Let $z\_1,z\_2,\ldots,z\_n$ be i.i.d. random variables in the unit circle. Consider the polynomial
$$
p(z)=\prod\_{i=1}^{n}{(t-z\_i)}=t^n+a\_{1}t^{n-1}+\cdots+a\_{n-2}t^2+a\_{n-1}t+a\_n
$$
where the $a\_i$ are the symmetric functions
$$
a\_{1}=(-1)\sum\_{i=1}^{n}{z\_{i}}\hspace{0.3cm},\quad a\_{2}=(-1)^2\sum\_{1\leq i< j\leq n}{z\_{i}z\_{j}}\hspace{0.3cm} \quad\ldots\quad a\_{n}=(-1)^{n}z\_{1}z\_{2}\ldots z\_{n}.
$$


How can we estimate the random variable $Z$ defined as
$$
Z=\sum\_{j=1}^{n}{|a\_{j}|}
$$
asymptotically as $n\to\infty$?


It is not very difficult to estimate $|\sum\_{j=1}^{n}{a\_{j}}|$ by estimating $\log p(1)$ via the CLT. However, $Z$ seems to be much more difficult. Any idea of what can work here? 



> 
> **Update:** If we look at the term at the central symmetric random variable
>  $a\_{\lfloor n/2 \rfloor}$ $$
> > a\_{\lfloor n/2 \rfloor}=\text{sum of
> > the products of $\lfloor n/2 \rfloor$
> > of different $z\_{i}$'s} $$ 
>  it is not hard to see that it
>  has uniform distributed phase in
>  $(-\pi,\pi]$. 
> 
> 
> However, its magnitude is blowing up extremely fast!
> 
> 
> Does anyone knows how to compute the limit distribution of 
>  $|a\_{\lfloor n/2 \rfloor}|$ under the appropriate normalizations? 
> 
> 
> 


Thanks!



Estimates for Symmetric Functions
OK, here is my argument (sorry for the delay).


First of all, $Z$ is essentially the maximum of the absolute value of the polynomial $P(z)=\prod\_j(1-z\_jz)$ on the unit circumference (up to a factor of $n$, but it is not noticeable on the scale we are talking about).


Second, the maximum of the absolute value of a (trigonometric) polynomial of degree $K$ can be read from any $AK$ uniformly distributed points on the unit circumference $\mathbb T$ (say, roots of unity of degree $AK$) with relative error of order $A^{-1}$.


Now let $\psi(z)=\log(1-z)$. We want to find the asymptotic distribution of the $\max\_z n^{-1/2}Re\sum\_j \psi(zz\_j)$ where $z\_j$ are i.i.d. random variables uniformly. Since it is the logarithm of $|P(z)|$, the maximum can be found using $10n$ points. 


Decompose $\psi(z)$ into its Fourier series $-\sum\_{k\ge 1}\frac 1kz^k$. Then, formally, we have $n^{-1/2}\sum\_j \psi(zz\_j)=-\sum\_k\left(n^{-1/2}\sum\_j z\_j^k\right)\frac{z^k}{k}$. It is tempting to say that the random variables $\xi\_n,k=n^{-1/2}\sum\_j z\_j^k$ converge to the uncorrelated standard complex Gaussians $\xi\_k$ by the CLT in distribution and, therefore, the whole sum converges in distribution to the random function $F(z)=\sum\_k\xi\_k\frac{z^k}k$, so $n^{-1/2}\log Z$ converges to $\max\_z\Re F(z)$ (the $-$ sign doesn't matter because the limiting distribution is symmetric). This argument would be valid literally if we had a finite sum in $k$ but, of course, it is patently false for the infinite series (just because if we replace $\max$ by $\min$, we get an obvious nonsense in the end). Still, it can be salvaged if we do it more carefully.


Let $K$ run over the powers of $2$. Choose some big $K\_0$ and apply the above naiive argument to $\sum\_{k=1}^{K\_0}$. Then we can safely say that the first $K\_0$ terms in the series give us essentially the random function $F\_{K\_0}(z)$ which is the $K\_0$-th partial sum of $F$ when $n$ is large enough. 


