This question just came in my mind and I don't know whether this or similar questions have been considered in the literature. Consider a complexvalued polynomial $$p(z)=a_0+a_1z+\dots+a_nz^n$$ of degree n with with given coefficients. If $z$ is chosen uniformly in the circle(disc) $z \leq 1$ ,then $Y=p(z)$ is obviously a random variable when z is uniformly chosen in the unit disc $z \leq 1$. Is it possible to find out the pdf of $Y$ or even some of the statistical measures like mean, variance etc. as functions of $n$ and of the coefficients or would that be too hard analytically? I have no idea as to how to proceed. Could someone kindly point out the relevant results in complex analysis that would bear something on this problem? I would greatly appreciate any hints/suggestions/links in this regard.
PS:of course the distribution of $Y$ will change as we vary the coefficients. So precisely speaking I am talking about the distribution of $Y$ given the coefficient vector $(a_0,a_1,...a_n)$. If you think prescribing some distribution to the coefficients will make it more interesting and meaningful, kindly write your response to that problem as well

$\begingroup$ doesn't the average of p(z) exist for a given vector of coefficients $(a_0,a_1,...,a_n)$?If we take the coefficients as per some distribution say,$N(0,1)$ would the problem make sense in that case ? $\endgroup$– AgnostMysticCommented Jan 14, 2022 at 11:09

$\begingroup$ i need the average in terms of n and the coefficients which are assumed to b fixed and given,more specifically I need the distribution of $Y=p(z)$ given $(a_0,a_1,..,a_n)$ $\endgroup$– AgnostMysticCommented Jan 14, 2022 at 12:00

2$\begingroup$ OK, so the even moments may be computed by integrating over the ball in polar coordinates. $\endgroup$– CloudscapeCommented Jan 14, 2022 at 13:55

$\begingroup$ @AlgebraicsAnonymous could you kindly explain how the moments can be computed? $\endgroup$– AgnostMysticCommented Jan 14, 2022 at 16:11

1$\begingroup$ Not the one you consider, but a wellstudied quantity is the Mahler measure of a polynomial $p$, which is the geometric mean of $p$ over the unit torus. There is also the zeta Mahler function $Z(p,s)$. $\endgroup$– François BrunaultCommented Jan 14, 2022 at 21:24
1 Answer
$\newcommand\ol\overline$As pointed out by AlgebraicsAnonymous, it is not a problem to find all evenorder moments of $Y$.
However, Mathematica cannot find $EY$ even when $n=1$, $a_0=1$, and $a_1=2$. So, it is highly unlikely that a closed form expression for $EY$ exists in general, let alone higher oddorder moments.
Here is the image of a corresponding Mathematica notebook:
Remark 1: The OP asked for details on how the above expression for mod (?) in the Mathematica notebook was derived. So, here is the derivation: for $z=re^{it}$ with $r\ge0$ and $t\in[0,2\pi)$, \begin{equation*} \begin{aligned} & p(z)^2=\Re(p(z)^2) \\ &=\Re(p(z)\overline{p(z)}) \\ &=\Re\Big(\sum_{j=0}^n a_j z^j\,\sum_{k=0}^n \ol{a_k}\, \ol z^k\Big) \\ &=\Re\Big(\sum_{j,k=0}^n a_j \ol{a_k} z^j\, \ol z^k\Big) =\sum_{j,k=0}^n \Re(a_j \ol{a_k} z^j\, \ol z^k) \\ &=\sum_{j,k=0}^n \Re(a_j \ol{a_k} r^{j+k} e^{i(jk)t)}) =\sum_{j,k=0}^n a_j a_k r^{j+k}\cos((jk)t) \end{aligned} \end{equation*} if the $a_j$'s are real numbers (which is so in the case considered in the Mathematica notebook).
Of course, one can bound the oddorder moments of $Y$ using the logconvexity of $m_p:=EY^p$ in $p$: \begin{equation*} m_{2q+2}^{3/2}m_{2q+4}^{1/2}\le m_{2q+1}\le m_{2q}^{1/2}m_{2q+2}^{1/2} \tag{1} \end{equation*} for $q=0,1,\dots$.
Remark 2: The OP asked for details on how the logconvexity of $m_p$ in $p$ is used to get (1). Here we go: the logconvexity of $m_p$ in $p$ means that $l_p:=\ln m_p$ is convex in $p$, that is, \begin{equation*} l_{p_t}\le (1t)l_{p_0}+tl_{p_1}, \tag{2} \end{equation*} where $p_0$ and $p_1$ are any nonnegative real numbers, $t\in[0,1]$, and $p_t:=(1t)p_0+tp_1$. Using now (2) with $p_0=2q+1$, $p_1=2q+4$, and $t=1/3$ (and some simple algebra), we get the first inequality in (1). Using (2) with $p_0=2q$, $p_1=2q+2$, and $t=1/2$ (and some simple algebra), we get the second inequality in (1).
Note also the special case of the second inequality in (1) with $q=0$: \begin{equation*} m_1\le m_2^{1/2}, \tag{3} \end{equation*} which is also an instance of the CauchySchwarz, Lyapunov, and Hölder inequalities.
As for the evenorder moments, using the multinomial expansion of $Y^{2q}=p(z)^q\overline{p(z})^q$ for $q=0,1,\dots$ and integrating the resulting expansion termwise in polar coordinates, we get $$m_{2q}=\frac{(q!)^2}{2\pi}\, \sum_{s=0}^{nq}\frac1{s+1} \sum\nolimits_s\prod_{j=0}^n\frac{a_j^{q_j}\,\overline{a_j}^{\,t_j}}{q_j!\, t_j!},$$ where $\sum\nolimits_s$ denotes the summation over all $(n+1)$tuples $(q_0,\dots,q_n)$ and $(t_0,\dots,t_n)$ of nonnegative integers such that $\sum_{j=0}^n j q_j=\sum_{j=0}^n j t_j=s$.

$\begingroup$ @losif Panels, could you kindly explain where the expression for m in the code comes from,I mean how you derive it $\endgroup$ Commented Jan 15, 2022 at 5:14

$\begingroup$ @sajjadveeri : There is no m in the code. I guess, you meant mod instead of m. Anyhow, I have added a detailed derivation for mod. Also, my last name is Pinelis, not Panels. $\endgroup$ Commented Jan 16, 2022 at 1:00

$\begingroup$ @losif Pinelis,apologies for misspelling ypur name and thank you for the wonderful and lucid explanation.Could you kindly explain how you actually used the logconvexity in bounding the odd moments and why you divided by $\pi $ and not the area of the circle?Second ,I did some calculations and I found that $p(z)^2$ can be computed in a closed form,since $\int_0^{2 \pi}cos(jk)t dt=0$ when j is not equal to k .In the rest of case we will be left with terms of the form $ \int_o^1 a_j^2 r^{2j}dr$ which are readily computable. $\endgroup$ Commented Jan 16, 2022 at 15:41

$\begingroup$ Am I right here?If yes ,can we bound average of $p(z)$ using average of $p(z)^2$ ?thank you and regards $\endgroup$ Commented Jan 16, 2022 at 15:41

$\begingroup$ @sajjadveeri : I have provided details on the use of the logconvexity. Also, I have noted the inequality $m_1\le m_2^{1/2}$ (now inequality (3)) as a special case. $\endgroup$ Commented Jan 16, 2022 at 16:11