Let $S_n$ be a random variable with the PDF of the gamma distribution when $\lambda = 1$: $$ f_{S_n}(s_n) = \frac{x^{n-1}}{\Gamma(n)}e^{-s_n}. $$ My goal is to find the joint distribution of the random variable $\frac{S_i}{S_n}$ where $i < n$. My thought is that I could do this using Bayes theorem:
$$ f_{S_n}(S_n|S_i) = \frac{f_{S_iS_n}(s_i,s_n)}{f_{S_i}(s_i)}. $$ But I am not exactly sure what $f_{S_n}(S_n|S_i)$ would be. I know that $$ f_{S_n}(s_n) = f_{S_i}(s_i)\cdot \frac{x^{n-i}}{\Gamma(n)/\Gamma(i)}e^{-s_n+s_i}. $$ Any suggestions would be much appriciated.
Bayes' Theorem says:- $$f_{S_n\mid S_i}(s_n\mid s_i)=\dfrac{f_{S_i,S_n}(s_i,s_n)}{f_{S_i}(s_i)}$$
However, you seek the ratio distribution $$\begin{align}f_{S_i/S_n}(r)=\mathbf 1_{0<r}\,\int_0^\infty s\,f_{S_i,S_n}(rs,s)\,\mathrm d s \end{align}$$
Yet, to find that, you do still need to identify how, if they are not independent, these gamma distributed random variables are related? Ie what is their joint distribution?
For instance, if $(X_k)$ is a sequence of independent and identically exponential distributed random variables (with rate parameter $1$), then their sum is Gamma distributed.$$\sum_{k=1}^j X_k\sim \mathcal{Gamma}(j, 1)$$
If this is your case, then you may relate $S_i$ and $S_n$, since $S_i$ and $S_n{-}S_i$ are independent.$$\begin{align}f_{S_i,S_n}(t,s)&=f_{S_i}(t)~f_{S_n-S_i}(s-t)\\[1ex]&=\dfrac{{t}^{i-1}\mathrm e^{-t}}{\Gamma(i)}\cdot\dfrac{(s-t)^{n-i-1}\mathrm e^{-(s-t)}}{\Gamma(n-i)}\\[1ex]&=\dfrac{{t}^{i-1}(s-t)^{n-i-1}\mathrm e^{-s}}{\Gamma(i)\Gamma(n-i)} \end{align}$$