Mixing
Need help understanding why $Var(max(X,Y)) = Var(min(X,Y))$ if $X$ and $Y$ are iid $N(0,1)$
$begingroup$
$Var(max(X,Y)) = Var(min(X,Y))$ if $X$ and $Y$ are iid $N(0,1)$
The book solution says
$(X,Y)$ has the same distribution as $(-X,-Y)$ since $X$ and $Y$ are iid $N(0,1)$
it follows that $Var(max(X,Y)) = Var(min(X,Y))$
It make sense intuitively but I'm having trouble seeing it.
It make sense to me that $max(X,Y) = -min(-X,-Y)$ so for example the $max(1,4) = -min(-1,-4)$
Then take the variance of both sides $Var(max(X,Y)) = Var(-min(-X,-Y))$
Then take out the negative of the RHS and we have $Var(max(X,Y)) = Var(min(-X,-Y))$
But from here I'm stuck. Thanks for your help and patience.
probability variance
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
$endgroup$
add a comment |
$begingroup$
$Var(max(X,Y)) = Var(min(X,Y))$ if $X$ and $Y$ are iid $N(0,1)$
The book solution says
$(X,Y)$ has the same distribution as $(-X,-Y)$ since $X$ and $Y$ are iid $N(0,1)$
it follows that $Var(max(X,Y)) = Var(min(X,Y))$
It make sense intuitively but I'm having trouble seeing it.
It make sense to me that $max(X,Y) = -min(-X,-Y)$ so for example the $max(1,4) = -min(-1,-4)$
Then take the variance of both sides $Var(max(X,Y)) = Var(-min(-X,-Y))$
Then take out the negative of the RHS and we have $Var(max(X,Y)) = Var(min(-X,-Y))$
But from here I'm stuck. Thanks for your help and patience.
probability variance
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
$endgroup$
1
$begingroup$
With $X$ and $Y$ iid $N(0,1)$ variables, $-X$ and $-Y$ are also $N(0,1)$ variables.
$endgroup$
– kimchi lover
Jan 28 at 1:34
$begingroup$
@kimchilover thanks for your help. It makes sense that $-X,-Y$ are also $N(0,1)$. It seems true to me that $(X,Y) neq (-X,-Y)$ and I get stuck at #3 in my argument above.
$endgroup$
– HJ_beginner
Jan 28 at 2:11
1
$begingroup$
Of course $(X,Y)ne(-X,-Y)$ but they do have the same distribution, and so do any functions of them, too. The probability law of $min(X,Y)$ is the same as that of $min(-X,-Y)$, and so on.
$endgroup$
– kimchi lover
Jan 28 at 2:13
$begingroup$
Ahhh good point... I have no issue believing $Var(X) = Var(Y) = 1$ if $X$ and $Y$ are independent $N(0,1)$ but of course $X neq Y$ . Thanks for your help
$endgroup$
– HJ_beginner
Jan 28 at 2:17
add a comment |
$begingroup$
$Var(max(X,Y)) = Var(min(X,Y))$ if $X$ and $Y$ are iid $N(0,1)$
The book solution says
$(X,Y)$ has the same distribution as $(-X,-Y)$ since $X$ and $Y$ are iid $N(0,1)$
it follows that $Var(max(X,Y)) = Var(min(X,Y))$
It make sense intuitively but I'm having trouble seeing it.
It make sense to me that $max(X,Y) = -min(-X,-Y)$ so for example the $max(1,4) = -min(-1,-4)$
Then take the variance of both sides $Var(max(X,Y)) = Var(-min(-X,-Y))$
Then take out the negative of the RHS and we have $Var(max(X,Y)) = Var(min(-X,-Y))$
But from here I'm stuck. Thanks for your help and patience.
probability variance
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
$endgroup$
$Var(max(X,Y)) = Var(min(X,Y))$ if $X$ and $Y$ are iid $N(0,1)$
The book solution says
$(X,Y)$ has the same distribution as $(-X,-Y)$ since $X$ and $Y$ are iid $N(0,1)$
it follows that $Var(max(X,Y)) = Var(min(X,Y))$
It make sense intuitively but I'm having trouble seeing it.
It make sense to me that $max(X,Y) = -min(-X,-Y)$ so for example the $max(1,4) = -min(-1,-4)$
Then take the variance of both sides $Var(max(X,Y)) = Var(-min(-X,-Y))$
Then take out the negative of the RHS and we have $Var(max(X,Y)) = Var(min(-X,-Y))$
But from here I'm stuck. Thanks for your help and patience.
probability variance
probability variance
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
asked Jan 28 at 1:19
HJ_beginnerHJ_beginner
8951415
8951415
1
$begingroup$
With $X$ and $Y$ iid $N(0,1)$ variables, $-X$ and $-Y$ are also $N(0,1)$ variables.
$endgroup$
– kimchi lover
Jan 28 at 1:34
$begingroup$
@kimchilover thanks for your help. It makes sense that $-X,-Y$ are also $N(0,1)$. It seems true to me that $(X,Y) neq (-X,-Y)$ and I get stuck at #3 in my argument above.
