How can I find the characteristic function of a bivariate gaussian distribution?

[steven187]

Hi all,

Im currently researching into Multivariate distributions, in particular I am trying to derive the characteristic function of the bivariate distribution of a gaussian. While knowing that a gaussian density function cannot be integrated how is it possible to find the characteristic function. I have been working on it but I keep bumping into a dead end. The following is what I did in the most recent dead end: (does anybody know why the latex thing aint working? I havnt been on here for a while)

so I have a double integral of the

exp{i*x*t_x+i*y*t_y}*(joint density function)

then I did the substitutions
u=x-m_x
and
v=y-m_y

I then simplified it down such that I get the following in the exponential expression

u^2*s_y^2-2*(si)*u*v*s_x*s_y+v^2*s_x^2

where si is the correlation coefficient, the problem is I can't complete the square because of the existence of si, would anybody know what I would need to do?

I had looked at Wolfram Mathworld website, its very interesting how it uses Eulers formula but isn't there any other way of doing it?

Regards

Steven

 

[mathman]

let w=au+bv and z=bu-av, where a^2+b^2=1. There will be some value of a where there will not be a cross product term for wz. Good luck!

 

[tacman]

To find the characteristic function of a bivariate Gaussian distribution, you can use the definition of the characteristic function, which is the expected value of the exponential of a complex-valued random variable. In this case, the random variable is (x,y), and the characteristic function is given by:

φ(t) = E[e^(itx+ity)] = ∫∫e^(itx+ity)f(x,y)dxdy

Where f(x,y) is the joint density function of the bivariate Gaussian distribution, and t is a complex number.

To solve this integral, you can use the properties of the Gaussian distribution, which states that the joint density function can be written as:

f(x,y) = (1/(2πσ_xσ_y√(1-ρ^2)))e^(-(1/(2(1-ρ^2)))(x^2/σ_x^2+y^2/σ_y^2-2ρxy/(σ_xσ_y)))

Where σ_x and σ_y are the standard deviations of x and y, and ρ is the correlation coefficient.

Substituting this into the integral, we get:

φ(t) = (1/(2πσ_xσ_y√(1-ρ^2)))∫∫e^(itx+ity-e^(-(1/(2(1-ρ^2)))(x^2/σ_x^2+y^2/σ_y^2-2ρxy/(σ_xσ_y))))dxdy

To solve this integral, you can use the properties of the exponential function and the fact that the integral of e^(-ax^2+bx+c) is √(π/a)e^(b^2/4a+c).

After some algebraic manipulation, you should be able to get the characteristic function in the form:

φ(t) = e^(im_xt+im_yt-1/2(t^2σ_x^2+t^2σ_y^2+2ρt^2σ_xσ_y))

Where m_x and m_y are the means of x and y. This is the characteristic function of a bivariate Gaussian distribution, and it can also be written as:

φ(t) = e^(-(1/2)(t^TΣt))

Where Σ is the covariance matrix of the bivariate Gaussian distribution, given by:

Σ = [σ_x^2