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I don't get how to calculate a conditional expected value. Here's the setting: $v_s$ and $v_b$ are independent and both uniformly distributed on [0,1]. Furthermore, $p_s(v_s)=a_s +c_sv_s$ and $p_b(v_b)=a_b+c_bv_b$.

Now I am asked to calculate $E[p_s(v_s)|p_s(v_s)≤p_b(v_b)]$ and $E[p_b(v_b)|p_b(v_b)≥p_s(v_s)]$.

Can anyone please help me?

I would really appreciate any help!

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    you can simplify this notation drastically by observing that $p_s(v_s)$ and $p_b(v_b)$ are independent uniform $[a_s,a_s+c_s]$ and $[a_b,a_b+c_b]$ respectively. (assuming the $c's$ are greater than 0). – 2017-01-16
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    Thanks for your answer! Can you also help me with the conditional expectation? Unfortunately, I haven't computed conditional expectation yet. – 2017-01-16
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    also, if you don't get how to compute a conditional expected value, this is a horrible problem to start on. Try computing $E(U_1 | U_1 < U_2)$ for $U_1,U_2$ independent $U(0,1)$'s – 2017-01-16
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    The solution states that $E[p_s(v_s)|p_s(v_s)≤p_b(v_b)]=(a_s+p_b)/2$ and $E[p_b(v_b)|p_b(v_b)≥p_s(v_s)]=(p_s+a_b+c_b)/2$ but I don't see how to get there. Could you please explain to me how to get there? – 2017-01-16
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    Posted an answer to the first problem. I would say unless there are additional constraints on the $a$'s and $c$'s, the solution is oversimplified. But if the conditions are there, it matches case 3 in my answer below. – 2017-01-16

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If you have a uniform random variable $V$ on the interval $[a,b]$ and you want to compute the conditional expected value of $V$ given $V

First, if $C>b,$ then the inequality $V

If $C

If $a\le C\le b$ then you know that $a < V < C$ so $V$ is conditionally uniform in $[a,C]$ which means that the conditional expected value is $ (a+C)/2.$

So in your original problem, if the constants are picked so that $a_s < p_b < a_s+c_s$ always hold regardless of what $v_b$ is, then the third case gives $E(p_s | p_s < p_b) = (a_s+p_b)/2.$