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A Gaussian distribution characteristic function

  1. Jan 5, 2017 #1
    Hello, guys. I am trying to solve for characteristic function of normal distribution and I've got to the point where some manipulation has been made with the term in integrands exponent. And a new term of t2σ2/2 has appeared. Could you be so kind and explain that to me, please.

    [tex]=Ae^{it\mu}\int_{-\infty}^{\infty}e^{-\frac{1}{c^2}(\alpha^{2}-i2t\sigma ^{2}\alpha)}d\alpha=Ae^{(it\mu-\frac{t^{2}\sigma^{2}}{2})}\int_{-\infty}^{\infty}e^{-\frac{(\alpha-it\sigma^{2})^{2}}{c^{2}}} d\alpha [/tex]
     
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  3. Jan 5, 2017 #2

    Charles Link

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    What was done is called "completing the square". e.g. take ## x^2+6x ##. You divide the x coefficient by 2 and square that result to complete the square: ## x^2+6x=x^2+6x+9-9 =(x+3)^2-9 ##. ( ## (6/2)^2=9 ##). You add it and also subtract it from the expression. editing ... In the case you presented, ## \alpha=x ##. ## \\ ## Additional editing: On closer inspection, the term should have ## \sigma^4 ## and not ## \sigma^2 ##, and it should have a ## c^2 ## in the denominator instead of a 2. No wonder it puzzled you !!!
     
    Last edited: Jan 5, 2017
  4. Jan 6, 2017 #3
    Nice! Thank you very much!
     
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