$Y (t)$ is the number of events occurring in $[0,1]$ where for each $t> 0$, $Y (t)~\sim\operatorname{Poi} (\lambda t)$ and $X$ measures the time taken for the $r$th event to occur.

Am I right in saying that the event $(X \le t) = (Y(t) \ge t)$?

Also, how can I write the cdf of $X$ as the sum of poisson probabilities using the above?

  • $\begingroup$ If $X$ is indeed the waiting time until the $r$-th event, then $X$ is a sum of $r$ independent exponentially distributed random variables with parameter $\lambda$. $\endgroup$ – André Nicolas Mar 14 '14 at 0:51
  • 1
    $\begingroup$ Perhaps you are looking for the details about the distribution of $X$. In that case, search under "gamma distribution." $\endgroup$ – André Nicolas Mar 14 '14 at 1:05

The $r$th event occurs before time $t$ if and only if the number of events before time $t$ is at least $r$.

So $[X<t] = [Y(t)\ge r]$.


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