Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A, B, and C are random variables uniformly distributed over (0,1).

Question

Suppose that $A, B,$ and $C$ are independent random variables, each being uniformly distributed over $(0,1)$. What is the probability that $Ax^2 + Bx + C$ has real roots?

First, I set $P(B^2 - 4AC \ge 0)$

Then I am told that $$\begin{align} \int_0^1 \int_0^1 \int_{\min\{1, \sqrt{4ac}\}}^1 1 \;\text{d}b\,\text{d}c\,\text{d} &a= \int_0^1 \int_0^{\min(1, \frac{1}{4a})}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a\\ &= \int_0^{\frac{1}{4}} \int_0^1 \int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a + \int_{\frac{1}{4}}^1 \int_0^{\frac{1}{4a}}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a \end{align}$$

why the middle integrate from $0$ to $\min(1, \frac{1}{4a})$ from the second integral...where does $\frac{1}{4a}$ come from? why the min{...} does not go to the front integral? why they break up into last step like this (I refer to one integral + another integral) ?

Thanks a lot

Answer 1

To resolve $\min\{1,\sqrt{4ac}\}$, we need to figure out which of the two arguments is smaller. If it's $1$, the integral is $0$ and thus doesn't contribute. For it to be $\sqrt{4ac}$, we need to have $\sqrt{4ac}\le1$, and thus $c\le1/(4a)$. This is the answer to your question where $1/(4a)$ comes from.

I'm not sure how to answer your question why the $\min$ doesn't go to the front integral. My counter-question would be how you'd propose to move it to the front integral such that the result is equivalent. Unless you can come up with such a proposal, I'd suggest to concentrate on why the form written here is equivalent to the one in the previous step. Generally speaking, it makes sense that resolving a minimum in a limit of the innermost integral affects the limits of the immediately enclosing integral and not of a more remote enclosing integral.

The breaking up into two integrals is again the direct result of resolving the minimum in the upper limit of the second integral. That minimum is $1$ if $1\le1/(4a)$, i.e. if $a\le\frac14$, and is $1/(4a)$ otherwise; thus we have to split the outer integral over $a$ into two parts according as $a\lessgtr\frac14$.