Find $P(x_2/x_3 \leq a)$ where $x_1,x_2,\dots,x_n $ are iid $\operatorname{Uniform}(0,1)$

Question

Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.

How to find $P(x_2/x_3 \leq a)$?

---

Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below

$$f(x_2, x_3) = 1$$

$$P(\frac {x_2}{x_3} \leq a) = \int _0^{1} \int _0^{a*x_3} dx_2 dx_3 = \frac{a}{2} $$

---

Edit, I need to find the probability by conditioning. S my approach is not right.

---

Update: after reviewing the comments below, I summarized them into:

$$P\left(\frac {x_2}{x_3} \leq a\right) = \begin{cases} \int_0^1 \int_0^{a*x_3} dx_2 dx_3 = a/2 && a \le 1\\ \int_0^1 \int_1^{x_2/a} dx_3 dx_2 = 1 - \frac{1}{2*a} && a > 1 \end{cases} $$

these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results

I'm still thinking of the best way to get these results using conditional probability

Answer 1

Hint: You should reason that (for positive a) $$P\left(\frac {x_2}{x_3} \leq a\right) = \int _0^{1} \min(a \,x_3, 1) \, dx_3 = \begin{cases} \int_0^1 a \, x_3 \, dx_3 && a \le 1\\ \int _0^{\frac1a} a \, x_3 \, dx_3 + \int _{\frac1a}^{1} 1 \, dx_3 && a > 1 \end{cases} $$

Answer 2

For $x\in\left(0,1\right)$ we have by independency:

$P\left\{ X_{1}/X_{2}\leq a\mid X_{2}=x\right\} =P\left\{ X_{1}\leq ax\right\} $

So if function $f$ is defined by $f\left(x\right)=\begin{cases} 1 & \text{if }ax\geq1\\ 0 & \text{if }ax\leq0\\ ax & \text{otherwise}\end{cases}$

Then:

$P\left\{ X_{1}/X_{2}\leq a\mid X_{2}\right\} =f\left(X_{2}\right)$ and $P\left\{ X_{1}/X_{2}\leq a\right\} =\mathbb{E}f\left(X_{2}\right)=\int_{0}^{1}f\left(x\right)dx$