The probability that a polynomial has complex roots Find the probability that $x^2 - 2ax + b$ has complex roots if the coefficients $a$ and $b$ are independent random variables with the common density


*

*uniform, that is $1/h$, and

*exponential, that is $\alpha e^{-\alpha x}$


This comes down to finding $P(a^2 \lt b)$. But since $a$ and $b$ are both random variables, would it be $P(a^2\lt b) = P(x\lt k)P(y \lt k^2)$? That doesn't seem particularly correct.
 A: I have seen several similar questions here. The idea is to use the joint density function of $a$ and $b$, which is (assuming independence),
(1) $f(x,y)=\frac1{h^2}$ in the square $[0,h]^2$ and 0 otherwise;
(2) $f(x,y)=\lambda^2e^{-\lambda(x+y)}$ in the first quadrant and 0 otherwise (I replace the parameter $a$ by $\lambda$ because it's confusing with a r.v. $a$.)
In both cases:
$$P(a^2<b)=\iint_{\{(x,y):x^2<y\}} f(x,y)dxdy.$$
I only solve (1) when $h\le 1$. The case when $h>1$ should be similar.
$$P(a^2<b)=\int_0^h\int_{x^2}^h dydx=h^2-\frac{h^3}3,$$
which is the area of the region within the square $[0,h]^2$ and above the parabola $y=x^2$.
Case (2) can be solved similarly with a different integral, and I'll leave it to you.
A: I'm guessing that you need to add "uncorrelated" to the description of the random variables a and b. In that case, you will have to integrate over the possible values. For the uniform case this is easy, the probability is proportional to the area. For the other, I'm going to let you think about it and see what you come up with.
A: The polynomial  has complex roots if and only if (2a)^2 - 4b <0.  That means $a^2 -b <0.$  Now all you have to calculate $P(A^2 < B)$ with both of those distributions.
That I will leave to you (or another poster) to do.
