Law of Unconscious statistician - Proof on wiki wrong?
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So I don't understand the last part of how wikipedia proves the LOTUS theorem. Here is the link.
After we prove that,
$F_Y(y) = F_X(g^{-1}(y))$
It says by chain rule we have,
$F_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
But didn't we get this by differentiating. This should be,
$dF_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
Then the second last equation would be
$displaystyleint_{-infty}^{infty}g(x) f_X(x) dx = displaystyleint_{-infty}^{infty} y,. dF_Y(y).dy$
Which makes sense. By definition the expected value of a function of random variable should be equal to the integral of the product of the value of the function of the random variable i.e. $y$ and the PDF of the function of the random variable $f_Y(y)$.
statistics probability-distributions expected-value
asked Nov 14 at 8:43
gallickgunner
1155
add a comment |
up vote
0
down vote
favorite
So I don't understand the last part of how wikipedia proves the LOTUS theorem. Here is the link.
After we prove that,
$F_Y(y) = F_X(g^{-1}(y))$
It says by chain rule we have,
$F_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
But didn't we get this by differentiating. This should be,
$dF_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
Then the second last equation would be
$displaystyleint_{-infty}^{infty}g(x) f_X(x) dx = displaystyleint_{-infty}^{infty} y,. dF_Y(y).dy$
Which makes sense. By definition the expected value of a function of random variable should be equal to the integral of the product of the value of the function of the random variable i.e. $y$ and the PDF of the function of the random variable $f_Y(y)$.
statistics probability-distributions expected-value
asked Nov 14 at 8:43
gallickgunner
1155
Yes, the wiki page has mistakes.
– Kavi Rama Murthy
Nov 14 at 8:47
Ok thanks just wanted to confirm it.
– gallickgunner
Nov 14 at 13:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I don't understand the last part of how wikipedia proves the LOTUS theorem. Here is the link.
After we prove that,
$F_Y(y) = F_X(g^{-1}(y))$
It says by chain rule we have,
$F_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
But didn't we get this by differentiating. This should be,
$dF_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
Then the second last equation would be
$displaystyleint_{-infty}^{infty}g(x) f_X(x) dx = displaystyleint_{-infty}^{infty} y,. dF_Y(y).dy$
Which makes sense. By definition the expected value of a function of random variable should be equal to the integral of the product of the value of the function of the random variable i.e. $y$ and the PDF of the function of the random variable $f_Y(y)$.
statistics probability-distributions expected-value
asked Nov 14 at 8:43
gallickgunner
1155
So I don't understand the last part of how wikipedia proves the LOTUS theorem. Here is the link.
After we prove that,
$F_Y(y) = F_X(g^{-1}(y))$
It says by chain rule we have,
$F_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
But didn't we get this by differentiating. This should be,
$dF_Y(y) = f_X(g^{-1}(y)) * 1/g^{'}(g^{-1}(y))$
Then the second last equation would be
$displaystyleint_{-infty}^{infty}g(x) f_X(x) dx = displaystyleint_{-infty}^{infty} y,. dF_Y(y).dy$
Which makes sense. By definition the expected value of a function of random variable should be equal to the integral of the product of the value of the function of the random variable i.e. $y$ and the PDF of the function of the random variable $f_Y(y)$.
statistics probability-distributions expected-value
statistics probability-distributions expected-value
asked Nov 14 at 8:43
gallickgunner
1155
asked Nov 14 at 8:43
gallickgunner
1155
asked Nov 14 at 8:43
gallickgunner
1155
asked Nov 14 at 8:43
gallickgunner
1155
asked Nov 14 at 8:43
gallickgunner
1155
1155
Yes, the wiki page has mistakes.
– Kavi Rama Murthy
Nov 14 at 8:47
Ok thanks just wanted to confirm it.
– gallickgunner
Nov 14 at 13:14
add a comment |
Yes, the wiki page has mistakes.
– Kavi Rama Murthy
Nov 14 at 8:47
Ok thanks just wanted to confirm it.
– gallickgunner
Nov 14 at 13:14
Yes, the wiki page has mistakes.
– Kavi Rama Murthy
Nov 14 at 8:47
Yes, the wiki page has mistakes.
– Kavi Rama Murthy
Nov 14 at 8:47
Ok thanks just wanted to confirm it.
– gallickgunner
Nov 14 at 13:14
Ok thanks just wanted to confirm it.
– gallickgunner
Nov 14 at 13:14
add a comment |
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Yes, the wiki page has mistakes.
– Kavi Rama Murthy
Nov 14 at 8:47
Ok thanks just wanted to confirm it.
– gallickgunner
Nov 14 at 13:14