Given $\ A_{(1)}, \ A_{(2)}, \ A_{(3)}, \ A_{(4)}, A_{(5)}\ $ are the order statistics of i.i.d random variables $A_i$ where $f(a_i|\alpha) = \beta^{-1}\ exp(\beta^{-1} (a_i-\alpha))$. Let $Y_{i} = (n-i+1)(A_{(i)} - A_{(i-1)})$ for $\ 1\leq i\leq 4\ $ and $\ A_{(0)} = 0$. Define $\ S = Y_{2} + Y_{3} + Y_{4} + Y_{5}$.
(a) Compute the marginal density of $(Y_1, S)$.
(b) Show that $Y_1+na\sim exp(\frac{1}{\beta})$, and $Y_1, T_{n-1}$ are independent.
(c) Define $U_1= Y_2$, and $U_{j} = U_{j-1} + Y_{j+1}$ for $2\leq j\leq 3$. Compute the joint density $f(Y_2+Y_3, Y_2+Y_3+Y_4|Y_1, Y_2+\ldots Y_5)$.
My attempt:
(a) We observe that $f(y_1, s) = f(s|y_1)f(y_1) = f(s|y_1)\ \beta\exp(\beta(y-na))$. Now, since the joint distribution of $Y_1, Y_2,..., Y_5$ is $g(y_1, y_2, \ldots, y_5) =\beta^{-1}exp(-\frac{y_{1}-n\alpha}{\beta})\beta^{-1}exp(-\frac{y_{2}}{\beta}) \ldots\beta^{-1}exp(-\frac{y_{n}}{\beta})$, we get: $Y_2, \ldots, Y_n$ are i.i.d with distributions $exp(\frac{1}{\beta})$, and $Y_1$ follows a distribution $\frac{1}{\beta} exp(-\frac{y_{1}-n\alpha}{\beta})$. So $S$ follows $Gamma(4, \frac{1}{\beta})$, but how do we show $f(s|y_1) = f(s)$?
(b) We have: $F_{Y_1+n\alpha}(y) = P(Y_1 + n\alpha\leq y) = \int_{y - n\alpha}^{2\alpha} \frac{1}{\beta}e^{-\frac{y_1-n\alpha}{\beta}}dy$. This results in $F_{Y_1+n\alpha}(y) = -exp(\frac{y+2n\alpha}{\beta}) + e^{-\frac{n\alpha}{\beta}}$.
Taking derivative with respect to $y$, we end up with: $f_{Y_1+n\alpha}(y) = -\frac{1}{\beta} exp(\frac{2n\alpha + y}{\beta})$, which is indeed the pdf of $exp(\frac{1}{\beta})$.
(c) I am completely stucked on this part for several hours. Could anyone please help me resolve this difficult part?