Let $X_i \sim Exp(1)$ be independent. I need to show that $((X_1,...,X_n)|X_1+\dots+X_n=t)$ is uniformly distributed over all nonnegative vectors that sum to t.
What does "over all nonnegative vectors that sum to t" mean? What do I need to show exactly?
I assume I have to use the cumulutive distribution function and the indepence. Then I get $$F_X(x) = F_{X_1}(x_1)\cdot ...\cdot F_{X_n}(x_n) = (1-e^{-x_1})\cdot...\cdot(1-e^{-x_n})$$
What do I have to do next?
If we weren't told $\sum_{i=1}^nX_i=t$, the joint PDF of the $X_i$ would be $\exp\left(-\sum_i x_i\right)$ on $[0,\,\infty)^n$. Over the $X_i\ge0$ region of the hyperplane plane $\sum_iX_i=t$ for some $t>0$, this PDF is $\exp(-t)$; and this region is a simplex of measure $\frac{t^n}{n!}$, so the probability density of $t$ is $\frac{t^n}{n!}\exp(-t)$ on $[0,\,\infty)$ (which, as a sanity check, indeed integrates to $1$). We obtain the conditional density of $\vec{X}$ by noting that, if $A$ is a subset of the above simplex,$$P(\vec{X}\in A|\sum_iX_i=t)=\frac{p(\vec{X}\in A)}{p(\sum_i X_i=t)},$$where the $p$ are probability densities. This is$$\frac{\int_A\exp\left(-\sum_ix_i\right)d^nx}{\frac{t^n}{n!}\exp(-t)}=\frac{\int_A\exp(-t)d^nx}{\frac{t^n}{n!}\exp(-t)}=\frac{n!}{t^n}\mu(A).$$Therefore, the density on an infinitesimal region in the hyperplane is $\frac{n!}{t^n}$.