The probability of a uniform random variable and its square

Question

For the random variables $X \sim U[-1,1]$ and $Y=X^2$. I already showed that their covariance = 0 because $E(XY) = 0$.

My problem is that I can't understand how to find the probabilities $P(X > \frac{1}{2} , Y > \frac{1}{4} )$ and $P(X > \frac{1}{2})P(Y > \frac{1}{4} )$?

Also what would I conclude if both probabilities are equal or not? Thanks for help..

Answer 1

If $X>\frac{1}{2}$, then $Y$ is certainly greater than $\frac{1}{4}$, so the first probability is $1$.

For the second probability, $P(X>\frac{1}{2})$ is $\frac{1}{4}$, and $P(Y>\frac{1}{4})$, which corresponds to $P(|X|>\frac{1}{2})$, is $\frac{1}{2}$. So the probability is $\frac{1}{8}$.

Answer 2

First:

$P(X > \frac{1}{2} , Y > \frac{1}{4} ) = P(Y > \frac{1}{4} | X > \frac{1}{2}) P(X > \frac{1}{2}) = P(X^2 > \frac{1}{4} | X > \frac{1}{2}) \frac{1}{4} = \frac{1}{4} $

Second:

$P(X > \frac{1}{2})P(Y > \frac{1}{4} ) = \frac{1}{4}P(X^2 > \frac{1}{4} ) = \frac{1}{4}P(|X| > \frac{1}{2} ) =\frac{1}{4}P(X > \frac{1}{2} or X < -\frac{1}{2} ) = \frac{1}{4}( P(X > \frac{1}{2})+ P(X < -\frac{1}{2} )) = \frac{1}{4} \frac{1}{2} = \frac{1}{8}$