Given a sequence of independent and identically distributed random variables $(X_n)_{n \in \mathbb{N}}$ with $\mathbb{E}(|X_1|) < \infty$ and the power series $$\sum_{n=1}^{\infty} X_n z^n$$
In previous exercises I have shown that the radius of convergence is almost surely constant (using Kolmogorov 0-1 law). But now I am struggeling to determine the radius of convergence, because there is no specific distribution given.
Do you have any hints for me?
Thanks!
Best Answer
If $|z|\lt 1$, then $$\mathbb E\left[\sum_{n=1}^N \left|X_n\right| \left|z\right|^n \right] = \mathbb E\left[ \left|X_1\right|\right]\sum_{n=1}^N \left|z\right|^n\leqslant \mathbb E\left[ \left|X_1\right|\right]\sum_{n=1}^\infty \left|z\right|^n$$ and by monotone convergence, the random variable $\sum_{n=1}^\infty \left|X_n\right| \left|z\right|^n$ has a finite expectation, hence is finite almost everywhere. Consequently, the series $\sum_{n=1}^{\infty} X_n z^n$ converges almost everywhere. The radius of convergence is thus bigger or equal to $1$.
Now, let $r\gt 1$. The series $\sum_{n=1}^{\infty} \left|X_n\right| r^n$ cannot be convergent (unless $X_n=0$ a.s.): by the $0$-$1$ law, it should be almost everywhere convergent, hence $\left|X_n\right| r^n\to 0$ almost surely, hence in probability hence $\mathbb P\left\{\left|X_1\right|\gt r^{-n}\right\}\to 0$ which entails $\mathbb P\left\{\left|X_1\right|\gt 0\right\}=0$.
In conclusion: if $X_1=0$ a.s., the radius of convergence is almost surely infinite, otherwise, it is $1$.