0
$\begingroup$

From my book:

Let $W$ denote the waiting time until the first occurrence during the observation of a Poisson process in which the mean number of occurrences in the unit interval is $\lambda$.

...Thus, when $w>0$, the pdf of $W$ is

$$F'(w) = f(w) = \lambda e^{-\lambda w}$$

My question is, why does it say "when $w>0$ as opposed to "when $1>w>0$? Since we are on a unit interval, I would've thought $W$ is bounded above by $1$.

$\endgroup$
  • 1
    $\begingroup$ The following thoughts might relate to your understanding about the unit interval. No matter what the mean intensity over that interval is, there is some chance the count will be zero. (That chance is $e^{-\lambda}$.) This means there is a positive (nonzero) chance that nothing happens during that interval. Since by definition what happens in the next interval of time is independent of what just happened, there is a positive chance that nothing happens in the next interval, either. Proceeding in this fashion, we see there is a small but positive chance of waiting any arbitrarily long time. $\endgroup$ – whuber Jun 8 '17 at 18:58
2
$\begingroup$

Read the wording extra carefully! The text states that lambda is the "...mean number of occurrences in the unit interval...". It does not state that our distribution is bounded by the unit interval.

We could define lambda over any interval we wanted. This way though lambda is directly interpretable as the rate of occurrence, so it's a convenient one to use.

$\endgroup$
1
$\begingroup$

We are not on the unit interval. The waiting time $W$ for the first occurrence can have any nonnegative real number value; it is the number of occurrences $N$ in $(0,1]$ that is a Poisson random variable with mean $\lambda$ and hence probability mass function $$p_N(k) = e^{-\lambda}\frac{\lambda^k}{k!}, ~ k = 0, 1, 2, 3, \ldots$$ Note that $\lambda$ is also known as the parameter of the Poisson random variable $N$. More generally, the number of occurrences in $(0,T]$ is a Poisson random variable $N_{(0,T]}$ with parameter $\lambda T$. Thus, the probability that the waiting time $W$ for the first arrival is greater than $T$ is $$P\{W > T\} = P\{N_{(0,T]} = 0\} = e^{-\lambda T}.$$ But, $P\{W > T\} = 1 - F_{W}(T)$ and so $F^\prime(T) = f_W(T) = \lambda e^{-\lambda T}$ for $T > 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.