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!


1 Answer 1


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 $(0,X_2)$.

  • $\begingroup$ 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, 2019 at 21:03

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