Finding Moments of Normal Distribution with Unknown Constant

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Homework Statement


You have i.i.d. random variables X1,...,Xn ~ Normal (\Theta,c^2*\Theta^2), where "c" is a known positive constant (relative variability = std. dev(X)/E[X]) and \Theta is an unknown positive constant. Find the first four moments. E.g E[Xj] where j=1,2,3,4.

Homework Equations

The Attempt at a Solution


So i know the pdf of a normal distribution to be, and what i did was input (c*\Theta)2 as the standard deviation into this. Then I took the expected value of that to get the first order moment. Is this the proper way to do it? Does having (c\Theta)2 change how you integrate the expectation of a normal distribution relative to how you would for the usual N(\mu,\sigma) case?
 
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Homework Statement


You have i.i.d. random variables X1,...,Xn ~ Normal (\Theta,c^2*\Theta^2), where "c" is a known positive constant (relative variability = std. dev(X)/E[X]) and \Theta is an unknown positive constant. Find the first four moments. E.g E[Xj] where j=1,2,3,4.


Homework Equations




The Attempt at a Solution


So i know the pdf of a normal distribution to be, and what i did was input (c*\Theta)^2 as the standard deviation into this. Then I took the expected value of that to get the first order moment. Is this the proper way to do it? Does having (c\Theta)^2 change how you integrate the expectation of a normal distribution relative to how you would for the usual N(\mu,\sigma) case?
 
Sorry the first one i formatted poorly, and couldn't figure it out. New to this forum, just figuring stuff out.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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