splitting of poisson variables

Question

this is some lecture slides from my school. I don't understand why we need to sum up all from $m=0$ to $m=+ \infty$. I think $m$ is a fixed number which represents the number of type $2$ event? plz help me! many thanks in advance!

Answer 1

Recall the rule of average conditional probabilities:

For a partition $B_1,\dotsc,B_n$ of an outcome space $\Omega$, $$P(A) = P(A|B_1)+\dotsb+P(A|B_n)P(B_n).$$ In your case \begin{align*} P(X_1 = n) &= P(X_1 = n|X = n+0)P(X = n+0)\\ &\qquad\qquad+P(X_1 = n|X = n+1)P(X = n+1)\\ &\qquad\qquad+P(X_1 = n|X = n+2)P(X = n+2)+\dotsb\\ &= \sum_{m=0}^\infty P(X_1 = n|X=n+m)P(X = n+m). \end{align*} Then, just plug in the given information.

Answer 2

It is essential that:$$\{X_1=n\}=\bigcup_{m=1}^{\infty}\{X_1=n\wedge X=m\}$$ Note that the events on RHS are disjoint. So this gives: $$P(X_1=n)=\sum_{m=0}^{\infty}P(X_1=n\wedge X=m)$$ Here: $$P(X_1=n\wedge X=m)=P(X_1=n\mid X=m)P(X=m)$$

For $m<n$ the events are empty so you can start with $m=n$.

Finally: $$P(X_1=n)=\sum_{m=n}^{\infty}P(X_1=n\mid X=m)P(X=m)=$$$$\sum_{m=0}^{\infty}P(X_1=n\mid X=n+m)P(X=n+m)$$