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Homework Help: The mean of the Probability Density Function

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data
    The mean of a function is as follows:
    $${1 \over {a - b}}\int_b^a {f(x)\,dx} $$

    So why is the mean of the PDF as follows:
    $$\int_{ - \infty }^\infty {xf(x)\,dx} $$

    I thought it would have been this way:
    $$\lim \,b \to - \infty \,{1 \over { - b}}\int_b^0 {f(x)\,dx\,} + \,\,\lim \,a \to \infty {1 \over a}\int_0^a {f(x)\,dx = 0 + 0 = 0} $$
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 14, 2012 #2

    I like Serena

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    Hi p75213! :smile:

    The difference is that the mean of the function is an average of y-values.
    The mean of the PDF is a weighed average of x-values.
  4. Nov 14, 2012 #3

    Ray Vickson

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    The integral $$\int_{ - \infty }^\infty {xf(x)\,dx} $$ is not the "mean of the pdf"; it is the mean of the random variable X for which f is the density function. Sometimes you will see authors use rather sloppy language and say something like "let μ be the mean of the pdf", but when they say that they do not mean it literally: they mean that μ is the mean corresponding to the pdf.

  5. Nov 14, 2012 #4
    Thanks guys. I've done some more reading and investigation. Rather than the mean I prefer to think of it as "expected value".
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