- #1

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## Homework Statement

Hi all, I have some difficulty understanding the following problem, help is greatly appreciated!

Let ##U_1, U_2, U_3## be independent random variables uniform on ##[0,1]##. Find the probability that the roots of the quadratic ##U_1 x^2 + U_2 x + U3## are real.

## Homework Equations

## The Attempt at a Solution

From the determinant, the following must hold for real solutions

$$U_2 ^2 \geq 4U_1 U_3$$

And the corresponding probability to compute is

$$P(U_1 \leq \frac{U_2 ^2}{4U_3} )$$

I fixed ##u_2## which gives me the following function,

$$P(U_1 \leq \frac{u_2 ^2}{4U_3} | U_2 = u_2)$$

which I solved for by integrating over the domain specified by inequality, the result was

$$u_2 ^2 /4 - \frac{u_2 ^2}{2} \ln (u_2 / 2)$$

Now I want the 2nd equation from the top, which gives the total probability of having ##U_1 \leq \frac{U_2 ^2}{4U_3} ##. My intuition was to integrate the result from ##u_2 \in [0,1]##, which according to my solutions manual turns out to be true. But doesn't this imply that

$$P(U_1 \leq \frac{U_2 ^2}{4U_3} ) = \int P(U_1 \leq \frac{u_2 ^2}{4U_3} | U_2 = u_2) \ du_2$$

which is something I've not really seen before. But it leads to the answer, so my question is, why would this step be correct?

I have tried looking around the net for something relating a conditional CDF to a joint CDF, and I found this,

$$F(x|y) f_y (y) = \frac{dF(x,y)}{dy}$$

could it be that

$$F(x,y) = \int_{-\infty}^{y} F(x|u) f_y (u) \ du $$

$$P(X\leq x) = \lim_{y \to \infty} F(x,y) = \int_{-\infty}^{\infty} F(x|u) f_y (u) \ du $$

in my case ##f_y## would simply be 1, and so would the upper limit of my integral. Is this sound?

Many thanks in advance for any assistance!