# Price distribution: Negative values?

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?

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

## 1 Answer

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.

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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? – denesp Apr 23 '15 at 5:11