0
$\begingroup$

My professor said in the class that for "Additive White noise, the noise in pixels adjacent to each other are independent". How?

This is what I have got so far: White noise implies that PSD (Power Spectral Density) is flat which implies Covariance matrix is an impulse function i.e.

$$\begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0\end{bmatrix}$$

How does this imply that noise at adjacent pixels is independent? At max, this says that noise is uncorrelated.

This question asks the same thing, though that part of the question is unanswered.

Improve this question
$\endgroup$
7
  • $\begingroup$ a covariance matrix will have a bit of a hard time being an impulse function. The covariance matrix should be a diagonal matrix in this case, so the matrix you've shown us is a very bad example. $\endgroup$
    – Marcus Müller
    Commented Feb 5, 2019 at 16:34
  • $\begingroup$ Yeah, I realize that if the noise is independent, then covariance matrix should be diagonal. But when I took inverse 2D DFT of a matrix of ones (Flat PSD) in MatLab, this is what I got. It confused me more. Please explain if I have understood something wrong. Thanks! $\endgroup$
    – Nagabhushan S N
    Commented Feb 5, 2019 at 16:58
  • $\begingroup$ Your 2D-DFT of a single constant 2D image has nothing to do with the covariance matrix of a set of variables! $\endgroup$
    – Marcus Müller
    Commented Feb 5, 2019 at 17:07
  • $\begingroup$ Sorry. I meant 2D IDFT. Taking Inverse DFT of PSD gives Autocorrelation right? $\endgroup$
    – Nagabhushan S N
    Commented Feb 5, 2019 at 17:14
  • $\begingroup$ yes, but your constant-entry matrix is not a PSD. $\endgroup$
    – Marcus Müller
    Commented Feb 5, 2019 at 17:21

1 Answer 1

Reset to default
2
$\begingroup$

How did the professor state this? Did they say that the noise is iid? That means that the samples are independent, identically distributed. That means they are independent by definition (they are assumed to be so).

White noise, while a little more vague, also makes the assumption of independence. Again, this means that the noise is assumed to be independent.

However, if all that is said is that the samples are uncorrelated with the given covariance matrix, then you are correct: this does not imply independence.

Improve this answer
$\endgroup$
4
  • 4
    $\begingroup$ maybe the independence implication comes from the usual assumption of joint Gaussian distribution. $\endgroup$
    – AlexTP
    Commented Feb 5, 2019 at 14:09
  • $\begingroup$ @Peter No. All he mentioned was that "noise is additive and white. As an implication, noise in adjacent pixels are independent" $\endgroup$
    – Nagabhushan S N
    Commented Feb 5, 2019 at 16:29
  • $\begingroup$ @AlexTP can you explain a bit more? $\endgroup$
    – Nagabhushan S N
    Commented Feb 5, 2019 at 16:30
  • 2
    $\begingroup$ @NagabhushanSN "jointly normal distributed" and "uncorrelated" imply "independent". That's a special property of the normal distribution, which means that if jointly normal variables (e.g. pixel values) are independent, if they have "zero correlation" / are "uncorrelated" – which is what "white" means, in some definition. In other definitions, "white" doesn't only mean uncorrelated, but "independent". It's all a game of knowing which definitions your professor use. We can't help you more than Peter did in his answer. $\endgroup$
    – Marcus Müller
    Commented Feb 5, 2019 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.