Are uniformly distributed random variables $X$, $X^2$ (in)dependent?
$begingroup$
We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.
However I am not sure how to show that they are (in)dependent.
probability uniform-distribution
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
$endgroup$
add a comment |
$begingroup$
We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.
However I am not sure how to show that they are (in)dependent.
probability uniform-distribution
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
$endgroup$
$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53
add a comment |
$begingroup$
We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.
However I am not sure how to show that they are (in)dependent.
probability uniform-distribution
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
$endgroup$
We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.
However I am not sure how to show that they are (in)dependent.
probability uniform-distribution
probability uniform-distribution
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
edited Dec 20 '18 at 8:14
Björn Friedrich
2,66961831
2,66961831
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
asked Dec 19 '18 at 15:36
dwarf_dzdwarf_dz
1
1
$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53
add a comment |
$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53
$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53
$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
begin{align}
P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
end{align}
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
$endgroup$
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
add a comment |
$begingroup$
They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
begin{align}
P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
end{align}
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
Hint:
begin{align}
P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
end{align}
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
Hint:
begin{align}
P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
end{align}
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
Hint:
begin{align}
P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
end{align}
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 19 '18 at 15:42
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
add a comment |
add a comment |
$begingroup$
For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
$endgroup$
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
add a comment |
$begingroup$
For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
$endgroup$
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
add a comment |
$begingroup$
For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
$endgroup$
For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
edited Dec 20 '18 at 7:57
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
answered Dec 19 '18 at 15:46
drhabdrhab
101k545136
101k545136
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
add a comment |
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
For proving independence?
$endgroup$
– NCh
Dec 20 '18 at 1:11
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
$begingroup$
@NCh Thanks for attending me. Repaired.
$endgroup$
– drhab
Dec 20 '18 at 7:57
add a comment |
$begingroup$
They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
$endgroup$
add a comment |
$begingroup$
They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
$endgroup$
add a comment |
$begingroup$
They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
$endgroup$
They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
answered Dec 19 '18 at 15:45
J.G.J.G.
26.9k22742
26.9k22742
add a comment |
add a comment |
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$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53