4
$\begingroup$

Let $a_1,a_0$ be i.i.d. real random variables with uniform distribution in $[-1,1]$. I'm interested in the random zeros of the polynomial $$p(x) = x^2 + a_1x + a_0. $$

One thing (between many) thing I'm particularly interested is in computing the expected value of it's zeros. It's possible to write exactly what are the zeros $z_1,z_2$ of $p(x)$, they are $$ z_1 = \frac{-a_1+\sqrt{a_1^2-4a_0}}{2} $$ and $$ z_2 = \frac{-a_1-\sqrt{a_1^2-4a_0}}{2}. $$

Therefore what I want is to calculate $$E[z_1] = E\Bigg[\frac{-a_1+\sqrt{a_1^2-4a_0}}{2}\Bigg] = \frac{-E[a_1] + E\Big[{\sqrt{a_1^2-4a_0}}\Big]}{2} = \frac{E\Big[{\sqrt{a_1^2-4a_0}}\Big]}{2}.$$

Note that I only need to calculate $E[z_1]$, for $E[z_2] = \overline{E[z_1]}$. Also, doing some experimentation with Matlab, it looks like $E[z_1] = i$.

I want to confirm that result with a mathematical proof, if possible. Thank you.

Edit: From my own computation, we can see that the correct relation between the expected values should be $E[z_2] = -E[z_1]$, not $E[z_2] = \overline{E[z_1]}$.

$\endgroup$
4
  • $\begingroup$ I don't think it's possible for your result to be correct, since $\operatorname{Re}(E\Big[{\sqrt{a_1^2-4a_0}}\Big])=E\Big[\operatorname{Re}(\sqrt{a_1^2-4a_0})\Big]>0$. $\endgroup$
    – f''
    Jul 12, 2016 at 4:00
  • $\begingroup$ How do you get that inequality? It doesn't look obvious at all. $\endgroup$
    – Integral
    Jul 12, 2016 at 4:05
  • 1
    $\begingroup$ The real part of the expectation is the same as the expectation of the real part. But the real part of the square root is sometimes positive and never negative, so its expectation must be positive. $\endgroup$
    – f''
    Jul 12, 2016 at 4:10
  • $\begingroup$ Depending on your purpose, you might get more insight separating the case or real roots from the case of complex roots. The real root (positive square root) case is a density on the segment $[-1/2, (1/2)(1+\sqrt{5})]$. The complex root (positive square root) case is a density on the part of the unit disk with positive imaginary part and real part in $[-1/2,1/2]$. Looking at the negative square roots gives the negative interval and the conjugate region, respectively. $\endgroup$ Jul 12, 2016 at 6:22

2 Answers 2

6
$\begingroup$

$$E[z_1]=\frac{E\Big[{\sqrt{a_1^2-4a_0}}\Big]}2\\=\frac18\left(\int_{-1}^1\int_{-1}^1\sqrt{x^2-4y}dydx\right)\\=\frac1{48}\left(\int_{-1}^1(x^2+4)^{\frac32}-(x^2-4)^{\frac32}dx\right)\\=\frac1{24}\left(\int_0^1(4+x^2)^\frac32dx+i\int_0^1(4-x^2)^\frac32dx\right)\\=\frac{1}{24}\left(\left(\frac{11\sqrt5}4+6\sinh^{-1}\frac12\right)+i\left(\frac{9\sqrt3}4+\pi\right)\right)\\\approx0.3765+0.2933i$$

$\endgroup$
2
$\begingroup$

Let the roots be the random variables $U$ and $V$. Then by symmetry $E(U)=E(V)$. But $U+V=-a_1$, so $U+V$ has mean $0$. Thus $E(U)=E(V)=0$.

$\endgroup$
8
  • $\begingroup$ I'm sorry, but what symmetry are you talking about? $\endgroup$
    – Integral
    Jul 12, 2016 at 3:39
  • $\begingroup$ There is no symmetry, because one root is defined with the positive square root and the other is defined with the negative square root. In particular, $\operatorname{Im}(z_1)\ge0$. $\endgroup$
    – f''
    Jul 12, 2016 at 3:40
  • $\begingroup$ OK, let's use the quadratic formula. By the linearity of expectation, $E((-a_1+\sqrt{D})/2 +(-a_1+\sqrt{D})/2))=E(-a_1)$. $\endgroup$ Jul 12, 2016 at 3:49
  • $\begingroup$ I agree that $U+V$ has mean $0$ but can't see how $E(U) = E(V)$. $\endgroup$
    – Integral
    Jul 12, 2016 at 3:53
  • 2
    $\begingroup$ @AndréNicolas But you don't randomly choose one of them. They are defined so that $\operatorname{Re}(z_1)\ge\operatorname{Re}(z_2)$ and $\operatorname{Im}(z_1)\ge\operatorname{Im}(z_2)$ $\endgroup$
    – f''
    Jul 12, 2016 at 3:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .