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In classical probability, adding two independent random variables corresponds to adding their cumulant generating functions, i.e. the logarithms of their Fourier transforms.

In Voiculescu's free probability, adding two freely independent random variables corresponds to adding their $R$-transforms.

However, in Voiculescu's free probability there is a separate notion used in the situation of multiplying two freely independent variables, called the $S$-transform.

My question is, why is a separate transform needed for multiplication in the free case, and why is it not needed in the classical (commutative) case?

keej
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  • In the commutative setting you can pass between addition and multiplication using exponentials and logarithms, which you can't do in the noncommutative setting. – Qiaochu Yuan Aug 23 '17 at 04:28
  • Why can't you take exponentials and logs in the noncommutative setting? It seems like you can just use the taylor series as usual. Or is it because exp(A+B) = exp(A)exp(B) doesn't hold in the noncommutative setting? – keej Aug 24 '17 at 23:55
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    Yes, you can still take exponentials and logs, but they no longer relate addition and multiplication. – Qiaochu Yuan Aug 25 '17 at 00:04

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