For random variables
$$N_1 \sim P(\lambda_1)$$ $$N_2 \sim P(\lambda_2)$$ $$N \sim P(\lambda)$$ where $N=N_1+N_2$ and $N_1$, $N_2$ are independent.
Why is the following statement true?
$$P(N_1=n_1)=P(N=n,N_2=n-n_1)$$
In other words, why is the $probability$ of $(N_1 = n_1)$ equal to the $joint$ $probability$ of $(N=n \cap N_2=n-n_1)$. I know that algebraically they are true when treating them as 'normal' variables.
This is not true.
We indeed have: $$ \big \{\omega : (N(\omega) = n) \wedge (N_2(\omega) = n - n_1)\big\} = \big \{\omega : (N_1(\omega) = n_1) \wedge (N_2(\omega) = n - n_1) \big\}, $$ because: $$ \Big( (N(\omega)=n) \wedge (N_2(\omega) = n - n_1) \Big) \Rightarrow \Big( (N_1(\omega)=n_1) \wedge (N_2(\omega) = n - n_1) \Big), $$ and because: $$ \Big( (N_1(\omega)=n_1) \wedge (N_2(\omega) = n - n_1) \Big) \Rightarrow \Big( (N(\omega)=n) \wedge (N_2(\omega) = n - n_1) \Big). $$
We thus have: $$ \mathbb{P}[(N=n)\wedge(N_2=n-n_1)] = \mathbb{P}[(N_1=n_1)\wedge(N_2=n-n_1)], $$ and then: $$ \mathbb{P}[(N=n)\wedge(N_2=n-n_1)] = \mathbb{P}[N_1=n_1]\mathbb{P}[N_2=n-n_1], $$ since $N_1$ and $N_2$ are independent.
