Convergence of Pólya Urns and Expanding Real Polynomials – Markov Chains

Question

I'm reading these notes, and in the proof of this Proposition 2, which states:

Let $d\ge 2$ and $S\ge 1$ be integers. Let also $(\alpha_1,\dots,\alpha_d)\in\mathbb{N}^d \setminus\{ 0 \}$. Let $(P_n)_{n\ge0}$ be the $d$-color Pólya urn random process having $S\cdot Id$ as replacement matrix and $(\alpha_1,\dots,\alpha_d)$ as initial composition. Then, almost surely and in any $L^t$, $t\ge 1$,
$$\frac{P_n}{nS}\to V\ \ \text{as}\ n\to\infty$$ where $V$ is a $d$-dimensional Dirichlet-distributed random vector, with parameters $(\frac{\alpha_1}S,\dots,\frac{\alpha_d}S)$.

I'm stuck near the end of the proof, when it says "Besides, expanding real polynomials $X^p=X_1^{p_1}\dots X_d^{p_d}$ in the basis $(\Gamma_p)_{p\in\mathbb{N}^d}$, one gets formulae $$X^p=S^{|p|}\Gamma_p+\displaystyle\sum_{{k\in\mathbb{N}^d}\atop {|k|\le |p|-1}}a_{p,k}\Gamma_k{X}\tag{1}$$ where the $a_{p,k}$ are rational numbers, $p=(p_1,\dots,p_d)\in\mathbb{N}^d$, $|p|=\sum_{k=1}^d p_k$, $S\ge 1$ is an integer and $\Gamma_p(v):=\prod_{k=1}^d\frac{\Gamma\big(\frac{v_k}S+p_k\big)}{\Gamma\big(\frac{v_k}{S}\big)}$ for $v=\sum_{k=1}^dv_ke_k$, where $(e_k)_{k\ge 1}$ denotes the canonical basis in $\mathbb{R}^d$.

Main question: So the main question here is, what does it mean expanding the polynomial in the basis $\Gamma_p$? I would like to do this explicitely, but I don't understand what I can do, what is this basis? I thought of computing $\Gamma_p(e_k)$ for $k=1,\dots,d$ but this does not lead me anywhere.

The results just before this statement are:

Consequently, after a direct induction, for any $p\in\mathbb{N}^d$, $$\mathbb{E}[\Gamma_p(P_n)]=\frac{\Gamma(\frac{\alpha}{S}+n+|p|)}{\Gamma(\frac{\alpha}{S}+n)}\cdot\frac{\Gamma(\frac{\alpha}{S})}{\Gamma(\frac{\alpha}{S}+|p|)}\cdot\Gamma_p(P_0)$$ so that, when $n$ tends to infinity, by Stirling’s formula, $$\mathbb{E}[\Gamma_p(P_n)]=n^{|p|}\cdot\frac{\Gamma(\frac{\alpha}{S})}{\Gamma(\frac{\alpha}{S}+|p|)}\cdot\Gamma_p(P_0)\cdot \big(1+O\big(\frac1n\big)\big)$$

Second question: as a consequence of the previous question I should get (I don't know how since I don't understand firstly what is above)

Consequently, when $n$ tends to infinity, one gets the asymptotics $$\mathbb{E}\bigg(\frac{J_n}{\alpha+nS}\bigg)^p=\frac{\Gamma(\frac{\alpha}{S})}{\Gamma(\frac{\alpha}{S}+|p|)}\cdot\Gamma_p(P_0)\cdot \big(1+O\big(\frac1n\big)\big)\tag{2}$$

Here I suppose they use again Stirling's approximation and that they mean $\mathbb{E}\bigg[\bigg(\frac{J_n}{\alpha+nS}\bigg)^p\bigg]$ rather than $\mathbb{E}\bigg[\frac{J_n}{\alpha+nS}\bigg]^p$, and I don't know what is $J_n$ (In particular, it is showed that the martingale $\big(\frac{P_n}{\alpha+nS}\big)$ converges to a random variable $V$ a.s. and earlier in the notes, the only time they define $J_n$ is saying "$J_k(n)$ is the number of leaves at time $n$ of the $k$-th subtree." and further "is exactly distributed like the composition vector at time $(n−1)$ of an $(S+1)$-color Pólya urn process", so I guess they meant to write $P_n$ in this proof). Even if i find formula $(1)$, I don't undesrtand how to get $(2)$.

Answer 1

The second question from the first:

I also agree that $J_n$ should be $P_n$. Apply expectation to $P_n^p$, by using (1). Except for the first term, all other terms end up $O(n^{p-1})$. Then it becomes $O(1/n)$ after dividing $(\alpha + nS)^p$.

The first question:

Formula (1) on the right side, $X$ must be removed.

Note that for $X=(x_1,\ldots, x_d)$, $$\Gamma_p(X)= \left(\frac{x_1}S \cdot (\frac{x_1}S+1) \cdots (\frac{x_1}S+p_1-1)\right) \cdots \left(\frac{x_d}S \cdot (\frac{x_d}S+1) \cdots (\frac{x_d}S+p_d-1)\right)$$ by repeatedly applying $\Gamma(s+1)=s\Gamma(s)$.

Thus, expanding the polynomial in the basis $\Gamma_p$ means writing monomials $X^p$ in terms of $\Gamma_p$ polynomials as above. This can be done through Stirling numbers of the second kind $S(p,k)$. It is enough to have for single variable $x$ and $p\in\mathbb{N}$, $$ x^p=(-1)^p\sum_{k=0}^p S(p,k) (-1)^k x(x+1)\cdots (x+k-1). $$ Putting $x/S$ in $x$, we have $$ x^p=(-1)^pS^p\sum_{k=0}^p S(p,k) (-1)^k \frac xS (\frac xS+1) \cdots (\frac xS+k-1). $$