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Moment Generating Function of normally distributed variable

  1. Jan 10, 2012 #1
    Hi guys,

    I need to find the moment generating function for X ~ N (0,1) and then also the MGF for X2 . I know how to do the first part but i'm unsure for X2.

    do i use the identity that if Y = aX then

    MY(t) = E(eY(t)) = E(e(t)aX)

    or do i just square 2pi-1/2e x2/2 and then solve as normal for an MGF?
     
  2. jcsd
  3. Jan 11, 2012 #2

    chiro

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    Hey johnaphun and welcome to the forums.

    For the chi-square distribution, are you given the pdf function or do you have to derive it from the normal function?

    If you can use the definition of the pdf, then you do exactly the same thing as you did for the normal. If you need to derive the pdf for the chi-square distribution then you need to use a transformation method.

    You don't need to use any identity, just the pdf of the appropriate distribution as well as the MGF definition.
     
  4. Jan 11, 2012 #3
    Hi thanks for the reply,

    I've not been given any pdf function, i've only been told that Xi is standardised normally distributed and to thus find the MGF for Xi (which i can do) and (Xi)2.

    I assume it's asking for me to find an MGF for a random variable Xi with chi distribution which is (1 − 2 t)−1/2, i'm just not sure how to go about proving that.
     
    Last edited: Jan 11, 2012
  5. Jan 11, 2012 #4

    chiro

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    That seems a little odd because at the very least you should be given the normal pdf function.

    The way I would do it (i.e. the second part) is to use the chi-square pdf and then use the mgf and do some algebra to get the final mgf function.

    The thing though is that if you have to derive the chi-square pdf from the normal pdf, then you may lose marks. It doesn't explicitly say you have to do this though, so you are probably in all likelihood, safe to just use the chi-square pdf.

    Based on this information, do you know how to go about solving this problem?
     
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