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My question is pretty simple: I have two equations which are supposedly true (from a published paper) but I have no idea how to arrive at the solution myself. So it would be great if someone could a) tell me whether the equations are correct and b) how to get to the solution or what specific statistic topic I need to learn to understand the solution.

This question has an economics background, so let me first tell you the story. (But you can also ignore the story and only look at the maths)

Suppose people have the choice to work either in their home country $a$ or in a foreign country $b$. If they stay at home they can expect wage $z_a$ which is randomly distributed between 0 and 1. If they go abroad they get wage $z_b$ which is also randomly distributed between 0 and 1; however; $z_b$ is multiplied by a constant $y$, $y \in [0,1]$.

Obviously, if a person knows its respective values of $z_a, z_b$ and $y$ it will only go abroad if $y z_b > z_a$.

Now, I am interested in the EX ANTE probability $p$ of an individual person going abroad. And this is were my problem starts. According to the paper I'm reading, this probability is calculated as follows:

1) ---> $p = \int_0^y P(z_a < y z_b) \;dz_a = \frac{y}{2}$

So we only consider the integral over those with a $z_a < y$ because the others would never go abroad, and thus calculate $P(z_a < y z_b)$ over said integral. I get that. But not the result. Wouldn't $\frac{y}{2}$ imply that $P(z_a < y z_b)$ = $\frac{1}{2}$ ? Or am I just too dumb to work with probabilities?

In a next step I want to know the average wage rate $w_n$ of those who do NOT go abroad. It is calculated as

2) ---> $w_n = \frac{1}{1 - p} [ \int_0^y z_a P(z_a \geq y z_b) \;dz_a + \int_y^1 z_a \;dz_a ] = \frac{3 - y^2}{6 - 3y}$

So we take the average $\frac{1}{1 - p}$ (remember that $p = \frac{y}{2}$ as calculated above) over the expected wage $z_a$ of all those who stay at home because $P(z_a \geq y z_b)$ plus those who stay because their $z_a$ is $> y$ anyways. Again, I think I get the setup. But how does this add up to $\frac{3 - y^2}{6 - 3y}$?

My question is not about the setup of the equations. I'd just like to know whether they are right, and if so, maybe a hint how to get the solution myself... Any help would be appreciated.

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You are conditioning on the wage rate in the home country. That is why the integral is $dz_a$. $P(z_a < yz_b \vert z_a) = 1-\frac {z_a}y, y < $ etc. If you put this or $1-$ this resp. in for the probabilities in the integral you should get their results (I haven't checked the second). As to your specific criticism of the first you should get $P(z_a < yz_b) = \frac y 2$ which is what they are calculating by conditioning on $z_a$. –  mike Apr 24 '12 at 19:28
    
thx mike, feeling a bit stupid but also relieved, have tried to work this out for some time now... –  hiasei Apr 24 '12 at 20:02
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