Answer

We are asked to find the CDF of $$V = \frac{X}{Y}$$ where $$X, Y \sim \text{Uniform}[0,1]$$ and independent. Start with the definition: $$F_V(v) = P\left(\frac{X}{Y} \leq v\right) = P(X \leq vY)$$. Since $$X$$ and $$Y$$ are independent and uniform on $$[0,1]$$, the joint PDF is 1 over the unit square, so $$F_V(v) = \int_0^1 \int_0^{\min(vy,1)} dx\,dy$$. Split into two cases. If $$0 \leq v \leq 1$$, then $$vy \leq 1$$ for all $$y \in [0,1]$$, so $$F_V(v) = \int_0^1 vy\,dy = \frac{v}{2}$$. If $$v > 1$$, split the integral: for $$y \leq \frac{1}{v}$$, upper limit is $$vy$$; for $$y > \frac{1}{v}$$, it’s 1. Then $$F_V(v) = \int_0^{1/v} vy\,dy + \int_{1/v}^1 1\,dy = \frac{1}{2v} + \left(1 - \frac{1}{v}\right) = 1 - \frac{1}{2v}$$. Final result: $$F_V(v) = \begin{cases} \frac{v}{2}, & 0 \leq v \leq 1 \\ 1 - \frac{1}{2v}, & v > 1 \end{cases}$$.