Let $X_k$ be independent exponential random variables of parameter $\lambda>0$ such that
$$f_{X_k}(x)=\begin{cases} \lambda e^{-\lambda x} & x>0\\ 0 & x\leq0 \end{cases}$$
and let $S_n$ be the sum of the $n$ first $X_k$, that is,
$$S_n=\sum_{k=1}^\infty X_k=X_1+X_2+...+X_n.$$
Here, I am looking for the conditional density of $X_1/t$ given that $S_n = t$ in order to demonstrate that it is given by
$$f_{X_1/t|S_n}(z,t) =(n-1)(1-z)^{n-2},$$
a Beta distribution with parameters $\alpha = 1$ and $\beta = n-1$ where $1>z>0$.
Here's what I did to get the wrong answer:
First, I looked for the conditional density of $X_1$ given $S_n$ and for that, I defined a new random variable as follows
$$Z_n = \sum_{k=2}^\infty X_k = S_n - X_1$$
such that I can write
$$f_{S_n | X_1}(t|x) = f_{Z_n}(t - x)$$
knowing that the probability of $S_n = t$ given that $X_1=x$ is simply the probability of $Z_n = t - x$. I know that the density of the sum of $n$ exponential random variables with parameter $\lambda$ is given by
$$f_n(t) = \frac{\lambda^{n}}{\Gamma\left(n\right)}e^{-\lambda t}t^{n-1}$$
for $t>0$, I can deduce that we must have
$$f_{S_{n}}\left(t\right)=\begin{cases} \frac{\lambda^{n}}{\Gamma\left(n\right)}e^{-\lambda t}t^{n-1} & t>0\\ 0 & t\leq0 \end{cases}$$
and
$$f_{Z_{n}}\left(t\right)=\begin{cases} \frac{\lambda^{n-1}}{\Gamma\left(n-1\right)}e^{-\lambda t}t^{n-2} & t>0\\ 0 & t\leq0 \end{cases}.$$
According to Bayes formula, we get
$$f_{X_1 | S_n}(x|t) = \frac{f_{S_n | X_1}(t|x) f_{X_1}(x)}{f_{S_n}(t)}= \frac{f_{Z_{n}}\left(t-x\right)f_{X_{1}}\left(x\right)}{f_{S_{n}}\left(t\right)}$$
which then gives us the result
$$f_{X_{1}|S_{n}}\left(x|t\right)=\begin{cases} \left(n-1\right)\frac{\left(t-x\right)^{n-2}}{t^{n-1}} & t>x>0\\ 0 & \text{otherwise} \end{cases}.$$
From this, I can get the conditional density of $X_1/t = z$ given that $S_n = t$ since we deduce we must have
$$f_{X_1/t}(z) = f_{X_1}(zt)$$
and therefore
$$f_{X_1/t | S_n}(z, t) = f_{X_1 | S_n}(zt | t)$$
knowing that, of course, $X_1/t = z$ and $X_1 = zt$ are equivalent expressions. From what we got earlier, we get
$$f_{X_{1}/t |S_{n}}\left(z|t\right)=\begin{cases} \left(n-1\right)\frac{\left(1-z\right)^{n-2}}{t} & 1>z>0\\ 0 & \text{otherwise} \end{cases}$$
which is a promising result, but still has an extra factor of $t$ in the denominator. I surely overlooked something and hopefully someone can tell me where this factor comes from and how I can get rid of it to get the answer I am looking for.
