Is this combinatorial identity known? (of interest for random matrix theory)

Question

While playing around with random matrices and I arrived at a different formula for the mean of the limiting normal distribution for a spectral CLT for sample covariance matrices. More precisely I have the formula \begin{align*} & \sum\limits_{b=1}^{r-1} c^{b} \left(A(r,b) - {r \choose b}^2 \, b\right) \end{align*} for the expression (1.23) from Bai and Silverstein's famous paper 'CLT for Linear Spectral Statistics of Large-Dimensional Sample Covariance Matrices', where \begin{align*} A(r,b) :=& \sum\limits_{m=1}^{\min(b,r-b+1)} (2m-1) {r \choose b-m}{r \choose b+m-1}\\ & + \sum\limits_{m=1}^{\min(b,r-b)} (2m+1) {r \choose b-m}{r \choose b+m} \ . \end{align*} In order to show that my expression is equal to that of Bai and Silverstein, I need to show the equality $$ A(r,b) = \frac{1}{2} {2r \choose 2b} + \left( b - \frac{1}{2} \right) {r \choose b}^2 $$ for all $r \in \mathbb{N}$ and $b<r$. I am not too well versed with combinatorial identities and so far my efforts have lead nowhere. Does someone have an idea how to prove this?

Testing with the computer shows this to be true for at least all $b<r \leq 1000$.

Technically I could just use the fact that my formula and Bai and Silverstein's formula describe the same object to show this identity, which seems rather crude unless this identity is not jet known.

Answer 1

Firstly, exploit the finite support to simplify the limits of the sums. Secondly, split the second sum. We get

$$\begin{align*}A(r,b) =& \sum\limits_{m=1}^{r} (2m-1) {r \choose b-m}{r \choose b+m-1} \\ & + \sum\limits_{m=1}^{r} 2m {r \choose b-m}{r \choose b+m} \\ & + \sum\limits_{m=1}^{r} {r \choose b-m}{r \choose b+m} \end{align*}$$

The first two are mechanical using Wilf-Zeilberger, so ask your favourite CAS (e.g. Wolfram Alpha or Sage) to get

$$\begin{align*}A(r,b) = & \frac{b(r-b+1)\binom{r}{b-1}\binom{r}{b} + (b-2r-1)(b+r)\binom{r}{b-r-1}\binom{r}{b+r}}{r} \\ & + \frac{(b+1)(r-b+1)\binom{r}{b-1}\binom{r}{b+1} + (b-2r-1)(b+r+1)\binom{r}{b-r-1}\binom{r}{b+r+1}}{r} \\ & + \sum_{m=1}^r \binom{r}{b-m} \binom{r}{b+m} \end{align*}$$

For the third sum, we apply $$\sum_k \binom{r}{m+k} \binom{s}{n-k} = \binom{r+s}{m+n}$$ (see e.g. identity (5.22) of Concrete Mathematics by Graham, Knuth and Patashnik). So $$\sum_{m=1}^r \binom{r}{b-m} \binom{r}{b+m} = \frac{\sum_{m=-r}^r \binom{r}{b-m} \binom{r}{b+m} - \binom{r}{b}^2}{2} = \frac{\binom{2r}{2b} - \binom{r}{b}^2}{2}$$

I've left you some cancellation, which I fully expect to be routine.