Uniform distribution- independent random variable max and min












-1












$begingroup$


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 ?






random-variables uniform-distribution







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$endgroup$












  • $begingroup$
    Intuitively, does knowing the value of $U$ tell you any information about $V$?
    $endgroup$
    – angryavian
    Dec 3 '18 at 22:16










  • $begingroup$
    I think, that yes.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:19










  • $begingroup$
    I think, that $U,V$ are dependent, but I have problem, because I trying find an example.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:27
















-1












$begingroup$


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 ?






random-variables uniform-distribution







share|cite|improve this question









$endgroup$












  • $begingroup$
    Intuitively, does knowing the value of $U$ tell you any information about $V$?
    $endgroup$
    – angryavian
    Dec 3 '18 at 22:16










  • $begingroup$
    I think, that yes.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:19










  • $begingroup$
    I think, that $U,V$ are dependent, but I have problem, because I trying find an example.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:27














-1












-1








-1





$begingroup$


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 ?






random-variables uniform-distribution







share|cite|improve this question









$endgroup$




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 ?






random-variables uniform-distribution




random-variables uniform-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 22:13









PabloZ392PabloZ392

686




686












  • $begingroup$
    Intuitively, does knowing the value of $U$ tell you any information about $V$?
    $endgroup$
    – angryavian
    Dec 3 '18 at 22:16










  • $begingroup$
    I think, that yes.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:19










  • $begingroup$
    I think, that $U,V$ are dependent, but I have problem, because I trying find an example.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:27


















  • $begingroup$
    Intuitively, does knowing the value of $U$ tell you any information about $V$?
    $endgroup$
    – angryavian
    Dec 3 '18 at 22:16










  • $begingroup$
    I think, that yes.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:19










  • $begingroup$
    I think, that $U,V$ are dependent, but I have problem, because I trying find an example.
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:27
















$begingroup$
Intuitively, does knowing the value of $U$ tell you any information about $V$?
$endgroup$
– angryavian
Dec 3 '18 at 22:16




$begingroup$
Intuitively, does knowing the value of $U$ tell you any information about $V$?
$endgroup$
– angryavian
Dec 3 '18 at 22:16












$begingroup$
I think, that yes.
$endgroup$
– PabloZ392
Dec 3 '18 at 22:19




$begingroup$
I think, that yes.
$endgroup$
– PabloZ392
Dec 3 '18 at 22:19












$begingroup$
I think, that $U,V$ are dependent, but I have problem, because I trying find an example.
$endgroup$
– PabloZ392
Dec 3 '18 at 22:27




$begingroup$
I think, that $U,V$ are dependent, but I have problem, because I trying find an example.
$endgroup$
– PabloZ392
Dec 3 '18 at 22:27










1 Answer
1






active

oldest

votes


















3












$begingroup$

$P(V > 0) = P(X > 0 wedge Y > 0) = 0.5 times 0.5 = 0.25$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:41












  • $begingroup$
    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.
    $endgroup$
    – ConMan
    Dec 3 '18 at 22:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$P(V > 0) = P(X > 0 wedge Y > 0) = 0.5 times 0.5 = 0.25$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:41












  • $begingroup$
    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.
    $endgroup$
    – ConMan
    Dec 3 '18 at 22:49
















3












$begingroup$

$P(V > 0) = P(X > 0 wedge Y > 0) = 0.5 times 0.5 = 0.25$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:41












  • $begingroup$
    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.
    $endgroup$
    – ConMan
    Dec 3 '18 at 22:49














3












3








3





$begingroup$

$P(V > 0) = P(X > 0 wedge Y > 0) = 0.5 times 0.5 = 0.25$



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.






share|cite|improve this answer









$endgroup$



$P(V > 0) = P(X > 0 wedge Y > 0) = 0.5 times 0.5 = 0.25$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 22:30









ConManConMan

7,6121324




7,6121324












  • $begingroup$
    $P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:41












  • $begingroup$
    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.
    $endgroup$
    – ConMan
    Dec 3 '18 at 22:49


















  • $begingroup$
    $P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
    $endgroup$
    – PabloZ392
    Dec 3 '18 at 22:41












  • $begingroup$
    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.
    $endgroup$
    – ConMan
    Dec 3 '18 at 22:49
















$begingroup$
$P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
$endgroup$
– PabloZ392
Dec 3 '18 at 22:41






$begingroup$
$P(V > 0) = P(X > 0 color{red}{wedge} Y > 0) = 0.5 times 0.5 = 0.25$. Is this correct?
$endgroup$
– PabloZ392
Dec 3 '18 at 22:41














$begingroup$
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.
$endgroup$
– ConMan
Dec 3 '18 at 22:49




$begingroup$
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.
$endgroup$
– ConMan
Dec 3 '18 at 22:49


















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