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

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.

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

Quote:

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) ?

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

Quote:

Originally Posted by

**askhwhelp** 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 see you asked this question fresh in a new thread. Bad form. Please keep 1 thread per question.

To answer your question.

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?