$endgroup$
– HJ_beginner
Jan 28 at 2:11
1
$begingroup$
Of course $(X,Y)ne(-X,-Y)$ but they do have the same distribution, and so do any functions of them, too. The probability law of $min(X,Y)$ is the same as that of $min(-X,-Y)$, and so on.
$endgroup$
– kimchi lover
Jan 28 at 2:13
$begingroup$
Ahhh good point... I have no issue believing $Var(X) = Var(Y) = 1$ if $X$ and $Y$ are independent $N(0,1)$ but of course $X neq Y$ . Thanks for your help
$endgroup$
– HJ_beginner
Jan 28 at 2:17
add a comment |
1
$begingroup$
With $X$ and $Y$ iid $N(0,1)$ variables, $-X$ and $-Y$ are also $N(0,1)$ variables.
$endgroup$
– kimchi lover
Jan 28 at 1:34
$begingroup$
@kimchilover thanks for your help. It makes sense that $-X,-Y$ are also $N(0,1)$. It seems true to me that $(X,Y) neq (-X,-Y)$ and I get stuck at #3 in my argument above.
$endgroup$
– HJ_beginner
Jan 28 at 2:11
1
$begingroup$
Of course $(X,Y)ne(-X,-Y)$ but they do have the same distribution, and so do any functions of them, too. The probability law of $min(X,Y)$ is the same as that of $min(-X,-Y)$, and so on.
$endgroup$
– kimchi lover
Jan 28 at 2:13
$begingroup$
Ahhh good point... I have no issue believing $Var(X) = Var(Y) = 1$ if $X$ and $Y$ are independent $N(0,1)$ but of course $X neq Y$ . Thanks for your help
$endgroup$
– HJ_beginner
Jan 28 at 2:17
1
1
$begingroup$
With $X$ and $Y$ iid $N(0,1)$ variables, $-X$ and $-Y$ are also $N(0,1)$ variables.
$endgroup$
– kimchi lover
Jan 28 at 1:34
$begingroup$
With $X$ and $Y$ iid $N(0,1)$ variables, $-X$ and $-Y$ are also $N(0,1)$ variables.
$endgroup$
– kimchi lover
Jan 28 at 1:34
$begingroup$
@kimchilover thanks for your help. It makes sense that $-X,-Y$ are also $N(0,1)$. It seems true to me that $(X,Y) neq (-X,-Y)$ and I get stuck at #3 in my argument above.
$endgroup$
– HJ_beginner
Jan 28 at 2:11
$begingroup$
@kimchilover thanks for your help. It makes sense that $-X,-Y$ are also $N(0,1)$. It seems true to me that $(X,Y) neq (-X,-Y)$ and I get stuck at #3 in my argument above.
$endgroup$
– HJ_beginner
Jan 28 at 2:11
1
1
$begingroup$
Of course $(X,Y)ne(-X,-Y)$ but they do have the same distribution, and so do any functions of them, too. The probability law of $min(X,Y)$ is the same as that of $min(-X,-Y)$, and so on.
$endgroup$
– kimchi lover
Jan 28 at 2:13
$begingroup$
Of course $(X,Y)ne(-X,-Y)$ but they do have the same distribution, and so do any functions of them, too. The probability law of $min(X,Y)$ is the same as that of $min(-X,-Y)$, and so on.
$endgroup$
– kimchi lover
Jan 28 at 2:13
$begingroup$
Ahhh good point... I have no issue believing $Var(X) = Var(Y) = 1$ if $X$ and $Y$ are independent $N(0,1)$ but of course $X neq Y$ . Thanks for your help
$endgroup$
– HJ_beginner
Jan 28 at 2:17
$begingroup$
Ahhh good point... I have no issue believing $Var(X) = Var(Y) = 1$ if $X$ and $Y$ are independent $N(0,1)$ but of course $X neq Y$ . Thanks for your help
$endgroup$
– HJ_beginner
Jan 28 at 2:17
add a comment |
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1
$begingroup$
With $X$ and $Y$ iid $N(0,1)$ variables, $-X$ and $-Y$ are also $N(0,1)$ variables.
$endgroup$
– kimchi lover
Jan 28 at 1:34
$begingroup$
@kimchilover thanks for your help. It makes sense that $-X,-Y$ are also $N(0,1)$. It seems true to me that $(X,Y) neq (-X,-Y)$ and I get stuck at #3 in my argument above.
$endgroup$
– HJ_beginner
Jan 28 at 2:11
1
$begingroup$
Of course $(X,Y)ne(-X,-Y)$ but they do have the same distribution, and so do any functions of them, too. The probability law of $min(X,Y)$ is the same as that of $min(-X,-Y)$, and so on.
$endgroup$
– kimchi lover
Jan 28 at 2:13
$begingroup$
Ahhh good point... I have no issue believing $Var(X) = Var(Y) = 1$ if $X$ and $Y$ are independent $N(0,1)$ but of course $X neq Y$ . Thanks for your help
$endgroup$
– HJ_beginner
Jan 28 at 2:17