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Uniformity of Poisson arrivals in random interval

hemanth

New member
May 31, 2012
9
Given that an Poisson arrival has occurred in an interval [0,t], where t is geometric with mean (alpha).
Is it true that the arrival instant is uniform in [0,t]?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Given that an Poisson arrival has occurred in an interval [0,t], where t is geometric with mean (alpha).
Is it true that the arrival instant is uniform in [0,t]?
In a Poisson process with mean $\lambda$ the probability that n events occurred in [0,t] is...

$$ P \{ N(t)-N(0) = n \} = e^{- \lambda\ t}\ \frac{(\lambda\ t)^{n}}{n!}\ (1)$$

Once You know that one event occurred at the time $\tau$ with $0 < \tau < t$, the $tau$ is uniformly distributed in [0,t]...

Kind regards

$\chi$ $\sigma$
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,795
Given that an Poisson arrival has occurred in an interval [0,t], where t is geometric with mean (alpha).
Is it true that the arrival instant is uniform in [0,t]?
Let $T$ be the time of the first arrival, let $\tau$ be an arbitrary time between 0 and t, and let the Poisson distribution have a mean of $\lambda$ arrivals per unit of time.

Then, from the definition of conditional probability:
$$P(T < \tau \ |\ T < t) = \frac{P(T < \tau \wedge T < t)}{P(T<t)} = \frac{P(T < \tau)}{P(T<t)} \qquad (1)$$

From the Poisson distribution we know that:
$$P(T < t) = P(\text{at least 1 arrival in }[0,t]) = 1 - P(\text{0 arrivals in }[0,t]) = 1 - \frac{e^{-\lambda t}(\lambda t)^0}{0!} = 1 - e^{-\lambda t} \qquad (2)$$

So:
$$P(T < \tau \ |\ T < t) = \frac{P(T < \tau)}{P(T<t)} = \frac{1 - e^{-\lambda \tau}}{1 - e^{-\lambda t}} \qquad (3)$$

This is not a uniform distribution.$\qquad \blacksquare$
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Given that an Poisson arrival has occurred in an interval [0,t], where t is geometric with mean (alpha).
Is it true that the arrival instant is uniform in [0,t]?
I apologize for the fact that at first I didn't realize that t is a geometric r.v. and not a deterministic r.v., so that I have to better specify my answer. In general in a stationary Poisson process with mean $\lambda$ the probability that n events occur in a time between $\tau$ and $\tau + t$ is...


$$P \{ N(\tau + t) - N(\tau) = n\} = e^{- \lambda\ t}\ \frac{(\lambda\ t)^{n}}{n!}\ (1)$$

... and, very important detail, the probability is independent from $\tau$. That means that in the case of one event [n=1], setting $\tau=0$, the event time $t_{0}$ is uniformely distributed in [o,t]. But t is not a deterministic but a geometric variable and that means that we are in the same situation described in...

http://www.mathhelpboards.com/f19/transformation-random-variable-5079/#post23090

Kind regards

$\chi$ $\sigma$