I am new to Statistics. I am studying Poisson process, I have certain questions to ask.

A process of arrival times in continuous time is called a Poisson process of rate $\lambda$ if the following two conditions hold:

  • The number of arrivals in an interval of length $t$ is $\text{Pois}(\lambda t)$ random variable.
  • The number of arrivals that occur in disjoint time intervals are independent of each other.

Let $X_1$ denote the time of first arrival in a Poisson process of rate $\lambda$. Let $X_2$ denote the time elapsed between the first arrival and the second arrival. We can find the distribution of $X_1$ as follows: $$\mathbb{P}(X_1>t)=\mathbb{P}\left(\text{No arrivals in }[0,t]\right)=\mathrm{e}^{-\lambda t}$$ Thus $\mathbb{P}(X_1\le t)=1-\mathrm{e}^{-\lambda t}$, and hence $X_1\sim\text{Expo}(\lambda)$.

Suppose we want to find the conditional distribution of $X_2$ given $X_1$. I found the following discussion in my textbook.

$\begin{equation}\begin{split}\mathbb{P}(X_2>t\mid X_1=s) & = \mathbb{P}\left(\text{No arrivals in }(s,s+t] \mid \text{Exactly one arrival in [0,s]} \right) \\ & =\mathbb{P}\left(\text{No arrivals in }(s,s+t]\right)\\ &=\mathrm{e}^{-\lambda t}\end{split}\end{equation}$.

Thus, $X_1$ and $X_2$ are independent, and $X_2\sim\text{Expo}(\lambda)$.

However, I have the following questions regarding the above discussion.

  1. Since $X_1$ is a continuous random variable, $\mathbb{P}(X_1=k)=0$ for every $k\in\mathbb{R}$. Thus, $\mathbb{P}(X_1=s)=0$. In other words, we are conditioning on an event with zero probability. But when I studied conditional probability, conditioning on events with zero probability was not defined. So in this case, is conditioning on an event with zero probability valid?

  2. Second, assuming that conditioning on $X_1=s$ is valid, what we have found is the conditional distribution of $X_2$ given $X_1=s$. In other words, the conditional distribution of $X_2$ given $X_1$ is $\text{Expo}(\lambda)$, not the distribution of $X_2$ itself. But the author claims that $X_2\sim\text{Expo}(\lambda)$. Why is this true?


2 Answers 2


If the conditional distribution of $X_2$ given the event $X_1=s$ is the same for all values of $s$, then the marginal (i.e. not conditional) distribution of $X_2$ is also that same distribution, and they are independent.

If the conditional distribution of $X_2$ given $X_1=s$ depended on $s$, then the distribution of $X_2$ would be a weighted average of those conditional distributions, with weights given by the distribution of $X_1$. But if all of those conditional distributions are the same, then you're taking a weighted average of things that are all the same.

How to define conditioning on an event of probability $0$ is somewhat more delicate; maybe I'll say more about that later.

  1. Conditioning on an event with probability 0 is Ok, this is a property of continuous random variables.

  2. $X_2$ has an exponential distribution of mean $\lambda$ begun at time $s$. So $\mathbb{P}(\mbox{no arrivals in }(s,s+t]) = e^{-\lambda (t+s-s)} = e^{-\lambda t}$. Think of this as having two exponential clocks. As soon as the first one rings you start the second one, by the independence property the likelihood of the second one ringing after it has been going for a time $t$ is independent of when the previous clock rang. The distribution of $X_2$ should be $EXP(\lambda/2)$, i.e if we ignore the first clock ringing, the intensity should decrease by a factor of two. That is, the superposition of two intensity $\lambda/2$ Poisson process is an intensity $\lambda$ PP (look up Bernoulli splitting of Poisson processes)


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