# Uniform distribution- independent random variable max and min

Let $$X,Y \sim U_{[-1,1]}$$ are independent random variables. Let us define $$U=\max (X,Y)$$ and $$V=\min(X,Y)$$. Are random variables $$U,V$$ independent ?

• Intuitively, does knowing the value of $U$ tell you any information about $V$? – angryavian Dec 3 '18 at 22:16
• I think, that yes. – PabloZ392 Dec 3 '18 at 22:19
• I think, that $U,V$ are dependent, but I have problem, because I trying find an example. – PabloZ392 Dec 3 '18 at 22:27

$$P(V > 0) = P(X > 0 \wedge Y > 0) = 0.5 \times 0.5 = 0.25$$
However, $$P(V > 0 | U = 0) = 0$$, since $$U = 0 \implies X \leq 0 \wedge Y \leq 0$$.
i.e. the distribution of $$V | U$$ is not the same as the unconditional distribution of $$V$$, so they are not independent.
• $P(V > 0) = P(X > 0 \color{red}{\wedge} Y > 0) = 0.5 \times 0.5 = 0.25$. Is this correct? – PabloZ392 Dec 3 '18 at 22:41
• Yes - the event $min(X, Y) > 0$ is the same as the event $X, Y > 0$, i.e. the minimum is greater than zero if both of the values are greater. – ConMan Dec 3 '18 at 22:49