There are two IID standard uniform random variable ${X_1},{X_2}$. assume $Z = \frac{X_1}{X_2}$ ,what is the probability that $\lfloor Z\rfloor$ (floor function) is odd?
my solution is as follow:
${p_Z}(u) = \frac{1}{2}[\frac{{(H(u - 1))}}{{{u^2}}} + H(1 - u)]$
$\int {} {p_Z}(u)du = {P_Z}(u) = \frac{{{u^2} - {{(u - 1)}^2}H(u - 1)}}{{2u}}$
note that H is Heaviside step function. we define ${P_{odd}}(u) = P(2u) - P(2u - 1) = \frac{{4{u^2} - {{(2u - 1)}^2}H(2u - 1)}}{{4u}} - \frac{{{{(2u - 1)}^2} - {{(2u - 2)}^2}H(2u - 2)}}{{2(2u - 1)}}$
we can show that ${P_{odd}}(1)$ is the probability of $Z$ being the first odd number and so on. using wolfram alpha I found out
$\sum\limits_{i \in N} {{P_{odd}}(i)} = \frac{{\ln (4)}}{4}$
I tested this answer using Monte Carlo simulation and it seemed legit,I'm wondering is there any other simpler ways to answer this question? for example without using infinite sum.
For convenience, instead of $X_1, X_2$, let us use $(X,Y)$ and observe their joint distribution is uniform on the unit square. Then $Y/X$ is the slope of a line through the origin, and the set of all $(X,Y)$ such that $\lfloor Y/X \rfloor$ is odd, is the union of the set of triangles $A_k$ with vertices $$\{(0,0), (\tfrac{1}{2k-1}, 1), (\tfrac{1}{2k}, 1)\}$$ excluding the edge joining $(0,0)$ and $(\tfrac{1}{2k}, 1)\}$. So the total probability is simply the sum of the areas of each $A_k$, and since they all have common height $1$, the answer is simply $$\frac{1}{2} \sum_{k=1}^\infty \frac{1}{2k-1} - \frac{1}{2k} = \frac{\log 2}{2}.$$