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

#### hemanth

##### New member
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
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

$\chi$ $\sigma$