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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<x_2<x_3) \\ &=\exp(-\tau x_3)\mathbb I(x_1<x_2<x_3) \end{align*} 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!

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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 $(0,X_2)$.

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  • $\begingroup$ Thanks for the reply! I'll look into Poisson processes (with which I'm not very familiar), but a quick look into Wikipedia suggests that your answer makes much sense! Do you know of any way to simulate the conditional distribution of, say, the second variable in the sequence from repeated draws of the whole sequence? $\endgroup$ – baruuum May 24 at 19:17
  • $\begingroup$ Since it was my fault to ask two questions in the same post, I'll edit the question (dropping the simulation part) and accept your answer. Thanks! $\endgroup$ – baruuum May 24 at 21:03

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