[Math] Poisson events distributed uniformly in a given time

poisson distributionprobability

It is given that $4$ Poisson events occur between $12:00$ to $13:00 $ (interval denoted by T).
Intuitively, Why the probability of each event to occur at time $t \in T$ is uniform across $T$?

I'm well aware of the calculation for that:
Let's calculate the probability of $3$ (out of $ 4$) events to occur between $12:30$ to $13:00$ ($\frac{1}{2}$ an hour).
Let the following random variables:

  1. $X$ – the $4 $ events from $12:00$ to $13:00$.
  2. $X_1$ – the single event that has happened between $12:00 $ to $ 12:30$ .
  3. $X_2$ – the 3 events which happened between $12:30$ to $13:00$.

So their distributions are:
$$
\begin{align*}
& X \sim Poi(\lambda \cdot t) = Poi(4 \cdot 1) = Poi(4) \\
& X_1, X_2 \sim Poi(4 \cdot \frac{1}{2}) = Poi(2)
\end{align*}
$$

Given that a total of 4 events occurred at $T$, then:

$$
\begin{align*} P(X_2 = 3 \,|\, X = 4) &= \\
& = \frac{P(X_2 = 3 \cap X = 4)}{P(X = 4)} \\
& = \frac{P(X_2 = 3 \cap (X_1 = 1 \cup X_2 = 3))}{P(X = 4)} \\
& = \frac{P(X_2 = 3 \cap X_1 = 1)}{P(X = 4)} \\
& = \frac{P(X_2 = 3) \cdot P(X_1 = 1)}{P(X = 4)} \\
& = \frac{\frac{e^{-2} \cdot 2^3}{3!} \cdot \frac{e^{-2} \cdot 2^1}{1!}}{\frac{e^{-4} \cdot 4^4}{4!}} \\
& = \frac{4!}{3! 1!} \cdot \left(\frac{2}{4}\right)^1 \cdot \left(\frac{2}{4}\right)^3\\
& = \frac{2^4 \cdot 4!}{3! \cdot 4^4}\\
& = 4 \cdot \left(\frac{1}{2}\right)^4\\
& = 0.25
\end{align*}
$$
(Note that at the end we get a binomial distribution)

But I just can't understand this intuitively.

Edit:

The root of my confusion:

Indeed, it is given that in an hour there are precisely 4 events.
Therefore the total 4 events are NOT distributed like Poission.
But it is not given how much time each event takes to occur, so maybe the second, third and fourth events are distributed like Poission? Perhaps the time that each event takes to occurs is distributed exponentially (which indicates on a Poission distribution)?

Best Answer

It has to do with the fact that it is given that there are exactly 4 events in the hour, whereas the Poisson distribution allows any number of events in a given time period. Both $X_1$ and $X_2$ are both greater than or equal to $1$ (there is at least one event in both half hours), so we need to divide the remaining two events over $X_1$ and $X_2$. This distribution is no longer Poisson. The probability that the final two events both fall in $X_2$ is simply $0.5^2 = 0.25.$

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