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Mean and varince of Log(X) Where X~U[1,0]

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    find the mean and varince of Log(X) Where X~U[1,0] (X is continuous Random variable)

    2. Relevant equations
    [tex] \mathbb{E}(X) = \int_{-\infity}^{\infity}{x f_X(x)} dx [/tex]

    [tex] \mathbb{E}(X^2) = \int_{-\infity}^{\infity}{x^2 f_X(x)} dx [/tex]

    [tex] Var(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2 [/tex]

    3. The attempt at a solution
    [tex] \mathbb{E}[log(x)] = \int_0^1{xlog(x)} = \frac{-1}{4} [/tex]
    [tex] \mathbb{E}[log(x)^2] = \int_0^1{x^2log(x)} = \frac{-1}{9} [/tex]
    [tex] Var(X) = \mathbb{E}[log(x)^2] - \mathbb{E}[log(x)]^2 = \frac{-25}{144} [/tex]

    but i know variance cant be negative
     
  2. jcsd
  3. Nov 10, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    you're right, variance can't be negative

    not too sure what you've done, but i think you've assumed [itex] f_{log(X)}(log(X)=x) = log(x)[/itex] which doesn't make any sense... in fact over the given domain log(X) is negative, which doesn't make sense for a pdf, and isn't normalised

    So first, define the random variable Y, related to X by
    Y = log(X)

    you then need to find the probabilty distrubution g(y).dy. It can be related to the original f(x), by

    g(y).dy = f(x).dx
    where the dy & dx are related

    once you have that
    [tex] E[log(X)] = E[Y] = \int{g(y)}dy [/tex]

    and so on, note the lmits of the pdf of y, g(y) wll be realte to the orginal x values, but not necessarily the same
     
    Last edited: Nov 10, 2009
  4. Nov 10, 2009 #3
    i think i get it now
    let:[tex] f(x) = 1 \mbox{(the pdf for U[0,1]), }g(y) = e^y = x \Rightarrow \frac{dx}{dy} = e^y, f(g(y)) = 1 [/tex]

    this gives you [tex]f_x(x) = \int_0^1{1 dx} = \int_{log(0)}^{log(1)}{e^y dy} = \int^0_{-\infty}e^y dy[/tex]
    get the pdf fo y from that and use the def/formula for mean and varaince.

    it seems so simple thanks
     
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