Conditional probability in Kaplan, Menzio (2014)

Question

This is question about Kaplan and Menzio's shopping time model.

Pages 7,8: Unemployed search once or twice (for a seller).

• $\psi_u$:probability of searching twice, searching once with prob $1-\psi_u$
• $\nu$ is the probability of finding a seller.
• Searches are independent. An unemployed who searches twice, hence has the probability of finding two sellers of $\nu^2$.

Now here's the problem, on page 10, they look at it from a seller's point of view. Conditional of the seller of being matched with a buyer, what is the probability of the buyer being matched with another seller?

$$ Prob(\text{being matched second seller} | \text{being matched with first seller}) = \frac{Prob(\text{being matched with first and second seller})}{Prob(\text{being matched with first seller})} \\ = \frac{\text{search twice and find both times}}{\text{search once and find a seller or search twice and find one or two sellers }} \\ = \frac{\psi_u\nu^2}{((1-\psi_u) * \nu) + (\psi_u)*(\nu + \nu)}\\ = \frac{\psi_u\nu}{1+\psi_u}\\ $$

However, what they get is

$$ \frac{2\psi_u\nu}{1+\psi_u}$$

They compute some "intermediate probabilities" on page 8, but I don't see how they help get their result. How does one get their result?

Answer 1

The first / second seller terminology can be confusing here (does it relate to time or to conditionality?). It's safer to focus on a particular seller.

Let the probability that a buyer finds a particular seller, $s$, via single search be $\nu_s$. The probability that a buyer is matched with $s$ and (either before or after) another seller is then (assuming that $\nu_s$ is small so that the probability of finding $s$ twice in at most 2 searches can be ignored):

$$\psi_u(\nu_s\nu+\nu\nu_s)= 2\psi_u\nu_s\nu$$

The probability that a buyer is matched with $s$ via either a single or a double search is (again ignoring the probability, given a double search, of finding $s$ twice):

$$(1-\psi_u)\nu_s + \psi_u(\nu_s + \nu_s)=\nu_s(1+\psi_u)$$

The conditional probability of a buyer being matched with $s$ and another seller, conditional on $s$ being matched with that buyer, is therefore:

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

But the latter formula does not depend on $\nu_s$. Hence the same formula applies to any seller.

Answer 2

Here's another way, using the same notation as Adam, to get to the same result:

$$P(\text{another match | being matched}) = P(\text{searching twice | being matched})\cdot P(\text{another match | searching twice}) $$

Now,

$$P(\text{searching twice | being matched}) = \frac{P(\text{searching twice and being matched})}{P(\text{being matched})} \\ = \frac{\psi_u 2 \nu_s}{(1+\psi_u)\nu_s} $$

$$P(\text{another match | searching twice}) = \frac{P(\text{another match and searching twice})}{P(\text{searching twice})} \\ = \frac{\psi_u \nu}{\psi_u} $$

So,

$$P(\text{another match | being matched}) = \frac{\psi_u 2 \nu_s}{(1+\psi_u)\nu_s}\frac{\psi_u \nu}{\psi_u} = \frac{2\psi_u\nu}{(1+\psi_u)}$$