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

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

<p>Now, </p>

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

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

<p>So, </p>

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

CC BY-SA 3.0
2015-05-29T20:18:44.003Z
2015-05-29T20:18:44.003Z
FooBar
<p>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.</p>

<p>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):</p>

<p>$$\psi_u(\nu_s\nu+\nu\nu_s)= 2\psi_u\nu_s\nu$$</p>

<p>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):</p>

<p>$$(1-\psi_u)\nu_s + \psi_u(\nu_s + \nu_s)=\nu_s(1+\psi_u)$$</p>

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

<p>$$\frac{2\psi_u\nu_s\nu}{\nu_s(1+\psi_u)}=\frac{2\psi_u\nu}{1+\psi_u}$$</p>

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

CC BY-SA 3.0
2015-03-20T13:46:27.953Z
2015-03-20T13:46:27.953Z
Adam Bailey
<p>This is question about Kaplan and Menzio's <a href="http://ideas.repec.org/p/nbr/nberwo/18777.html">shopping time model</a>.</p>

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

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

<p>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?</p>

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

<p>However, what they get is</p>

<p>$$ \frac{2\psi_u\nu}{1+\psi_u}$$</p>

<p>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?</p>

<probability>
Conditional probability in Kaplan, Menzio (2014)
FooBar
# Conditional probability in Kaplan, Menzio (2014)

(Asked by: username_0 on 2015-03-19)

This is question about Kaplan and Menzio's [shopping time model](http://ideas.repec.org/p/nbr/nberwo/18777.html).

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?

**Tags:** probability


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## 2 answers

### Answer by username_1 on 2015-03-20. Score: 4 | Accepted answer

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.


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### Answer by username_0 on 2015-05-29. Score: 2

Here's another way, using the same notation as username_1, 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,

$$P(\text{another match | being matched}) = \frac{\psi\_u 2 \nu\_s}{(1+\psi\_u)\nu\_s}\frac{\psi\_u \nu}{\psi\_u} = \frac{2\psi\_u\nu}{(1+\psi\_u)}$$


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