How to Derive Normal Probability Function

The normal probability function, also called the Gaussian distribution after its originator, Johann Carl Friedrich Gauss, is invaluable to the study of statistics. If you have just two measurements, the mean of the data, mu, and the variance, sigma^2, this function allows you to find out what portion of the data falls under a given part of the curve. The function is: f(x) = (1/sqrt(2*pi*sigma^2))*e^(-(x-mu)^2/(2*sigma^2)). You can derive it if you are familiar with basic rules for taking derivatives.

Instructions

    • 1

      Ignore the constant out front for the time being (1/sqrt(2*pi*sigma^2)). It doesn't affect the process of taking the derivative, so wait until the end to put it back in.

    • 2

      Take the derivative of the outside function first using the chain rule, and then multiply by the derivative of the nested function. Recall that the derivative of e^x is ee^x, so the derivative of e^(-(x-mu)^2/(2*sigma^2)) is e^(-(x-mu)^2/(2*sigma^2)) times the derivative of -(x-mu)^2/(2*sigma^2).

    • 3

      Take the derivative of -(x-mu)^2/(2*sigma^2). X is the only variable here, and it is acted upon by a power, so use the power rule. Take the power, 2, and put it out front, then reduce the power by 1. As the power is now 2 -1 = 1, we no longer need to write a power. Note that you can use the chain rule again on the nested expression (x-mu), but the derivative of this expression is 1, so you would only end up multiplying everything by 1 anyway. Result: -2*(x-mu)/(2*sigma^2)*e^(-(x-mu)^2/(2*sigma^2)).

    • 4

      Return the constant we ignored to its rightful place at the front of the equation, and you have the final result of deriving the normal probability function:

      (1/sqrt(2*pi*sigma^2))*(-2)*(x-mu)/(2*sigma^2)*e^(-(x-mu)^2/(2*sigma^2))

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