Determining independence of a joint continuous probability distribtuion
$begingroup$
I'm just confused on determining if a joint pdf has independence between it's two variables X and Y.
The theorem given to me is f(x,y) = fx(x)*fy(y) as a way to test independence, it's easy enough to understand
But another theorem was given to me as an alternative
f(x,y) = g(x)*h(y) where g(x) and h(y) are non-negative function of only x or y, respectively.
How do I determine g(x) and h(y)? Is it just a factored form of f(x,y)? And if that's the case, is independence just determined on whether or not I can factor f(x,y)?
An easy example: f(x,y) = 2y......so g(x)=y, h(y)=2? Likewise, is g(x)=4y and h(y) =.5 viable (although probably unnecessary)?
probability statistics probability-distributions
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
$endgroup$
add a comment |
$begingroup$
I'm just confused on determining if a joint pdf has independence between it's two variables X and Y.
The theorem given to me is f(x,y) = fx(x)*fy(y) as a way to test independence, it's easy enough to understand
But another theorem was given to me as an alternative
f(x,y) = g(x)*h(y) where g(x) and h(y) are non-negative function of only x or y, respectively.
How do I determine g(x) and h(y)? Is it just a factored form of f(x,y)? And if that's the case, is independence just determined on whether or not I can factor f(x,y)?
An easy example: f(x,y) = 2y......so g(x)=y, h(y)=2? Likewise, is g(x)=4y and h(y) =.5 viable (although probably unnecessary)?
probability statistics probability-distributions
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
$endgroup$
add a comment |
$begingroup$
I'm just confused on determining if a joint pdf has independence between it's two variables X and Y.
The theorem given to me is f(x,y) = fx(x)*fy(y) as a way to test independence, it's easy enough to understand
But another theorem was given to me as an alternative
f(x,y) = g(x)*h(y) where g(x) and h(y) are non-negative function of only x or y, respectively.
How do I determine g(x) and h(y)? Is it just a factored form of f(x,y)? And if that's the case, is independence just determined on whether or not I can factor f(x,y)?
An easy example: f(x,y) = 2y......so g(x)=y, h(y)=2? Likewise, is g(x)=4y and h(y) =.5 viable (although probably unnecessary)?
probability statistics probability-distributions
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
$endgroup$
I'm just confused on determining if a joint pdf has independence between it's two variables X and Y.
The theorem given to me is f(x,y) = fx(x)*fy(y) as a way to test independence, it's easy enough to understand
But another theorem was given to me as an alternative
f(x,y) = g(x)*h(y) where g(x) and h(y) are non-negative function of only x or y, respectively.
How do I determine g(x) and h(y)? Is it just a factored form of f(x,y)? And if that's the case, is independence just determined on whether or not I can factor f(x,y)?
An easy example: f(x,y) = 2y......so g(x)=y, h(y)=2? Likewise, is g(x)=4y and h(y) =.5 viable (although probably unnecessary)?
probability statistics probability-distributions
probability statistics probability-distributions
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
asked Dec 9 '18 at 23:30
ajdawgajdawg
32
32
add a comment |
add a comment |
1 Answer
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$begingroup$
If $f(x,y) =g(x)h(y)$ (with $g$ and $h$ non-negative and measurable) integrating w.r.t. $x$ and then w.r.t. $y$ we get $1=int int f(x,y)dxdy=int (int g(x) dx) h(y)dy=(int g(x) dx)(int h(y) dy)$ $,, $ (1).
Now let $g_1=frac g {int g(x)dx}$ and $h_1=frac h {int h(x)dx}$. It follows from (1) that $g_1$ and $h_1$ are both density functions and $f(x,y)=g_1(x)h_1(y)$. It follows that $g_1$ and $h_1$ are the marginals if $f$; independence follows from this.
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $f(x,y) =g(x)h(y)$ (with $g$ and $h$ non-negative and measurable) integrating w.r.t. $x$ and then w.r.t. $y$ we get $1=int int f(x,y)dxdy=int (int g(x) dx) h(y)dy=(int g(x) dx)(int h(y) dy)$ $,, $ (1).
Now let $g_1=frac g {int g(x)dx}$ and $h_1=frac h {int h(x)dx}$. It follows from (1) that $g_1$ and $h_1$ are both density functions and $f(x,y)=g_1(x)h_1(y)$. It follows that $g_1$ and $h_1$ are the marginals if $f$; independence follows from this.
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
$begingroup$
If $f(x,y) =g(x)h(y)$ (with $g$ and $h$ non-negative and measurable) integrating w.r.t. $x$ and then w.r.t. $y$ we get $1=int int f(x,y)dxdy=int (int g(x) dx) h(y)dy=(int g(x) dx)(int h(y) dy)$ $,, $ (1).
Now let $g_1=frac g {int g(x)dx}$ and $h_1=frac h {int h(x)dx}$. It follows from (1) that $g_1$ and $h_1$ are both density functions and $f(x,y)=g_1(x)h_1(y)$. It follows that $g_1$ and $h_1$ are the marginals if $f$; independence follows from this.
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
$begingroup$
If $f(x,y) =g(x)h(y)$ (with $g$ and $h$ non-negative and measurable) integrating w.r.t. $x$ and then w.r.t. $y$ we get $1=int int f(x,y)dxdy=int (int g(x) dx) h(y)dy=(int g(x) dx)(int h(y) dy)$ $,, $ (1).
Now let $g_1=frac g {int g(x)dx}$ and $h_1=frac h {int h(x)dx}$. It follows from (1) that $g_1$ and $h_1$ are both density functions and $f(x,y)=g_1(x)h_1(y)$. It follows that $g_1$ and $h_1$ are the marginals if $f$; independence follows from this.
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
If $f(x,y) =g(x)h(y)$ (with $g$ and $h$ non-negative and measurable) integrating w.r.t. $x$ and then w.r.t. $y$ we get $1=int int f(x,y)dxdy=int (int g(x) dx) h(y)dy=(int g(x) dx)(int h(y) dy)$ $,, $ (1).
Now let $g_1=frac g {int g(x)dx}$ and $h_1=frac h {int h(x)dx}$. It follows from (1) that $g_1$ and $h_1$ are both density functions and $f(x,y)=g_1(x)h_1(y)$. It follows that $g_1$ and $h_1$ are the marginals if $f$; independence follows from this.
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 10 '18 at 0:04
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
55.4k42057
add a comment |
add a comment |
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