Prove that that PDF can be “factorized”, if its log satisfies a homogeneous PDE of the second order
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The following problem is taken from "The Advanced Theory of Statistics, Kendall & Stuart, Volume 1, Second Edition, excercise 1.17:
Let $f(x,y)$ be a probability density function. Prove that the linear transformation, that transforms variables to the independent ones, exists if and only if $f(x,y)$ satisfies the following equation:
$$
left( A frac{ partial^2 }{partial x^2} +
2 H frac{ partial^2 }{partial x partial y} +
B frac{ partial^2 }{partial y^2} right) log f = 0,
$$
where $A$, $H$ and $B$ are some constants.
Variables $xi$, $phi$ are called independent if $f(xi, phi) = f_{1}(xi)f_{2}(phi)$, where $f_{1}(xi)$, $f_{2}(phi)$ are density functions.
I'd like to prove the "if" part. The most obvious way is to separate the variables, assuming $log f = u(x,y) = X(x)Y(y)$, though two quesions immidiately arise:
- Is it possible, that there are other solutions, that can't be represented in the form of $X(x)Y(y)$?
- Are there any simpler methods that don't use PDE technique? Kendall & Stuart is a statistics texbook, after all, so it's rather suspicious that it requires the knowledge of PDEs.
analysis statistics pde linear-transformations density-function
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
$endgroup$
add a comment |
$begingroup$
The following problem is taken from "The Advanced Theory of Statistics, Kendall & Stuart, Volume 1, Second Edition, excercise 1.17:
Let $f(x,y)$ be a probability density function. Prove that the linear transformation, that transforms variables to the independent ones, exists if and only if $f(x,y)$ satisfies the following equation:
$$
left( A frac{ partial^2 }{partial x^2} +
2 H frac{ partial^2 }{partial x partial y} +
B frac{ partial^2 }{partial y^2} right) log f = 0,
$$
where $A$, $H$ and $B$ are some constants.
Variables $xi$, $phi$ are called independent if $f(xi, phi) = f_{1}(xi)f_{2}(phi)$, where $f_{1}(xi)$, $f_{2}(phi)$ are density functions.
I'd like to prove the "if" part. The most obvious way is to separate the variables, assuming $log f = u(x,y) = X(x)Y(y)$, though two quesions immidiately arise:
- Is it possible, that there are other solutions, that can't be represented in the form of $X(x)Y(y)$?
- Are there any simpler methods that don't use PDE technique? Kendall & Stuart is a statistics texbook, after all, so it's rather suspicious that it requires the knowledge of PDEs.
analysis statistics pde linear-transformations density-function
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
$endgroup$
1
$begingroup$
Using a PDE is not that weird, is it? It measures the rate of change of one variable when another one changes. Which is more or less measuring dependence.
$endgroup$
– Benjamin Lindqvist
Dec 28 '18 at 21:24
add a comment |
$begingroup$
The following problem is taken from "The Advanced Theory of Statistics, Kendall & Stuart, Volume 1, Second Edition, excercise 1.17:
Let $f(x,y)$ be a probability density function. Prove that the linear transformation, that transforms variables to the independent ones, exists if and only if $f(x,y)$ satisfies the following equation:
$$
left( A frac{ partial^2 }{partial x^2} +
2 H frac{ partial^2 }{partial x partial y} +
B frac{ partial^2 }{partial y^2} right) log f = 0,
$$
where $A$, $H$ and $B$ are some constants.
Variables $xi$, $phi$ are called independent if $f(xi, phi) = f_{1}(xi)f_{2}(phi)$, where $f_{1}(xi)$, $f_{2}(phi)$ are density functions.
I'd like to prove the "if" part. The most obvious way is to separate the variables, assuming $log f = u(x,y) = X(x)Y(y)$, though two quesions immidiately arise:
- Is it possible, that there are other solutions, that can't be represented in the form of $X(x)Y(y)$?
- Are there any simpler methods that don't use PDE technique? Kendall & Stuart is a statistics texbook, after all, so it's rather suspicious that it requires the knowledge of PDEs.
analysis statistics pde linear-transformations density-function
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
$endgroup$
The following problem is taken from "The Advanced Theory of Statistics, Kendall & Stuart, Volume 1, Second Edition, excercise 1.17:
Let $f(x,y)$ be a probability density function. Prove that the linear transformation, that transforms variables to the independent ones, exists if and only if $f(x,y)$ satisfies the following equation:
$$
left( A frac{ partial^2 }{partial x^2} +
2 H frac{ partial^2 }{partial x partial y} +
B frac{ partial^2 }{partial y^2} right) log f = 0,
$$
where $A$, $H$ and $B$ are some constants.
Variables $xi$, $phi$ are called independent if $f(xi, phi) = f_{1}(xi)f_{2}(phi)$, where $f_{1}(xi)$, $f_{2}(phi)$ are density functions.
I'd like to prove the "if" part. The most obvious way is to separate the variables, assuming $log f = u(x,y) = X(x)Y(y)$, though two quesions immidiately arise:
- Is it possible, that there are other solutions, that can't be represented in the form of $X(x)Y(y)$?
- Are there any simpler methods that don't use PDE technique? Kendall & Stuart is a statistics texbook, after all, so it's rather suspicious that it requires the knowledge of PDEs.
analysis statistics pde linear-transformations density-function
analysis statistics pde linear-transformations density-function
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
edited Dec 28 '18 at 22:16
Simeon Y.
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
asked Dec 28 '18 at 21:10
Simeon Y.Simeon Y.
184
184
1
$begingroup$
Using a PDE is not that weird, is it? It measures the rate of change of one variable when another one changes. Which is more or less measuring dependence.
$endgroup$
– Benjamin Lindqvist
Dec 28 '18 at 21:24
add a comment |
1
$begingroup$
Using a PDE is not that weird, is it? It measures the rate of change of one variable when another one changes. Which is more or less measuring dependence.
$endgroup$
– Benjamin Lindqvist
Dec 28 '18 at 21:24
1
1
$begingroup$
Using a PDE is not that weird, is it? It measures the rate of change of one variable when another one changes. Which is more or less measuring dependence.
$endgroup$
– Benjamin Lindqvist
Dec 28 '18 at 21:24
$begingroup$
Using a PDE is not that weird, is it? It measures the rate of change of one variable when another one changes. Which is more or less measuring dependence.
$endgroup$
– Benjamin Lindqvist
Dec 28 '18 at 21:24
add a comment |
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$begingroup$
Using a PDE is not that weird, is it? It measures the rate of change of one variable when another one changes. Which is more or less measuring dependence.
$endgroup$
– Benjamin Lindqvist
Dec 28 '18 at 21:24