Uniformity of Poisson arrivals in random interval

[hemanth]

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]

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

$\chi$ $\sigma$

 

[I like Serena]

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.$\qquad \blacksquare$

 

[chisigma]

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

$\chi$ $\sigma$

 

[blue_raver22]

No, it is not necessarily true that the arrival instant is uniform in [0,t]. The Poisson process is a stochastic process where events occur randomly and independently over time, according to a Poisson distribution. This means that the arrival instant of a Poisson arrival is not determined by the length of the interval, but rather by the probability of an arrival occurring within that interval. Therefore, the arrival instant can vary and is not necessarily uniform in [0,t]. The mean value of t being geometric with mean (alpha) simply indicates the average length of the interval, but it does not determine the distribution of arrival instants within that interval.