# Inter-arrival time distribution of a Poisson process

I'm actually looking at a rigorous way to prove that if $$\{N_t\}_{t\geq 0}$$is a Poisson process of rate $$\lambda >0$$, then the inter-arrival times $$(H_n)_{n\in \mathbb{N}}$$ are independently and identically distributed exponential random variable with rate parameter $$\lambda$$.

Note that for this I use the two definitions of Poisson process that does not use the above statement as a definition that is (the equivalence between the two definition is already proved):

Def 1: A Poisson process $$\{N_t\}_{t\geq 0}$$ of rate $$\lambda >0$$ is a stochastic process with values in $$\mathbb{N}$$ satisfying:

1. $$N_0=0$$ a.s
2. The increments are independent, that is given $$n\in \mathbb{N}$$ and $$0\leq t_0<\dots $$N_{t_0}, N_{t_1}-N_{t_0},\dots,N_{t_{n-1}}-N_{t_n}$$ are independent.
3. $$P(N_t-N_s=k)=P(N_{t-s}=k), t\geq s$$
4. $$P(N_{t+\delta}-N_t=1)=\lambda\delta+o(\delta)$$
5. $$P(N_{t+\delta}-N_t \geq 2)=o(\delta)$$

Def 2: A Poisson process $$\{N_t\}_{t\geq 0}$$ of rate $$\lambda >0$$ is a stochastic process with values in $$\mathbb{N}$$ satisfying:

1. $$N_0=0$$ a.s
2. The increments are independent, that is given $$n\in \mathbb{N}$$ and $$0\leq t_0<\dots $$N_{t_0}, N_{t_1}-N_{t_0},\dots,N_{t_{n-1}}-N_{t_n}$$ are independent.
3. $$P(N_t-N_s=k)=P(N_{t-s}=k), t\geq s$$
4. $$N_t \sim Poi(\lambda t)$$

What I want to avoid is what seems to be the classic proof that consider object like $$P(H_2>t|H_1=t_1)$$ which has no sense for me as $$H_1$$ is continuous ( namely exponential) and $$H_2$$ could be (at this step) any type of random variable. (and hence $$P(H_1=t)=0 \forall t$$)

Thank's by advance to the one that will propose me a good proof for this.

• I'm not sure what the issue is with conditioning on $H_1 = t_1$. The fact that $H_1$ is continuous only means that it can take on any positive real value. Once it is realized/observed it will still have to take some value. If you are not willing to condition on continuous random variables, there will be many things in probability that you won't be able to do. Nov 23 '20 at 14:07
• Maybe my background is too light but my issue is just that $P(H_1=t)=0$ so I can't use the classic formula of conditioned probability by an event. Nov 23 '20 at 14:10
• The details of the proof will critically depend on the definition you are using, so it would be nice if you can fix one such version. Nov 23 '20 at 14:24
• Okay I'll add it to the initial post right now. Thanks for answering. Nov 23 '20 at 14:26
• You can still condition on the event $H_1 = t$. If you condition on it, it has already happened. You can easily simulate from exponential distributions, and any such simulation will give you specific real values, each of which technically has zero probability to appear. If you need to do a Bayes-like computation, relax $H_1 = t$ to $H_1 \in\left[t,t+dt\right]$ and take $dt \to 0$. You will see that $dt$ cancels out from the numerator and denominator, so there is no issue of dividing by zero. Nov 23 '20 at 17:57

## 1 Answer

We will assume that $$N$$ has right-continuous sample paths. Define the firth arrival time $$T_1$$ by

$$T_1 = \inf \{ t \geq 0 : N_t \geq 1 \}.$$

By the right-continuity, this implies that $$N_{T_1} = 1$$. Then we will prove the following claim, which is a special case of the strong Markov property:

Claim. Define a new stochastic process $$\tilde{N} = ( \tilde{N}_t )_{t\geq 0}$$ by $$\tilde{N}_t = N_{T_1 + t} - N_{T_1} = N_{T_1 + t} - 1.$$ Then we have the following:

1. $$T_1$$ and $$\tilde{N}$$ are independent.
2. $$\tilde{N}$$ is a Poisson process of rate $$\lambda$$.
3. $$T_1$$ is an exponential random variable of rate $$\lambda$$.

Proof. Fix $$0 < t_0 < t_1 < \cdots < t_n$$. Also, let $$\delta > 0$$ be sufficiently small so that $$0 < t_0 - \delta$$ and $$t_{i-1}+\delta < t_i-\delta$$ holds for all $$i = 1, \dots, n$$.

