Unable to compute conditional probability

$\bf{Setup}:$ Suppose I have a sequence of random variables $(X_n)_{n\geq 1}$ from $(\Omega,\mathcal{F},\mathbb{P}) \rightarrow$ $(E,\mathcal{E})$ such that each $X_n$ has some probability measure $\lambda$ as their common distribution. Suppose we have a Poisson random variable $K$ with mean $c\in(0,\infty)$, i.e. $\mathbb{P}(K=k)=\frac{e^{-c} c^k}{k!}$ for every $k\in{0,1,...}$Further, suppose $K,X_1,X_2,...$ are independent.

Define $N_\omega(A)=N(\omega,A)=\sum_{n=1}^{K(\omega)}\mathbb{I}_A(X_n(\omega))$, $A\in\mathcal{E}$, where $\mathbb{I}$ is the indicator function.

$\mathbf{Question}$: Letting $\{A,...,B\}$ be a finite measurable partition of $E$, in view of the assumptions on $K$ and the $X_n$, we have for $i,...,j\;$ in $\mathbb{N}\equiv\{0,1,2,...\}$ with $i+...+j=k$,

\begin{align*} \mathbb{P}\{N(A) =i,...,N(B)=j\} = \mathbb{P}\{K=k\}\mathbb{P}\{N(A)=i,...,N(B)=j |K=k \} =\frac{e^{-c}c^k}{k!} \cdot \frac{k!}{i!...j!}\lambda(A)^i...\lambda(B)^j \end{align*}

I do not know how to obtain the last equality. Can someone help me out?

Solution 1:

First, to clarify things, I would recommend the following change in notation: for any $(k,m)\in\mathbb{N}^2$, $A_1,...,A_m$ is a partition of $E$ and $(i_1,...,i_m)\in\mathbb{N}^k$ is such that $\sum_{j=1}^m i_j=k$. Now, the trick is to recognize that conditional on $K=k$, $(N(A_1),...,N(A_m))$ follows a multinomial distribution $\mathcal{M}(k,\lambda(A_1),...,\lambda(A_m))$. Thus, $$\mathbb{P}(N(A_1)=i_1,...,N(A_m)=i_m)=\frac{k!}{i_1!...i_m!} \prod_{j=1}^m \lambda(A_j)^{i_j}.$$

Why is it the case that $(N(A_1),...,N(A_m))\sim \mathcal{M}(k,\lambda(A_1),...,\lambda(A_m))$? Well, for each $n\in\{1,...,k\}$, $X_n$ belongs to only one set $A_j$ (as the $(A_j)_{j=1...m}$ forms a partition), and does so with probability $\lambda(A_j)$. So $(\mathbb{1}_{A_1}(X_n),...,\mathbb{1}_{A_m}(X_n))\sim \mathcal{M}(1,\lambda(A_1),...,\lambda(A_m))$. The result follows by independence between the $(X_n)_{n\ge 1}$.