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I'm working on an exercise related to Poisson processes and have encountered a step in the solution that confuses me. The exercise is as follows:

Consider two independent Poisson processes $X_1(t)$ and $X_2(t)$, where $X_i(t)$ represents the number of arrivals for process $i$ during the interval $[0, t]$. The average number of arrivals per unit time for the two processes are $\lambda_1 = 1$ and $\lambda_2 = 1$, respectively.

Compute $P[X_1(1) + X_2(1) = 2 | X_1(2) = 1]$.

I understand that to solve this, we need to condition on the arrival of $X_1$ before or after the first unit of time, which gives us:

$P[X_1(1) + X_2(1) = 2 | X_1(1) = 1] \cdot P[X_1(1) = 1 | X_1(2) = 1] + P[X_1(1) + X_2(1) = 2 | X_1(1) = 0] \cdot P[X_1(1) = 0 | X_1(2) = 1]$

However, the solution then states:

$P[X_1(1) = 1 | X_1(2) = 1] = \frac{1}{2} = P[X_1(1) = 0 | X_1(2) = 1]$

This is justified by using the Binomial Distribution $B(1, \frac{1}{2})$ to obtain $\frac{1}{2}$ for both probabilities. My confusion arises here for two main reasons:

  1. Shouldn't $P[X_1(1) = 0 | X_1(2) = 1]$ also be equal to $P[X_1(1) = 1] = e^{-1}$? This seems logical since we can have no arrivals in the first interval only if we have an arrival in the second interval.
  2. Given that it's a Poisson distribution, not a uniform one, why does the probability split evenly between the two intervals, i.e., $\frac{1}{2}$? Shouldn't the distribution of arrivals correspond more to the exponential distribution of interarrival times rather than being evenly split?

Could someone help clarify these points for me? I'm particularly interested in understanding the logical reasoning behind the equal probability split given the properties of the Poisson process.

Vlad Ionescu
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    $X_1(2)=1$ says there was one arrival in the interval $[0,2]$. So either there was one arrival in $[0,1]$ (probability $P(X_1(1)=1)=\lambda e^{-\lambda}$) and zero arrivals in $[1,2]$ (probability $P(X_1(2)-X_1(1)=0)= e^{-\lambda}$) with probability $P(X_1(1)=1, P(X_1(2)=1)=\lambda e^{-2\lambda}$, or there were zero arrivals in $[0,1]$ (probability $P(X_1(1)=0)=e^{-\lambda}$) and one arrival in $[1,2]$ (probability $P(X_1(2)-X_1(1)=1)=\lambda e^{-\lambda}$) with probability $P(X_1(1)=0, P(X_1(2)=1)=\lambda e^{-2\lambda}$, the same. That makes $P(X_1(1)=1 \mid P(X_1(2)=1)=\frac12$. – Henry Feb 11 '24 at 01:04
  • I think I got it, so you're saying since they have equal probabilities and sum up to 1 in the $P(X_1(2) = 1)$ probability space, that means that each one should be 1/2 when conditioned on it? – Vlad Ionescu Feb 11 '24 at 22:54
  • Yes. A symmetry argument is the quick way of getting the result. Doing the conditional analysis as $\dfrac{\lambda e^{-2\lambda}}{2\lambda e^{-2\lambda}}=\dfrac12$ confirms it. – Henry Feb 12 '24 at 01:42

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