Distribution of inter arrival times in a Poisson process

Question

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.

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

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

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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?

Answer 1

The discussion in your book is not phrased correctly in some aspects, but first let me address your question about conditioning on an event of probability $0$; something that is explicitly forbidden in the definition of conditional probability in the earlier chapter of your book.

For jointly continuous random variables $X$ and $Y$ with joint pdf $f_{X,Y}(u,v)$, the conditional pdf of $Y$ given that $X = x$ is defined to be $$f_{Y\mid X}(v\mid X = u) = \begin{cases} \displaystyle \frac{f_{X,Y}(u,v)}{f_{X}(u)}, & \text{if }~f_{X}(u)>0,\\0, &\text{otherwise.}\end{cases}$$ where $f_X(u)$ is the (marginal) pdf of $X$. The conditional complementary CDF is $$1-F_{Y\mid X}(t\mid X = u) = P\{Y > t\mid X = u\} = \int_t^\infty f_{Y\mid X}(v\mid X = u) \,\mathrm dv$$ Now, in your application, $P\{X_2 > t\mid X_1 = s\}$ can be calculated directly since we are told that the first arrival occurred at $s$ and are being asked for the conditional probability that no arrivals have occurred in $(s,s+t]$. But, what happens in $(s,s+t]$ is independent of what happened in $(0,s]$ since the time intervals are disjoint. That is, $P\{\text{no arrivals in} ~ (s,s+t]\mid X_1=s\}$ is the same regardless of whether we assume that there was an arrival at $s$ or the first arrival occurred before time $s$, and so $$P\{X_2 > t\mid X_1 = s\} = P\{\text{no arrivals in} ~ (s,s+t]\} = e^{-\lambda t}.$$ and thus we get that the conditional pdf $f_{X_2\mid X_1 = s}(v\mid X_1 = s)$ is the same as the unconditional pdf $f_{X_2}(v) = \lambda e^{-\lambda v}, v > 0$. Conditionally or unconditionally, the distribution of $X_2$ is exponential with parameter $\lambda$. Furthermore, \begin{align} f_{X_2}(v) = f_{X_2\mid X_1 = s}(v\mid X_1 = s) = \displaystyle \frac{f_{X_1,X_2}(s,v)}{f_{X_1}(s)} \implies f_{X_1,X_2}(s,v) = f_{X_1}(s)f_{X_2}(v) \end{align} showing that $X_1$ and $X_2$ are independent (exponential random variables with parameter $\lambda$).

The answers to our specific questions are hidden somewhere in the above.