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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?

cyberic
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  • Are you sure that's your definition for $X_i$? If $X_i=t_i-t_{i-1}$, then $X_3-X_2$ is not independent from $X_2$. – Graham Kemp Mar 09 '18 at 01:08

1 Answers1

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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

$$\mathsf P(X_3-X_2<c\mid X_2<c)=\dfrac{\mathsf P(X_2<c,X_3<X_2+c)}{\mathsf P(X_2<c)}=\dfrac{\int_0^c\int_0^{c+s} f_{X_2}(s)f_{X_3}(t)\mathsf dt~\mathsf ds}{\int_0^c f_{X_2}(s)\mathsf d s}$$

Graham Kemp
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  • Just to make sure I understand this concept correctly, the$ P(X_3 < c | X_2 >c) = P(X_3 < c)$ isn't it? – cyberic Mar 09 '18 at 02:15
  • Yes. The waiting time between the second and third arrivals is independent from the waiting time between the first and second arrivals. – Graham Kemp Mar 09 '18 at 02:25