Let the parameter $\gamma$ be in $(0, 1)$ and $I_1, I_2, \dots$ a random sequence of uniformly and independently chosen closed subintervals of $[0, 1]$ of length $\gamma$ apiece. Let $X_{\gamma}$ be the minimum $n$ such that $$\bigcup_{k=1}^n I_k$$ has Lebesgue measure 1, and $Y_{\gamma}$ the minimum $n$ such that $$\bigcap_{k=1}^n I_k$$ has Lebesgue measure 0.
What is the joint probability mass function of $X_{\gamma}$ and $Y_{\gamma}$?
Thoughts
- $f_{\gamma}(0, y) = f_{\gamma}(x, 0) \equiv 0$ by the definitions of union/intersection over no sets.
- $f_{\gamma}(1, y) = f_{\gamma}(x, 1) \equiv 0$: the measure of the first set is $\alpha\not\in\{0, 1\}$.
- Extending the above in one direction, $f_{\gamma}(x, y) \equiv 0$ when $x \leq \frac{1}{\gamma}$ because the union of $x$ of the subintervals has measure bounded above by $\frac{x}{\gamma}$.
- $\displaystyle\lim_{x^2 + y^2\rightarrow \infty} f(x, y) = 0$. Because ... it's a pmf and its integral has to be finite. (Also, from the side of the definition of the distribution ... um, for grad prob theory reasons I currently forget but promise to look up.)
Here's where I get confused. By definition of pmf, $\sum_{m, n\geq 0} f(m, n) = 1$, so it must be non-zero somewhere. But ... where? For the union of the subintervals to have measure 1, it is necessary that the minimum of all left sides of the intervals be 0. But the probability is zero that that is the case after any finite number of selections, so f(x, y) must seem to be identically zero restricted to a fixed $x$. What gives?