Writing a random variable as the sum of independent random variables

Question

Let $X_i$ be independent random variables with $X_i \sim Po(1)$. We know (for instance by looking at characteristic functions) that then $\sum_{i=1}^{n} X_i \sim Po(n)$. I am interested in the converse of that implication.
Let $Y \sim Po(n)$. Can we find independent random variables $Y_1, …, Y_n$ with $Y_i \sim Po(1)$ such that $\sum_{i=1}^{n} Y_i=Y$?

Answer 1

First sample $X\sim\text{Poisson}(n)$. Then toss a $n$-faced fair die $X$ times independently, letting $Y_k$ denote the total number of times face number $k\in\{1,\ldots,n\}$ has showed up. It is clear that $$X=\sum_{k=1}^nY_k$$ whatever the random realization of $X$ may be. Further, $Y_1,\ldots,Y_n$ are independent $\text{Poisson}(1)$ r.v. Indeed, for any $a_1,\ldots,a_n\in\Bbb Z_+$ we have, with $N:=a_1+\cdots+a_n$, \begin{align*} \Bbb P\Bigl(Y_1=a_1,\ldots,Y_n=a_n\Bigr)&=\Bbb P\Bigl(Y_1=a_1,\ldots,Y_n=a_n\:\Big|\: X=N\Bigr)\cdot\Bbb P(X=N)\\[.4em] &=\binom{N}{a_1,\ldots,a_n}\left(\frac1n\right)^{\!N}\cdot{\mathrm e}^{-n}\frac{n^N}{N!}\\[.4em] &=\left(\frac{\mathrm e^{-1}}{a_1!}\right)\cdots\left(\frac{\mathrm e^{-1}}{a_n!}\right)\\[.4em] &=\Bbb P\Bigl(\text{Poisson}(1)=a_1\Bigr)\cdots\Bbb P\Bigl(\text{Poisson}(1)=a_n\Bigr). \end{align*}

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Formal construction of $(Y_1,\ldots,Y_n)$. As stated in my earliest comment, we need a bit of extra independent randomness for this construction. So, suppose there exists a Uniform$(0,1)$ random variable $U$ independent of $X$. (If it does not exist, say we augment the probability space by taking its product with $((0,1),\mathcal B[0,1),\text{Leb})$.)

Then $(Y_1,\ldots,Y_n):=f(X,U)$ where $f\colon\Bbb Z_+\times(0,1)\to\mathbb(\Bbb Z_+)^n$ is the measurable map such that for any $y_1,\ldots,y_n\in\mathbb Z_+$, we have $(y_1,\ldots,y_n)=f(x,u)$ if, for all $1\le k\le n$, $y_k$ is the number of occurrences of digit $k$ in the first $x$ digits of $u$ when written in base $n$.