How to show that the distribution of $Y_{1}, Y_{2}, \ldots, Y_{n-1}$ is uniform on the simplex in $\mathbb{R}^{n-1}$

Question

I need to show that the distribution of $Y_{1}, Y_{2}, \ldots, Y_{n-1}$ is uniform on the simplex in $\mathbb{R}^{n-1}$, that is, the pdf is $(n-1)!$ in a suitable support.

Here, for $i = 1, \ldots, n-1$, $Y_{i}$ is defined as $$ Y_{i}=\frac{X_{i}}{X_{1}+\ldots+X_{n}}, i=1, \ldots, n-1, $$ where $X_{1}, X_{2}, \ldots, X_{n}$ are independent distributed random variables, each with a standard exponential distribution.

Answer 1

To prove that result, you can integrate your distribution over $T=\sum_1^n{X_i}$

$$ f(y_1,y_2,...,y_{n-1})dy_1dy_2...dy_{n-1}=\int_{t=0}^\infty{g(x_1,x_2,...,x_n)}dx_1dx_2...dx_{n-1}dt $$

where g is the distribution over $X_1,...,X_n$

you can notice that $dx_1dx_2...dx_{n-1}=t^{n-1}dy_1dy_2...dy_{n-1}$

and $g(x_1,x_2,...,x_n) = \lambda^n.e^{-\lambda.t}$

so $$ f(y_1,y_2,...,y_{n-1})=\int_{t=0}^\infty{t^{n-1}\lambda^ne^{-\lambda.t}dt} $$

with $z=\lambda t$, you get $$ f(y_1,y_2,...,y_{n-1})=\int_{t=0}^\infty {z^{n-1}e^{z}dz}=(n-1)! $$