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I want to lower bound the expected value of the square root of a randomly chosen eigenvalue of a Wishart matrix.

To get the bound I want I need a lower bound on

$$T_n = \int_0^\infty\sqrt{x}e^{-x}L_n(x)^2dx,$$

for $n>1$ where $L_n(x)$ is the $n$-th Laguerre polynomial.

Mathematica convinced me that $T_n > \sqrt{n+1}$ but I wasn't able to find a proof for this.

Does anyone have an idea of how to estimate this integral?

Note: if the $\sqrt{x}$ is replaced by $x$ then Wikipedia has a closed form solution for it.

Thanks!

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  • $\begingroup$ I include $n>1$ because the bound fails for $n=1$. $\endgroup$ – Afonso S. Bandeira Aug 10 '13 at 18:15
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    $\begingroup$ Your integral can be evaluated by expanding the Laguerre polynomial in $x$, squaring the result, and then the individual terms are integrable as gamma functions. For example, for $n=3$ I get $\frac{687\sqrt{\pi}}{512}$. I agree with your bound, but I have not been able to see how to obtain it even with the exact evaluation for given $n$ values. Maybe this will help you see a way! - Tom $\endgroup$ – Tom Dickens Aug 12 '13 at 0:45
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Here is a proof.

First evaluate the integral. I use the generating function for squares of Laguerre polynomials ($I_0$ the modified Bessel function):

$$ \sum_{n=0}^{\infty}L_{n}(x)^2 z^n = \frac{1}{1-z} \exp\left(-\frac{2 x z}{1-z}\right) I_{0}\left(\frac{2 x \sqrt{z}}{1-z}\right) $$

for $|z|<1$ (All the formulas I use, one can find it in e.g. Gradshteyn and Ryzhik.)

This gives for the integral

$$ T_{n} = \int_{0}^{\infty} dx \sqrt{x} e^{-x} L_{n}(x)^2 = [z^n] \frac{1}{1-z} \int_{0}^{\infty} dx \sqrt{x} \exp\left(-x \frac{1+z}{1-z}\right) I_{0}\left(x \frac{2 \sqrt{z}}{1-z}\right) $$

$[z^n]$ means, as usual, "take the coefficient of $z^n$ in the (formal) power expansion in $z$ of the following expression".

The integral can be evaluated (it is in Gradshteyn and Ryzhik) and the result is

$$ T_{n} = [z^n] \frac{\sqrt{\pi}}{2} (1-z)^{-1} P_{1/2}\left(\frac{1+z}{1-z}\right) $$

with $P_{1/2}(x)$ the Legendre Function.

Expanding the Legendre function and the prefactor in powers of $z$ gives for the coefficient of $[z^n]$

$$ T_{n} = \frac{1}{8\sqrt{\pi}}\sum_{m=0}^{n}\frac{\Gamma\left(m-\frac{1}{2}\right) (2(n-m)+1)!}{4^{n-m} m!^2 (n-m)!^2} $$

Now I was lazy and asked Mathematica to simplify, which gives

$$ T_{n}= \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} {_3}F_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right) $$

with the generalized hypergeometric function ${_3}F_{2}$ evaluated at $1$.

With the exact solution at hands we are almost done. We observe that the sum defining the hypergeometric function

$$ {_3}F_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right) =\sum_{m=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{m}^{2}(-n)_m}{m!^2 \left(-n-\frac{1}{2}\right)_{m} } $$

has only positive terms (the minus signs in the Pochhammer symbols $(-n)_m$ and $(-n-\frac{1}{2})_m$ cancel.). We get a lower bound by taking only the first two expansion terms of ${_3}F_{2}$ and the well known estimate for the Gamma functions in front

$$ \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} \geq n^{1/2} $$

and find

$$ T_{n} \geq n^{1/2} \left(1+\frac{n}{2(n+1)}\right) $$

which is larger than $\sqrt{n+1}$ for $n\geq 2$.

Btw, the asymptotic ($n\rightarrow \infty$) value of $T_n$ is

$$ T_{n}\sim \frac{4}{\pi} n^{1/2} $$

I am pretty sure that the calculation can be done more elegantly, but had not the time to dig deeper.

Edit: With the same trick one finds more generally:

$$ \int_{0}^{\infty} dx \, x^{\nu} e^{-x} L_{n}(x)^2 = \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {_3}F_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {_2}F_{1}(-\nu,-\nu;1;1)) (1+ O(n^{-1})) $$ The last term simplifies to: $$ \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} \frac{\Gamma(2 \nu +1)}{\Gamma(\nu+1)^2} (1+ O(n^{-1})) $$

for $\nu > 0$.

Edit: After some massaging with Mathematica I could determine the coefficient of $n^{-1}$ in the above expansion: $$ \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {_3}F_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {_2}F_{1}(-\nu,-\nu;1;1)) (1 + \frac{\nu}{2 n} + O(n^{-1-2 \nu})) $$ The determination of the next order coefficient as well as the exact exponent is rather cumbersome. One could do better than $O(n^{-1-2 \nu})$.

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  • $\begingroup$ That is a very nice result, and teaches me a new way to look at such problems. Thanks! $\endgroup$ – Tom Dickens Aug 14 '13 at 14:13
  • $\begingroup$ Thanks! This is great! I think there is a small typo, the second term of the series should be $\frac{n}{2(2n-1)}$ but the idea works and I really like it. Thanks! $\endgroup$ – Afonso S. Bandeira Sep 4 '13 at 2:29
  • $\begingroup$ Actually, I have a follow-up question. In the edit, is there a known bound on the constant in that $O(n^{-1})$ term? $\endgroup$ – Afonso S. Bandeira Sep 4 '13 at 20:00
  • $\begingroup$ @Alfonso Sorry for not answering for so long. I am quite busy currently, but did not forget about your follow up question. $\endgroup$ – Johannes Trost Sep 17 '13 at 14:07

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