Question:
Consider three random variables A, B, C, jointly distributed such that all vectors (A, B, C) $\in [0, 1] \times [0,2] \times [0, 3]$ are equally likely. Find the joint density of A, B, C. Find the marginal densities. Are A, B, C independent? Consider the quadratic equation $Ax^2 + Bx + C = 0$. What is the probability that the roots of the equation are real?
My Attempt at Solution:
Since the vector (A, B, C) is uniformly distributed in the box $[0, 1] \times [0,2] \times [0, 3]$, the joint density is a uniform distribution.
$f_{A, B, C} = \frac{1}{6}$ for (A, B, C) $\in [0, 1] \times [0,2] \times [0, 3]$ and $f_{A, B, C} = 0$ otherwise.
Then to get the marginal densities, I simply indtegrated this joint distribution with respoect to the other variables, giving:
$$f_A = 1, f_B = \frac{1}{2}, f_C = \frac{1}{3}$$ over their respective intervals and $0$ otherwise.
My trouble is really with the last part of the question. For the roots of the quadratic to be real, we are looking for the probability:
$$P\left(B^2 - 4AC \geq 0 \right) $$
I was wondering how to compute this. My first thought was that I have to consider the product of two uniformly distributed random variables, or let $Z=AC$. Then find the density for $Z$, or $f_Z(z)$. Then, Find the density for $4Z$. Then find the density for $B^2$. Then, finally we can get the density for the whole expression $B^2 - 4AC$, which we can use to evaluate the probability. I am very confused as to how to get this result. Any suggestions?
Since the joint density is of the form $f_A(a)f_B(b)f_C(c)$ it follows that $A,B,C$ are independent. Hence $P(B^{2} \geq 4AC)=P(A \leq \frac {B^{2}} {4C})=\mathbb E P(\frac {B^{2}} {4C}I_{B^{2} \leq 4C})$. We can write this as $\frac 1 3 \int_0^{3} \mathbb E(\frac {B^{2}} {4t}I_{B^{2} \leq 4t})dt$. But $\int_0^{3} \frac {B^{2}} {4t}I_{B^{2} \leq 4t})dt=\frac 1 4 (\ln t)|_{B^{2}/4} ^{3}$. Now use the density of $B$ to finish.