Find the probability that $P( \frac{1}{\sqrt{2}}\leq (\frac{X^2}{Y^2})\leq\sqrt{2})$

Question

PROBLEM: Given that $X$, $Y$ are indepedent uniformly distributed r.vs's over range $[0,1]$ with their joint pdf given by: $f_{X,Y}(x,y)=1$

find $P( \frac{1}{\sqrt{2}}\leq (\frac{X^2}{Y^2})\leq\sqrt{2})=?$

MY ATTEMPT:

Solving the inequality $\frac{1}{\sqrt{2}}\leq (\frac{X^2}{Y^2})\leq\sqrt{2}$
gives us the following result: $(2^{-\frac{1}{4}} \cdot x \leq y\leq 2^{\frac{1}{4}} \cdot x).$

Hence, $P( \frac{1}{\sqrt{2}}\leq (\frac{X^2}{Y^2})\leq\sqrt{2})=P(2^{-\frac{1}{4}} \cdot x \leq y\leq 2^{\frac{1}{4}} \cdot x)=P(y\leq 2^{\frac{1}{4}} \cdot x)-P(2^{-\frac{1}{4}} \cdot x \leq y)$

From here I don't know how to proceed further. I know joint pdf will be used but I dont know how. Kindly help

Answer 1

Correction: $$P( \frac{1}{\sqrt{2}}\leq (\frac{X^2}{Y^2})\leq\sqrt{2})$$ $$=P(2^{-\frac{1}{4}} \cdot y \leq x\leq 2^{\frac{1}{4}} \cdot y)$$ $$=P(x\leq 2^{\frac{1}{4}} \cdot y)-P(x<2^{-\frac{1}{4}} \cdot y ).$$

Now $P(x\leq 2^{\frac{1}{4}} \cdot y)=\int_0^{2^{-1/4}}\int_0^{2^{1/4}y} dx dy +\int_{2^{-1/4}} ^{1} \int_0^{1} dxdy$ and $P(x<2^{-\frac{1}{4}} \cdot y )=\int_0^{1} \int_0^{2^{-1/4}y} dx dy$.

For the first probability I have split the integral into two parts because $2^{1/4}y$ may exceed $1$ (in which case you have to integrate w.r.t $x$ from $0$ to $1$.

I will let you compute the integrals.