I'm given two uniform random variables $V,U \sim\mathbf{U}(-1,1)$. I also get the function: $$h(s)=s^2+Us+V$$ I'm interested to answer queations like for which values $h(s)$ has only: zero /one/ two roots and what is the probability for that.
So of course I have to use this: $$s_{1,2}=\frac{-U\pm \sqrt{U^2-4V}}{2}$$
One root: $U^2=4V$, now correct me if I'm wrong, but since there is no inequality sign this is not an accumulating event and therefore its probability is $0$? Now I have the feeling I'm terribly wrong but I have no way to prove myself wrong.
Two roots: $$U^2>4V \rightarrow P(U^2>4V)=1-P(U^2<4V) = 1-P(-2\sqrt{V}<U<2\sqrt{V})$$ And again I get to a phrase I'm not 100% sure is correct, let along I'm not sure how to calculate. Can anyone hint / guide how to proceed from here? what kind of method should I generally use in problems like this?
The conclusion is right. The intuition is also basically right, though the expression "accumulating event" is unknown to me. The correct term would be "a zero measure" event.
Basically: because $U,V$ are uniform and independent on the unit square, the probability of any event $E$ that depends on some (in)equality on $U,V$, is given by the area corresponding to that event on the $U,V$ plane. Because in this case the condition is $V=U^2/4$ , the region is a curve (a parabola) , and its area is zero.
In general (roughly) : inequalities like $a < g(U,V) < b$ will give regions corresponding to areas (hence positive probability... unless the region doesn't intersect the unit square) and equalities like $g(U,V)=0$ will give you a curve (hence zero probability... unless the equation is degenerate , like $0 U + 0 V = 0$).
The second part corresponds to the inequality case: to compute $P(U^2<4V)$ you need to integrate the joint density over that region, which in this case correponds to computing its area:
$$P(U^2<4V)=\int_{u^2<4v}f_{U,V}(u,v) du dv $$
To compute the integral you need to graph the region and compute its area by using the proper integrals.