Special problem: Let $G(z)$ be a probability generating function(pgf, the $z$ can be seen as real number or complex number), that is $$G(z) = \sum\limits_{i = 0}^\infty {{p_i}{z^i}} ,(\left| z \right| \leqslant 1;\sum\limits_{i = 0}^\infty {{p_i} = 1;} 0 \leqslant {p_i} \leqslant 1).$$ the logarithm of $G(z)$ is $$\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}} ,(\left| z \right| \leqslant 1).$$ If ${p_0} > {p_1} > {p_2} > \cdots$ , then show that ${{q_i}}$ is absolutely convergent, namely $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|} < \infty $, or give a counterexample.

I want to use Faa Di Bruno Formula (http://en.wikipedia.org/wiki/Faà_di_Bruno's_formula) to get the expression of $q_i$. then to estimate $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|}$.

Remark: The problem above is a special case of my Problem which is connected to probability generating function:

Let $G(z)$ be a probability generating function, that is $$G(z) = \sum\limits_{i = 0}^\infty {{p_i}{z^i}} ,(\left| z \right| \leqslant 1;\sum\limits_{i = 0}^\infty {{p_i} = 1;} 0 \leqslant {p_i} \leqslant 1).$$

Problem 1. Under which necessary and sufficient condition on $p_i$, the logarithm of $G(z)$ $$\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}} ,(\left| z \right| \leqslant 1).$$ is absolutely convergent, namely $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|} < \infty $?

My conjecture is that if $G(z)$ has no zeros then $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|} < \infty $. If this conjecture is not ture, can somebody give me a counterexample.

The Problem 1 is similar to an open problem:

Problem 2. Let $f(t) = \sum\limits_{k = 0}^\infty {{a_k}{e^{kit}}}$, we use the norm $\left\| f \right\| = \sum\limits_{k = 0}^\infty {\left| {{a_k}} \right| } $. Under which conditions on $f$ is the sequence $\left\| {{f^n}} \right\|$ bounded?

This problem are firstly proposed by Beurling(1938), quoted by Henry(1953). Hedstrom(1966) call the problem "Norms of powers of absolutely convergent Fourier series". It relates to complex-valued probabilities, see Baishanski(1999).

Problem 3. If some $q_i$ are negative, under which necessary and sufficient condition on $q_i$, $$\exp (\sum\limits_{i = 0}^\infty {{q_i}{z^i}} ) = \exp [\sum\limits_{i = 1}^\infty {{q_i}({z^i}} - 1)]$$ is a pgf?

Lévy(1937) prove that $P(z) ={e^{\sum\limits_{i = 1}^m {{q _i}({z^i}-1)} }}{\rm{, }}(\left| z \right| \le 1)$ is a pgf when a term with a sufficiently small negative coefficient is preceded by one term with positive coefficient and followed by at least two terms with positive coefficients as well(see Johnson(2005), p393-394), namely ${q_1} > 0,{q_{m - 1}} > 0,{q_m} > 0$. For $m=4$, van Harn(1987) give four inequalities to ensure $$P(z) = {e^{a(z-1) - b({z^2}-1) + c({z^3}-1) + d({z^4}-1)}}{\rm{, }}(\left| z \right| \le 1)$$ is a pgf, namely $a,b,c,d > 0$ and $ b \le \min \{ \frac{{{a^2}}}{3},\frac{c}{a},\frac{{ad}}{{2c}},\frac{{{c^2}}}{{3d}}\} $.

Problem 4. The $\frac{1}{k}($ for each $ k \in \mathbb{N})$ power of $G(z)$ is $$\sqrt[k]{{G(z)}} = \sum\limits_{i = 0}^\infty {q_i^{(k)}{z^i}} ,(\left| z \right| \le 1).$$ If some ${q_i^{(k)}}$ are negative, under which necessary and sufficient condition on $p_i$, $\sqrt[k]{{G(z)}}$ is absolutely convergent, namely $\sum\limits_{i = 0}^\infty {\left| {q_i^{(k)}} \right|} < \infty $?

When ${p_0} \ge {p_1} \ge {p_2} \ge \cdots \ge 0$ in Problem 4, this is Szekely's Discrete Convex Theorem (See Theorem 2.3.1 of Kerns(2004), page29). This problem appear firstly in Székely(2005). Székely(2005) discuss the condition that $\sqrt[k]{{G(z)}}$ is absolute convergence, when $G(z)$ is pgf of Bernoulli distibution. It is related to Negative probability(http://en.wikipedia.org/wiki/Negative_probability). Since the absolute convergence of $\sqrt[k]{{G(z)}}$ and $\ln G(z)$ are not equiavalent, so Problem 1 and Problem 4 are different.

