Consider the two independent random variables $X$ and $Y$. Assume both are uniformly distributed on $[0,1]$. I want to calculate the probability that $X$ lies between some "low" linear transformation of $Y$, say $\delta_1Y$, and some "higher" linear transformation of $Y$, say $\delta_2Y$, where $\delta_1<\delta_2\Leftrightarrow\delta_1Y<\delta_2Y$. Further we know that $\delta_1>0$ and $\delta_2\geq1$. Thus, the probability I want to calculate is $\mathbb{P}(\delta_1Y<X<\delta_2Y)$.
I know that the probability that $X$ is less than $Y$ can be calculated by:
$$\mathbb{P}[X<Y]=\int_{y=0}^{1}\int_{x=0}^{y}P_{X}(x)P_{Y}(y)dxdy$$
where $P_X(x)$ is the PDF of $X$ and $P_Y(y)$ is the PDF of $Y$. A simple solution would be if you simply could calculate the probability as:
$$\mathbb{P}[\delta_{1}Y<X<\delta_{2}Y]=\mathbb{P}[\delta_{1}Y<X]\cdot\mathbb{P}[X<\delta_{2}Y]$$
But since this implicitly would assume that the two probabilities are independent, I'm pretty sure this is not a viable option. Does anyone have an idea how to solve this?
You could start with:
$$\begin{aligned}P\left(\delta_{1}Y<X<\delta_{2}Y\right) & =\mathbb{E}\mathbf{1}_{\delta_{1}Y<X<\delta_{2}Y}\\ & =\int_{0}^{1}\int_{0}^{1}\mathbf{1}_{\delta_{1}y<x<\delta_{2}y}dxdy\\ & =\int_{0}^{1}\int_{\max\left(0,\delta_{1}y\right)}^{\min\left(1,\delta_{2}y\right)}dxdy\\ & =\int_{0}^{1}\min\left(1,\delta_{2}y\right)-\max\left(0,\delta_{1}y\right)dy \end{aligned} $$
If you already know that $\delta_1\geq0$ then $\max(0,\delta_1y)=\delta_1y$ and things become a bit easyer.
Go on with discerning cases.