Why does this combinatorial sum vanish?

Question

We define the coefficients $c_{k,k-i}$ of ${n \choose k}$ by the following easy expansion: \begin{align*} & {n \choose k} = \frac{1}{k!} n(n-1) \dots (n-k+1) = \frac{1}{k!} \prod\limits_{t=0}^{k-1} (n-t)\\ & = \sum\limits_{i=0}^{k} n^{i} \frac{(-1)^{k-i}}{k!} \underbrace{\sum\limits_{\substack{0 \leq x_1 < \dots < x_{k-i} < k}} x_1 \dots x_{k-i}}_{=: c_{k,k-i}} \end{align*} For given $L \in \mathbb{N}$ and $i,j \in \mathbb{N}$ with $i+j \leq L$ the sum \begin{align*} & S_{L,i,j} := \sum\limits_{b=i}^\infty \sum\limits_{w=j}^\infty (-1)^{b+w} {L-1 \choose b+w-2} {b+w-2 \choose b-1} \frac{c_{b,b-i}}{b!} \frac{c_{w,w-j}}{w!} \end{align*} vanishes whenever $(i+j)$ and $L$ have the same parity, i.e. \begin{align*} & (L+i+j) \, \operatorname{mod} \, 2 = 0 \ \ \Rightarrow \ \ S_{L,i,j} = 0 \ . \end{align*}

Question: Is there a nice way to see from the form of $S_{L,i,j}$ why this property holds?

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The motivation behind this question is the fact that the trace moments of $XX^*$, where $X$ is a $(p \times n)$ random matrix with iid standard complex normal entries, are given by \begin{align*} & \mathbb{E}\big[ \operatorname{tr}((XX^*)^{L}) \big] = L! \sum\limits_{i,j=0}^\infty (-1)^{i+j} p^i n^j S_{L,i,j} \end{align*} and some other theoretical arguments show that the influence of summands with the wrong parity of $i+j$ do not contribute. The aim of the question is to gain a better understanding of $S_{L,i,j}$ in the non-vanishing cases by firstly understanding why it sometimes vanishes.

Answer 1

1. Consider two variables $x,y$. Then we have $$(-1)^{i+j}S_{L,i,j}=[x^iy^j]\sum_{b,w}{L-1\choose b+w-2}{b+w-2\choose b-1}{x\choose b}{y\choose w}.$$ where as usual $[M]F$ denotes a coefficient of monomial $M$ in the polynomial (or power series etc) $F$. Here ${x\choose b}:=\frac{x(x-1)\ldots(x-b+1)}{b!}$ is considered as a polynomial in $x$. So, denote $$A(x,y):=\sum_{b,w}{L-1\choose b+w-2}{b+w-2\choose b-1}{x\choose b}{y\choose w}.$$ We need to prove $$A(-x,-y)=(-1)^{L+1}A(x,y).\tag{1}$$

2. Introduce two new variables $t,s$ and consider Laurent series in $t,s$. Then $$A(x,y)=[1] t^{-1}s^{-1}(1+t^{-1}+s^{-1})^{L-1}(1+t)^x(1+s)^y, \tag{2}$$ where $(1+t)^x:=\sum {x\choose b}t^b$ is a power series in $t$, and polynomials in $x$ are its coefficients.

3. It remains to prove (1) for the polynomial (2) (which already looks like something more pleasant than long combinatorial sums. Also, it is quite a general statement in a sense that $L$, $x$, $y$ are arbitrary, and the general identities are usually easier to prove that very specific ones.)

The idea is a suitable change of variables. Forget about $s$ for a moment and think about power series in $t$. We have $[1] h(t)=[t^{-1}]t^{-1}h(t)={\rm res}\, t^{-1}h(t)$, and the residue (which may be viewed a circuit integral) enjoys the change of variables formula. We should have some connection between $(1+t)^x$ and $(1+t)^{-x}$, a natural change of variables for this goal is the involutive function $t=-\frac{\tau}{1+\tau}$. We have $$ [1]h(t)={\rm res}\, t^{-1}h(t)=\oint t^{-1}h(t)dt=\oint -\frac{1+\tau}{\tau}h\left(-\frac{\tau}{1+\tau}\right)-\frac{d\tau}{(1+\tau)^2}\\= {\rm res}\, \tau^{-1}(1+\tau)^{-1}h\left(-\frac{\tau}{1+\tau}\right)=[1](1+\tau)^{-1}h\left(-\frac{\tau}{1+\tau}\right). \tag{3} $$ [remarks: a) what we do here is a partial case of the Lagrange–Bürmann formula which is a change of variables in generating functions context. b) The formula (3) itself may be proved just by checking it for $h(t)=t^m$ for every integer $m$, but then it could seem more enigmatic than it deserves to be.]

4. So, now we denote $t=-\tau/(1+\tau)$, $s=-\xi/(1+\xi)$ and apply (3) for (2). We get $$A(x,y)\\=[1](1+\tau)^{-1}(1+\xi)^{-1}\frac{(1+\tau)(1+\xi)}{\tau \xi}(-1-\tau^{-1}-\xi^{-1})^{L-1}(1+\tau)^{-x}(1+\xi)^{-y}\\=(-1)^{L-1}A(-x,-y)$$ as desired.