0
$\begingroup$

I have the following polynom $x^2+Ux+V$ where U, V independent and uniformly distributed on $[0,1]$. Since I'm looking for a root, I'm using the solution formula:

$x_{1,2}=-\frac{U}{2}\pm\sqrt{\left(\frac{U}{2}\right)^2-V}$.

Furthermore I only look for real roots. Therefore I am looking for the following probability: $P(\tfrac{U^2}{4}>V)$.

I try to find the density of the random variable $X:=U^2/4$. I am using the following mapping: $g(x)=x^2/4$. The derivation is $g'(x)=x/2$ and the inversion formula is $f(x)=2\sqrt{x}$. Because of that I get as a density for $X$: $f_X(x)=\begin{cases}\frac{1}{\sqrt{a}} & \text{for }\frac{1}{4}\geq x>0\\ 0 & \text{else}\end{cases}$.

Now I can calculate $P(\tfrac{U^2}{4}>V)$:

$P(X>V)=\int_{0}^{\frac{1}{4}}\int_0^xf_V(y)dyf_X(x)dx=\int_{0}^{\frac{1}{4}}x\cdot\frac{1}{\sqrt{x}}dx=\int_{0}^{\frac{1}{4}}\sqrt{x}dx=\left[\frac{2}{3}\cdot x^{\frac{3}{2}}\right]_0^{\frac{1}{4}}$

I am very thankful for any help.

Sincerely, Hypertrooper

$\endgroup$

1 Answer 1

0
$\begingroup$

For fixed $c\in\Bbb R$, $$ P(U^2>c)=\begin{cases}1&\text{if }c\le 0\\ 0&\text{if }c\ge 1\\ 1-\sqrt c&\text{otherwise}\end{cases}$$ Now $$P(U^2>4V)=\int_0^1P(U^2>4v)\,\mathrm dv =\int_0^{\frac14}(1-\sqrt {4v})\,\mathrm dv.$$

$\endgroup$
1
  • $\begingroup$ Thank you! This is a more elegant approach to the problem. But at the end, the chain of equation of integrals was correct since both give the same value. It helped a lot to calm down a heated discussion. $\endgroup$
    – user239922
    Aug 20, 2017 at 15:35

You must log in to answer this question.