# $X_1, X_2, …, X_n \sim Exp(\lambda)$, 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?

• The joint PDF of $Y_1^*$, $Y_2^*$, $\ldots$, $Y_{n-1}^*$ is $$f_{Y_1^*,...,Y_{n-1}^*}(y_1,..., y_{n-1})=(n-1)!f(y_1)\cdots f(y_{n-1}) = (n-1)! \frac{1}{t^{n-1}} \mathbb 1(0\leq y_1\leq \ldots \leq y_n\leq t).$$ – NCh Mar 24 at 11:20

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 $$(Y_1^*, \ldots, Y_{n - 1}^*)$$.

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.)