assume that X and Y are independent with X ~ UNIF(-1,1) and Y~UNIF(0,1).

Question

I am trying to find the probability that the roots of the equation h(t)=0 are real, where h(t)=t^2+2Xt+Y of the given data.

I know that I need to look at Uniform continuous distributions but I am not sure where to start

Answer 1

We want the discriminant to be $\ge 0$, so we want $4X^2-4Y\ge 0$, that is, $Y\le X^2$.

Draw the parabola $y=x^2$. The joint density function is $\frac{1}{2}$ on the rectangle $-1\le x\le 1$, $0\le y\le 1$. Let $A$ be the area of the part of the rectangle that is below the parabola. Then our required probability is $\frac{A}{2}$.

We can save ourselves the trouble of dividing by $2$ by finding the area of the region below $y=x^2$, above the $x$-axis, from $x=0$ to $x=1$.

Answer 2

First, you need to find where on the rectangle, $(X,Y)\in[-1,1]\times[0,1]$, the quadratic equation $t^2+2Xt+Y=0$ has real roots, and which values of $X,Y$ make the roots complex.

Hint: Examine the square-root term in the quadratic root formula. When is this not imaginary?

Next you need to measure the probability of this, either algebraically or graphically. The graphical method involves simply measuring the portion of the rectangle where the roots are real.

$$\newcommand{\hidden}[1]{}\hidden{ \begin{align} \mathbb P\bigg(\frac{-2X\pm\sqrt{4X^2-4Y}}{2} \in\mathbb R\bigg) & = \mathbb P(Y\leq X^2) \\ & = \int_{-1}^1 \int_0^{x^2} \frac 1 2\operatorname d y\operatorname d x \\ & =\frac 1 2 \int_{-1}^1 x^2 \operatorname d x \\ & = \frac 1 3 \end{align} }$$