# Radius of convergence of a series of random variables

Let $X_n$ be i.i.d. and (a.s.) bounded random variables.(none of them identically zero) Prove that the radius of convergence of the series with coefficients $X_n$, $f(\omega,t)=\sum_{n=0}^{+\infty}X_n(\omega)t^n$ is exactly $1$. (Hint:use Borell-Cantelly lemma.)

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Hint: Which characterization of the radius of convergence of a (deterministic) series do you know? – Did Sep 23 '12 at 7:07
If $X_n=0$ for each $n$, the radius is infinite. So I guess you mean the radius of convergence is at least $1$. – Davide Giraudo Sep 23 '12 at 8:50
@Davide Giraudo: $X_n$ have to be i.i.d., which cannot be the case if they are all zero. – Ahriman Sep 23 '12 at 9:56
@Ahriman Every almost surely constant random variable is independent of everything else hence, yes, if the random variables are almost surely constant then they are independent. – Did Sep 23 '12 at 10:52
@Mathfollower For i.i.d. bounded random variables, the only case when $R\ne1$ is when the random variables are almost surely zero. – Did Sep 23 '12 at 10:54

Let $0\le t<1$, then $\forall \omega \in \Omega$ $$\sum_{i=0}^n |X_i(\omega)| t^n \le M \sum_{i=0}^n t^n \le \frac{M}{1-t}$$ Thus the series is absolutely convergent (increasing and bounded) with a radius of convergence $R\ge 1$.
 @Mathfollower : I assumed "bounded" meant any random variable $X_n$ is bounded by a unique $M$, which is a strong hypothesis. Please detail yours in your question if it differs. – vanna Sep 23 '12 at 10:15