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A Moments of normal distribution

  1. Jan 6, 2017 #1
    I have calculated characteristic function of normal distribution [tex]f_{X}(k)=e^{(ika-\frac{\sigma ^{2}k^{2}}{2})}[/tex] and now I would like to find the moments, so I know that you could expand characteristic function by Taylor series

    [tex]f_{X}(k)=exp(1+\frac{1}{1!}(ika - \frac{\sigma^2k^2}{2})+\frac{1}{2!}(ika - \frac{\sigma^2k^2}{2})^2+\frac{1}{3!}(ika - \frac{\sigma^2k^2}{2})^3+...)[/tex]

    [tex]f_{X}(k)=exp(1+\frac{(ik)}{1!}\left \langle X^1 \right \rangle+\frac{(ik)^2}{2!}\left \langle X^2 \right \rangle+\frac{(ik)^3}{3!}\left \langle X^3 \right \rangle+...)[/tex]

    and the moments will be
    [tex]\left \langle X^n \right \rangle[/tex]

    Now the problem is that I completely forgot how to evaluate Taylor series.
    Could you help me to calculate for example second moment? I know what the answer should be, but I couldn't get it right.
     
    Last edited: Jan 6, 2017
  2. jcsd
  3. Jan 6, 2017 #2

    jambaugh

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    Take the n-th derivative of the taylor series and evaluate at zero.
    BTW your exponential function should not appear within the taylor expansion... this expansion IS the exponential funcition.
     
  4. Jan 6, 2017 #3
    I am trying to calculate first derivate of term <x1>

    [tex]\frac{d}{dk}(ika-\frac{\sigma^2k^{2} }{2})=(ia - \sigma ^{2}k)[/tex]

    now I am evaluating it at 0

    [tex]f_{X}(0)= ia[/tex]

    And what will happen with a i term?
     
  5. Jan 6, 2017 #4

    mathman

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    [itex]f'_X(0)=iE(X)[/itex], not E(X).
     
  6. Jan 7, 2017 #5
    Thank you!
     
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