Timeline for Why is free probability a generalization of probability theory?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2017 at 16:27 | comment | added | Terry Tao | @Henry.L Consider for instance the random $2 \times 2$ matrix $X := \begin{pmatrix} x & 0 \\ 0 & -x \end{pmatrix}$, where $x$ is drawn uniformly from $[0,1]$. Using the classical expectation structure ${\bf E}$, $X$ has non-zero mean $\begin{pmatrix} 1/2 & 0 \\ 0 & -1/2 \end{pmatrix}$. But using the noncommutative trace structure $\frac{1}{2} {\bf E} \mathrm{tr}$, $X$ has mean zero (or trace zero, if one prefers). | |
| Apr 25, 2017 at 5:07 | comment | added | Chill2Macht | This answer was very helpful to me on several levels. As particular examples, it made clear to me how I was misinterpreting the notion of localizable measurable space, as well as emphasizing the distinction between non-scalar classical random variables and scalar non-commutative random variables, and making me aware of non-commutative non-scalar random variables (which I did not even know existed). I appreciate your taking the time to comment on this, and hope that I did not misrepresent or misinterpret your notes in any way you would find objectionable. | |
| Apr 25, 2017 at 4:59 | vote | accept | Chill2Macht | ||
| Apr 24, 2017 at 18:57 | comment | added | Henry.L | Fabulous, I learnt a lot, especially for the last comment clarified my misunderstanding in one place of your answer. My appreciation! Can you also give an example to the statement in your post "...but this is a rather different structure (note now that the expectation is a matrix rather than a scalar) and should not be confused with the scalar noncommutative probability structure one can place here." I think it might be very helpful! | |
| Apr 24, 2017 at 18:06 | comment | added | Terry Tao | For instance, much as how classical probability theorems such as the law of large numbers or the central limit theorem can fail if one deals with variables of infinite mean or variance, free probability theorems such as the free law of large numbers or the free central limit theorem can similarly fail if certain traces are now undefined or infinite. | |
| Apr 24, 2017 at 18:01 | comment | added | Terry Tao | (c) One could consider noncommutative probability in which traces are not necessarily defined, much as one can consider the commutativity (or lack thereof) of operators on a Hilbert space that are not necessarily of trace class. For instance, one could define a notion of freeness of variables $X,Y$ that are not of finite trace, but nevertheless have a functional calculus, by declaring $f(X), g(Y)$ to be free for any bounded $f,g$ in this calculus. But one has to take care as some free probability theorems may now fail in this unbounded setting. | |
| Apr 24, 2017 at 17:58 | comment | added | Terry Tao | (b) Classical exchangeability is related to classical independence via de Finetti's theorem, and there is a notion of free exchangeability that is similarly related to free independence via a free version of de Finetti's theorem (see e.g. arxiv.org/abs/0807.0677). So I would view the concept of exchangeability (both classical and free) as being orthogonal to the classical probability / free probability distinction; also, as pointed out in the previous post, there are also many other ways to modify the notion of exchangeability. | |
| Apr 24, 2017 at 17:57 | comment | added | Terry Tao | (a) Standard probability spaces only appear in classical probability theory; noncommutative probability theory does not have any underlying probability space (though, through tools such as the GNS construction, one can create a Hilbert space that is somewhat analogous to the classical probability space, in much the same way that the Hilbert space of quantum states is analogous to classical phase space). ... | |
| Apr 24, 2017 at 17:19 | comment | added | Henry.L | Thanks for a very clarifying answer! Two questions, If we are discussing within a std prob space, what is the difference between your "noncommutative rv" and "inexchangeable rv"? And is it really necessary to have an integral representation (expectation) to discuss commutativity within free prob? | |
| Apr 24, 2017 at 16:58 | history | edited | Terry Tao | CC BY-SA 3.0 |
added 46 characters in body
|
| Apr 24, 2017 at 16:52 | history | edited | Terry Tao | CC BY-SA 3.0 |
added 265 characters in body
|
| Apr 24, 2017 at 16:46 | history | edited | Terry Tao | CC BY-SA 3.0 |
added 34 characters in body
|
| Apr 24, 2017 at 16:41 | history | edited | Terry Tao | CC BY-SA 3.0 |
added 280 characters in body
|
| Apr 24, 2017 at 16:33 | history | answered | Terry Tao | CC BY-SA 3.0 |