Characteristic function and Fourier transform

We were trying to prove that Linear combination of I.I.D Gaussian Rvs produce Gaussian Rvs( not necessarily independent though).

The MGF of Gaussian distribution is as follows:




The usual trick is to calculate the MGF of the sum of the IID Gaussian RVs and after a bit of re-arranging, it can be shown that this MGF has the form of that of MGF of a Gaussian RV.Since the MGF of the sum has the form of that of a Gaussian RV, he concluded that the sum is indeed a RV. I was not quite convinced as to how the inverse would be unique. I asked my prof if he could give some intuition as to how the pdf-MGF pair was unique. He said that one can look at the characteristic function as the Fourier transform of the pdf. Oh boy ! I had never looked at this way. I feel this interpretation is cool. You can read more abt this at

http://homepages.inf.ed.ac.uk/rbf/CVonline/LOCAL_COPIES/SHUTLER3/node2.html

Undergraduation in ECE is not entirely useless after all ! :-)