I'm having some trouble understanding how to derive the variance of the Fourier transform. This is for an image, i.e., it's a 2D transform.
The variance is $|\hat{I}(\xi,\eta)|^2$, the absolute square of the transform of the image; $\xi$ and $\eta$ are the coordinates in the frequency domain. $$\hat{I}(\xi,\eta) = \iint \mathbf{I}(x,y)\,e^{-i2\pi(x\xi + y\eta)}\,dx\,dy,$$ where $I(x, y)$ is the function representing the original image and the coordinates $x,y$ are in the spatial domain.
To calculate the variance, then, I want to do $$|\hat{I}(\xi,\eta)|^2 = \iint \mathbf{I}(x,y)\,e^{-i2\pi(x\xi + y\eta)}\,dx\,dy\,\iint \bar{\mathbf{I}}(x,y)\,e^{i2\pi(x\xi + y\eta)}\,dx\,dy$$ $$= \iint \mathbf{I}(x,y)\,\bar{\mathbf{I}}(x,y)\,e^{i2\pi(x\xi + y\eta) \,- \,i2\pi(x\xi + y\eta)}\,dx\,dy$$
My questions are:
- how do I deal with the conjugate of the function $\mathbf{I}$, i.e., $\bar{\mathbf{I}}(x,y)$? My inclination is that it doesn't change because all its values are real;
- should the exponential term really cancel? As written now, it will go to $e^0 = 1$. What am I missing about conjugating functions?
The variance should be proportional to $\frac{1}{\xi^2 + \eta^2}$, which is why I suspect something is going wrong with the exponential term.
Edited to clarify: The purpose of this is to show that the variance of $\hat{I}(\xi, \eta)$ is some scalar multiple of $\frac{1}{\xi^2 + \eta^2}$, i.e., $$ E_p[|\hat{I}(\xi, \eta)|^2] = s\frac{1}{\xi^2 + \eta^2}. $$ I read here that "the power of the process, which is expressed by its variance, is equal to the integral of the spectral density function over the interval $[−\pi, \pi]$".
The power is expressed as $|\hat{I}(\xi,\eta)|^2$. Shouldn't I just take the absolute square of $\hat{I}$?