# The average value of the modulus $|f(z)|$ of a polynomial with real coefficients

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

• 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 ? Commented Jan 14, 2022 at 11:09
• 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 $(a_0,a_1,..,a_n)$ Commented Jan 14, 2022 at 12:00
• OK, so the even moments may be computed by integrating over the ball in polar coordinates. Commented Jan 14, 2022 at 13:55
• @AlgebraicsAnonymous could you kindly explain how the moments can be computed? Commented Jan 14, 2022 at 16:11
• 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 $Z(p,s)$. Commented Jan 14, 2022 at 21:24

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

• @losif Panels, could you kindly explain where the expression for m in the code comes from,I mean how you derive it Commented Jan 15, 2022 at 5:14
• @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. Commented Jan 16, 2022 at 1:00
• @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. Commented Jan 16, 2022 at 15:41
• Am I right here?If yes ,can we bound average of $|p(z)|$ using average of $|p(z)|^2$ ?thank you and regards Commented Jan 16, 2022 at 15:41
• @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. Commented Jan 16, 2022 at 16:11