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Completing The Square

  1. Jun 16, 2007 #1
    Hi all

    A basic question here:

    I am deriving the equation for the expectation of a lognormal variable. This involves rearranging the contents of the integral

    [itex]
    \int_{-\infty}^{+\infty} e^x e^{-(x-\mu)^2/2\sigma^2}dx
    [/itex]

    A proof I have seen completes the square like so:

    [itex]
    x-\frac{(x-\mu)^2}{2\sigma^2} = \frac{2\sigma^2x-(x-\mu)^2}{2\sigma^2}
    [/itex]

    [itex]
    = \frac{(x-(\mu+\sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2}
    [/itex]

    So, trying this (ignoring the 2\*sigma^2 denominator for now):

    [itex]
    2\sigma^2x-(x-\mu)^2 = 2\sigma^2x-(x^2-2\mu x +\mu^2)
    [/itex]

    [itex]
    =-x^2+(2\mu+2\sigma^2)x-\mu^2
    [/itex]

    [itex]
    \Rightarrow x^2-(\mu+2\sigma^2)x =\mu^2
    [/itex]

    [itex]
    x^2-(2\mu+2\sigma^2)x+(-\mu-\sigma^2)^2 = -\mu^2 + (-\mu-\sigma^2)^2
    [/itex]

    When attempt to express the LHS as a square:

    [itex]
    (x-(\mu+\sigma^2))^2 = -\mu^2 + (\mu^2 + 2\sigma^2\mu + \sigma^4)
    [/itex]

    [itex]
    \Rightarrow (x-(\mu+\sigma^2))^2 = 2\sigma^2\mu+\sigma^4)
    [/itex]

    Bringing over the RHS terms, and factoring in the 2*sigma^2 denominator:

    [itex]
    \Rightarrow \frac{(x-(\mu+\sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2} = 0
    [/itex]

    [ok , got it now. thanks to the poster below]
     
    Last edited: Jun 16, 2007
  2. jcsd
  3. Jun 16, 2007 #2

    arildno

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    Dearly Missed

    1st line mistake: When expanding the square parenthesis into the full parenthesis, there shall be a + in front of [itex]\mu^{2}[/itex], not - as you have done.
     
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