From: Tim Wescott on 26 Apr 2010 12:49 ombz wrote: (top posting fixed)>> On Apr 26, 11:00=A0am, "ombz" >> wrote: >>> Hi all. >>> I'd like to rotate a 2D-Gaussian bump. I.e. parametrize the 2D Gaussian >>> distribution with the rotation angle phi. I first thought I could > simply >>> apply a rotation matrix on the covariance matrix. But my MATLAB output >>> tells me that I'm wrong. How do I make it correctly? >>> Thanks for your help >>> Andreas >> Hello Andreas, >> >> >> Since the 2D gaussian has rotational symmetry, why do you want to >> rotate it? If you are saying you have a 2D object that you want to >> rotate in 3D, then write up the rotation equations and mathematically >> apply them to the gaussian. You may even use Euler angles for this. >> Let us know what you really need. >> >> IHTH, >> Clay >> > I didn't actually mean the symmetric "gaussian", sorry. I mean a 2D > multivariate normal distribution with covariance matrix > > Cov = [sigma_x^2 0 > 0 sigma_y^2] > > where sigma_x != sigma_y, i.e. no rotational symmetry. > > I was applying a rotation matrix R = [ cos(phi) -sin(phi); sin(phi) > cos(phi) ]. But it did not rotate it. > > The complete cov. matrix from 2D multivariate distribution follows > > Cov = [sigma_x^2 rho*sigma_x*sigma_y; > rho*sigma_x*sigma_y; sigma_y^2] > > Where rho is the correlation coefficient. Maybe all I'd have to do is > parametrize rho = rho(phi) = phi/pi if phi in [-pi,pi]? Yeah, I'll > test > this ASAP... Think. You're applying the rotation to x: xr = R * x. But by definition, if x is in n(0, Cov_x), then Cov_x is the expected value of x times itself: Cov_x = E{x * x^T}. Also by definition Cov_xr is Cov_xr = E{xr * xr^T}. See where this is going? Tell me this isn't homework* (by telling me what you're really doing) and I'll give you the rest. Tell me it _is_ homework and I'll give you enough help that you understand it, but not so much that you forget it before the final. * Sorry if my attitude strikes you as coy, but it's that time of year. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com From: dvsarwate on 26 Apr 2010 13:12 On Apr 26, 10:43 am, "ombz" wrote:> I didn't actually mean the symmetric "gaussian", sorry. I mean a 2D > multivariate normal distribution with covariance matrix > > Cov = [sigma_x^2     0 >           0      sigma_y^2] > > where sigma_x != sigma_y, i.e. no rotational symmetry. > > I was applying a rotation matrix R = [ cos(phi) -sin(phi); sin(phi) > cos(phi) ]. But it did not rotate it. > > The complete cov. matrix from 2D multivariate distribution follows > > Cov = [sigma_x^2              rho*sigma_x*sigma_y; >        rho*sigma_x*sigma_y;   sigma_y^2] > > Where rho is the correlation coefficient. Maybe all I'd have to do is > parametrize rho = rho(phi) = phi/pi if phi in [-pi,pi]? Yeah, I'll test > this ASAP... The reverse problem -- how to rotate an arbitrary 2D Gaussian distribution with correlated Gaussian RVs so as to get independent Gaussian RVs -- is stated in (http://courses.ece.illinois.edu/ECE313/spring09/homework/PS13.pdf) and solved in (http://courses.ece.illinois.edu/ECE313/spring09/homework/HW13.pdf) Hope this helps. --Dilip Sarwate From: ombz on 27 Apr 2010 05:34 @Dilip: thanks for the links - I think they'll help me to solve the problem. @Tim: no - this is no homework. I think the rotated distribution might help me in a tracking problem as I'd need to have properly rotated distributions if I want to better cope with rotation. But in this moment I think I might even add the rotation to the distribution itself...But aah, I was never too good in statistics.;) Final question: how about making the distribution asymmetric? I think I'd have to change to another type of distribution - no? Anyway. Thanks for your help. I think I should be able to handle it now on my own. Best, Andreas >On Apr 26, 10:43=A0am, "ombz" >wrote: >> I didn't actually mean the symmetric "gaussian", sorry. I mean a 2D >> multivariate normal distribution with covariance matrix >> >> Cov =3D [sigma_x^2 =A0 =A0 0 >> =A0 =A0 =A0 =A0 =A0 0 =A0 =A0 =A0sigma_y^2] >> >> where sigma_x !