$N = Poisson(\lambda)$, $\{U_i\}$ iid $\implies (N_1, N_2) = Po(\lambda p_1)$x $Po(\lambda p_2)$

Question

Let $\{N\}\cup\{U_i\}$ be independent random variables.

$N = $ Poisson$(\lambda)$

$\{U_i\}$ iid, taking values in $\{1,2\}$, $\mathbb{P}[U_i = 1] = p_1$ and $\mathbb{P}[U_i = 2] = p_2$, $p_1 + p_2 =1$.

Define $N_j = \sum_{i=1}^N\textbf{1}_{[U_i =j]}$, $j \in \{1,2\}$.

Then $(N_1, N_2) = Po(\lambda p_1)$x $Po(\lambda p_2)$, that is,

\begin{align*} \mathbb{P}[N_1 = k_1, N_2 = k_2] = \frac{e^{\lambda p_1}(\lambda p_1)^{k_1}}{k_1!}\frac{e^{\lambda p_2}(\lambda p_2)^{k_2}}{k_2!} \end{align*}

My attempt:

\begin{align*} &[N_1 = k_1, N_2 = k_2] = \big[\sum_{i=1}^{k_1 + k_2} \textbf{1}_{[U_i =1]} = k_1, \sum_{i=1}^{k_1 + k_2} \textbf{1}_{[U_i = 2]} = k_2 \big]\\ &= \big[ \exists i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2} \in \{1, \dots, k_1 + k_2\} : U_{i^1_l} =1, l \in \{1, \dots, k_1\}, U_{i^2_m} = 2, m \in \{1, \dots, k_2\} \big] \\ &= \cup_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \big[ [U_{i^1_1} = 1]\cap\dots\cap[U_{i^1_{k_1}} = 1]\cap[U_{i^2_1} = 2]\cap\dots\cap[U_{i^2_{k_2}} = 2]\big]] \end{align*}

Then, $\mathbb{P}[N_1 = k_1, N_2 = k_2]$ = $\mathbb{P}\Big[\cup_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \big[ [U_{i^1_1} = 1]\cap\dots\cap[U_{i^1_{k_1}} = 1]\cap[U_{i^2_1} = 2]\cap\dots\cap[U_{i^2_{k_2}} = 2]\big] \Big]$ = $\sum_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \mathbb{P}\Big[[U_{i^1_1} = 1]\cap\dots\cap[U_{i^1_{k_1}} = 1]\cap[U_{i^2_1} = 2]\cap\dots\cap[U_{i^2_{k_2}} = 2]\Big]$ = $\sum_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \prod_{l=1}^{k_1}\mathbb{P}\big[U_{i^1_l} = 1]\prod_{m=1}^{k_2}\mathbb{P}\big[U_{i^2_m} = 1]$ = $\sum_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}}\prod_{l=1}^{k_1}p_1\prod_{m=1}^{k_2} p_2 $

Answer 1

Note that for fixed $n$, $X_n := \sum_{i = 1}^n I(U_i = 1) \sim \text{Bin}(n, p_1)$, and $\sum_{i = 1}^n I(U_i = 2) = n - X_n$.

Thus for $k_1, k_2 \in \{0, 1, \ldots\}$, by law of total probability, we have \begin{align} & P[N_1 = k_1, N_2 = k_2] \\ = & \sum_{n = 0}^\infty P[N_1 = k_1, N_2 = k_2 \mid N = n]P[N = n] \\ = & \sum_{n = 0}^\infty P\left[\sum_{i = 1}^n I(U_i = 1)= k_1, \sum_{i = 1}^n I(U_i = 2) = k_2 \mid N = n\right]P[N = n] \\ = & \sum_{n = 0}^\infty P\left[X_n = k_1, n - X_n = k_2\right]P[N = n] \quad\text{ by independence}\\ = & P[X_{k_1 + k_2} = k_1]P[N = k_1 + k_2] \quad\text{ the summand is not zero if and only if $n = k_1 + k_2$.}\\ = & \binom{k_1 + k_2}{k_1}p_2^{k_2}p_1^{k_1}e^{-\lambda}\frac{\lambda^{k_1 + k_2}}{(k_1 + k_2)!} \\ = & \frac{e^{-\lambda p_1}(\lambda p_1)^{k_1}}{k_1!}\times \frac{e^{-\lambda p_2}(\lambda p_2)^{k_2}}{k_2!}. \end{align}