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" (*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info at wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 5481, 158]*) (*NotebookOutlinePosition[ 6319, 185]*) (* CellTagsIndexPosition[ 6275, 181]*) (*WindowFrame->Normal*) Notebook[{ 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\)\)\)\)\)], "DisplayFormula"], Cell["is", "Text"], Cell[BoxData[ \(TraditionalForm \`\[ExponentialE]\^\(- \(\[Sum]\+\(i = 1\)\%n \((x - \[Mu]\_i)\)\^2\/\(2\ \[Sigma]\_i\%2\)\)\)\/\(\((2\ \[Pi]) \)\^\(n/2\)\ \(\[Product]\+\(i = 1\)\%n \[Sigma]\_i\)\)\)], "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\)\)], "DisplayFormula"], Cell["Using pattern-matching, we complete the square", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm \`\(x\^2\/2\) \(\[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\) /. a_\ x\^2 + b_\ x + c_ \[Rule] a\ \((x + b\/\(2 a\))\)\^2 - b\^2\/\(4 a\) + c\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(-\(\((\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2)\)\^2\/\(2\ \(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2\)\)\)\) + 1\/2\ \((\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2)\)\ \((x - \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2 \)\/\(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2\))\)\^2 + 1\/2\ \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\%2\/\[Sigma]\_i\%2\)\)], "Output"] }, Open ]], Cell["Hence the general product in simplest terms reads", "Text"], Cell[BoxData[ \(TraditionalForm \`\(\[ExponentialE]\^\(\((\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2) \)\^2\/\(2\ \(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2\)\) - 1\/2\ \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\%2\/\[Sigma]\_i\%2 \)\)\/\(\((2\ \[Pi])\)\^\(n/2\)\ \(\[Product]\+\(i = 1\)\%n \[Sigma]\_i\)\)\) \[ExponentialE]\^\(\(-\(1\/2\)\)\ \((\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2)\)\ \((x - \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2 \)\/\(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2\))\)\^2 \)\)], "Input"], Cell[TextData[{ "Comparing the last exponential term with ", Cell[BoxData[ \(TraditionalForm \`\[ExponentialE]\^\(-\(\((x - \[Mu])\)\^2\/\(2\ \[Sigma]\^2\)\)\)\)]], " we see that" }], "Text"], Cell[BoxData[ \(TraditionalForm \`\(\[Sum]\+\(i = 1\)\%n 1\/\[Sigma]\_i\%2 \[Rule] 1\/\[Sigma]\^2, \)\)], "DisplayFormula"], Cell["and", "Text"], Cell[BoxData[ \(TraditionalForm \`\(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\/\[Sigma]\_i\%2 \[Rule] \[Mu]\/\[Sigma]\^2, \)\)], "DisplayFormula"], Cell["and hence the product can be written as", "Text"], Cell[BoxData[ \(TraditionalForm \`\(\[ExponentialE]\^\(\(1\/2\) \[Mu]\^2\/\[Sigma]\^2 - 1\/2\ \(\[Sum]\+\(i = 1\)\%n \[Mu]\_i\%2\/\[Sigma]\_i\%2 \)\)\/\(\((2\ \[Pi])\)\^\(n/2\)\ \(\[Product]\+\(i = 1\)\%n \[Sigma]\_i\)\)\) \(\[ExponentialE]\^\(- \(\((x - \[Mu])\)\^2\/\(2\ \[Sigma]\^2\)\)\)\[CenterDot]\)\)], "DisplayFormula"], Cell["\<\ Note that, in general, the product is no longer correctly \ normalized.\ \>", "Text"] }, FrontEndVersion->"Macintosh 3.0", ScreenRectangle->{{0, 800}, {0, 580}}, WindowSize->{795, 532}, WindowMargins->{{Automatic, 1}, {Automatic, 1}}, MacintoshSystemPageSetup->"\<\ 00<0001804P000000c at 2@?oeooH3?`990dL5N`?P0080001804P000000`d26P01 0000I00000400 at 410?l00BL?00400@0000000000000006P801T1T0000000h000 00000000007oo`00003oo`000o800000\>" ] (*********************************************************************** Cached data follows. 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