Finding the cumulative distribution function of the sum of two functions of two independent random variables

Question

Since we want two real roots, then we're looking for a positive discriminant.

Let $Z$ be the random variable representing the value taken by the discriminant.

So, $$Z = 4(U^2 - V)$$

However, I don't know where to take it from here.

Any Hints would be really appreciated.

Thanks

Answer 1

We want to find $\mathbb P(Z>0)$, or $\mathbb P(U^2>V)$. For $0<t<1$ we have $$ \mathbb P(U^2\leqslant t) = \mathbb P(U\leqslant t^{\frac12}) = t^{\frac12} $$ and so $U^2$ has density $$ f(t) = \frac12 t^{-\frac12}\mathsf 1_{(0,1)}(t). $$ We know already that $V$ has density $g(t) = \mathsf 1_{(0,1)}(t)$, so integrating over the joint density we have \begin{align} \mathbb P(U^2>V) &= \int_0^1 \int_t^1 \frac12 s^{-\frac12} \ \mathsf ds\ \mathsf dt\\ &= \int_0^1 (1-t^{\frac12})\ \mathsf dt\\ &= \frac13. \end{align}

Answer 2

$P\{U^{2} >V\}=EP(\{U^{2} >V\}|U\})=EU^{2}=\int _0^{1} u^{2}\, du =\frac 1 3$.

Answer 3

The positive discriminant implies $Z=4(U^2-V)>0$, hence you need to find $\mathbb P(U^2-V>0)$. Refer to the graph:

$\hspace{1cm}$

The required probability is the green area: $$\mathbb P(U^2-V>0)=\mathbb P(V<U^2)=\int_0^1 U^2 dU=\frac{U^3}{3}\bigg{|}_0^1=\frac13.$$