I have the following problem -
$a$ and $b$ are independently drawn from uniform distribution. $a$ is drawn from uniform distribution $\{m, 1+m\}$ and $b$ is drawn from uniform distribution $\{0,1\}$. where, $0<m<1$
I have to compute expectation of a provided that $a>b$ ie. E(a|a>b) -
This will be done piecewise -
$Pr (0<b<m)\times E(a) + Pr( m<b<1)\times E(a|a>b)$
Using order statistics -$$ m\times \frac{2m+1}{2} + (1-m) \int_{m}^{1+m} 2a^2 da $$ = $$ \frac{1}{6} (-12 m^3 + 6 m^2 + 11 m + 4)$$
Vs Expectation using old fashioned integration -
$$ m \frac{\int_0^m \int_m^{1 + m} a da db}{\int_0^m \int_m^{1 + m} 1da db} + (1 - m) \frac{ \int_m^1 \int_b^{1 + m} a da db} { \int_m^1 \int_b^{1 + m} 1da db} = \frac{2 m^3 + 3 m^2 + 9 m + 4}{6 m + 6} $$
Both behave same at end points, m=0 and m=1. But expectation rises more quickly when calculated using order statistics than the integration way.
Which one is correct? Why are they different? Is there a problem with calculation?
I am re-raising the query. The solution of @Egor Larionov seemed correct until I reversed the role. If we calculate the expectation of b when b > a using @Egor Larionov formulation, I get following -
$$\frac{\int_{m}^{1} b\, db \int_{b}^{1} da}{\int_{m}^{1} \, db \int_{m}^{b} da}= \frac{2-m-m^2}{3(1-m)}$$
But expectation is undefined at m=1, which isn't right.
Please help anyone