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MathGroup Archive 1998

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Re: Problem to be solved (Product of Normal Distributions)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14166] Re: Problem to be solved (Product of Normal Distributions)
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Wed, 30 Sep 1998 02:04:15 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

>Imagine I have something like ax^2+bx+c. Now I want to know how to make
>Mathematica return me something in the form (x+d)^2, where d depends on
>a, b and c.
>
>This is the simple form. In reality I am trying to determine the sigma
>and mu of a product of the form
>
>        f_i=1/(sqrt(2 pi) sigma_i) Exp( -(x-mu_i)^2/(2 sigma_i^2)  )
>
>So I want to know what is the sigma and mu of f_1 times f_2.
>
>Now that you know it, I can also add that I tried Collect in every way,
>and some substituion rules.

Attached is one (semi-automatic) approach using pattern-matching and
taking advantage of Mathematica's typesetting.

Cheers,
	Paul

"NormalDistribution.nb"

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Cell[TextData[{
  "The product of ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " (normalized) normal distributions," }], "Text"],

Cell[BoxData[
    \(TraditionalForm\`\(\
    \(1\/\(\ at \(2\ \[Pi]\)\ \[Sigma]\_i\)\)
      \[ExponentialE]\^\(-
          \(\((x - \[Mu]\_i)\)\^2\/\(2\ \[Sigma]\_i\%2\)\)\)\)\)],
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          \(\[Sum]\+\(i = 1\)\%n
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  "DisplayFormula"],

Cell["The (negative) argument of the exponential function is", "Text"],

Cell[BoxData[
    \(TraditionalForm
    \`\[Sum]\+\(i = 1\)\%n\((x - \[Mu]\_i)\)\^2\/\(2\ \[Sigma]\_i\%2\)
      \[Equal] \(x\^2\/2\) \(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2\) -
        x \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2\) +
        \(1\/2\) \(\[Sum]\+\(i = 1\)\%n
\[Mu]\_i\%2\/\[Sigma]\_i\%2\)\)],
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Cell["Using pattern-matching, we complete the square", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(TraditionalForm
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        x \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2\) +
        \(1\/2\) \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\%2\/\[Sigma]\_i\%2\) /.
      a_\ x\^2 + b_\ x + c_ \[Rule]
        a\ \((x + b\/\(2  a\))\)\^2 - b\^2\/\(4  a\) + c\)], "Input"],

Cell[BoxData[
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        \((x - \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2
                      \)\/\(\[Sum]\+\(i = 1\)\%n
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      1\/2\ \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\%2\/\[Sigma]\_i\%2\)\)],
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Cell["Hence the general product in simplest terms reads", "Text"],

Cell[BoxData[
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                    \)\^2\/\(2\
                  \(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2\)\) -
              1\/2\ \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\%2\/\[Sigma]\_i\%2
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            \(\[Product]\+\(i = 1\)\%n \[Sigma]\_i\)\)\)
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          \((\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2)\)\
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    \(TraditionalForm
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Cell[BoxData[
    \(TraditionalForm
    \`\(\[ExponentialE]\^\(\(1\/2\) \[Mu]\^2\/\[Sigma]\^2 -
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                      \)\)\/\(\((2\ \[Pi])\)\^\(n/2\)\
            \(\[Product]\+\(i = 1\)\%n \[Sigma]\_i\)\)\)
      \(\[ExponentialE]\^\(-
            \(\((x - \[Mu])\)\^2\/\(2\
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Note that, in general, the product is no longer correctly \ normalized.\
\>", "Text"]
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