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I would like to find the pdf for the $n-1$ interarrival times $w$ of $n$ iid events from an exponential distribution of time constant $\tau$. Using the fact that the first event follows an exponential distribution of time constant $\tau/n$, and using memorylessness, I can show that the pdf for interarrival time $j$ is an exponential distribution of time constant $\tau/(n-j)$. In other words, for fixed $n$ and $j$, the distribution of waiting time $j$ follows an exponential distribution of time constant $\tau/(n-j)$ - I have simulated this and obtained consistent results.

However, I don't see how to get to the overall distribution of the interarrival times. In other words, for fixed $n$, what is the pdf of the distribution of all of the $w_k$, regardless of $j$?

I would have thought the pdf of $w$ is just the average of the pdfs of $w_i$, in other words a linear combination of exponentials, but I have tried some simulations, and the result looks more like a single exponential.

The application is I have experimental data coming from several realizations of $n$ events from the same exponential distribution, and I am trying to reproduce the overall distribution of interarrival times.

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  • $\begingroup$ If by "overall" you mean the pdf of a randomly chosen interarrival time (choosing the $w_j$ with probability $1/(n-1)$, then yes, the pdf is just a mixture of each pdf. But perhaps you mean something else by "overall"? $\endgroup$ Jun 4, 2017 at 20:06
  • $\begingroup$ I mean the following: if I draw $n$ iid events from an exponential distribution of time constant $\tau$, I can determine $n-1$ interarrival times. If I now repeat this many times, and plot the distribution of all of those waiting times $w$ irrespective of if they are $w_i$ or other, what distribution does that follow? I would have thought that the probability of picking $w_i$ is just $1/(n-1)$ as you write, so the pdf would be $pdf(w)=H(w)\frac{1}{\tau} \sum_{i=1}^{n-1} \frac{n-i}{n-1}e^{-\frac{(n-i)w}{\tau}}$, but my simulations don't look like that for now (am double-checking). $\endgroup$
    – Mister Mak
    Jun 4, 2017 at 21:34

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It turns out my calculation was correct, but my simulation wasn't (wrong normalization --- ooops!). When $n$ iid events are drawn from an exponential distribution of time constant $\tau$, the pdf of the $n-1$ waiting times is indeed $pdf(w)=H(w)\frac{1}{\tau} \sum_{i=1}^{n-1} \frac{n-i}{n-1}e^{-\frac{(n-i)w}{\tau}}$. I attach a figure to illustrate this. Wait times of exponential distribution

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Let $X_1, ..., X_n$ be the waiting times, each with parameter $\tau$. As you say, each iid $X_i \sim Exp(\tau)$. Let $W_1, ..., W_n$ be the arrival times, so that $W_k = X_1 + \cdots + X_k$. You can find the distribution of the $W_i$s via moment generating functions. Recall their multiplicative property:

$$M_{W_k}(t) = W_{\sum_{i = 1}^k X_i}(t) = \prod_{i = 1}^k M_{X_i}(t)$$

And so we can see that actually

$$W_k \sim \text{Gamma}(k, \tau)$$

I hope I answered your question properly. I only inferred what you want to know based on your phrasing. I do not know where the $\tau/n$s come into your argument.

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  • $\begingroup$ Thanks, but not sure I understand your answer (also note our notations for $w$ are not the same): I know the distribution of waiting time $w_k$ for a given $k$, but am looking for the distribution of all waiting times $w$ regardless of $k$. Ie if I generate $m$ realizations of $n$ events from an identical exponential distribution, what will the overall distribution of the $n-1$ wait times look like? (not just the distribution of $w_k$ for a given $k$) $\endgroup$
    – Mister Mak
    Jun 4, 2017 at 19:16

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