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Let $z_{1},z_{2},\ldots,z_{n}$ be i.i.d random points in the unit circle ($|z_i|=1$) with uniform distribution of their angles. Consider the random polynomial $P(z)$ given by $$ P(z)=\prod_{i=1}^{n}(z-z_i). $$

Let $m$ be the maximum absolute value of $P(z)$ on the unit circle $m=\max\{|P(z)|:|z|=1\}$.

How can I estimate $m$? More specifically, I would like to prove that there exist $\alpha>0$ such that the following holds almost surely as $n\to\infty$ $$ m\geq e^{\alpha\sqrt{n}}. $$

Any idea of what can be useful here?

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    You want a lower bound on the maximum?2011-04-27
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    Take all $z_i=0$. Then $P(z)=z^n$ with the maximum $m=1$. So your bound cannot be true...2011-04-27
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    Did you consider any special cases such as (1) $z_1=z_2=\ldots =z_n = -1 + \varepsilon$, $\varepsilon>0$ or (2) $z_k=e^{i\dot 2\pi k/n}$, $k=1,\ldots,n$?2011-04-27
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    @Fabian, $z_i$ need to be on the circle, $0$ is not on the unit circle.2011-04-27
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    The points are on the circle2011-04-27
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    @Fabian you are right let me restate the question to be more specific.2011-04-27
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    @Thomas: Why they are on the circle? (I read in the circle)2011-04-27
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    @gatu: why do you think the inequality should hold?2011-04-27
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    @Fabian: I did some simulations that suggest that the inequality should hold.2011-04-27
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    @gatu: Nice problem! I think I can get exponent $\frac{1}{2} +\epsilon$ for any positive $\epsilon$, where you are looking for $\frac{1}{2}$.2011-04-28
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    @User6312: That's good! What is the idea behind?2011-04-28
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    This reminds me of the *Kahane polynomials* - these are polynomials of degree $n$ having coefficients on the unit circle and a sup norm like $\sqrt{n} + \varepsilon$ (Note the $\ell^2$- norm of such polynomial is $\sqrt{n}$). The difference in this case is, of course, that we are looking at the zeros here...2011-04-28
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    @gatu: There was a serious mistake in my argument. Perhaps it can be fixed, but I am not optimistic.2011-04-28
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    To me, this problem doesn't really seem probabilistic. Isn't it the same thing to simply say "for any choice of $n$ points on the unit circle, $\max_{|z|=1} |\prod(z-z_i)| \ge e^{\alpha\sqrt{n}}$"?2011-04-28
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    @Hans: no, "almost surely as $n\to\infty$" is not the same "always for finite $n$". @ght: There seems to be a recurring misunderstanding of your phrase "in the unit circle" as "on the unit circle" -- perhaps it would be preferable to write "in the unit disk"?2011-04-28
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    @joriki: Hm, I didn't notice the phrase "as $n\to\infty$". Not that it can immediately tell what difference it makes, but perhaps if I brush up my probability skills a little... :-) And yes, "in the unit disk" or "inside the unit circle" is much clearer than "in the unit circle" (I, for example, thought it meant $|z|=1$).2011-04-28
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    @Hans: As an example, if you throw $n$ coins, you will almost surely throw tails at least once as $n\to\infty$, but it's not true for any finite $n$ that you will always throw tails at least once.2011-04-28
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    @Hans and @joriki: I want the points $z_{1},z_{2},\ldots,z_{n}$ to be in the unit circle, meaning that $|z_{i}|=1$ for all $i$. Let me clarify this further in the question.2011-04-28
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    @ght: Then "on the unit circle" would probably be the clearest.2011-04-28
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    @joriki: In the case at hand, the estimate is false if you look at any fixed $n$, since $P(z)=z^n+1$ has its roots on the unit circle and has maximum modulus equal to 2 (independently of $n$). So if I understand correctly, one would need to show (roughly speaking) that the probability of coming close to such a symmetric arrangement of the roots (or to any arrangement which violates the desired estimate) tends to zero as $n \to \infty$?2011-04-29
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    @Hans: Exactly.2011-04-29

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