# Random variable power series

Let $X_1,X_2,...$ be independent but not necessarily identically distributed random variables. Consider the power series $$\sum_{n=0}^{\infty} X_n z^n$$. Show that for each $z$, the series either converges almost surely or diverges almost surely.

When the random variables are i.i.d. then the result follows from zero-one laws. I think in this case also it will come from zero-one laws.

Let's denote the event that series $\sum_{n=0}^{\infty} X_n z^n$ converges as $E_z$.

Then for every $z$ event $E_z$ is a so-called tail event and all tail events have a probability in $\{0,1\}$.

Let $k$ denote some positive integer and observe that series $\sum_{n=0}^{\infty}X_nz^n$ converges if and only if series $\sum_{n=k}^{\infty}X_nz^n$ converges.

This for every positive integer $k$ and that is enough to conclude that the event of convergence is an element of $\sigma$-algebra:$$\bigcap_{n=0}^{\infty}\sigma(\{X_n,X_{n+1},\dots\})$$

For every event $E$ belonging to that $\sigma$-algebra it can be shown that $\mathsf P(A\cap E)=\mathsf P(A)\mathsf P(E)$.

Then $A=E$ leads to $\mathsf P(E)=\mathsf P(E)^2$ and consequently $\mathsf P(E)\in\{0,1\}$.

So here we have $\mathsf P(E_z)\in\{0,1\}$ for every $z$.

Which of the values - $\mathsf P(E_z)=0$ or $\mathsf P(E_z)=1$ - also depends on $z$.

• Isn't convergence of series to a tail event depends on z value? It will be true for $\sum_{n=k}^{\infty} X_n$ but here we have a power series. – Nikki Oct 31 '17 at 19:40
• No. Essential is the fact that convergence completely depends upon the values taken by the $X_i$ with $i\geq k$, and this for every positive integer $k$. That fact determines that it is a tail event. Power series (or not) is not relevant. – drhab Oct 31 '17 at 20:52
• What I mean to say is that - If $E_z$ denotes the event that the power series converges - then it is a tail event for every $z$. So its probability will be $0$ or $1$. Which of those values does depend also on $z$ though. – drhab Nov 1 '17 at 7:29