Distribution of inter arrival times in a Poisson process 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.

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

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?
 A: 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.
