Matlab and Mathematica can't do this integration

• MATLAB
• ay0034
In summary: If it is multidimensional, Monte Carlo integration is the way to go.In summary, the conversation is about a researcher struggling to integrate a complicated PDF in order to obtain a CDF. They have tried using MATLAB and Mathematica, but have not been successful. They also considered changing the order of integration, but it did not yield good results. The researcher has also tried calculating the mean and variance, but without a closed form PDF, it seems impossible. They have also discussed the possibility of using polar coordinates and transforming the variables, but it is not feasible in this case. The conversation also mentions the option of using numerical methods such as quadrature or simulation to calculate the desired values.
ay0034
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}}\{ \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 \} }$

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.

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.

With respect to which variable is the integration to be performed? $dz$ ?, $d\theta$ ?

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

Let $h(z,\theta) =$ 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:

$$\int_{-\infty}^{\infty} f_Z(z)dz = \lim_{a \rightarrow \infty} \int_{-a}^{a} f_Z(z) dz$$

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

$$= \int_{0}^{2\pi} \left( \lim_{a \rightarrow \infty} \int_{-a}^{a} h(z,\theta)\ dz \right) \ d\theta$$

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.

ay0034 said:
Since the integration does not have a closed form.

Which integration does not have a closed form?
$$\int z e^{C_1 z^2 + C_2 z} dz$$ ?

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

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 ?

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.

wrt z you could get the answer in terms of the error-function, if you want that.

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.

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.)

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 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.

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.

ay0034 said:
...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)

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.

1. Can't Matlab and Mathematica do any type of integration?

No, Matlab and Mathematica have limitations when it comes to certain types of integration. They may not be able to solve integrals with complex or highly oscillatory functions, or those that involve special functions.

2. Why can't Matlab and Mathematica solve some integrals?

Matlab and Mathematica use algorithms and numerical methods to solve integrals. These methods may not be able to handle certain types of functions or integrals that require more advanced techniques.

3. Are there any alternatives to Matlab and Mathematica for solving integrals?

Yes, there are other software programs and programming languages that specialize in solving integrals, such as Maple, Maxima, and Python. It may be worth exploring these alternatives if Matlab and Mathematica cannot solve a particular integral.

4. Is it possible to improve the integration capabilities of Matlab and Mathematica?

Yes, it is possible to improve the integration capabilities of Matlab and Mathematica by using different algorithms and techniques, or by adjusting the parameters of the integration function. However, this may require a deeper understanding of the underlying mathematics and programming.

5. Can I get help with solving a difficult integral using Matlab or Mathematica?

Yes, there are online communities and forums dedicated to discussing and solving difficult integrals using Matlab and Mathematica. You can also seek help from experts in these software programs or consider hiring a professional mathematician or programmer for assistance.

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