Probability one random variable exceeds another by ratio k

Question

I have constants $j,k \in [0,1]$ . I take two independent draws $x,y$ from a distribution uniform over 0,1. What is the probability that $x*j> y * k$ ?

Answer 1

Hint :

That is same as probability of $x$ being greater than $\frac{k}{j}$ times $y$.

This is same as picking $x$ from $[0,1]$ and $y$ from $[0,\frac{k}{j}]$ and finding probability that $x$ is bigger than $y$.

This is $$\int_0^1\int_0^{k/j}I_{x>y}.\frac{j}{k} dxdy$$

Where $I$ is the indicator function. Note that $\frac{j}{k}$ is multiplies to the integral since the density function changed.

This integral evaluates to $\frac{j}{2k}$

Answer 2

Hint:

It equals integral:$$\mathsf E1_{\{Xj>Yk\}}=\int_0^1\int_0^1[xj>yk]dxdy$$where $[xj>yk]$ denotes the function $[0,1]^2\to\mathbb R$ prescribed by $\left\langle x,y\right\rangle \mapsto\begin{cases} 1 & \text{if }xj>yk\\ 0 & \text{otherwise} \end{cases}$