Hi,

I have a problem in integrating the FB5 distribution. The FB5's probability density function is given by f(x)=\frac{1}{c(\kappa, \beta)}\exp[\kappa \gamma_1 x +\beta[(\gamma_2 x)^2 - (\gamma_3 x)^2]]. We know that \gamma_1 represents the mean direction of the distribution and that \gamma_2 and \gamma_3 represent the major and minor axes respectively. The three columns \gamma_1, \gamma_2 and \gamma_3 form an orthogonal (rotation) 3x3 matrix.

I would like to integrate this PDF and therefore I need to express its exponent in polar coordinates. I know that \gamma_1 x can be expressed as: \gamma_1 x= \cos(\theta) \cos(\theta_o) + \sin(\theta) \sin(\theta_o) \cos(\phi - \phi_o) since it is the vector dot product of two 3d vectors. In the above \gamma_1 = [\sin(\theta_o) \cos(\phi_o)        \sin(\theta_o) \sin(\phi_o)        \cos(\theta_o) ] and x= [\sin(\theta) \cos(\phi)       \sin(\theta) \sin(\phi)        \cos(\theta) ].

How can we express the \gamma_2 x and \gamma_3 x in a similar way?

Do you think that the following form would be correct? : \gamma_2 x= \cos(\theta) \cos(\theta major) + \sin(\theta) \sin(\theta major) \cos(\phi - \phi major)...same for \gamma_3 x ???? . However, I have read that one needs to define only one angle to specify the major and minor axes. This is very confusing to me. As a result, I am not sure if I can express \gamma_2 x and \gamma_3 x as stated above. Please advise!!!!!!!!


Many Thanks

BR

Alex.