Poisson Process conditional probability intersection?

Question

I've come across this type of problem a few times now so I understand what the correct steps are but I am struggling to understand the intuition behind this. The example problem states:

Consider a Poisson process on the interval [0, T] with rate λ > 0, and let 0 < τ < T. Define X1 to be the number of counts during [0, τ ], X2 to be the number of counts during [τ, T], and X to be the total number of counts during [0, T]. Let i, j, n be nonnegative integers such that n = i + j. Express the following probabilities in terms of n, i, j, τ, T, and λ, simplifying your answers as much as possible: Find P(X1 = i|X = n).

Steps to solve seem to be:

1. P(X1=i|X=n) = P(X1=i,X=n)/P(X=n)
2. P(X1=i,X-X1=n-i)/P(X=n)
3. P(X1=i,X2=j)/P(X=n) ...

I don't understand why step 2 is true. Why is the intersection of X1=i and X=n equivalent to X1=1 intersected with X-Xi=n-1?

Answer 1

The event that there were $6$ arrivals before time $\tau$ and a total of $9$ arrivals is the same as the event that there were $6$ arrivals before time $\tau$ and $3$ arrivals after that time.

Answer 2

Why is the intersection of X1=i and X=n equivalent to X1=1 intersected with X-Xi=n-1?

It is not. You are confusing some $i$ and $1$. (Wh1ch m1ght just be a typ0, but st1ll, be careful.)

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Because $X$ is the total of counts over both periods, so $X=X_1+X_2$.   Thus $X_2=X-X_1$.

When $X_1=i$ and $X_1+X_2=n$ then $X_2=n-i$. When $X_1=i$ and $X_2=n-i$ then $X_1+X_2=n$.

Therefore the event of $X_1=i$ and $X=n$ is the event of $X_1=i$ and $X_2=n-i$.

Since they are the same event, their probabilities are equal.

$$\begin{align}\mathsf P(X_1=i, X=n) ~&=~ \mathsf P(X_1=i, X_2=n-i)\\ \end{align}$$