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Math Help - Deriving the formula for the normal distribution (exponential derivative)

  1. #1
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    Deriving the formula for the normal distribution (exponential derivative)

    Hi everyone,

    I'm trying to get my head around the formula for the normal distribution:

    \dfrac{1}{\sigma \sqrt(2\pi)}  e^-{\dfrac{(x-\mu)^2}{2\sigma^2}}

    where:

    \dfrac{1}{\sigma \sqrt(2\pi)}

    Should be a normalising constant so that the area below the curve is always equal to 1. If I am not mistaken, this normalising constant is the derivative of the exponential term. I've looked at the rules for taking derivatives and seem to be unable to get this as an answer for the derivative of the exponential.

    In specific, I fail to see where the square root and pi come from..

    I was hoping someone here could give me a clue. Many thanks in advance!

    Arthur

    Can anyone give me
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Arthur View Post
    Hi everyone,

    I'm trying to get my head around the formula for the normal distribution:

    \dfrac{1}{\sigma \sqrt(2\pi)}  e^-{\dfrac{(x-\mu)^2}{2\sigma^2}}

    where:

    \dfrac{1}{\sigma \sqrt(2\pi)}

    Should be a normalising constant so that the area below the curve is always equal to 1. If I am not mistaken, this normalising constant is the derivative of the exponential term.
    You are mistaken. The "normalising" constant is chosen to "normalise" the probability function- to make the probability of getting any number whatsoever (and we must get some number) equal to 1. It has nothing to with the derivative but is equal to an integral. It is one over \int_{-\infty}^\infty e^{-\frac{(x-\mu)^2}{2\sigma^2}}

    I've looked at the rules for taking derivatives and seem to be unable to get this as an answer for the derivative of the exponential.

    In specific, I fail to see where the square root and pi come from..

    I was hoping someone here could give me a clue. Many thanks in advance!

    Arthur

    Can anyone give me
    Follow Math Help Forum on Facebook and Google+

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