F,t+1} + \delta I_{V,t+1} \Big] \\ I_{V,t} &= -\kappa + \frac{1}{1+r}(1-f(u_{t-1},v_t))I_{V,t+1} + f(u_{t-1},v_t) \Big[\frac{1}{1+r} \Big((1-\delta)I_{F,t+1} + \delta I_{V,t+1}\Big) \Big] \end{align*} $y$ is firm output (exogenous), $w$ is the wage (for now assume exogenous), $\delta$ is the exogenous probability that the firm-worker match breaks, and $\kappa$ is the cost of opening a vacancy (exogenous). In order to simplify the problem, let $w=0$. Finally, $f$ is a function that captures the probability of a match between a worker and a firm occurring, and depends on the unemployment rate from the previous period as well as the vacancy rate this period. Assume that in every period, free entry holds (this is a standard assumption in economics, and it simply means that $I_{V,t} = 0 \forall t$). Then \begin{align*} I_{F,t} &= y + \frac{1-\delta}{1+r} I_{F,t+1}\\ \kappa &= f(u_{t-1},V_t) \frac{1-\delta}{1+r} I_{F,t+1} \end{align*} My main goal: solve for vacancies every period (sitting inside my $f()$ function). But what do I do with this $I_{F,t+1}$? How do I calculate that? This is my problem. Here is my solution, and I'm hoping someone has some insight into whether this is a valid way of dealing with this:
I can calculate \begin{align*} I_{F,0}^j &= y + \frac{1}{1+r}(1-\delta)I_{F,1}\\ I_{F,1}^j &=y + \frac{1}{1+r}(1-\delta)I_{F,2} \end{align*} And so on. Plugging in at time zero: \begin{align*} I_{F,0} &= y \Big( 1+ \frac{1}{1+r}(1-\delta)\Big) + \Big(\frac{1}{1+r}(1-\delta)\Big)^2 I_{F,2}\\ \\ &=y \sum_{k=0}^\infty \Big(\frac{1}{1+r}(1-\delta)\Big)^k + \lim_{k \to \infty} \Big(\frac{1}{1+r}(1-\delta)\Big)^k I_{F,k}\\ &=y \sum_{k=0}^\infty \Big(\frac{1}{1+r}(1-\delta)\Big)^k \\ \\ &= y\Big(\frac{1+r}{\delta+r} \Big) \end{align*}
And in steady state (i.e. in the long run, as $t \to \infty$), $I_{F,t} = I_{F,t+1} = I_F \ \forall t$, giving me \begin{align*} I_F &= y + \frac{1-\delta}{1+r} I_F\\ I_F &= y\frac{1+r}{r+\delta} \end{align*}
I have, then, that \begin{align*} I_{F,0} &= y\frac{1+r}{\delta+r}\\ \lim_{t \to \infty} I_{F, t} &= y\frac{1+r}{\delta + r} \end{align*} Can I then treat $F$ as a constant, equal to this value, for every $t$? If so, why? As in, how do I know it never deviates from this value? If not, why not?
Thanks in advance to anyone who can offer me some insights.