I am interested in finding the following problem:
Let $\tau_1$ and $\tau_2$ are ordered statistics from a set of 2 independent uniform $(0,t)$ R.V. and let $Y_1,Y_2,Y_3$ are nonnegative iid R.V. that are also independent of $\tau_1 \& \tau_2$. Then we want to show that
$P(Y_1<\tau_1, Y_1+Y_2<\tau_2|Y_1+Y_2+Y_3=t)=\frac{1}{3}$.
For a fixed values of $\tau_1=u_1 \& \tau_2=u_2$ we have,
$P(Y_1<\tau_1, Y_1+Y_2<\tau_2|Y_1+Y_2+Y_3=t, \tau_1=u_1, \tau_2=u_2)=\\ P(Y_1<\tau_1, t-Y_3<\tau_2|Y_1+Y_2+Y_3=t, \tau_1=u_1, \tau_2=u_2)=\\ P(Y_1<\tau_1|Y_1+Y_2+Y_3=t, \tau_1=u_1)P(Y_3>t-\tau_2|Y_1+Y_2+Y_3=t, \tau_2=u_2)=\\ P(Y_1<\tau_1|Y_1+Y_2+Y_3=t, \tau_1=u_1)P(Y_3<\tau_2|Y_1+Y_2+Y_3=t, \tau_2=u_2)=\\ u_1/t*u_2/t=u_1u_2/t^2$
Then taking expectaion with respect to joint distribution of $(\tau_1,\tau_2)$ which is $f(\tau_1,\tau_2)=2/t^2, 0<\tau_1<\tau_2<t $, we have $P(Y_1<\tau_1, Y_1+Y_2<\tau_2|Y_1+Y_2+Y_3=t)=E(\tau_1\tau_2/t^2)=\frac{1}{4} $
which is not desired answer 1/3. I am not sure if my justification is correct. I appreciate any help for this problem.