In their equation (5), Kaplan and Menzio claim that the price distribution in their Burdett-Judd market is given by

$$ F(p, u) = \{u \cdot A_1 \left[1 - \left(1 - B_1(u)\right)\frac{(r-c)p}{(p-c)r}y_u\right] \\ + (1-u) \cdot A_2 \left[1 - (1 - B_2(u))\frac{(r-c)p}{(p-c)r}*w(u)\right] \}/C$$

For positive $A_i$, $B_i$, $C$, where $u$ denotes the unemployment rate and $p$ denotes the price. They continue claiming that

  • $F$ is continuous
  • has connected support

$c$ is the households outside option, $r$ is the reservation price, hence the distribution should only give positive mass to prices between $[c, r]$.

$$ B_1(u) = 2\nu(u)\frac{\psi_u}{1+\psi_u}$$

where $\nu(\sigma(u)) = \frac{s}{b} = \frac{1-u}{1+u(\psi_u - \psi_e)}$. In their calibration: $\psi_e = 0.02$, $\psi_u = 0.27$. Hence

$$ B_1(u) = 2\frac{1-u}{1+0.25u}\frac{0.27}{1+0.27}$$

The Issue

For example, at an unemployment rate of $0.05$, we have $B_1(0.05) = .38$. However, for $p = c + \epsilon$ (for small $\epsilon$), the denominator $p-c$ becomes very small small.

This means that the product of $1-B_1(0.05)\cdot (r-c)\cdots$ becomes very large. One minus that is an very large negative number. The denominator $C$ is positive. A similar phenomenon happens with $B_2(0.05)$.

They call $F(p, u)$ the distribution. I assume this means the pdf. Can a pdf have negative values? Or what am I missing here?

  • 1
    $\begingroup$ On the side $F(p,u)$ is a probability, i.e. a cumulative distribution function (CDF), constrained in $[0,1]$. $\endgroup$ Apr 23 '15 at 23:50

In Lemma 1 they say that the support $\left[\underline{p}_t, \bar{p}_t \right]$ is such that $ c < \underline{p}_t $. So even though the support is connected, it does not extend to $c$, hence the $p \to c$ problem never arises.

  • 1
    $\begingroup$ Your current argument seems strange to me. The equilibrium strategy is such that that $F(\underline{p}_t) = 0$. This is how $\underline{p}_t$ is determined. The authors state this in Appendix A, Claim 5, page 41. This limits how close $p$ can be to $c$. Unless there is another condition that $\underline{p}_t$ has to fulfill I don't see a contradiction? $\endgroup$
    – Giskard
    Apr 23 '15 at 5:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.