Recent content by ay0034

Yes, the PDF I wrote down is already transformed to a polar coordinates. And z you referred to came from the fact that dxdy is transformed to r*drd\theta. As you know, this is kind of a function of r.v., I began with a CDF, transformed it to a polar coordinates, and took derivative of the CDF...

You're right. The thing is I have to integrate that error function with respect to theta. I almost gave up to do this double integration, and am trying to calculate mean and variance. But without a closed form PDF, that looks impossible as well.

Unfortunately, I can't. If so, my PDF would be a Rayleigh distribution and I can take advantage of existing information out there. That is my problem.

Both of them. With respect to theta and with respect to z.

Oh i forgot to put that. It's dtheta, not dz. And I have tried what you are talking about, and I got no good results. Since the integration does not have a closed form, replacing it with a and using limit function was not helpful.

mu's and sigma's can be any number. And in this case, mu1 and mu2 are different, and sig1 and sig2 are different as well.

Hello, While doing a research, I obtained the following PDF: f_{Z}(z)=\frac{1}{2\pi\sigma^{2}_{1}\sigma^{2}_{2}}e^{-\frac{1}{2} ( \frac{\mu^{2}_{1}}{\sigma^{2}_{1}} + \frac{\mu^{2}_{2}}{\sigma^{2}_{2}})}\int^{2\pi}_{0}ze^{-\frac{1}{2\sigma^{2}_{1}\sigma^{2}_{2}}\{...

thank you so much. I'll look it up.

For a non-central chi distribution, means can be different. But can variances be different as well? As far as I know, variances must be 1.

Plz help!

Hello all, I've been working on error analysis of the system, and I finally faced a big problem. Let X~N(mu1, sigma1^2) and Y~N(mu2, sigma2^2), and Z=sqrt( X^2 + Y^2 ) For Z to be a Chi, mu's should be zero and sigma's should be 1, to be a Rayleigh, mu's should be zero and two...