Mixing
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Independent Transformations of the same random variable
Clash Royale CLAN TAG#URR8PPP
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For simplicity let $X$ be a uniformly distributed random variable on $[0,1]$. Does there exist a pair of measurable functions $f$ and $g$ such that $f(X)$ and $g(X)$ are independent random variables?
Building off of Parcly's answer, if $xin [0,1)$ is expressed in base 2, $x=0.b_1b_2dots_2$, then $f(x) = 0.b_1b_3b_5dots$ and $g(x) = 0.b_2b_4b_6dots$ are both limits of independant random variables so they at least measurable. In particular, $f_n = 0.b_1b_3...b_2n+1uparrow f$ and similarly $g_nuparrow g$. Thus for every $y$ and $z$,
beginequation
P(f<ytextrm and g<z)=lim P(f_n<ytextrm and g_n<z)=lim P(f_n<y)P(g_n<z)=(lim P(f_n<y))(lim P(g_n<z))=P(f<y)P(g<z)
endequation
where the first and last equalities are due to the regularity of the measure. Additionally, each $f$ and $g$ are bijections of the interval with itself and so for any other measurable functions $h(x)$ and $j(x)$, we have $hcirc f$ and $jcirc g$ independent random variables. What's more, $hcirc f$ and $hcirc g$ are independent while it is not true that $h(x)$ and $j(x)$ are independent when mappings of $X$ alone.
probability random-variables
asked Jul 19 at 0:19
itheypsi
385
add a comment |Â
up vote
2
down vote
favorite
For simplicity let $X$ be a uniformly distributed random variable on $[0,1]$. Does there exist a pair of measurable functions $f$ and $g$ such that $f(X)$ and $g(X)$ are independent random variables?
Building off of Parcly's answer, if $xin [0,1)$ is expressed in base 2, $x=0.b_1b_2dots_2$, then $f(x) = 0.b_1b_3b_5dots$ and $g(x) = 0.b_2b_4b_6dots$ are both limits of independant random variables so they at least measurable. In particular, $f_n = 0.b_1b_3...b_2n+1uparrow f$ and similarly $g_nuparrow g$. Thus for every $y$ and $z$,
beginequation
P(f<ytextrm and g<z)=lim P(f_n<ytextrm and g_n<z)=lim P(f_n<y)P(g_n<z)=(lim P(f_n<y))(lim P(g_n<z))=P(f<y)P(g<z)
endequation
where the first and last equalities are due to the regularity of the measure. Additionally, each $f$ and $g$ are bijections of the interval with itself and so for any other measurable functions $h(x)$ and $j(x)$, we have $hcirc f$ and $jcirc g$ independent random variables. What's more, $hcirc f$ and $hcirc g$ are independent while it is not true that $h(x)$ and $j(x)$ are independent when mappings of $X$ alone.
probability random-variables
asked Jul 19 at 0:19
itheypsi
385
Sure,e.g. $f=g$ being a constant function.
– Clement C.
Jul 19 at 0:51
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For simplicity let $X$ be a uniformly distributed random variable on $[0,1]$. Does there exist a pair of measurable functions $f$ and $g$ such that $f(X)$ and $g(X)$ are independent random variables?
Building off of Parcly's answer, if $xin [0,1)$ is expressed in base 2, $x=0.b_1b_2dots_2$, then $f(x) = 0.b_1b_3b_5dots$ and $g(x) = 0.b_2b_4b_6dots$ are both limits of independant random variables so they at least measurable. In particular, $f_n = 0.b_1b_3...b_2n+1uparrow f$ and similarly $g_nuparrow g$. Thus for every $y$ and $z$,
beginequation
P(f<ytextrm and g<z)=lim P(f_n<ytextrm and g_n<z)=lim P(f_n<y)P(g_n<z)=(lim P(f_n<y))(lim P(g_n<z))=P(f<y)P(g<z)
endequation
where the first and last equalities are due to the regularity of the measure. Additionally, each $f$ and $g$ are bijections of the interval with itself and so for any other measurable functions $h(x)$ and $j(x)$, we have $hcirc f$ and $jcirc g$ independent random variables. What's more, $hcirc f$ and $hcirc g$ are independent while it is not true that $h(x)$ and $j(x)$ are independent when mappings of $X$ alone.
probability random-variables
asked Jul 19 at 0:19
itheypsi
385
For simplicity let $X$ be a uniformly distributed random variable on $[0,1]$. Does there exist a pair of measurable functions $f$ and $g$ such that $f(X)$ and $g(X)$ are independent random variables?
