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Standard Deviation in terms of Probability Function

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    I've still yet to learn Latex since I'm pretty good with words equation editor, so here's the question typed out in words.
    P4fFhXl.jpg

    2. Relevant equations
    I really don't know what to do here.


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 27, 2013 #2

    Simon Bridge

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    Have you tried starting by expanding out the brackets in the sum?
     
  4. Jan 27, 2013 #3
    I did:

    3waloSA.jpg
     
  5. Jan 27, 2013 #4

    Simon Bridge

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    There's a 1/N outside the sum.
     
  6. Jan 27, 2013 #5
    But I can't really see how to turn mu^2/n into P(x_j)
     
  7. Jan 27, 2013 #6

    Simon Bridge

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    $$\frac{1}{N}\sum_{i=1}^N \mu^2 = ?$$
    ... though you seem to be changing between n and N in the sums.
     
  8. Jan 27, 2013 #7
    Yeah, N is the number of trials, and n is the number of possible outcomes from each trial. and also, considering mu is also a summation, I don't really know how to calculate mu^2.
     
  9. Jan 27, 2013 #8

    Ray Vickson

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    The expression
    [tex] \mu = \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^N x_i [/tex]
    is correct, but the expression
    [tex] \mu = \lim_{N \to \infty} \sum_{j=1}^n x_j P(x_j)[/tex] is false. Do you see why?
     
  10. Jan 27, 2013 #9

    Simon Bridge

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    So what is ##\mu## the mean of?
     
  11. Jan 27, 2013 #10

    Ray Vickson

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    A lot of your difficulties are, I suspect, a result of using bad notation, so first, let's change it to something sensible. You have a random sample ##X_1, X_2, \ldots, X_N## drawn independently from a distribution ##\{ v_j, P(v_j), \: j = 1,2,\ldots n \}##. That is the possible values of each ##X_i ## are ##v_1, v_2, \ldots, v_n## with probabilities ##P(v_1), P(v_2), \ldots, P(v_n).## Now by definition, ##\mu = \sum_{j=1}^n P(v_j) v_j ## and ##\sigma^2 = \sum_{j=1}^n P(v_j)(v_j - \mu)^2.## Note that there are no N's or limits, or anything like that in these two expressions!

    You are being asked to show that μ and σ2 are also given by the limiting expressions wiritten in the question. (To be technical, these are "almost-sure" equalities, but never mind than for now.)
     
    Last edited: Jan 27, 2013
  12. Jan 27, 2013 #11
    μ is supposed to be the average value of measured x.

    And also, the question is given to us like that, but I understand what you mean since I do believe that's what it should be. And I've been wondering about the N and limit myself since at the end there are no N values in the expression.
     
  13. Jan 27, 2013 #12

    Ray Vickson

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    No, ##\mu## is not supposed to be the average of the measured x; it is supposed to be the mean of the distribution---given by the formula in my previous post.

    Yes, there are no N values and limits in the end, because what you are being asked to show is the the large-N limit equals something, and that 'something' is just a number, not a function of N. For example, ##\lim_{N \to \infty} 1 + (1/N) = 1,## and at that point the '1' does not have any N's in it, or any 'lim', or anything like that: it is just the number '1'.
     
  14. Jan 27, 2013 #13
    Ok I realize this. But I'm still not sure how to move on. For one thing,
    xrR9jUA.jpg
     
  15. Jan 27, 2013 #14

    Ray Vickson

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    The number ##\mu^2## is just the square of the number ##\mu##---no, I am not kidding! It is a number, and has nothing at all to do with ##X_1, X_2, \ldots, X_N##.
     
  16. Jan 27, 2013 #15
    This is where I am currently stuck at:
    CR7J0NY.jpg

    If μ is just a number then I don't see a way to move on.
     
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