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Standard Deviation

  1. Jun 15, 2007 #1
    Hi all.

    If I have a function V(x1,x2,x3,x4) and I want to calculate it's standard deviation when x1,x2,x3,x4 are standard normal and their variances are small, then formula (8) on this page
    http://www.devicelink.com/mem/archive/99/09/003.html is an approximation to the standard deviation. Can anyone offer me a proof or tell me where I can read more about this formula?

    -Anders
     
  2. jcsd
  3. Jun 15, 2007 #2

    EnumaElish

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  4. Jun 15, 2007 #3
    Hi.

    Hmm, I think it's the clever part I don't understand then. Must I taylor expand V and then compute the standard deviation? And also, I can't see the connection between "difference" and "variance". I'd appreciate a few formulas. Thanks.

    -Anders
     
  5. Jun 15, 2007 #4

    EnumaElish

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    I thought it would follow from dV = Sum[(partial V/partial xi) dxi, i=1,2,3,4]. The "d" operator is similar to "deviation" (e.g., from the mean). Let's say you have only 2 x's. Then dv = (Dv/Dx1) dx1 + (Dv/Dx2) dx2 ==> dv^2 = (Dv/Dx1)^2 dx1^2 + (Dv/Dx2)^2 dx2^2 + ignored term* approx. equal to (Dv/Dx1)^2 dx1^2 + (Dv/Dx2)^2 dx2^2. (I've used capital D for "partial.")

    In fact, you don't even need the link I posted.

    *This is the interaction term 2(Dv/Dx1)(Dv/Dx2) dx1 dx2. If you think that for a given "random draw" either of dx1 or dx2 (but not necessarily both, and you don't know which) is likely to be "very small" then you can assume 2(Dv/Dx1)(Dv/Dx2) dx1 dx2 = 0.
     
    Last edited: Jun 16, 2007
  6. Jun 15, 2007 #5
    Okay. The only part I don't understand now is why dx is the same as a standard deviation ?
     
  7. Jun 15, 2007 #6

    EnumaElish

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    Last edited: Jun 15, 2007
  8. Jun 17, 2007 #7
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