Let $\{X(t) : t \geq 0\}$ be a Poisson process with rate $\lambda$, and let $W_n$ denote the waiting time for the $n$-th event. For $s \geq 0$, determine $P( W_{X(t)} \leq t+s)$ and $P( W_{X(t)+2} \leq t+s)$.
Here's what I've tried.
For $W_{X(t)}$ to be less than $t+s$, the $X(t)$-th event must occur before time $t+s$. This implies that at least $X(t)$ many events must occur before time $t+s$, so we have
$$
P( W_{X(t)} \leq t+s)
= P( X(t+s) \geq X(t))
= P( X(t+s) – X(t) \geq 0)
= 1.
$$
Applying similar logic to the second one,
\begin{align*}
P(W_{X(t)+2} \leq t + s)
&= P(X(t+s) \geq X(t) + 2) \\
&= 1 – P(X(t+s) – X(t) < 2) \\
&= 1 – P(X(t+s) – X(t) = 0) – P(X(t+s) – X(t) = 1) \\
&= 1 – e^{-\lambda t} – \lambda t e^{-\lambda t}.
\end{align*}
Is what I've done ok? I'm getting the niggling feeling that I should be using the uniform distribution somehow. Please advise, thanks!
Best Answer
After much discussion with another student, we've determined that what I did is correct except for a typo in the last line. The answer is $$ P(W_{X(t)+2} \leq t + s) = 1 - e^{-\lambda s} - \lambda s e^{-\lambda s}. $$ Rather than focusing on the Poisson process on its own, it's better to think of things as a renewal process and consider the special case of Poisson processes.