Slides06-2010

# Slides06-2010 - Derivation of the Normal Distribution...

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Derivation of the Normal Distribution: Lecture VI Charles B. Moss September 2, 2010 Charles B. Moss () Derivation of the Normal Distribution: Lecture VI September 2, 2010 1 / 16

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Outline 1 Derivation of the Normal Distribution Function 2 Expected Values Charles B. Moss () Derivation of the Normal Distribution: Lecture VI September 2, 2010 2 / 16
Derivation of the Normal Distribution Function Derivation of the Normal Distribution Function The order of proof of the normal distribution function is to start with the standard normal: f ( x )= 1 2 π e x 2 2 (1) First, we need to demonstrate that the distribution function does integrate to one over the entire sample space, which is −∞ to .Th i s is typically accomplished by proving the constant. Let us start by assuming that I = Z −∞ e y 2 2 dy (2) Squaring this expression yields I 2 = Z −∞ e y 2 2 dy Z −∞ e x 2 2 dx Z y 2 + x 2 (3) Charles B. Moss () Derivation of the Normal Distribution: Lecture VI September 2, 2010 3 / 16

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Derivation of the Normal Distribution Function The trick to this integration is changing the variables into a polar form. Polar Integration : The notion of polar integration is basically one of a change in variables. SpeciFcally, some integrals may be ill-posed in the traditional Cartesian plane, but easily solved in a polar space.
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## This note was uploaded on 07/15/2011 for the course AEB 6180 taught by Professor Staff during the Spring '10 term at University of Florida.

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Slides06-2010 - Derivation of the Normal Distribution...

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