What is the probability that the roots of $P(x) = \frac{1}{4} x^2 +Ux+V^2$ are real?

Question

Assume that $U$ and $V$ are independent normally distributed random variables, each with mean $0$ and variance $1$.

Find the probability that the roots of the polynomial $P(x) = \frac{1}{4} x^2 +Ux+V^2$ are real.

This is a question I had to work out on my intro probability midterm. It's clear to see by looking at the discriminant that $$\begin{align} \Bbb P(\text{roots of }P(x)\text{ are real}) &=\Bbb P(U^2-V^2\ge0) \\ &=\Bbb P(U^2\ge V^2) \\ &=\Bbb P(U\ge V\text{ or }-U\ge -V), \end{align}$$ but what next?

I appreciate any help.

Thanks in advance!

So, if you know the distribution of $U$ and $V$, can you write down an expression for the probability that $U\ge V$? — Gerry Myerson, Jul 14 '13 at 07:19
No need to do any calculation. Just uses the fact your probability distribution is symmetric under $U \leftrightarrow V$. — achille hui, Jul 14 '13 at 07:22
@achillehui note that you also need to know the probability that $U=V$ — Mark Bennet, Jul 14 '13 at 07:31
Your condition is not quite expressed correctly - easier to write $|U|\ge|V|$. — Mark Bennet, Jul 14 '13 at 07:35

2'5 9'2 · Answer 1 · 2013-07-15T07:14:14.563

0

Since the variables are independent and Normally distributed, the codensity function of $u$ and $v$ is very symmetric about $(0,0)$ in the $(u,v)$-plane. You are looking for $P(|U|\geq |V|)$. Sketching the region "$|U|\geq |V|$" in the $(u,v)$-plane, we have two sectors that are each one-quarter of the plane. All the symmetry granted in the setup allows us to conclude that the probability is $\frac12$.

edited Jul 15 '13 at 07:14

answered Jul 14 '13 at 07:48

2'5 9'2

54,717

Why not following @achille.hui's suggestion instead of delving into irrelevant specifics of the distribution? The couples (U,V) and (V,U) are identically distributed and [U=V] has probability zero, QED. – Did Jul 14 '13 at 08:31
@Did Is it not more relevant that $[|U|=|V|]$ has measure $0$? OP does not want a region containing, say $u=1,v=-2$, even though $U>V$ there. Also I intended to appeal to symmetry; I though I was avoiding details, only giving enough details to justify the symmetry. Thanks for the second glance though- I had some incorrect inequalities. – 2'5 9'2 Jul 15 '13 at 07:19
The new version is better although, oddly enough, neither the argument guaranteeing a symmetry nor the exact symmetry involved are mentioned. Since one could write down a full solution without even using the word "normal", I am not sure (once again) that to rely on the specifics of the standard gaussian distribution yields the best possible answer. – Did Jul 15 '13 at 07:58

What is the probability that the roots of $P(x) = \frac{1}{4} x^2 +Ux+V^2$ are real?

1 Answers1