0
$\begingroup$

Let $U,V$ be random numbers chosen independently from the interval [0,1]. Find the cumulative distribution and density for the the random variables:

$$Y=max(U,V)$$ and $$Y=min(U,V)$$

I am not sure how to go about this with the max and min terms. I have the solutions to these but am not sure how they got to them or even where to start.

$\endgroup$

1 Answer 1

1
$\begingroup$

Hint:

$$Pr(\max(U,V) \le t) = Pr(U \le t, V \le t)$$

$$Pr(\min(U,V) \le t) = 1- Pr(\min(U,V) > t)$$

$\endgroup$
4
  • $\begingroup$ I get this but the answer for the first one is $F(x)=x^2$ and $f(x)=2x$ on [0,1]. how did they get to that? why $x^2$ and not something else? $\endgroup$
    – Derek
    Nov 6, 2017 at 2:50
  • 1
    $\begingroup$ You use that $F_U(x) = F_V(x) = x$ on $(0,1)$. The $x^2$ comes from the fact that $U$ and $V$ are independent $\endgroup$
    – amsmath
    Nov 6, 2017 at 2:52
  • 1
    $\begingroup$ Okay. So I think that makes sense to me but then the Min is $F(x)=2x-x^2$. This is because $F_U(x)+F_V(x)=2x$? $\endgroup$
    – Derek
    Nov 6, 2017 at 2:58
  • $\begingroup$ Are you able to simplify $Pr(\min(U,V) > t)$? $\endgroup$ Nov 6, 2017 at 3:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .