Moment Generating Function of normally distributed variable

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SUMMARY

The moment generating function (MGF) for a standard normally distributed variable X ~ N(0,1) is derived using the definition of the MGF, which is E(e^(tX)). For the chi-square distribution, the MGF can be expressed as (1 - 2t)^(-k/2) for k degrees of freedom. In this discussion, the focus is on finding the MGF for the square of a standard normal variable, X², which follows a chi-square distribution with 1 degree of freedom. The participants emphasize using the appropriate probability density function (pdf) and the MGF definition without unnecessary identities.

PREREQUISITES
  • Understanding of moment generating functions (MGF)
  • Knowledge of probability density functions (pdf) for normal and chi-square distributions
  • Familiarity with the properties of the chi-square distribution
  • Basic calculus for evaluating expectations
NEXT STEPS
  • Study the derivation of the moment generating function for the chi-square distribution
  • Learn how to transform a standard normal variable to derive its chi-square counterpart
  • Explore the applications of moment generating functions in statistical inference
  • Review the properties and applications of the chi-square distribution in hypothesis testing
USEFUL FOR

Statisticians, data analysts, and students studying probability theory who need to understand moment generating functions and their applications in statistical distributions.

johnaphun
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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?
 
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johnaphun said:
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?

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.
 
chiro said:
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.

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:
johnaphun said:
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.

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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