Calculating the Moments of the Marchenko-Pastur Distribution

Question

I'm trying to follow a proof of the Marchenko-Pastur theorem. In particular I'm trying to show that the kth moment of the Marchenko-Pastur distribution: $$ a_k = \int_{(1-\sqrt{\gamma})^2}^{(1+\sqrt{\gamma})^2}x^k \frac{1}{2\pi\gamma x}\sqrt{((1+\sqrt{\gamma})^2-x)(x-(1-\sqrt{\gamma})^2)}dx $$ is equal to $$ \sum_{r=0}^{k-1} \frac{\gamma^r}{r+1}\binom{k}{r}\binom{k-1}{r} $$

Where I've got to so far:

I understand how we can use a substitution and rewrite the kth moment as: $$ \frac{1}{2\pi}\int_{-2}^2(\sqrt{\gamma}y+1+\gamma)^{k-1}\sqrt{4-y^2}dy $$ Then the notes I'm working through says you should expand the k-1 power, use the formula for the moments of the semicircle law (i.e. the Catalan number formula) and use Vandermonde's identity. Expanding the k-1 power we can write this as: \begin{align} \frac{1}{2\pi}\int_{-2}^2\sum_{r=0}^{k-1} \binom{k-1}{r} (\sqrt{\gamma} y)^r (1+\gamma)^{k-1-r}\sqrt{4-y^2}dy \\ =\sum_{r=0}^{k-1} \binom{k-1}{r} \gamma^{r/2} (1+\gamma)^{k-1-r}\frac{1}{2\pi}\int_{-2}^2 y^r\sqrt{4-y^2}dy \\ =\sum_{r=0, r\text{ even}}^{k-1} \binom{k-1}{r} \gamma^{r/2} (1+\gamma)^{k-1-r}C_{r/2} \end{align} where $C_i$ is the ith Catalan Number...

But I'm not sure how we can get from here to the result - or where we're going to use Vandermonde's identity? Any help is appreciated.

Answer 1

$$ \frac{1}{2\pi}\int_{-2}^2(\sqrt{\gamma}y+1+\gamma)^{k-1}\sqrt{4-y^2}dy \\ = \sum_{i+j\le k-1} {k-1 \choose i;j;k-1-i-j} \gamma^{i/2+j}\cdot \frac{1}{2\pi} \int_{-2}^2 y^{i}\sqrt{4-y^2} dy \\ = \sum_{2l+j\le k-1} {k-1 \choose 2l;j;k-1-2l-j} \gamma^{l+j}\cdot C_{l} \\= \sum_{2l+j\le k-1} \frac{(k-1)!}{(2l)!j! (k-1-2l-j)!}\cdot\frac {(2l)!}{l!(l+1)!}\cdot \gamma^{l+j} = \big| l+j = r\big|\\ = \sum_{l+r\le k-1}\frac{(k-1)!}{l!(r-l)! (l+1)!(k-1-l-r)!} \cdot \gamma^{r}\\ = \sum_{r = 0}^{k-1}\frac{(k-1)!}{r!(k-r)!}\cdot\gamma^r \sum_{l=0}^{k-1-r} {r\choose l} \cdot {k-r \choose k-1-r-l} \\ = \sum_{r = 0}^{k-1} \frac{(k-1)!}{r!(k-r)!}\cdot\gamma^r \cdot {k \choose k-r-1} = \sum_{r = 0}^{k-1} \frac{1}{r+1}{k \choose r} \cdot {k-1 \choose r}\cdot\gamma^r. $$