Problem 1 can be seen as inverse problem of Problem 3, Problem 4 can be seen as inverse problem of Problem 2.

[1]Baishanski, B. (1999). Norms of powers and a central limit theorem for complex-valued probabilities. In Analysis of Divergence (pp. 523-543). Birkhäuser Boston.

[2]Beurling, A. (1938, August). Sur les intégrales de Fourier absolument convergentes et leur applicationa une transformation fonctionnelle. In Ninth Scandinavian Mathematical Congress (pp. 345-366).

[3]Hedstrom, G. W. (1967). Norms of powers of absolutely convergent Fourier series in several variables. The Michigan Mathematical Journal, 14(4), 493-495.

[4]Helson, H., & Beurling, A. (1953). Fourier-Stieltjes transforms with bounded powers. Mathematica Scandinavica, 1, 120-126.

[5]Kerns, G. J. (2004). Signed Measures in Exchangeability and Infinite Divisibility (Doctoral dissertation, Bowling Green State University).

[6]Johnson, N. L., Kemp, A. W., Kotz S. (2005). Univariate Discrete Distributions, 3ed. Wiley, New Jersey.

[7]Lévy, P. (1937). Sur les exponentielles de polynômes et sur l’arithmétique des produits de lois de Poisson. Annales scientifiques de l’École Normale Supérieure, 54, 231–292.

[8]Székely, G. J. (2005). Half of a coin: negative probabilities. Wilmott Magazine, 66-68.

[9]van Harn K.(1978). Classifying infinitely divisible distributions by functional equations. Amsterdam: Mathematisch.

  • 4
    $\begingroup$ There cannot be zeros in $|z|<1$ or even $|z| \leq 1$ because the hypothesis $p_0 > p_1 > p_2 > \cdots > 0$ implies that in the expansion $$ (1-z) \, G(z) = p_0 - \sum_{j=1}^\infty (p_j - p_{j-1}) \, z^j $$ all the coefficients $p_j - p_{j-1}$ are positive and their sum telescopes to $p_0$, so $$ \left| \sum_{j=1}^\infty (p_j - p_{j-1}) \, z^j \right| < p_0$ $$ for all $z \neq 1$ such that $|z| \leq 1$. $\endgroup$ Mar 18 '14 at 3:06
  • 4
    $\begingroup$ 14 edits by author, in under 24 hours. Hard to hit a moving target. $\endgroup$ Mar 18 '14 at 4:46
  • 2
    $\begingroup$ Waouh, 24 edits now! This must be a world record. $\endgroup$
    – abx
    Mar 18 '14 at 13:02
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    $\begingroup$ Noam has essentially answered it. All we need to notice is that the trick he mentioned allows one to show that the real part of $(1-z)G(z)$ is positive in the closed unit disk except for the point $z=1$, so $G(z)$ cannot be $0$ or negative real (Except, maybe, at $1$? Nah, it equals $1$ there!). Hence, $\log$ has an analytic brunch in some neighborhood ($\mathbb C\setminus(-\infty,0]$) of the compact $G( \text{Clos}\mathbb D )$, so Wiener's theorem finishes the story in no time. $\endgroup$
    – fedja
    Mar 30 '14 at 9:44
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    $\begingroup$ As to homework, the answer is "Yes", if the person took a decent course on Banach algebras recently and "No" otherwise. Our (or, at least, my) memory is short and rusty and our education is patchy, so, Noah, I guess I can kill you with an elementary for me question and, most likely, you can return the favor :-). Let us, hence, assume (unless it is obvious otherwise) that whoever asks a question asks it in good faith and for a good reason ;-). $\endgroup$
    – fedja
    Mar 30 '14 at 9:53

Not a complete solution, only a remark: assume $G(z)$ has no zeroes in the closed unit disk. If the series for $G(z)$ has radius of convergence >1 the series for $\log(G(z)$ will converge absolutely in $z=1$. (Because $G^\prime(z)/G(z)$ is then regular in a disk slightly larger than the unit disk.) Hence counterexamples (if any) must have a singularity in $z=1$.


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