Timeline for Moment/mgf of cosine of directional vectors?
Current License: CC BY-SA 4.0
23 events
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| Jan 31, 2023 at 9:17 | history | edited | User1865345 | CC BY-SA 4.0 |
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Mar 8, 2017 at 22:48 | comment | added | Henry.L | @YaroslavBulatov It is hard to say whether it is a coincidence without a neat calculation. It could be a conincidence or there is a deeper theoretical link that I do not know. Is mathematica doing symbolic calculation? (Sorry I know little about the computational side of stat beyond R.) | |
| Mar 8, 2017 at 18:13 | comment | added | Yaroslav Bulatov | btw, I checked the formula for 4 variables numerically (code in answer), and it is still within error boundaries (although Mathematica reported error boundaries for NIntegrate get quite large, 0.38) | |
| Mar 8, 2017 at 18:02 | comment | added | Yaroslav Bulatov | ah, so the nice formula for 3 dimensions is just a coincidence? That's unfortunate | |
| Mar 8, 2017 at 18:02 | history | bounty ended | Yaroslav Bulatov | ||
| Mar 8, 2017 at 3:03 | vote | accept | Yaroslav Bulatov | ||
| Mar 8, 2017 at 2:20 | history | edited | Henry.L | CC BY-SA 3.0 |
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| Mar 8, 2017 at 2:07 | comment | added | Henry.L | @YaroslavBulatov No, and I think its closed form involves a hypergeometric function according to my primary computation on bus. The projected normal distribution requires a bit complex technique than polar coordinates than [Mardia&Peter] claimed in 2-dim. Derivation is discussed in another post stats.stackexchange.com/questions/91303/… | |
| Mar 8, 2017 at 2:04 | comment | added | Henry.L | @Student001 I carouse the answer again, I think the first step is actually not necessary because the projected normal density is in its closed form for any form of $\Sigma$. So the answer only involves calculation of $\mathcal{PN}_k$ and a transformation formula. I do not see why orthogonality is important here...Thanks for your carefulness and can you see if the answer is clear to you now? | |
| Mar 8, 2017 at 2:01 | history | edited | Henry.L | CC BY-SA 3.0 |
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| Mar 7, 2017 at 23:51 | comment | added | KOE | Ah, OK. So $P$ is symmetric but not orthogonal then? If $PP = \Sigma$, you still seem to claim that $(Px)^TPy / \sqrt{x^TPPx y^TPPy} = x^T\Sigma y / \sqrt{x^T\Sigma x y^T \Sigma y} = x^Ty / \sqrt{x^Tx y^Ty}$, where $x$ and $y$ are indep. $N(0, I)$ and the last equality is in distribution. Could you provide a proof of this? | |
| Mar 7, 2017 at 21:54 | comment | added | Yaroslav Bulatov | so ... do you think my formula extends to more than 3 dimensions? | |
| Mar 7, 2017 at 17:33 | comment | added | Henry.L | @Student001 Maybe I should use better notation but I mean square root decomposition.en.wikipedia.org/wiki/Square_root_of_a_matrix | |
| Mar 7, 2017 at 16:32 | comment | added | KOE | No, if $P'\Lambda P$ is the spectral decomposition of $\Sigma$, then $PX$ as covariance matrix $\Lambda$, which need not be the identity, so at least that step doesn't justify $\Sigma = I$ w.l.o.g. Maybe your last comment does, I'm not sure. | |
| Mar 7, 2017 at 16:29 | history | edited | Henry.L | CC BY-SA 3.0 |
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| Mar 7, 2017 at 16:27 | comment | added | Henry.L | Even if we assumed only diagonal covariance, it only makes the projected normal distribution scaled on main axes and hence introduce only scalars into the density of $\mathcal{PN}_k$. And [Mardia&Peter] does not assume anything on the covariance matrix as I can see in the quote. Assuming identity means the projected image lies on a sphere which makes it easier to visualize. | |
| Mar 7, 2017 at 16:26 | comment | added | Henry.L | @Student001 If $\Sigma=P'\Lambda P$, then $PX$ have an identity covariance matrix. | |
| Mar 7, 2017 at 14:13 | comment | added | KOE | Could you provide a proof that assuming identity covariance matrix is w.l.o.g? It's not obvious to me. It's "easy" to show cardinal's claim that diagonal matrix is w.l.o.g, but how do you get rid of the eigenvalues? | |
| Mar 7, 2017 at 3:29 | comment | added | Henry.L | The answer I posted on MO is not exactly what the OP wanted because I was thinking that he is searching for the canonical angle. my bad. | |
| Mar 7, 2017 at 3:28 | history | edited | Henry.L | CC BY-SA 3.0 |
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| Mar 7, 2017 at 3:27 | comment | added | Henry.L | @YaroslavBulatov Hopefully this is well worth your bounty! | |
| Mar 7, 2017 at 3:17 | history | answered | Henry.L | CC BY-SA 3.0 |