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.
-
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?
-
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?
Best Answer
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.