1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Propagation of errors

  1. Jul 2, 2007 #1
    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]
    [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]
    [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]
    [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?
  2. jcsd
  3. Jul 3, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    When you are transforming from R^n to R^m and m < n, you can define n - m new random variables and set them equal to an identical number of the existing variables. Example: given transformation Y = X1 + X2, you can define Y1 = Y and Y2 = X2, at which point you will have a square matrix. I hope this is helpful.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?