# Thread: Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A, B,

1. ## Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A, B,

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, 1/4a\}}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a\\ &= \int_0^{1/4} \int_0^1 \int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a + \int_{1/4}^1 \int_0^{1/4a}\int_{\sqrt{4ac}}^1 1\;\text{d}b\,\text{d}c\,\text{d}a \end{align}

Could anyone first let me know whether this set up is right? If so, could you explain how each piece fit together. How to understand it?

Thanks a lot

2. ## Re: Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A,

It is set up correctly. As to why... well that's just too much typing sorry.

The first integral should be pretty clear. $4 A C >0$ and $B^2 \geq 4 A C \Rightarrow B \geq \sqrt{4 A C}$

but also $B \leq 1$ and so you end up with the Min function.

The rest of it is just carefully chasing those Min functions back through the integration chain.

3. ## Re: Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A,

Originally Posted by romsek

The rest of it is just carefully chasing those Min functions back through the integration chain.
Can you elaborate on this? This is what I get confused, especially why the middle integrate from 0 to min{1, 1/4a} from the second integral...where does 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) ?

4. ## Re: Understanding the setup for the probability that $Ax^2+Bx+C$ has real roots if A,

Can you elaborate on this? This is what I get confused, especially why the middle integrate from 0 to min{1, 1/4a} from the second integral...where does 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) ?

I assume you understand why we need $B\geq\sqrt{4AC}$ and that you understand that $B\leq 1$

So it should be clear why you integrate $B$ over $[\min(1,\sqrt{4AC}), 1]$

Now what does that min function require of $C$ ?

If $\sqrt{4AC}>1$ then $\min(1,\sqrt{4AC})=1$ and the last integral is 0. Thus $\sqrt{4AC} < 1 \Rightarrow C<\frac{1}{4A}$
And as before $C<1$ so our integration of $C$ is from $[0,\min(1,\frac{1}{4A})]$

Now what does this min function require of A?

If $\frac{1}{4A}>1 \Rightarrow A<\frac{1}{4}$ then $\min(1,\frac{1}{4A})=1$ otherwise $\min(1,\frac{1}{4A})=\frac{1}{4A}$

So for $0\leq A \leq \frac{1}{4}$ the upper limit of the integral over $C$ is $1$, for $\frac{1}{4}<A\leq 1$, the upper limit of the integral over $C$ is $\frac{1}{4A}$

Thus you split the integration of A over these two intevals with appropriate modification to the integration limits over C.

Is this clear?