Let $X_2$ and $X_3$ be the inter-arrival times in a Poisson process. That's $X_i = t_i - t_{i-1}$ where $t_i$ is the arrival of an event $i$. I'm trying to calculate the following probability:
$$P[X_3 - X_2 < c | X_2 < c] ..(1)$$
where $c$ is a constant. I calculated the CDF of the new R.V $Z = X_3 - X_2$:
$$1/2(1-exp(-\lambda t ))$$
Can I say that the $Z$ is also incrementally independent, so any information about $X_2$ doesn't affect the probability? In other words, $(1)$ can be written as:
$$P[Z<c]$$
if not, how can I find that probability? if so, how can I prove it?
Assuming $X_i$ is an inter-arrival time (between the $i-1$ and $i$ point events) of a Poisson process of rate $\lambda$ . Then $X_2, X_3$ are independent since a Poisson process is memoryless; they are also identically distributed. $f_{X_2}(x)=f_{X_3}(x)= \lambda\mathsf e^{-\lambda x}\mathbf 1_{x\in[0;\infty)}$
So just use the definition of conditional probability
$$\mathsf P(X_3-X_2<c\mid X_2<c)=\dfrac{\mathsf P(X_2<c,X_3<X_2+c)}{\mathsf P(X_2<c)}=\dfrac{\int_0^c\int_0^{c+s} f_{X_2}(s)f_{X_3}(t)\mathsf dt~\mathsf ds}{\int_0^c f_{X_2}(s)\mathsf d s}$$