# Transforming Poisson

Let $N, Y_n,n\in\mathbb{N}$ be independent random variables, with $N \sim P(\lambda), \lambda \lt \infty$ and $\mathbb{P}(Y_n = j)=p_j$, for $j=1,\dots,k$ and all $n$. Set $$N_j = \sum_{n=1}^{N}\mathbb{1}(Y_n=j).$$ Show that $N_1,\dots,N_k$ are independent random variables with $N_j \sim P(\lambda p_j)$ for all $j$.

For distribution I used that $N_j | N \sim Bin(N, p_j)$ and so $$\mathbb{P}(N_j=k)=\sum_{N=k}^{\infty}{{N}\choose{k}}p_j^k(1-p_j)^k\frac{\lambda^N}{N!}e^{-\lambda}=\dots=\frac{(p_j\lambda)^k}{k!}e^{-\lambda p_j}$$ Need help with independence.

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Let $T$, $T_1$, $\dots$, $T_k$ be indeterminates. Then the probability generating function of $N$ is $$f(T):={\Bbb E}[T^N]=e^{\lambda T}.$$ Now, if we condition on $N=n$, the joint distribution of the $N_j$'s will be multinomial and given by the coefficients of $(\sum_j p_j T_j)^n$. Therefore, setting $T=\sum_j p_j T_j$ in $f(T)$ will give the joint probability generating function of the $N_j$'s: $${\Bbb E}[\prod_j T_j^{N_j}]= f(\sum_j p_j T_j) = e^{\lambda (\sum_j p_j T_j)}.$$ This equals the product $$\prod_j e^{\lambda p_j T_j},$$ and so the $N_j$'s are independent with each $N_j\sim P(\lambda p_j)$.