- #1
MaartenB
- 1
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I want to know how to treat propagation of errors in general.
When the transformation of variables is a transformation of $R^n$ to $R^n$ it
simply involves a jacobian:
[tex]g(\vec{y}) = f(x(\vec{y}))|J|[/tex]
With
[tex]J_{ij} = \frac{\partial x_i}{\partial y_j} [/tex]
(see http://pdg.lbl.gov/2005/reviews/probrpp.pdf for instance)
But there are also situation of $R^n$ to $R^m$ possible.
This is how far I got:
(Much of this can be found in http://arxiv.org/abs/hep-ex/0002056 appendix A)
The characteristic function (CF) is defined as:
[tex]\phi_X(t) = E[e^{itX}] = \int e^{itx} f(x) dx[/tex]
The inverse (FT):
[tex]f(x) = \frac{1}{2\pi} \int e^{-itX} \phi_X(t) dt[/tex]
If we have a function Y=g(X) the pdf according to Kendal and Stuart (1943) is:
[tex]\phi_Y(t) = \int e^{itg(x)} f(x) dx[/tex]
Taking the inverse, and some rewriting
[tex]f(y) = \frac{1}{2\pi} \int e^{-ity} \phi_Y(t) dt = \int f(x)\delta(y-g(x))dx[/tex]
In vector form
[tex]f(y) = \int f(\vec{x})\delta(y-g(\vec{x}))d\vec{x}[/tex]
Take for instance two independent variables taken from the same distribution:
[tex]Y = g(\vec{X}) = g(X_1, X_2) = X_1 + X_2[/tex]
Then the resulting pdf is with:
[tex]f(y) = \int f(x_1,x_2)\delta(y-x_1-x_2)dx_1 dx_2 = \int f(x_1) f(y - x_1 - u)\delta(u)dx_1 du = \int f(x_1) f(y - x_1) dx_1 [/tex]
Using:
[tex]u = y - x_1 - x_2 [/tex]
[tex]x_2 = y - x_1 - u [/tex]
[tex]dx_2 = du [/tex]
Which is a simply a convolution, which is a known result,
see for instance: http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables
Or for $R^1$ to $R^1$:
[tex]f(y) = \int f(x)\delta(y-g(x))dx = \int f(g^{-1}(y-u))\frac{1}{g'(x)}\delta(u)du = f(g^{-1}(y))\frac{1}{g'(x)} = f(x(y)) \frac{dx}{dy}[/tex]
Using:
[tex]u = y - g(x)[/tex]
[tex]du = \frac{dg(x)}{dx}dx[/tex]
[tex]dx = \frac{1}{g'(x)}du[/tex]
[tex]x = g^{-1}(y-u)[/tex]
Also a general result, just a change of variable.
But how to do this for the case of $R^n$ to $R^m$?, I don't know how to evaluate
the dirac delta function in general.
I did find on wikipedia (http://en.wikipedia.org/wiki/Dirac_delta_function)
[tex] \int_V f(\mathbf{r}) \, \delta(g(\mathbf{r})) \, d^nr = \int_{\partial V}\frac{f(\mathbf{r})}{|\mathbf{\nabla}g|}\,d^{n-1}r [/tex]
But I couldn't find a reference where this is explained.
But then again, if y is a vector, how to solve this?:
[tex]f(\vec{y}) = \int f(\vec{x})\vec{\delta}(\vec{y}-g(\vec{x}))d\vec{x}[/tex]
Anyone got some hints? Or am I going the wrong direction with this?
When the transformation of variables is a transformation of $R^n$ to $R^n$ it
simply involves a jacobian:
[tex]g(\vec{y}) = f(x(\vec{y}))|J|[/tex]
With
[tex]J_{ij} = \frac{\partial x_i}{\partial y_j} [/tex]
(see http://pdg.lbl.gov/2005/reviews/probrpp.pdf for instance)
But there are also situation of $R^n$ to $R^m$ possible.
This is how far I got:
(Much of this can be found in http://arxiv.org/abs/hep-ex/0002056 appendix A)
The characteristic function (CF) is defined as:
[tex]\phi_X(t) = E[e^{itX}] = \int e^{itx} f(x) dx[/tex]
The inverse (FT):
[tex]f(x) = \frac{1}{2\pi} \int e^{-itX} \phi_X(t) dt[/tex]
If we have a function Y=g(X) the pdf according to Kendal and Stuart (1943) is:
[tex]\phi_Y(t) = \int e^{itg(x)} f(x) dx[/tex]
Taking the inverse, and some rewriting
[tex]f(y) = \frac{1}{2\pi} \int e^{-ity} \phi_Y(t) dt = \int f(x)\delta(y-g(x))dx[/tex]
In vector form
[tex]f(y) = \int f(\vec{x})\delta(y-g(\vec{x}))d\vec{x}[/tex]
Take for instance two independent variables taken from the same distribution:
[tex]Y = g(\vec{X}) = g(X_1, X_2) = X_1 + X_2[/tex]
Then the resulting pdf is with:
[tex]f(y) = \int f(x_1,x_2)\delta(y-x_1-x_2)dx_1 dx_2 = \int f(x_1) f(y - x_1 - u)\delta(u)dx_1 du = \int f(x_1) f(y - x_1) dx_1 [/tex]
Using:
[tex]u = y - x_1 - x_2 [/tex]
[tex]x_2 = y - x_1 - u [/tex]
[tex]dx_2 = du [/tex]
Which is a simply a convolution, which is a known result,
see for instance: http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables
Or for $R^1$ to $R^1$:
[tex]f(y) = \int f(x)\delta(y-g(x))dx = \int f(g^{-1}(y-u))\frac{1}{g'(x)}\delta(u)du = f(g^{-1}(y))\frac{1}{g'(x)} = f(x(y)) \frac{dx}{dy}[/tex]
Using:
[tex]u = y - g(x)[/tex]
[tex]du = \frac{dg(x)}{dx}dx[/tex]
[tex]dx = \frac{1}{g'(x)}du[/tex]
[tex]x = g^{-1}(y-u)[/tex]
Also a general result, just a change of variable.
But how to do this for the case of $R^n$ to $R^m$?, I don't know how to evaluate
the dirac delta function in general.
I did find on wikipedia (http://en.wikipedia.org/wiki/Dirac_delta_function)
[tex] \int_V f(\mathbf{r}) \, \delta(g(\mathbf{r})) \, d^nr = \int_{\partial V}\frac{f(\mathbf{r})}{|\mathbf{\nabla}g|}\,d^{n-1}r [/tex]
But I couldn't find a reference where this is explained.
But then again, if y is a vector, how to solve this?:
[tex]f(\vec{y}) = \int f(\vec{x})\vec{\delta}(\vec{y}-g(\vec{x}))d\vec{x}[/tex]
Anyone got some hints? Or am I going the wrong direction with this?