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Given a poisson process with on the avg "a" arrivals per unit time, find the probability that there will be no arrivals during a time interval of length t, namely, the probability that the waiting times between successive arrivals will be at least of length t.

What probability density would I use for this of which I could use to solve the question? Thanks!

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Hint: What is the distribution of the number of arrivals in $[0,t]$ for a Poisson process with $a$ arrivals (on average) per unit time? Now, what is the probability that the associated random variable takes on a value of zero? – cardinal Nov 12 '11 at 18:08
(I also think that whoever worded that question like that should really give it another shot; the part from namely... on could definitely be stated more clearly.) – cardinal Nov 12 '11 at 18:10
up vote 1 down vote accepted

You want to use the distribution for the waiting time between events of a poisson process. $W$ has exponential distribution with parameter $ \lambda$, where $\lambda$ is the average number of events per unit time for the Poisson process.

This isn't hard to see. If $X$ is the number of events occurring in the interval $[0,t]$, then $P[W>t]=P[X=0]= e^{-\lambda t}$. So the cdf of $W$ is $F_W(t)=1-e^{-\lambda t}, t\ge0$; whence the density of $W$ is $f_W(t)= \lambda e^{-\lambda t}, t\ge 0$.

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i don't get why P[X=0] = e^-λt. From the wiki, the probability is the λe^λx, how did you get that? Thanks. – nubela Nov 12 '11 at 19:01
$X$ has Poisson distribution with parameter $\lambda t$. So $X=0$ is the event that no events occur in the interval $[0,t]$. This is the same as $P[W>t]$, where $W$ is the waiting time. Note, I first found the cumulative density, $F_W$, of $W$ first, then differentiated to get the density (though I didn't explicitly say so). – David Mitra Nov 12 '11 at 19:04

The exponential distribution.

Exponential Distribution

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