I came across this question. If $\{X_n , n\geq 1\}$ are independent random variables, show that the radius of convergence of the power series $\sum_{n=1}^\infty X_n z^n$ is a constant with probability one, where $R^{-1} = \limsup |X_n|^{1/n}$.

PS: I guess this can be solved using second Borel-Cantelli lemma.


2 Answers 2


Consider $\sum_n X_n z^n$, where the $X_n=X_n(\omega)$ are independent random (complex) variables and $z$ is complex. (The real case is treated much the same way, but I want to discuss, however briefly, some topics that are inherently about complex series.)

First of all, the radius of convergence of the series (at a given $\omega$ in the underlying measure space) is $$ r(\omega)=(\limsup_{n\to\infty}|X_n(\omega)|^{1/n})^{-1}. $$ Note that $r$ is a measurable function of $\omega$, and its value does not depend on the values of a finite number of the $X_n$. Kolmogorov's zero-one law then gives us that $r(\omega)$ is a constant, say $R$, almost surely. This $R$ lies in $[0,\infty]$, and both $0$ and $+\infty$ are possible values, depending on the distribution of the $X_n$, though the most interesting cases to study are perhaps when $0<R<+\infty$.

Let me add some remarks. There is a nice book that presents the relevant theory:

Jean-Pierre Kahane. Some random series of functions. Second edition. Cambridge Studies in Advanced Mathematics, 5. Cambridge University Press, Cambridge, 1985. MR0833073 (87m:60119).

(The topic you want is "Random Taylor series", discussed in Chapter 4. What follows is based on Kahane's presentation. Kahane's book includes proofs of all the results below.)

This and similar questions were first considered by Borel, in

Emile Borel. Sur les séries de Taylor. C. r. hebd. Séanc. Acad. Sci., Paris 123, 1896, 1051-2.

(The journal is available here.)

Part of the problem was that at the time the concepts of probability theory were not quite formalized yet, so going from Borel's remarks to actual theorems took some time. Borel wrote

Si les coefficients sont quelconques, le cercle de convergence est une coupure.

What Borel is saying is that if the coefficients of a series $\sum_n X_n z^n$ are "arbitrary", then the circle of convergence is a natural boundary for the function. What this means is that there is no way to extend $F(z)=\sum_n X_n z^n$ analytically beyond the circle of convergence (because the singular points are dense on the boundary).

The first actual result in this regard is due to Steinhaus in 1929: If the $r_n$ are positive, and $0<\limsup_n r_n^{1/n}<\infty$, and the $\omega_n$ are independent random variables equidistributed on $[0,1]$, then $\sum_n r_ne^{2\pi i\omega_n}z^n$ has the circle of convergence as natural boundary, almost surely. A different formalization was found later by Paley and Zygmund, in 1932, in terms of Rademacher sequences.

On the other hand, Borel's statement cannot quite be translated as "the coefficients are independent random variables". Kahane's example is the series $$ \sum_n (2^n\pm1)z^n, $$ which has radius of convergence $1/2$, and $1/2$ is the only singular point on the circle of convergence.

Kahane mentions a conjecture of Blackwell that the general situation should be that one of the two scenarios above applies: Either

  1. $F(z)=\sum_n X_n z^n$ has the circle of convergence as natural boundary, or
  2. There is a series $\sum_n c_n z^n$ (the $c_n$ being constants, not random variables; Kahane calls it a sure series) that added to $F$ results on a (random) Taylor series with a strictly larger circle of convergence which is its natural boundary.

The conjecture was proved in 1953 by Ryll-Nardzewski.

  • 1
    $\begingroup$ Thanks a lot for your thorough and thoughtful answer $\endgroup$
    – ie86
    Commented Mar 2, 2013 at 1:41

Hints: For each $n$ let $\mathfrak F^n$ denote the sigma-algebra generated by $(X_k)_{k\geqslant n}$. Let $\mathfrak F^\infty$ denote the intersection of every $\mathfrak F^n$. Let $Y=\limsup|X_n|^{1/n}$. Then:

  • The random variable $Y$ is $\mathfrak F^\infty$-measurable.
  • The sigma-algebra $\mathfrak F^\infty$ is trivial in the sense that for every $A$ in $\mathfrak F^\infty$, $\mathbb P(A)$ is $0$ or $1$.

Is this enough to make you find a solution?

  • $\begingroup$ Are you using the Kolmogorov's zero-one law? How to say that the probability is 1 and not 0? $\endgroup$
    – ie86
    Commented Mar 1, 2013 at 17:20
  • $\begingroup$ Are you using the Kolmogorov's zero-one law? Yes. How to say that the probability is 1 and not 0? Sorry? Some sets in this sigma-algebra have probability 0, others have probability 1, hence I do not understand what you are asking. $\endgroup$
    – Did
    Commented Mar 1, 2013 at 17:22

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