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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.

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  • $\begingroup$ how should I read $b \land (r-b+1)$? the smallest of $b$ and $r-b+1$? $\endgroup$ Jul 19, 2022 at 11:35
  • $\begingroup$ Sorry, this was supposed to mean the minimum of b and (r-b+1). I've edited the question. $\endgroup$
    – Tardis
    Jul 19, 2022 at 11:45

1 Answer 1

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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.

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    $\begingroup$ Thank you! You made it look easy. I'm a bit embarrassed. $\endgroup$
    – Tardis
    Jul 19, 2022 at 16:15
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    $\begingroup$ It took a bit of experimentation and knowing where to look for known results. $\endgroup$ Jul 19, 2022 at 16:27

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