This is an extension of a question I previously asked.
Let $\gamma\in(0, 1)$ and $n\in\mathbb{N}$ be parameters, and let $I_1, I_2, \dots$ be independently and uniformly chosen subintervals of $[0, 1]$ of length $\gamma$ each (isomorphically: let the $I_k$ be independently and uniformly chosen points on the interval $[0, 1 - \gamma]$).
The random variable $x_{\gamma, n}$ is the Lebesgue measure of $\bigcup_{k=1}^n I_k$, and $y_{\gamma, n}$ is the Lebesgue measure of $\bigcap_{k=1}^n I_k$. What is the joint pdf $f_{\gamma, n}(x, y):[0, 1]^2\rightarrow\mathbb{R}$ of $(x_{\gamma, n}, y_{\gamma, n})$?
Thoughts
(I will omit subscripts when we can assume that $\gamma$ and $n$ are fixed and there is no risk of ambiguity.)
- $\big|\bigcup_{k=1}^n I_k\big|\in \big[\gamma, n\gamma\big]$, so $f(x, y) \equiv 0$ for $x\not\in \big[\gamma, n\gamma\big]$.
- $\big|\bigcap_{k=1}^n I_k\big|\leq \gamma$, so $f(x, y) \equiv 0$ for $y > \gamma$.
- When $n = 0$ or $1$, the distributions are very simple: $$ \begin{align} f_{\gamma, 0} = \delta_{0, 1} \end{align} $$ and $$ \begin{align} f_{\gamma, 1} = \delta_{\gamma, \gamma}. \end{align} $$
- When $n = 2$, $x+y = 2\gamma$ by the Inclusion-Exclusion Principle.
- Suppose that $\gamma < \frac{1}{2}$. Then as $n\rightarrow\infty$, the distribution "converges" (in a sense I want to make more precise) to $\delta_{1, 0}$.
Proof of #5: A necessary condition for $\cup I_n = 1$ is that the minimum of all left endpoints be zero and that the maximum of all right endpoints be 1. Choosing a small $t > 0$, the probability that the minimum left endpoint is greater than $t$ after $n$ trials is the probability that among $n$ uniformly and ind'ly chosen $t_n$ from $[0, 1 - \gamma]$ none is less than $t$. This probability is $\Big(\frac{t}{1-\gamma}\Big)^n$, which approaches zero as $n\rightarrow 0$. A symmetrical argument applies to the right endpoint (though there are independence issues to consider for a fully rigorous treatment).