I am trying to develop some intuition for problem 4.6.3a in Pitman (1993) (self study not HW) and need some help confirming my solution to a simpler analogous problem:
Simpler analogous problem: Let $U_1, U_2, U_3$ be iid discrete uniform random variables on the integers $\{0,1,2,3\}$; e.g., $P(U_1 = 2) = \frac{1}{4}$. Let $U_{(1)}, U_{(2)}, U_{(3)}$ be their orderings, with $U_{(1)}$ defined as $min(U_1, U_2, U_3)$. What is $P(U_{(1)} \ge 1, U_{(3)} \le 2)$?
Solution via counting: There are 20 possible values for $(U_{(1)}, U_{(2)}, U_{(3)})$. We know this from observing: if $U_{(1)} = 0 $, then $U_{(2)} \in \{0,1,2,3\}.$ Additionally:
- If $U_{(2)} = 0$, then $U_{(3)} \in \{0,1,2,3\}.$
- If $U_{(2)} = 1$, then $U_{(3)} \in \{1,2,3\}.$
- If $U_{(2)} = 2$, then $U_{(3)} \in \{2,3\}.$
- If $U_{(2)} = 3$, then $U_{(3)} = 3$.
So there are $4+3+2+1 = 10$ possible choices of $(U_{(1)}, U_{(2)}, U_{(3)})$ given $U_{(1)} = 0$. Similarly there are/(is) $6$, $3$, and $1$ possible choice(s) for $U_{(1)} = 1,2,3$, respectively, for a total of 20 possible events in our sample space.
There are only four possible choices of $(U_{(1)}, U_{(2)}, U_{(3)})$ that meet the criteria $U_{(1)} \ge 1, U_{(3)} \le 2$: {1,1,1}, {1,1,2}, {1,2,2}, {2,2,2}. Hence, $P(U_{(1)} \ge 1, U_{(3)} \le 2) = \frac{4}{20}$.
For the curious, I eventually want to do this via CDFs and then generalize this to the continuous case to show if $U_i$ ~ $U(0,1)$ and $0 \le x < y \le 1$, then
$$P(U_{(1)} > x, U_{(n)} < y) = P(x < U_{(1)} < y)\cdot P(x < U_{(2)} < y)\cdot…\cdot P(x < U_{(n)} < y) = (y-x)^n$$
This result doesn't make intuitive sense to me right now.
Best Answer
The random variables $U_{(1)},U_{(2)},U_{(3)}$ are dependent and do not have discrete uniform distribution, unlike $U_1,U_2,U_3$. So the cases you count are not equally probable. The inequalities $U_{(1)}\geq 1, U_{(3)}\leq 2$ means that all initial random variables $U_1$, $U_2$, $U_3$ lie in $\{1,2\}$. Since they are independent we can multiply probabilities: $$ \mathbb P(U_{(1)}\geq 1, U_{(3)}\leq 2)=\mathbb P(1\leq U_1\leq 2,\,1\leq U_2\leq 2,\,1\leq U_3\leq 2) $$ $$ =\mathbb P(1\leq U_1\leq 2)\cdot\mathbb P(1\leq U_2\leq 2)\cdot\mathbb P(1\leq U_3\leq 2) = \left(\frac14+\frac14\right)^3=\frac18. $$
For continuous uniform r.v. the last equation will be true if you replace order statistics in r.h.s. by unordered iid r.v.'s: for $0 \le x < y \le 1$ $$ \mathbb P(U_{(1)} > x, U_{(n)} < y) =\mathbb P(x < U_{1} < y)\cdot\mathbb P(x < U_{2} < y)\cdot\ldots\cdot \mathbb P(x < U_{n} < y) = (y-x)^n. $$