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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.

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  • $\begingroup$ It would help to have a precise statement for "$p(x)$ oscillates infinitely often as $x\rightarrow 1^{-}$" but it sound like one should be able to settle this using the central limit theorem instead of a law of large numbers. $\endgroup$ – Abdelmalek Abdesselam Apr 10 '17 at 16:24
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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.

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Edelman, Alan; Kostlan, Eric, How many zeros of a random polynomial are real?, Bull. Am. Math. Soc., New Ser. 32, No.1, 1-37 (1995). ZBL0820.34038.

Yes, this follows immediately from the Kac-Rice formula, see Edelman-Kostlan

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