Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
1  
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
1  
@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
add comment

1 Answer

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 $R\ge 1$.

share|improve this answer
    
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.