Even with the new Wolfram plugin for chatGPT, I'm struggling with the following question.
Suppose $X, Y \sim U(0, 2)$, i.e. both of them are independent and uniformly distributed. What would then be the probability that $\frac{A}{B} \in (1, 3)$?
This is what I've got so far and although I know the answer can't be right I simply don't see my mistake.
$\begin{align}&1 \leq \frac{A}{B} \leq 3, \newline &B\leq A\leq 3B,\end{align}$
whereby,
$\begin{align}&\int^2_{0}\left( \int^{3B}_{B}dA \right) dB, \newline &\int^2_{0}\left( 2B \right) dB, \newline &2^2-0^2 = 4\end{align}$,
would define the area that satisfies my condition. However, since the total parameter space is of size 4 this would imply that the required probability is 1 which doesn't make any sense.
Hopefully this question isn't too basic for this forum. Didn't know where else to turn.
Best,
Your suspicion is correct. When you sketch the event and sample space, it's clear that $\mathbb{P}\bigl(1<\frac{A}{B}<3\bigr) = \frac13$, as the triangle takes up one third of the square.
The problem with setting up the double integral the way you've suggested is that for each $B$ (vertical axis), the upper limit of integration for $A$ as a function of $B$ is piecewise: $$ \begin{cases} B < A < 3B, &\text{for } 0 < B < \tfrac23 \\ B < A < 2, &\text{for } \tfrac23 < B < 2 \end{cases} $$
The proper integral is then $$ \mathbb{P}\bigl(1<\frac{A}{B}<3\bigr) = \frac14 \Biggl( \int_0^{\frac23} \int_B^{3B} \mathrm{d}A \, \mathrm{d}B + \int_{\frac23}^2 \int_B^2 \mathrm{d}A \, \mathrm{d}B \Biggr) = \frac14 \biggl( \frac49 + \frac89 \biggr) = \frac13. $$
Of course, it's much cleaner to put $A$ on the outside of the integral, and for each $0 < A < 2$, consider $B$ such that $\frac13A < B < A$: $$ \mathbb{P}\bigl(1<\frac{A}{B}<3\bigr) = \frac14 \Biggl( \int_0^2 \int_{\frac13 A}^A \mathrm{d}B \, \mathrm{d}A \Biggr) = \frac14 \cdot \frac43 = \frac 13. $$