Is choosing the maximum and minimum of 2 random numbers in an interval [a, b] an independent event?

Question

I am unable to wrap my head around this. I was having a discussion with a friend who mentioned that suppose we have an interval $[1, 10]$ , and we pick $2$ real numbers randomly in that range. Let the larger of these $2$ real numbers be $A$ and the smaller of these $2$ real numbers be $B$. He mentioned that the joint PDF of $A$ and $B$ is $f_{A, B}(A, B) = f_A(A) \times f_B(B)$ where $A \sim \text{Uniform} (1, 10)$ and $B \sim \text{Uniform} (1, 10)$.

I don't really understand this because, since he wrote $f_{A, B}(A, B) = f_A(A) \times f_B(B)$, this implies that $A$ and $B$ are independent events, but intuitively to me, why should it be independent? Let me give an example. Suppose we know that $A = 8$, immediately, $B$ has a $\textbf{reduced sample space}$ since B can only take on values from $[1, 8]$. Doesn't this imply that they are not independent and hence we cannot simply apply $f_{A, B}(A, B) = f_A(A) \times f_B(B)$? If so, how should we find the joint PDF?

Answer 1

They are not independent. For example, $P(B>6|A<4)=0\neq P(B>6)$.

Answer 2

In fact, they are dependent, because the joint density of $(A,B)$ must have support $$\{(A, B) \in [1, 10] \times [1, 10] : A \le B\}.$$ Geometrically, in the Cartesian plane with abscissa $A$ and ordinate $B$, the support is the triangle with vertices at $(1,1)$, $(10,10)$, and $(1,10)$.

Thus, while the joint density is indeed constant when each of $A$ and $B$ is selected uniformly on the interval $[1,10]$, it is constant on the support; i.e., $$f_{A,B}(a,b) = \begin{cases} \frac{2}{81}, & 1 \le a \le b \le 10 \\ 0, & \text{otherwise}. \end{cases}$$