Our main task will be to show that the rest of the series cannot really change the maximum too much. More precisely, it contributes only a small absolute error with high probability. 


To this end, we need 


Lemma: Let $f(z)$ be an analytic in the unit disk function with $f(0)=0$, $|\Im f|\le \frac 12$. Then we have $\int\_{\mathbb T}e^{\Re f}dm\le \exp\left(2\int\_{\mathbb T}|f|^2dm\right)$ where $m$ is the Haar measure on $\mathbb T$.


Proof: By Cauchy-Schwartz, 
$$
\left(\int\_{\mathbb T}e^{\Re f}dm\right)^2\le \left(\int\_{\mathbb T}e^{2\Re f}e^{-2|\Im f|^2}dm\right)\left(\int\_{\mathbb T}e^{2|\Im f|^2}dm\right)
$$
Note that if$|\Im w|\le 1$, we have $e^{\Re w}e^{-|\Im w|^2}\le \Re e^w$. So the first integral does not exceed $\int\_{\mathbb T}\Re e^{2f}dm=\Re e^{2f(0)}=1$. Next, $e^s\le 1+2s$ for $0\le s\le\frac 12$, so $\int\_{\mathbb T}e^{2|\Im f|^2}dm\le 1+4\int\_{\mathbb T}|\Im f|^2dm\le 1+4\int\_{\mathbb T}|f|^2dm$. Taking the square root turns $4$ into $2$ and it remains to use that $1+s\le e^s$ 


The immediate consequence of Lemma 1 is a Bernstein type estimate for $G\_K(z)=\sum\_{k\in (K,2K]}\left(n^{-1/2}\sum\_j z\_j^k\right)\frac{z^k}{k}$
$$
P(\max|\Re G\_K|\ge 2T)\le 20Ke^{-T^2K/9}
$$
if $0\le TK\le \sqrt n$, say.


Indeed, just use the Bernstein trick on the independent random shifts of $g\_K(z)=\sum\_{k\in (K,2K]}\frac{z^k}{k}$:
$$
E e^{\pm t\Re G\_K(z)}\le \left(\int\_{\mathbb T}e^{\Re tn^{-1/2}g\_K}dm\right)^n\le e^{2t^2/K}
$$
for every $t\le \sqrt n/2$ (we used the Lemma to make the last estimate) and put $t=\frac{TK}{3}$. After that read the maximum from $10K$ points with small relative error and do the trivial union bound.


Choosing $T=K^{-1/3}$, we see that we can safely ignore the sum from $K=K\_0$ to $K=\sqrt n$ if $K\_0$ is large enough. Now we are left with 
$$
G\_K(z)=\sum\_{k\ge \sqrt n}\left(n^{-1/2}\sum\_j z\_j^k)\right)\frac{z^k}{k}
$$
to deal with. Recall that all we want here is to show that it is small at $10n$ uniformly distributed points. Again, if $g(z)=\sum\_{k\ge \sqrt n}\frac{z^k}{k}$, we have $|\Im g|\le 10$, say so we can use the same trick and get 
$$
P(\max\_{10n\text{ points}}|\Re G|\ge 2T)\le 20n e^{n^{-1/2}t^2-tT}
$$
if $0\le t\le \sqrt n/20$, say.
Here we do not need to be greedy at all: just take a fixed small $T$ and choose $t=\frac{2\log n}T$.


Now, returning to your original determinant problem, we see that the norm of the inverse matrix is essentially $Z/D$ where $D=\min\_i\prod\_{j:j\ne i}|z\_i-z\_j|$. We know the distribution of $\log Z$ and we have the trivial Hadamard bound $D\le n$. This already tells you that the typical $\lambda\_1$ is at most $e^{-c\sqrt n}$. The next logical step would be to investigate the distribution of $\log D$. 


