Timeline for Probability distribution of uAv…
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2014 at 10:03 | vote | accept | tam | ||
| Jun 2, 2014 at 13:59 | comment | added | tam | You are right, I have to use randn (normal distribution..) to generate the unitary matrices. but believe me, your expression works even for small n (n=3,4..): the distribution is exponential with mean $tr(AA^H)/n^2$ = mean given by Monte Carlo method! | |
| Jun 2, 2014 at 11:59 | comment | added | Carlo Beenakker | to generate unitary matrices that are uniformly distributed (with the Haar measure) you need to start with a normal distribution, a uniform distribution will not provide the proper result; I would advise you to double check that your U1 and U2 are really uniformly distributed, say by checking that the eigenphases lie uniformly on the unit circle. | |
| Jun 2, 2014 at 10:04 | comment | added | tam | When I use randn ( gaussian using matlab) to generate A, Q1 and Q2 (then i take the unitary matrices U1 and U2 using orth(Q1) and orth(Q2) resp.), it works well, even for small n!! but i think it is a particular case when i take randn instead of rand ? | |
| Jun 1, 2014 at 21:40 | comment | added | Carlo Beenakker | what are you taking for $A$? even if not unitary, it should not be sparse, you might draw $A$ from a Gaussian distribution, then the large-$n$ limit should hold. | |
| Jun 1, 2014 at 21:14 | comment | added | tam | I just want to say that, when A is not unitary and n>>1, the average using Monte Carlo method is not the same using your expression ( $tr(AA^H)/n^2$); ex: the first is 0.3 the second is 0.64 .//Anyway, Thank you for your clear and helpful explanation! | |
| Jun 1, 2014 at 18:43 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jun 1, 2014 at 18:12 | comment | added | Carlo Beenakker | added details at various steps of the derivation, and generalized it to the general case of nonunitary $A$. | |
| Jun 1, 2014 at 18:11 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jun 1, 2014 at 15:52 | comment | added | tam | "for non unitary A and n>>1 I would think the same exponential distribution will appear.." if you don't have the answer, I will be grateful if you give me the explanation for the case where A is unitary (and n>>1). | |
| Jun 1, 2014 at 15:44 | comment | added | Carlo Beenakker | in the mean time, you might check your numerics that for unitary $A$ you find $P(\xi)\propto(1-\xi)^{n-2}$, exactly for small $n$ (the exponential distribution only follows in the large-$n$ limit) | |
| Jun 1, 2014 at 15:42 | comment | added | Carlo Beenakker | hmm, I don't really have references, if you tell me for which step you would like to have further explanations, I can try to fill you in. | |
| Jun 1, 2014 at 15:28 | comment | added | tam | Could you please give me some references for the cases you have explained. thank you very much | |
| Jun 1, 2014 at 14:45 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jun 1, 2014 at 14:39 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jun 1, 2014 at 14:31 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |