I am having some problems to find the solution required in the next exercise:
$x^2-ax+b=0$ has real and not equal solutions and that b is a positive constant that such that $b \sim U(i,j)$ for certain i,j.
Find the $\mathbb{E}[X], \,Var[X]$ of the solutions.
What I have so far:
From the Quadratic Formula, it is known that:
And since these are real solutions,
$\implies a^2>4b$
$\implies \frac{a^2}{4}>b$
¿how do I use $b \sim U(i,j)$?
Well, simply by using the fact that since $b \sim U(0,a^2/4)$ then $\mathbb{E}(b) =4/a^2\int_{0}^{a^2/2}b\,db =\frac{a^2/4}{2} = a^2/8$.
Then, for $x_1$ for example, it is :
$$\mathbb{E}(x_1) = \mathbb{E}\left(\frac{a + \sqrt{a^2-4b}}{2}\right) = \frac{1}{2}\mathbb{E}\left(a+\sqrt{a^2-4b}\right)$$
But, $a$ is a constant (not following any distributions, as far as one can say per the problem hypothesis), thus $\mathbb{E}(a) = a$ and so :
$$\mathbb{E}(x_1) = \frac{1}{2}a + \frac{1}{2}\mathbb{E}\left(\sqrt{a^2-4b}\right) $$
Can you finish the calculation now ?
For the calculation of variance, again use the fact that since $b \sim U(0,a^2/4)$ then :
$$V(b) = \frac{1}{12}\left(\frac{a^2}{4}-0\right)^2= \frac{a^4}{12\cdot 16}$$
Be careful now, since if $a$ is a constant, then $V(a) = a^2$.