Building off of Parcly's answer, if $xin [0,1)$ is expressed in base 2, $x=0.b_1b_2dots_2$, then $f(x) = 0.b_1b_3b_5dots$ and $g(x) = 0.b_2b_4b_6dots$ are both limits of independant random variables so they at least measurable. In particular, $f_n = 0.b_1b_3...b_2n+1uparrow f$ and similarly $g_nuparrow g$. Thus for every $y$ and $z$,
beginequation
P(f<ytextrm and g<z)=lim P(f_n<ytextrm and g_n<z)=lim P(f_n<y)P(g_n<z)=(lim P(f_n<y))(lim P(g_n<z))=P(f<y)P(g<z)
endequation
where the first and last equalities are due to the regularity of the measure. Additionally, each $f$ and $g$ are bijections of the interval with itself and so for any other measurable functions $h(x)$ and $j(x)$, we have $hcirc f$ and $jcirc g$ independent random variables. What's more, $hcirc f$ and $hcirc g$ are independent while it is not true that $h(x)$ and $j(x)$ are independent when mappings of $X$ alone.
probability random-variables
asked Jul 19 at 0:19
itheypsi
385
edited Jul 19 at 15:29
asked Jul 19 at 0:19
itheypsi
385
asked Jul 19 at 0:19
itheypsi
385
asked Jul 19 at 0:19
itheypsi
385
385
Sure,e.g. $f=g$ being a constant function.
– Clement C.
Jul 19 at 0:51
add a comment |Â
Sure,e.g. $f=g$ being a constant function.
– Clement C.
Jul 19 at 0:51
Sure,e.g. $f=g$ being a constant function.
– Clement C.
Jul 19 at 0:51
Sure,e.g. $f=g$ being a constant function.
– Clement C.
Jul 19 at 0:51
add a comment |Â
1 Answer
1
active
oldest
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0
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For a given variate $x=0.b_1b_2b_3b_4dots_2$ from $X$, let $f(x)=b_1$ and $g(x)=b_2$. Since each bit in $x$ is independent of each other, it follows that $f(X)$ and $g(X)$ are two independent random variables with an equal probability of taking up the values 0 and 1.
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
1
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
For a given variate $x=0.b_1b_2b_3b_4dots_2$ from $X$, let $f(x)=b_1$ and $g(x)=b_2$. Since each bit in $x$ is independent of each other, it follows that $f(X)$ and $g(X)$ are two independent random variables with an equal probability of taking up the values 0 and 1.
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
1
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
add a comment |Â
up vote
0
down vote
accepted
For a given variate $x=0.b_1b_2b_3b_4dots_2$ from $X$, let $f(x)=b_1$ and $g(x)=b_2$. Since each bit in $x$ is independent of each other, it follows that $f(X)$ and $g(X)$ are two independent random variables with an equal probability of taking up the values 0 and 1.
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
1
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
For a given variate $x=0.b_1b_2b_3b_4dots_2$ from $X$, let $f(x)=b_1$ and $g(x)=b_2$. Since each bit in $x$ is independent of each other, it follows that $f(X)$ and $g(X)$ are two independent random variables with an equal probability of taking up the values 0 and 1.
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
For a given variate $x=0.b_1b_2b_3b_4dots_2$ from $X$, let $f(x)=b_1$ and $g(x)=b_2$. Since each bit in $x$ is independent of each other, it follows that $f(X)$ and $g(X)$ are two independent random variables with an equal probability of taking up the values 0 and 1.
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
edited Jul 19 at 1:12
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
answered Jul 19 at 0:30
Parcly Taxel
33.6k136588
33.6k136588
1
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
add a comment |Â
1
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
1
1
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
More simply, let $f(x) = b_1$ and $g(x) = b_2$.
– Mike Earnest
Jul 19 at 1:00
add a comment |Â
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Sure,e.g. $f=g$ being a constant function.
– Clement C.
Jul 19 at 0:51