Consider a polynomial $p(z)= \sum_0^n a_i z^i$. In the literature there are numerous bounds about the roots of $p(z)$. Now once we prescribe a certain distribution to the coefficients, the bound itself becomes a random variable. Generally, it might not be always possible to find the distribution of the bound. Suppose in certain cases we are able to find the distribution of the bound, say it is $F(y)$. Then can we assert the following :
For a positive number $c$, all the roots are less or equal to $c$ with a probability that is at least equal to $F(c)$?
I would be highly obliged for any help/clarification in this regard.
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4$\begingroup$ isn't your statement the definition of the cumulative distribution $F$? $\endgroup$– Carlo BeenakkerCommented Jan 25, 2023 at 15:07
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Not sure if this directly answers your question, but it might help in further analysis.
Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$
| variable | $P(z^{0})$ | $P(z^{1})$ | $P(z^{2})$ | $P(z^{3})$ |
|---|---|---|---|---|
| $x$ | $\frac{1}{4}$ | $\frac{3}{8}$ | $\frac{1}{8}$ | $\frac{1}{4}$ |
Which expands to the polynomial: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$
If we were to introduce multiple variables $x_{j}$ then the resulting distribution is the product of those polynomials
| variable | $P(z^{0})$ | $P(z^{1})$ | $P(z^{2})$ | $P(z^{3})$ |
|---|---|---|---|---|
| $x_{1}$ | $\frac{1}{4}$ | $\frac{3}{8}$ | $\frac{1}{8}$ | $\frac{1}{4}$ |
| $x_{2}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\frac{1}{9}$ | $\frac{16}{45}$ |
$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$
