Matlab and Mathematica can't do this integration


by ay0034
Tags: cdf, integration, mathematica, matlab, pdf
ay0034
ay0034 is offline
#1
Jun21-11, 04:01 PM
P: 11
Hello,

While doing a research, I obtained the following PDF:



[itex]f_{Z}[/itex][itex](z)=\frac{1}{2\pi\sigma^{2}_{1}\sigma^{2}_{2}}[/itex][itex]e^{-\frac{1}{2} ( \frac{\mu^{2}_{1}}{\sigma^{2}_{1}} + \frac{\mu^{2}_{2}}{\sigma^{2}_{2}})}[/itex][itex]\int^{2\pi}_{0}ze^{-\frac{1}{2\sigma^{2}_{1}\sigma^{2}_{2}}\{ \sigma^{2}_{2}z^{2}cos^{2}\theta+ \sigma^{2}_{1}z^{2}sin^{2}\theta-2\sigma^{2}_{2}\mu_{1}zcos\theta-2\sigma^{2}_{1}\mu_{2}zsin\theta \} }[/itex]



This integral won't be in a closed form. In addition to that, I have to integrate this PDF to get a CDF. Since this PDF is what I calculated, I want to check the CDF is going to be 1 as z goes to infinity.

However, both matlab and mathematica cannot integrate this PDF. Please help me with this annoying integration.

I appreciated it in advance.
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
ay0034
ay0034 is offline
#2
Jun21-11, 04:04 PM
P: 11
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.
Stephen Tashi
Stephen Tashi is offline
#3
Jun21-11, 07:45 PM
Sci Advisor
P: 3,173
With respect to which variable is the integration to be performed? [itex] dz [/itex] ?, [itex] d\theta [/itex] ?

Stephen Tashi
Stephen Tashi is offline
#4
Jun21-11, 09:51 PM
Sci Advisor
P: 3,173

Matlab and Mathematica can't do this integration


If your interest is only [itex] \int_{-\infty}^{\infty} f_Z(z) dz [/itex] have you tried changing the order of integration?

Let [itex] h(z,\theta) = [/itex] the messy function you are dealing with.

Taking some liberties with limits and the order of integration, which you would need to justify, we have:

[tex] \int_{-\infty}^{\infty} f_Z(z)dz = \lim_{a \rightarrow \infty} \int_{-a}^{a} f_Z(z) dz [/tex]

[tex] = \lim_{a \rightarrow \infty} \int_{-a}^{a} \int_{0}^{2\pi} h(z,\theta) d\theta\ dz [/tex]

[tex] = \int_{0}^{2\pi} \left( \lim_{a \rightarrow \infty} \int_{-a}^{a} h(z,\theta)\ dz \right) \ d\theta [/tex]
ay0034
ay0034 is offline
#5
Jun21-11, 10:48 PM
P: 11
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.
Stephen Tashi
Stephen Tashi is offline
#6
Jun21-11, 11:46 PM
Sci Advisor
P: 3,173
Quote Quote by ay0034 View Post
Since the integration does not have a closed form.
Which integration does not have a closed form?
[tex] \int z e^{C_1 z^2 + C_2 z} dz [/tex] ?
ay0034
ay0034 is offline
#7
Jun21-11, 11:53 PM
P: 11
Both of them. With respect to theta and with respect to z.
Stephen Tashi
Stephen Tashi is offline
#8
Jun22-11, 01:02 PM
Sci Advisor
P: 3,173
In the problem you are working, can you rescale the random variables so that the means of the rescaled variables are 0 and their standard deviations are 1 ?
ay0034
ay0034 is offline
#9
Jun22-11, 01:05 PM
P: 11
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.
disregardthat
disregardthat is offline
#10
Jun22-11, 01:41 PM
Sci Advisor
P: 1,686
wrt z you could get the answer in terms of the error-function, if you want that.
ay0034
ay0034 is offline
#11
Jun22-11, 01:46 PM
P: 11
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.
Stephen Tashi
Stephen Tashi is offline
#12
Jun22-11, 01:55 PM
Sci Advisor
P: 3,173
I'm trying to relate this problem to your thread on "Chi or Rayleigh or Ricean?". It looks like you are transforming to polar coordinates by X = z cos(theta) , Y = z sin(theta). Does the z in front of the exponential come from the volume element for polar coordinate integration? If so, isn't this calculation for a CDF rather than a PDF ? (I'm just guessing about your intentions.)
ay0034
ay0034 is offline
#13
Jun22-11, 02:07 PM
P: 11
Yes, the PDF I wrote down is already transformed to a polar coordinates. And z you refered to came from the fact that dxdy is transformed to r*drd[itex]\theta[/itex].

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 with respect to z so I get a PDF.

Matlab can numerically calculate values of PDF at each point, but cannot integrate it numerically nor symbolically.
winterfors
winterfors is offline
#14
Jun23-11, 03:56 AM
P: 71
As far as I can see, your integrand is only the product of two normal distributions over x and y, i.e. a bivariate normal distribution with zero covariance. Why do you want to express this in polar coordinates? The integral from from z=0 to infinity will indeed be equal to one, of course very easily demonstrated in cartesian coordinates.

If you absolutely want to express it in polar coordinates, the expression will be quite simple (with simple analytical expressions for the primitive functions of integrals) if you set mu_1 and mu_2 as the centre of your polar coordinates.
Stephen Tashi
Stephen Tashi is offline
#15
Jun23-11, 09:05 AM
Sci Advisor
P: 3,173
Quote Quote by winterfors View Post
Why do you want to express this in polar coordinates?
http://www.physicsforums.com/showthread.php?t=507750
winterfors
winterfors is offline
#16
Jun23-11, 04:43 PM
P: 71
Quote Quote by ay0034 View Post
...am trying to calculate mean and variance. But without a closed form PDF, that looks impossible as well.
Are you only after the mean and variance? In that case, I would not bother trying to calculate the distribution over Z, but just numerically evaluate the mean and variance of Z by integrating over X and Y (will work fine for any values of means and variances of the two normal distributions)
Pyrrhus
Pyrrhus is offline
#17
Jun28-11, 01:56 PM
HW Helper
Pyrrhus's Avatar
P: 2,280
I didn't check the integrand, but if it does not have a closed form, and you want a numeric value then your alternatives are quadrature or simulation. The integral is one-dimensional? then Quadrature should work fine.


Register to reply

Related Discussions
MATLAB and Mathematica help Math & Science Software 0
matlab vs mathematica Math & Science Software 20
how write the 'for' loop in mathematica Math & Science Software 11
how to write matlab code in mathematica Math & Science Software 0
Matlab/Mathematica help! Math & Science Software 1