Waiting time in Poisson process 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!

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.