Now, for any parameters $$s, s_1, \dots, s_n \geq 0$$, we consider

$$Y = sT_1 + \sum_{i=1}^{n} s_i (\tilde{N}_{t_i} - \tilde{N}_{t_{i-1}}).$$

Then, conditioned on $$\{T_1 \in ((k-1)\delta, k\delta] \}$$, we have

$$Y \geq s (k-1)\delta + \sum_{i=1}^{n} s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}})$$

and so,

$$\mathbb{E}[e^{-Y}] \leq \sum_{k=1}^{\infty} \mathbb{E}\left[ \exp\left\{ - s (k-1)\delta - \sum_{i=1}^{n} s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}}) \right\} \mathbf{1}_{\{T_1 \in ((k-1)\delta, k\delta] \}} \right]. \tag{1}$$

By noting that $$\{T_1 \in ((k-1)\delta, k\delta] \} = \{ N_{(k-1)\delta} = 0 \text{ and } N_{k\delta} \geq 1 \}$$, all of

$$\mathbf{1}_{\{T_1 \in ((k-1)\delta, k\delta] \}}, \quad N_{(k-1)\delta + t_1} - N_{k\delta + t_{0}}, \quad \dots, \quad N_{(k-1)\delta + t_n} - N_{k\delta + t_{n-1}}$$

are independent, and hence the bound $$\text{(1)}$$ reduces to

\begin{align*} \mathbb{E}[e^{-Y}] &\leq \sum_{k=1}^{\infty} e^{-s(k-1)\delta} \left( \prod_{i=1}^{n} \mathbb{E}\left[ e^{-s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}})} \right] \right) \mathbb{P} \left( T_1 \in ((k-1)\delta, k\delta] \right) \\ &= \sum_{k=1}^{\infty} e^{-s(k-1)\delta} \left( \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1} - \delta) (1 - e^{-s_i}) } \right) e^{-\lambda (k-1)\delta} (1 - e^{-\lambda \delta}) \\ &= \frac{1 - e^{-\lambda \delta}}{1 - e^{-(s+\lambda)\delta}} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1} - \delta) (1 - e^{-s_i}) }. \tag{2} \end{align*}

As $$\delta \to 0^+$$, the bound $$\text{(2)}$$ converges to

$$\mathbb{E}[e^{-Y}] \leq \frac{\lambda}{s+\lambda} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1}) (1 - e^{-s_i}) }. \tag{3}$$

By a similar computation applied to

$$Y \leq s k\delta + \sum_{i=1}^{n} s_i (N_{k\delta + t_i} - N_{(k-1)\delta + t_{i-1}}),$$

we can also obtain the reverse direction of $$\text{(3)}$$. So it follows that

$$\mathbb{E}[e^{-Y}] = \frac{\lambda}{s+\lambda} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1}) (1 - e^{-s_i}) } \tag{4}$$

holds for any choices of parameters $$s, s_1, \dots, s_n \geq 0$$. However, since the multidimensional Laplace transform uniquely determines the law of a given random vector, this proves that:

1. $$T_1$$, $$\tilde{N}_{t_1} - \tilde{N}_{t_{0}}$$, $$\ldots$$, $$\tilde{N}_{t_n} - \tilde{N}_{t_{n-1}}$$ are mutually independent,

2. $$\tilde{N}_{t_i} - \tilde{N}_{t_{i-1}} \sim \operatorname{Poisson}(\lambda(t_i - t_{i-1}))$$ for each $$i = 1, \dots, n$$, and

3. $$T_1 \sim \operatorname{Exp}(\lambda)$$.

Therefore the desired claim follows. $$\square$$

Now repeatedly applying this claim shows that inter-arrival times are independent exponential random variables of rate $$\lambda$$ as desired.

• Thank's for your contribution, now it clear ! Just a question, when you repeat you argument you will show that $T_1,T_2$ and the new (poisson) process will be independent right ? Nov 24 '20 at 8:45
• What I understand is that you prove that $\tilde{\tilde{N}}$ and $T_2$ have the good proprieties and after that you prove that $T_1$ is independent with $T_2$ and $\tilde{\tilde{N}}$ as they are function of random variable which are independent with $T_1$. Is that right ? Nov 24 '20 at 9:25
• @ArthurSerres, You are correct. The lemma tells that anything that is derived from $\tilde{N}$ is independent of $T_1$. (More precisely, the $\sigma$-algebra generated by $\tilde{N}$ is independent of $T_1$.) Since the 2nd inter-arrival time $T_2$ is the "first" arrival time for $\tilde{N}$ (hence is determined only by $\tilde{N}$), all of $T_1$, $T_2$, and $\tilde{\tilde{N}}$ will be mutually independent. This process will of course continue. Nov 24 '20 at 9:52
• Thanks a lot for this clarifications and the time you took for these explanations. Now it's clear for me :) Nov 24 '20 at 10:08