Random sample generated for i.i.d variables My attempt: Since $f(y;a) = \frac{1}{2a}exp(\frac{-y}{2a})$ for $y>0$, and $f(y;a) = \frac{1}{2a}exp(\frac{y}{2a})$ for $y<0$, it is easy to see $A_n$ is a sufficient statistics for a family $T$ of measure $\left\{f(y;a): x, a > 0\right\}$. This means when generating a sample that is equivalent to $Y_1, Y_2,\ldots Y_n$, we only care about the information of $\sum_{i=1}^{n} S_1+\ldots + S_n = A_n$. 
 A: Your last question is answered by saz in the comments.
Here is one approach to the problem. It's not really related to the hints though.
I take the problem to be sample from the conditional distribution of $Y_1,\dots,Y_n,$ given the sum $A_i=\sum_{i=1}^nY_i.$ I will treat this as a reasonable practical problem, where integrating an $n-1$-dimensional conditional pdf is too slow.
First recall (or read on Wikipedia) that $Y_i$ has the same distribution as $X_i-X'_i$ where $X_i,X'_i$ are independent $\mathrm{Exp}(1/a)$ variables. And $B_n=\sum_{i=1}^n X_i$ and $B'_n=\sum_{i=1}^n X'_i$ have independent $\mathrm{Erlang}(n, 1/a)$ distribution, also known as a gamma distribution.
Since $A_n=B_n-B'_n,$ we can think of $A_n$ as a difference of independent $\mathrm{Erlang}(n, 1/a)$ distributed variables. The conditional pdf of $B_n$ is given by
$$f_{B_n|A_n}(x)\propto x^{n-1}e^{-x/a} \cdot
(x-A_n)^{n-1}e^{-(x-A_n)/a} 
$$
for $x\geq A_n.$
This is not a common distribution. One way to sample from it would be to use the binomial expansion
$$x^{n-1}(x-A_n)^{n-1}=(x-A_n+A_n)^{n-1}(x-A_n)^{n-1}=\sum_{j=0}^{n-1}\binom{n-1}{j}(x-A_n)^{n+j}A_n^{n-1-j}$$
which reduces to sampling from a mix of Erlang (i.e. gamma) distributions. (More explanation below)
We can then sample $X_1,\dots,X_n$ conditioned on $B_i,$ which is much easier (see below). Similarly for $X'_1,\dots,X'_n.$ The differences $Y_i=X_i-X'_i$ will then have the correct distribution conditioned on $A_i.$
Sampling from i.i.d. exponentials conditioned on their sum
As mentioned in the comments, the joint pdf of $X_1,…,X_N$ conditioned on their sum $B_i$ is uniform on $\{X_1+\cdots+X_n=B_i\}$. This uniform distribution is homogeneous in $B_i$ - it doesn't depend on $B_i$ except for scaling - so you can just sample i.i.d exponential variables $X_1+\cdots+X_n$ and scale them to have sum $B_i$.
Sampling from a mixture of distributions
It is possible to efficiently sample from a pdf of the form
$$f(x) \propto \sum_{j=0}^{n-1} f_j(x)$$
as long as the integrals $Z_j=\int f_j(x)$ can be computed efficiently. For the case in question $Z_j$ can be expressed in terms of factorials and simple expressions of $a$ and $A_n.$ The marginal pdf of $j$ is given by $Z_j/(Z_0+\dots+Z_{n-1}).$ This involves sampling from a custom pdf, but it's just $n$ different values so can be computed quite quickly. Given $j$, the condition distribution of $x$ is given by $f(x\mid j)=f_j(x),$ e.g. a gamma distribution.
