$X_1, X_2, ..., X_n \sim Exp(\lambda)$, what's the joint distribution of $X_1, X_1+X_2, ..., X_1+X_2+...X_n$ and is it a uniform ordered distribution? To elaborate on the title, here is the entire problem:
Let $X_1, X_2, ..., X_n \thicksim Exp(\lambda)$ be an independent sample.
What's the joint distribution of the sequence of $X_1, X_1 + X_2, ..., X_1 + X_2 + ... + X_{n-1}$ with the condition of $X_1 + X_2 + ... + ... + X_n = t$ ?
And is this joint distribution equal to an $n-1$ member ordered uniform ($Unif[0,t]$) sample's joint distribution, meaning that:
If $Y_1, Y_2, ..., Y_{n-1} \thicksim Unif[0,t]$ independent sample, and we order them: $Y_1^*, Y_2^*, ..., Y_{n-1}^*$, then are these equal:
$$F_{X}(x_1,...,x_{n-1}) = \Bbb{P}(X_1 < x_1, X_1 + X_2 < x_2, ...,~~~ X_1 + X_2 + ... + X_{n-1} < x_{n-1} | X_1 + X_2 + ... + X_n = t) \stackrel{?}{=} \Bbb{P}(Y_1^* < x_1,  Y_2 < x_2, ..., Y_{n-1}^* < x_{n-1}) = F_{Y^*}(x_1,...,x_{n-1})$$
where $F_X$ is the joint distribution function of the $X_1, X_2, ...,X_1 + X_2 + ... + X_{n-1}$ sample with the condition of $\sum_{i=1}^n{X_i} = t$ and $F_{Y^*}$ is the joint distribution function of the $Y_1^*, Y_2^*, ..., Y_{n-1}^*$ sample.
If so, prove it; if not, disprove it.
The problem is...:
...that $X_1, X_1 + X_2, ..., X_1 + X_2 + ... + X_n$ aren't independent, so calculating the joint distribution function is hard, especially with a condition.
Ordered samples also follow a Beta distribution, which is generally tough to deal with:
$$\forall k \in \{1,...,n\}: \quad Y_k^* \thicksim \frac{1}{t}Beta(n,n-k+1)$$
Here is what I've tried so far:
1. Introduce new variables:
$$A_1 = X_1 \\
A_2 = X_1 + X_2 \\
\vdots \\
A_n = X_1 + X_2 + \dots + X_n$$
This way, we can write up the $X$'s like so:
$$X_1 = A_1 \\
X_2 = A_2 - A_1 \\
X_3 = A_3 - A_2 \\
\vdots \\
X_n = A_n - A_{n-1}$$
We can also calculate the individual distributions of these $A$'s:
$$\forall k \in \{1,...,n\} \quad A_k \thicksim Exp\left(\frac{\lambda}{k}\right)$$
But this didn't lead me much further, since we still can't write up the joint distribution functions of $A$'s or $X$'s since they're not independent.
2. I tried thinking outside the box: $X_1 + X_2 + ... + X_k$ could mean the arrival time of a truck, and if they're from an exponential distribution, then their arrival times are expected to be uniform. However, expected value says very little about joint distribution, plus this wouldn't be a very mathematically appropriate proof.
Can anyone lead me on the correct path?
 A: The discrete version of the problem is easier to solve.
Let $X_1, X_2, \dots, X_n$ be (positive) geometric random variables: if you flip a coin that lands heads with some probability $p>0$, the distribution measures the number of coinflips until the coin lands heads.
Then $X_1, X_1 + X_2, \dots, X_1 + X_2 + \dots + X_n$ measure the number of coinflips until the coin lands heads $1, 2, \dots, n$ times. If we condition on $X_1 + X_2 + \dots + X_n = T$, then we are conditioning on the fact that the $T^{\text{th}}$ coinflip, and exactly $n-1$ of the preceding coinflips, landed heads.
Here, each sequence of $T$ valid coinflips had the same probability $p^n (1-p)^{T-n}$ of occurring before we conditioned on $X_1 + X_2 + \dots + X_n = T$, and so the conditional distribution is the same as the distribution we get by picking, uniformly at random, $n-1$ distinct positions for the first $n-1$ coinflips to land heads.
We can approximate the continuous problem with the discrete one arbitrarily well. Divide the time period $[0,t]$ into $T = \frac{t}{\delta}$ steps, and on each one, flip a coin that lands heads with probability $\lambda\delta$. Let $X_i$ be the interval between the $(i-1)^{\text{th}}$ time and the $i^{\text{th}}$ time the coin lands heads, and condition on the $n^{\text{th}}$ time the coin lands heads being the $T^{\text{th}}$ step.
On one hand, for each $\delta > 0$, we obtain the discrete process described above with $p = \lambda\delta$ and $T = \frac{t}{\delta}$.
On the other hand, as $\delta \to 0$, the $(X_1, X_2, \dots, X_n)$ joint distribution converges in some reasonable sense to the exponential one, and the distribution of $(Y_1, Y_2, \dots, Y_{n-1})$, the distribution of $n-1$ uniformly random time steps, converted to real numbers in $[0,t]$, converges in the same sense to a joint uniform distribution.
(In particular, we can prove convergence of $F_X$ and $F_Y$, which is what we want.)
A: Let $S_i = X_1 + \ldots + X_i$.
$$F_{(S_1, ..., S_n)}(x_1, \ldots, x_n) = \\
\int_{-\infty}^{x_1} f_{X_1}(\tau_1)
 \int_{-\infty}^{x_2 - \tau_1} f_{X_2}(\tau_2) \cdots
 \int_{-\infty}^{x_n - \tau_1 - ... - \tau_{n - 1}} f_{X_n}(\tau_n)
 \, d\tau_n \cdots d\tau_1, \\
\frac {\partial^n} {\partial x_n \cdots \partial x_1}
 F_{(S_1, ..., S_n)}(x_1, \ldots, x_n) = \\
f_{X_1}(x_1) \frac {\partial^{n - 1}} {\partial x_n \cdots \partial x_2}
 \int_{-\infty}^{x_2 - x_1} f_{X_2}(\tau_2) \cdots
 \int_{-\infty}^{x_n - x_1 - \tau_2 - ... - \tau_{n - 1}} f_{X_n}(\tau_n)
 \, d\tau_n \cdots d\tau_2 = \ldots = \\
f_{X_1}(x_1) f_{X_2}(x_2 - x_1) \cdots f_{X_n}(x_n - x_{n - 1}).$$
For $f_{X_i}(x) = \lambda e^{-\lambda x} [0 < x]$, this gives
$$f_{(S_1, ..., S_n)}(x_1, \ldots x_n) =
\lambda^n e^{-\lambda x_n} [0 < x_1 < \ldots < x_n].$$
Next,
$$f_{S_n}(x) =
\mathcal L^{-1} {\left[ \left(
 \frac \lambda {p + \lambda} \right)^{\!n} \right]} =
\lambda^n e^{-\lambda x} \mathcal L^{-1}[p^{-n}] =
\frac {\lambda^n} {(n - 1)!} x^{n - 1} e^{-\lambda x} \, [0 < x], \\
f_{(S_1, ..., S_{n - 1}) \mid S_n = t}(x_1, \ldots, x_{n - 1}) =
\frac {f_{(S_1, ..., S_n)}(x_1, \ldots, x_{n - 1}, t)} {f_{S_n}(t)} = \\
\frac {(n - 1)!} {t^{n - 1}} \, [0 < x_1 < \ldots < x_{n - 1} < t],$$
which is the same as the pdf of the order statistic $(Y_1^*, \ldots, Y_{n - 1}^*)$.
