In [ ]:

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This note contains a bunch of calculations associated with the reading on directed search with incomplete information. Since $\lambda$ is s key word in sagemath, I'll switch the notation so that the probability with which each worker is a good type to be $\gamma$. Otherwise the notation is the same as in the reading in directed search with complete information.

We can start with the question about when we can find an equilibrium were good workers apply to the high wage firm for sure. Recall that the condition is that the good worker prefers to take a chance at the high wage firm if the other worker is also a good worker, or to get the low wage for sure (notice that the presumption is that the other worker will apply for sure at the high wage firm). The condition for this is $\gamma \frac{w_1}{2} + (1-\gamma)w_1 \ge w_2$

This depends in a sort of complicated way on $w_1$, $w_2$ and $\gamma$. The next calculation finds for all values of $w_1$ and $w_2$, the highest value that $\gamma$ can have before the good worker decides it is better just to apply to the low wage.

In [1]:

var('w_1,w_2,gamma,pi') indifference = gamma*w_1/2+(1-gamma)*w_1 == w_2 i_cond = solve(indifference,gamma) show(i_cond[0])

$\gamma = \frac{2 \, {\left(w_{1} - w_{2}\right)}}{w_{1}}$

Now, we can plot this for all values of $w_1 > \frac{1}{2}$ assuming that $w_2 = \frac{1}{2}$. The region we will be focussing on in what follows will be the region below the curve (in other words the region where it isn't so likely that other worker is a good worker).

We can express this in a slightly different way by figuring out how high $w_1$ has to be to support the equilibrium in which good workers apply there for sure. This is just a slight variant of what we just did.

In [7]:

w_cond = solve(indifference,w_1) show(w_cond[0])

$w_{1} = -\frac{2 \, w_{2}}{\gamma - 2}$

In words, this says that to support the equilibrium where the good workers apply to the high wage firm for sure, the wage $w_1$ must be higher than $\frac{2w_2}{2-\gamma}$.

In [3]:

plot(i_cond[0].rhs().substitute(w_2=1/2),(w_1,1/2,1))

Once we are in this region, we can work out what the bad workers will do. Assuming that $pi$ is the probability the bad workers apply to the high wage firm, we want them to be indifferent between the high and low wage firms. This happens when $(1-\gamma)(\pi\frac{w_1}{2}+(1-\pi)w_1) = \pi w_2+(1-\pi) \frac{w_2}{2}$ This kind of calculation is getting routine:

In [5]:

w_indiff = (1-gamma)*(pi*w_1/2+(1-pi)*w_1) == pi*w_2 +(1-pi)*w_2/2 prob = solve(w_indiff,pi) show(prob[0])

$\pi = \frac{2 \, {\left(\gamma - 1\right)} w_{1} + w_{2}}{{\left(\gamma - 1\right)} w_{1} - w_{2}}$

Notice in this formula what happens when $\gamma = 0$. It seems intuitively that if there is no chance the other worker is a good worker, then things reduce to the game we discussed before, and we should get the same formula as we did in the directed search game with complete information.

One interesting question is whether good workers are more likely to get jobs than bad workers are. This depends in a pretty complex way on $w_1$, $w_2$ and $\gamma$.

The probability a good worker gets a job is pretty straightforward, just $\frac{\gamma}{2}+(1-\gamma)$ The probability the bad worker gets a job is $\gamma(1-\pi)+(1-\gamma)(\pi(\frac{\pi}{2}+(1-\pi))+(1-\pi)(\pi+\frac{1-\pi}{2}))$

In [9]:

prob_good = gamma/2+(1-gamma) show(prob_good)

$-\frac{1}{2} \, \gamma + 1$

In [10]:

prob_bad = gamma*(1-pi)+(1-gamma)(pi*(pi/2+(1-pi))+(1-pi)*(pi+(1-pi)/2)) show(prob_bad)

$-\gamma {\left(\pi - 1\right)} + \frac{1}{2} \, {\left(\pi + 1\right)} {\left(\pi - 1\right)} + \frac{1}{2} \, {\left(\pi - 2\right)} \pi + 1$

In [12]:

form = prob_bad.substitute(pi=prob[0].rhs()) show(form)

$-\gamma {\left(\frac{2 \, {\left(\gamma - 1\right)} w_{1} + w_{2}}{{\left(\gamma - 1\right)} w_{1} - w_{2}} - 1\right)} + \frac{1}{2} \, {\left(\frac{2 \, {\left(\gamma - 1\right)} w_{1} + w_{2}}{{\left(\gamma - 1\right)} w_{1} - w_{2}} + 1\right)} {\left(\frac{2 \, {\left(\gamma - 1\right)} w_{1} + w_{2}}{{\left(\gamma - 1\right)} w_{1} - w_{2}} - 1\right)} + \frac{{\left(2 \, {\left(\gamma - 1\right)} w_{1} + w_{2}\right)} {\left(\frac{2 \, {\left(\gamma - 1\right)} w_{1} + w_{2}}{{\left(\gamma - 1\right)} w_{1} - w_{2}} - 2\right)}}{2 \, {\left({\left(\gamma - 1\right)} w_{1} - w_{2}\right)}} + 1$

This is way complicated, so we'll do some substitutions. First, as before we can let $w_2 = \frac{1}{2}$. Then we can take $\gamma$ to be equal to $\frac{1}{5}$. Then if you refer back to the previous plot, you can see that there are plenty of values for $w_1$ which will support an equilibrium of the kind we want.

In [19]:

plot(form.substitute(gamma=1/2,w_2=1/2),(w_1,1/2,1))+plot(prob_good.substitute(gamma=1/2),(w_1,1/2,1))

In [21]:

find_root(form.substitute(gamma=1/2,w_2=1/2) == prob_good.substitute(gamma=1/2), .5,1 )

0.6583592135001262

What this computation is saying is that whether bad workers are more or less likely to get a job depends on the difference between the low and high wage. The computation set $w_2 = \frac{1}{2}$ and $\gamma = \frac{1}{2}$. The numerical result says that if the high wage $w_1$ is lower than .65, then the bad workers will actually be more likely to find a job than the good workers will. Conversely, if $w_1 > .65$ then bad workers will be less likely to get a job. Notice they are less likely to get a job the higher is $w_1$.

In [37]:

var('y_1') F = function ('F') t = function('t') b = function('b') g = (y_1-w_1)*(1-(F(t(w_1,w_2))*(1-b(w_1,w_2)))^2) show(latex(g)) show(latex(g.differentiate(w_1)))

${\left({\left(b\left(w_{1}, w_{2}\right) - 1\right)}^{2} F\left(t\left(w_{1}, w_{2}\right)\right)^{2} - 1\right)} {\left(w_{1} - y_{1}\right)}$

${\left(b\left(w_{1}, w_{2}\right) - 1\right)}^{2} F\left(t\left(w_{1}, w_{2}\right)\right)^{2} + 2 \, {\left({\left(b\left(w_{1}, w_{2}\right) - 1\right)}^{2} F\left(t\left(w_{1}, w_{2}\right)\right) \mathrm{D}_{0}\left(F\right)\left(t\left(w_{1}, w_{2}\right)\right) \frac{\partial}{\partial w_{1}}t\left(w_{1}, w_{2}\right) + {\left(b\left(w_{1}, w_{2}\right) - 1\right)} F\left(t\left(w_{1}, w_{2}\right)\right)^{2} \frac{\partial}{\partial w_{1}}b\left(w_{1}, w_{2}\right)\right)} {\left(w_{1} - y_{1}\right)} - 1$