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In $\text{Shimer (2012)}^1$ the author describes the evolution of the unemployment rate as follows:

$$\dot{u}_{t+\tau} = \dot{u}_t^s(\tau)-\left(u_{t+\tau}-u_t^s(\tau)\right)f_t,$$

where

$f_t = -log(1-F_t \geq 0 \dots$ arrival rate of a Poisson process,

$F_t \dots$ job finding probability,

$u_t^s(\tau)\dots$ short term unemployment rate,

$\tau \dots$ time between passed between two measurement points in a panel data set with monthly interviews.

With $u_t^s(0)=0$ as an initial condition we should be able to come up for a solution for $u_{t+1}$ and $u^s_{t+1} \equiv u^s_t(1)$:

$$u_{t+1}=(1-F_t)u_t + u^s_{t+1}.$$

Trying to get there on my own, I struggle to find a proper solution approach for this kind of differential equations, since this resembles nothing of what I've learnt so far.

I would appreciate if you could hint me towards the right direction so that I may solve this problem eventually.

$^1 \text{Shimer, Robert (2012) Reassessing the ins and outs of unemployment, doi:10.1016/j.red.2012.02.001}$

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1 Answer 1

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First, note that derivative is with respect to $\tau$ and not $t$ (I got confused at the beginning). $t$ is a discrete variable from the reference you mention. Honestly, I believe that the notation is kind of akward, since $\tau$ dependence is sometimes written as $(\tau)$ and sometimes as subscript.

Anyway, define $v(\tau)=u_{t+\tau}-u_t^s(\tau)$ such that your equation becomes: $$ \dot{u}_{t+\tau} - \dot{u}_t^s(\tau)= - (u_{t+\tau}-u_t^s(\tau))f_t\implies \frac{dv(\tau)}{d\tau} = -v(\tau)f_t $$

This is a separable differential equation. Just divide by $v(\tau)$ and integrate both sides from $\tau=0$ to $\tau=1$ and get $$ \int_{v(0)}^{v(1)}\frac{dv}{v} = -\int_0^1f_td\tau = -f_t\implies \log\left(\frac{v(1)}{v(0)}\right) = -f_t = \log(1-F_t) $$ so that $$ v(1) = (1-F_t)v(0) $$ Now, $v(1) = u_{t+1}-u_t^s(1) = u_{t+1}-u_{t+1}^s$ since $u_t^s(1)=u_{t+1}^s$ by definition and $v(0)=u_t - u_{t}^s(0)=u_t$ from the initial condition $u_{t}^s(0)=0$. Thus, $$ v(1) = (1-F_t)v(0) \implies u_{t+1}-u_{t+1}^s = (1-F_t)u_t $$ which is now equivalent to what you needed.

Honestly, I believe the main issue with this problem is not the difficulty of the technique used (separation of variables is very standard) but the notation which prevents one to see right away what to do.

Hope this helps!

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  • $\begingroup$ Thank you so much for not only pointing out the fact that this differential equation is a function of $\tau$, but also for the detailed solution. I have some questions regarding your answer: Shouldn't it be $v(\tau) = u_{t+\tau}-u_t^s{tau}$ and $\log{\frac{v(1)}{v(0)}}=log(1-F_t) \Leftrightarrow v(1) = (1-F_t)v(0)$? $\endgroup$ Commented Jun 7, 2022 at 21:05
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    $\begingroup$ Sure, thats a typo in the definition of $v(\tau)$. I corrected it. Also another typo with the log. You are correct. Sorry! $\endgroup$ Commented Jun 8, 2022 at 9:48

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