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 complex-valued 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
Complex Variables – Average Value of the Modulus |f(z)| of a Polynomial with Real Coefficients
cv.complex-variablesprobability distributions
Best Answer
$\newcommand\ol\overline$As pointed out by AlgebraicsAnonymous, it is not a problem to find all even-order 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 odd-order 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(j-k)t)}) =\sum_{j,k=0}^n a_j a_k r^{j+k}\cos((j-k)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 odd-order moments of $Y$ using the log-convexity 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 log-convexity of $m_p$ in $p$ is used to get (1). Here we go: the log-convexity of $m_p$ in $p$ means that $l_p:=\ln m_p$ is convex in $p$, that is, \begin{equation*} l_{p_t}\le (1-t)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 $p_t:=(1-t)p_0+tp_1$. 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 Cauchy--Schwarz, Lyapunov, and Hölder inequalities.
As for the even-order 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 term-wise 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 $\sum_{j=0}^n j q_j=\sum_{j=0}^n j t_j=s$.