=3D sigma_y, i.e. no rotational symmetry. >> >> I was applying a rotation matrix R =3D [ cos(phi) -sin(phi); sin(phi) >> cos(phi) ]. But it did not rotate it. >> >> The complete cov. matrix from 2D multivariate distribution follows >> >> Cov =3D [sigma_x^2 =A0 =A0 =A0 =A0 =A0 =A0 =A0rho*sigma_x*sigma_y; >> =A0 =A0 =A0 =A0rho*sigma_x*sigma_y; =A0 sigma_y^2] >> >> Where rho is the correlation coefficient. Maybe all I'd have to do is >> parametrize rho =3D rho(phi) =3D phi/pi if phi in [-pi,pi]? Yeah, I'll te=>st >> this ASAP... > >The reverse problem -- how to rotate an arbitrary 2D >Gaussian distribution with correlated Gaussian RVs so >as to get independent Gaussian RVs -- is stated in > >(http://courses.ece.illinois.edu/ECE313/spring09/homework/PS13.pdf) > >and solved in > >(http://courses.ece.illinois.edu/ECE313/spring09/homework/HW13.pdf) > >Hope this helps. > >--Dilip Sarwate > > > > > From: Tim Wescott on 27 Apr 2010 11:18 ombz wrote: (top posting fixed)>> On Apr 26, 10:43=A0am, "ombz" >> wrote: >>> I didn't actually mean the symmetric "gaussian", sorry. I mean a 2D >>> multivariate normal distribution with covariance matrix >>> >>> Cov =3D [sigma_x^2 =A0 =A0 0 >>> =A0 =A0 =A0 =A0 =A0 0 =A0 =A0 =A0sigma_y^2] >>> >>> where sigma_x !=3D sigma_y, i.e. no rotational symmetry. >>> >>> I was applying a rotation matrix R =3D [ cos(phi) -sin(phi); sin(phi) >>> cos(phi) ]. But it did not rotate it. >>> >>> The complete cov. matrix from 2D multivariate distribution follows >>> >>> Cov =3D [sigma_x^2 =A0 =A0 =A0 =A0 =A0 =A0 =A0rho*sigma_x*sigma_y; >>> =A0 =A0 =A0 =A0rho*sigma_x*sigma_y; =A0 sigma_y^2] >>> >>> Where rho is the correlation coefficient. Maybe all I'd have to do is >>> parametrize rho =3D rho(phi) =3D phi/pi if phi in [-pi,pi]? Yeah, I'll > te= >> st >>> this ASAP... >> The reverse problem -- how to rotate an arbitrary 2D >> Gaussian distribution with correlated Gaussian RVs so >> as to get independent Gaussian RVs -- is stated in >> >> (http://courses.ece.illinois.edu/ECE313/spring09/homework/PS13.pdf) >> >> and solved in >> >> (http://courses.ece.illinois.edu/ECE313/spring09/homework/HW13.pdf) >> >> Hope this helps. > @Dilip: thanks for the links - I think they'll help me to solve the > problem. > > @Tim: no - this is no homework. I think the rotated distribution might help> me in a tracking problem as I'd need to have properly rotated distributions> if I want to better cope with rotation. But in this moment I think I might> even add the rotation to the distribution itself...But aah, I was never too> good in statistics.;) > > Final question: how about making the distribution asymmetric? I think I'd > have to change to another type of distribution - no? > > Anyway. Thanks for your help. I think I should be able to handle it now on> my own. I'm better than a lot of people at statistics, and it can still give me fits. A multivariate normal distribution is, by it's nature, symmetric around orthogonal axes. So yes, if you want an asymmetric distribution it's gotta be something else -- and then you have to be careful about what statistical tools you use, because so many of them only work right with normal distributions. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com From: glen herrmannsfeldt on 27 Apr 2010 14:16 ombz wrote:> I didn't actually mean the symmetric "gaussian", sorry. I mean a 2D > multivariate normal distribution with covariance matrix > Cov = [sigma_x^2 0 > 0 sigma_y^2] > where sigma_x != sigma_y, i.e. no rotational symmetry. > I was applying a rotation matrix R = [ cos(phi) -sin(phi); sin(phi) > cos(phi) ]. But it did not rotate it. > The complete cov. matrix from 2D multivariate distribution follows > Cov = [sigma_x^2 rho*sigma_x*sigma_y; > rho*sigma_x*sigma_y; sigma_y^2] > Where rho is the correlation coefficient. Maybe all I'd have to do is > parametrize rho = rho(phi) = phi/pi if phi in [-pi,pi]? Yeah, I'll test > this ASAP... It seems not so for off for small rotation angles. For larger angles, the normalization is wrong. I think it is that statisticians think about these problems in different ways than physicists do. The rotation form makes sense if you believe that it really is a two dimensional coordinate system that you are looking at from a different, rotated orthogonal, coordinate system. If there is no requirement that the axes be orthogonal, then that transform doens't apply. -- glen First  |  Prev  |  Pages: 1 2 Prev: Rotate 2D GaussianNext: Is there a "DIGITAL" equivalent of "Q"?