basics of uniform distribution

Question

I have a very very basic question

Let x and y be independent, uniformly distributed random variables over [0,1].a and R are pozitive real numbers where R>a. How can I find the probability of $$\mathbb{P}\left[x>\frac{(y)(a)}{R-a}\right]?$$

We can write it as $$1-\mathbb{P}\left[x<\frac{ya}{R-a}\right]$$ and then compute $$ 1-\mathbb{P}\left[x<\frac{ya}{R-a}\right] = 1-\left(\int_{y=0}^1\int_{x=0}^{x=ya/(R-a)} 1 dxdy\right) $$$$ =1-\left( \int_0^1\frac{ya}{R-a}dy\right) =1-\frac{a}{2(R-a)} $$

Thank you!!

Answer 1

This is rather a comment, but unfortunately, my reputation doesn't allow me to comment. I am somehow confused, whether the provided answer is correct, as $1-\frac{a}{2(R-a)}$ can take negative values. I would personally approach very differently.

$X$ and $Y$ are independent and uniformly distributed, thus their ratio $Z = \frac{x}{y}$ has a uniform ratio distribution, see here. The PDF is known: $$ f_Z(z) = \begin{cases} 1/2, & \text{if } 0 \le z \le 1 \\ 1/(2z^2), & \text{if } z > 1 \\ 0, & \text{otherwise}. \end{cases} $$

I would take it and integrate :

$$ P\biggl(\frac{X}{Y} > \frac{a}{R-a}\biggr) = \int_\frac{a}{R-a}^{\infty} f_Z(z) \ dz $$

Answer 2

This looks correct and the final answer would be $$1 - \frac{a}{2(R-a)}.$$ Another way to see that is direct:

$$ \begin{split} \mathbb{P}\left[x>\frac{(y)(a)}{R-a}\right] &= \int_{y=0}^{y=1} \int_{x=ya/(R-a)}^{x=1}1 dxdy \\ &= \int_0^1 \left[1 - \frac{ya}{R-a} \right]dy \\ &= 1 - \int_0^1 \frac{ya}{R-a} dy \\ &= 1 - \frac{a}{2(R-a)}. \end{split} $$

Of course, this argument only holds if $a/(R-a) < 1$. Otherwise, you need to substitute $$ \max\left\{1, \frac{ya}{R-a}\right\} $$