Joint Probability/Mutual Information Normalization

[Nahtix]

Homework Statement

Find the Corresponding normalization constants for the joint guassian

$$\\p\\(x,y) \alpha\ exp(-(ax^2/2) - (by^2/2) -cxy)$$

ie: find P(x), P(y), and P(X,Y) normalization constants


relevant equations:
Mutual Information: M(x,y)
$$\\M(x,y) = \sum\\p(x,y)ln(p(x,y)/(p(x)*p(y)))$$

The Attempt at a Solution

I've tried an integral over y then x but I end up with some messed up erf that I can't get rid of. My professor told me that the real point of this is that later we can use mutual information to get a ratio of the constants and that should equal $$\\1/2\\ln(1-c^2/ab)$$ but I haven't gotten anywhere close. I really just need a hint of where to start on this one as I have gotten nowhere with it any help is very appreciated.

 

[mighty2000]

To find the normalization constants for a joint Gaussian distribution, we can use the fact that the integral of the joint distribution over all possible values of x and y must equal 1. This is because the probability of any event occurring in a probability distribution must be equal to 1. Therefore, we can set up the following integral:

\\int\\int\\p(x,y)dx dy = 1

We can then substitute the given joint Gaussian distribution into this integral:

\\int\\int\\alpha exp(-(ax^2/2) - (by^2/2) -cxy) dx dy = 1

Next, we can use the properties of exponential functions to simplify this integral:

\\alpha\\int\\intexp(-(ax^2/2) - (by^2/2) -cxy) dx dy = 1

We can then use the fact that the exponential function can be written as the product of two exponential functions to split the integral into two separate integrals:

\\alpha\\int\\intexp(-(ax^2/2))exp(-(by^2/2))exp(-cxy) dx dy = 1

Now, we can use the fact that the integral of a product of functions is equal to the product of the integrals of each function to solve this integral:

\\alpha\\int\\intexp(-(ax^2/2))dx\\int\\intexp(-(by^2/2))dy\\int\\intexp(-cxy)dxdy = 1

The integrals of the exponential functions can be easily solved using known properties of Gaussian distributions. The integral of exp(-(ax^2/2))dx is equal to \\sqrt\\(2\\pi/a\\) and the integral of exp(-(by^2/2))dy is equal to \\sqrt\\(2\\pi/b\\). The integral of exp(-cxy)dxdy is equal to \\sqrt\\(2\\pi/c^2\\). Therefore, we can substitute these values into the equation and solve for the normalization constant \\alpha:

\\alpha\\sqrt\\(2\\pi/a\\)\\sqrt\\(2\\pi/b\\)\\sqrt\\(2\\pi/c^2\\) = 1

\\alpha = \\frac\\(1}{2\\pi\\sqrt\\(ab\\)\\sqrt\\(c^2\\)}

We can then use