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

hemanth
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Arrivals are Poisson distributed with parameter $$ \lambda$$.
Consider a system, where at the time of arrival of a tagged packet, it sees $$N_Q$$ packets.
Given that the tagged packet arrives at an instant $$t$$, which is uniform in [0, T],
what is the probability that all $$N_Q$$ packets arrived in [0,t]?
This is how i approached.

$$P\{N_Q \text{arrivals happened in} (0,t) |t\}= \frac{(\lambda \tau)^N_Q e^{-\lambda }}{N_Q!}$$
unconditioning on t, we get $$\frac{1}{T} \int _0^{T}\frac{(\lambda t)^N_Q e^{-\lambda t}}{N_Q!}dt$$Here $$N_Q$$ is a random variable in itself.
How do we get the expression independent of $$ N_Q$$?
 
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hemanth said:
Arrivals are Poisson distributed with parameter $$ \lambda$$.
Consider a system, where at the time of arrival of a tagged packet, it sees $$N_Q$$ packets.
Given that the tagged packet arrives at an instant $$t$$, which is uniform in [0, T],
what is the probability that all $$N_Q$$ packets arrived in [0,t]?
This is how i approached.

$$P\{N_Q \text{arrivals happened in} (0,t) |t\}= \frac{(\lambda \tau)^N_Q e^{-\lambda }}{N_Q!}$$
unconditioning on t, we get $$\frac{1}{T} \int _0^{T}\frac{(\lambda t)^N_Q e^{-\lambda t}}{N_Q!}dt$$Here $$N_Q$$ is a random variable in itself.
How do we get the expression independent of $$ N_Q$$?

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

$\chi$ $\sigma$
 
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