# Sequence of shifted exponential distributions has uniform conditionals?

Suppose we generate a sequence of random variables as follows. First, we let $$X_1 \sim Exp(\lambda)$$, where $$Exp(\lambda)$$ denotes the exponential distribution with rate parameter $$\lambda>0$$. For $$X_j$$, $$j>1$$, we let $$X_j \sim SExp(\lambda, X_{j-1})$$, where $$SExp(\lambda, \tau)$$ denotes the shifted (translated) exponential distribution with shift parameter $$\tau$$. The shifted exponential distribution has the density

$$f(x; \lambda,\tau) = \lambda e^{-a (x-\tau)}\mathbb I(x > \tau), \quad \lambda> 0, \tau > 0.$$

It is clear that $$X_j$$ has a marginal gamma distribution with shape parameter $$j$$ and rate parameter $$\lambda$$ (it is a sum of independent exponentially distributed random variables).

Now, when we look at the conditional distribution of $$X_j$$ for $$j>1$$, given the rest of the sequence, it turns out that it is uniform on $$(X_{j-1}, X_{j+1})$$, unless $$j$$ is the last term of the sequence.

To give a minimal example. Suppose that we generate three random variable according to the process described above. The joint density is then \begin{align*} p(x_1,x_2,x_3) &= p(x_1)p(x_2\vert x_1)p(x_3\vert x_2) \\ &\propto \exp(-\tau x_1)\exp(-\tau(x_2-x_1))\exp(-\tau(x_3-x_2)) \mathbb I(x_1 implying that the conditional density of $$X_2$$ is proportional to a constant.

Given the wide usage of the exponential distribution, is there an intuitive explanation for this?

Any suggestions would help! Thanks!

Yes, there ia a simple explanation for this. The $$X_j$$'s are the arrival times in a Poisson process with arrival rate $$\lambda$$, and it is a standard property of the Poisson process that given all the arrival times except the $$i$$-th, then the arrival time $$X_i$$ is uniformly distributed between $$X_{i-1}$$ and $$X_{i+1}$$. In fact, we don't even need to condition on all these arrival times; knowing just $$X_{i-1}$$ and $$X_{i+1}$$ suffices to give the conditional distribution of $$X_i$$ as uniform on the open interval between the two known arrival times.
Note also that given $$X_2$$, the conditional density of $$X_1$$ is uniform on $$(0,X_2)$$, that is, we don't really need to have an arrival at $$0$$. Given the time $$X_2$$ of the second arrival, the first arrival is (conditionally) uniformly distributed on the interval $$(0,X_2)$$.