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probability that all N_Q packets arrived in [0,t], in a Poisson process

hemanth

New member
May 31, 2012
9
Arrivals are Poisson distributed with parameter \(\displaystyle \lambda\).
Consider a system, where at the time of arrival of a tagged packet, it sees \(\displaystyle N_Q\) packets.
Given that the tagged packet arrives at an instant \(\displaystyle t\), which is uniform in [0, T],
what is the probability that all \(\displaystyle N_Q\) packets arrived in [0,t]?



This is how i approached.

\(\displaystyle P\{N_Q \text{arrivals happened in} (0,t) |t\}= \frac{(\lambda \tau)^N_Q e^{-\lambda }}{N_Q!}\)
unconditioning on t, we get \(\displaystyle \frac{1}{T} \int _0^{T}\frac{(\lambda t)^N_Q e^{-\lambda t}}{N_Q!}dt\)


Here \(\displaystyle N_Q\) is a random variable in itself.
How do we get the expression independent of \(\displaystyle N_Q\)?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Arrivals are Poisson distributed with parameter \(\displaystyle \lambda\).
Consider a system, where at the time of arrival of a tagged packet, it sees \(\displaystyle N_Q\) packets.
Given that the tagged packet arrives at an instant \(\displaystyle t\), which is uniform in [0, T],
what is the probability that all \(\displaystyle N_Q\) packets arrived in [0,t]?



This is how i approached.

\(\displaystyle P\{N_Q \text{arrivals happened in} (0,t) |t\}= \frac{(\lambda \tau)^N_Q e^{-\lambda }}{N_Q!}\)
unconditioning on t, we get \(\displaystyle \frac{1}{T} \int _0^{T}\frac{(\lambda t)^N_Q e^{-\lambda t}}{N_Q!}dt\)


Here \(\displaystyle N_Q\) is a random variable in itself.
How do we get the expression independent of \(\displaystyle N_Q\)?
The probability that each packet arrives in a time $\displaystyle 0 < \tau < t$ is...

$P \{0 < \tau < t\} = \frac{t}{T}\ (1)$

... and if the arrival times of all pachets are independent then the probability that all the pachets arrive in that time interval is...

$P_{N_{q}} = (\frac{t}{T})^{N_{q}}\ (2)$

Kind regards

$\chi$ $\sigma$