Let $X_1, X_2$ be two independent, non-negative random variables with finite means from the same distribution function $F(x)$ and density function $f(x)$.
Denote their order statistics with $0 < Y_1 < Y_2$.
I want to compute $E(Y_2 | Y_1 = y_1)$ or to be more specific I want to show for $y_1 > 0$:
\begin{align*}
E(Y_2 | Y_1 = y_1) = y_1 + \frac{\int_{y_1}^{\infty} 1 – F(u) du}{1 – F(y_1)}.
\end{align*}
The joint density function of $Y_1,Y_2$ is given by
\begin{align*}
f_{Y_1, Y_2}(y_1,y_2) = 2 f(y_1) f(y_2) \chi_{y_1 < y_2},
\end{align*}
where $\chi$ is the indicator function.
The conditional expectation above can be computed by
\begin{align*}
\frac{\int_0^{\infty} y_2 f_{Y_1,Y_2}(y_1,y_2) dy_2} {\int_0^{\infty} f_{Y_1,Y_2}(y_1,y_2) dy_2}
= \frac{\int_{y_1}^{\infty} y_2 f(y_1) f(y_2) dy_2} {\int_{y_1}^{\infty} f(y_1) f(y_2) dy_2}
= \frac{\int_{y_1}^{\infty} y_2 f(y_2) dy_2} {\int_{y_1}^{\infty} f(y_2) dy_2}
= \frac{\int_{y_1}^{\infty} y_2 f(y_2) dy_2} {1 – F(y_1)}
\end{align*}
At this point I am stuck.
Best Answer
\begin{align} \int_0^{y_1}(1-F(u))du&=\left(u(1-F(u)\right)\left|_{0}^{y_1}\right.+\int_0^{y_1}uf(u)du\\ &=y_1(1-F(y_1))+\int_0^{y_1}uf(u)du \end{align} further $\int_{0}^{\infty} (1-F(u))du=\int_0^{\infty}uf(u)du$ hence \begin{align} \int_{y_1}^{\infty} (1-F(u))du&=\int_0^{\infty}uf(u)du-\int_0^{y_1}uf(u)du\\ &=\int_0^{\infty}uf(u)du-y_1(1-F(y_1))-\int_0^{y_1}uf(u)du\\ &=\int_{y_1}^{\infty}uf(u)du-y_1(1-F(y_1)) \end{align}
implying that $$\frac{\int_{y_1}^{\infty} u f(u) du} {1 - F(y_1)} =\frac{\int_{y_1}^{\infty} (1-F(u))du+y_1(1-F(y_1))} {1 - F(y_1)}$$
as you are after.