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

  • $\begingroup$ 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 ? $\endgroup$ 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 $(a_0,a_1,..,a_n)$ $\endgroup$ Jan 14, 2022 at 12:00
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    $\begingroup$ OK, so the even moments may be computed by integrating over the ball in polar coordinates. $\endgroup$
    – Cloudscape
    Jan 14, 2022 at 13:55
  • $\begingroup$ @AlgebraicsAnonymous could you kindly explain how the moments can be computed? $\endgroup$ Jan 14, 2022 at 16:11
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    $\begingroup$ Not the one you consider, but a well-studied 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 $Z(p,s)$. $\endgroup$ Jan 14, 2022 at 21:24

1 Answer 1


$\newcommand\ol\overline$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:

enter image description here

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 $\sum_{j=0}^n j q_j=\sum_{j=0}^n j t_j=s$.

  • $\begingroup$ @losif Panels, could you kindly explain where the expression for m in the code comes from,I mean how you derive it $\endgroup$ 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. $\endgroup$ 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 log-convexity 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 $\int_0^{2 \pi}cos(j-k)t dt=0$ 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. $\endgroup$ 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 $\endgroup$ Jan 16, 2022 at 15:41
  • $\begingroup$ @sajjadveeri : I have provided details on the use of the log-convexity. Also, I have noted the inequality $m_1\le m_2^{1/2}$ (now inequality (3)) as a special case. $\endgroup$ Jan 16, 2022 at 16:11

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