Let $p(x) = \sum\_{k \geq 0} a\_k x^k$ where the $a\_k$'s are IID random variables taken from a mean-zero random variable taking finitely many values in $\mathbb{R}$; it clearly converges for $-1<x<1$. Is it a.s. true that the sign of $p(x)$ oscillates infinitely often as $x \rightarrow 1^-$? That is, is it the case (with probability 1) that there exist $x\_1 < x\_2 < x\_3 < \dots$ in (0,1) such that $p(x\_k)$ has the same sign as $(-1)^k$?


I imagine that this is well-known for $P(1) = P(-1) = 1/2$ (where $P(r)$ denotes the probability that $a\_k = r$); the case that most interests me is $P(1) = P(-1) = 1/4$ and $P(0) = 1/2$, but I'm sure that the same technique settles both cases.


I am also interested in knowing about the magnitude of the oscillations.



Sign-oscillations for power series with random coefficients
We can construct inductively a sequence $t\_n \to 1-$ and an increasing sequence $K\_n$ of positive integers
such that with probability $> 1 - 1/n^2$, $f(t\_n)$ has the same sign as
$V\_n = \sum\_{k=K\_{n-1}+1}^{K\_n} a\_k t\_n^k$. Since $\sum 1/n^2 < \infty$,
almost surely $f(t\_n)$ has the same sign as $V\_n$ for all but finitely many $n$. Almost surely the independent random variables $V\_n$ change sign infinitely often, so a.s. $f(t\_n)$ changes sign infinitely often.


