Substituting variables for probability distributions
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Let's say you have some probability distribution, $p(x)$. Then, let's say that $x = g(y)$, where $g$ is one to one and monotonic. You then want to find a probability distribution in terms of $y$. I don't understand why you cannot simply do $f(y) = p(g(y))$ to make this substitution and what goes wrong. I understand you need some stretching factor for integrals to evaluate to the same thing, but it astounds me that the $argmax(f(y))$ is NOT $g(argmax(p(x))$. Can someone please explain why this is wrong? If $p(x)$ is maximum for some $x_0$, why wouldn't a probability distribution in terms of $y$ be maximal at $g^{-1}(x_0)$?
probability probability-theory statistics probability-distributions
asked Jan 18 at 3:25
swedishfishedswedishfished
787
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add a comment |
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Let's say you have some probability distribution, $p(x)$. Then, let's say that $x = g(y)$, where $g$ is one to one and monotonic. You then want to find a probability distribution in terms of $y$. I don't understand why you cannot simply do $f(y) = p(g(y))$ to make this substitution and what goes wrong. I understand you need some stretching factor for integrals to evaluate to the same thing, but it astounds me that the $argmax(f(y))$ is NOT $g(argmax(p(x))$. Can someone please explain why this is wrong? If $p(x)$ is maximum for some $x_0$, why wouldn't a probability distribution in terms of $y$ be maximal at $g^{-1}(x_0)$?
probability probability-theory statistics probability-distributions
asked Jan 18 at 3:25
swedishfishedswedishfished
787
$endgroup$
$begingroup$
Relevant: math.stackexchange.com/questions/7605/…
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– nathan.j.mcdougall
Jan 18 at 3:30
add a comment |
$begingroup$
Let's say you have some probability distribution, $p(x)$. Then, let's say that $x = g(y)$, where $g$ is one to one and monotonic. You then want to find a probability distribution in terms of $y$. I don't understand why you cannot simply do $f(y) = p(g(y))$ to make this substitution and what goes wrong. I understand you need some stretching factor for integrals to evaluate to the same thing, but it astounds me that the $argmax(f(y))$ is NOT $g(argmax(p(x))$. Can someone please explain why this is wrong? If $p(x)$ is maximum for some $x_0$, why wouldn't a probability distribution in terms of $y$ be maximal at $g^{-1}(x_0)$?
probability probability-theory statistics probability-distributions
asked Jan 18 at 3:25
swedishfishedswedishfished
787
$endgroup$
Let's say you have some probability distribution, $p(x)$. Then, let's say that $x = g(y)$, where $g$ is one to one and monotonic. You then want to find a probability distribution in terms of $y$. I don't understand why you cannot simply do $f(y) = p(g(y))$ to make this substitution and what goes wrong. I understand you need some stretching factor for integrals to evaluate to the same thing, but it astounds me that the $argmax(f(y))$ is NOT $g(argmax(p(x))$. Can someone please explain why this is wrong? If $p(x)$ is maximum for some $x_0$, why wouldn't a probability distribution in terms of $y$ be maximal at $g^{-1}(x_0)$?
probability probability-theory statistics probability-distributions
probability probability-theory statistics probability-distributions
asked Jan 18 at 3:25
swedishfishedswedishfished
787
asked Jan 18 at 3:25
swedishfishedswedishfished
787
asked Jan 18 at 3:25
swedishfishedswedishfished
787
asked Jan 18 at 3:25
swedishfishedswedishfished
787
asked Jan 18 at 3:25
swedishfishedswedishfished
787
787
$begingroup$
Relevant: math.stackexchange.com/questions/7605/…
$endgroup$
– nathan.j.mcdougall
Jan 18 at 3:30
add a comment |
$begingroup$
Relevant: math.stackexchange.com/questions/7605/…
$endgroup$
– nathan.j.mcdougall
Jan 18 at 3:30
$begingroup$
Relevant: math.stackexchange.com/questions/7605/…
$endgroup$
– nathan.j.mcdougall
Jan 18 at 3:30
$begingroup$
Relevant: math.stackexchange.com/questions/7605/…
$endgroup$
– nathan.j.mcdougall
Jan 18 at 3:30
add a comment |
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Relevant: math.stackexchange.com/questions/7605/…
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– nathan.j.mcdougall
Jan 18 at 3:30