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If we have two random Variables that are independent and identically distributed, their sum's moment generating function is simply their product. But does the trick work if they are not identically distributed? For example, if one is Normally distributed, and the other is Exponentially distributed, can we still multiply the moment generating functions?

I can't see any reason it won't work, but have not been able to find any examples on the Internet, which is why I wanted to confirm this.

kjetil b halvorsen
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user4562262
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    The trick works – Dilip Sarwate Aug 12 '17 at 02:42
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    Wikipedia on [moment generating functions](https://en.wikipedia.org/wiki/Moment-generating_function#Linear_combination_of_independent_random_variables) gives a slightly more general result in an obvious place to look (what were you searching for?). Also see math.SE: [Moment generating function of X+Y using convolution of X and Y](https://math.stackexchange.com/questions/1868911/moment-generating-function-of-xy-using-convolution-of-x-and-y). – Glen_b Aug 12 '17 at 04:17
  • MGFs and characteristic functions are conceptually the same. Concerning the latter, see https://stats.stackexchange.com/questions/9617 and visit https://stats.stackexchange.com/a/72486 and https://stats.stackexchange.com/a/3684 for worked examples. – whuber Aug 12 '17 at 20:00
  • whuber, thanks - it looks like characteristic functions are useful if you want to invert to get the density function back. – user4562262 Aug 13 '17 at 14:14

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Answered in comments:

The trick works – Dilip Sarwate

More details:

Wikipedia on moment generating functions gives a slightly more general result in an obvious place to look (what were you searching for?). Also see math.SE: Moment generating function of $X+Y$ using convolution of $X$ and $Y$. – Glen_b

MGFs and characteristic functions are conceptually the same. Concerning the latter, see and visit 1 for worked examples. – whuber

kjetil b halvorsen
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