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Can anyone tell me what the meaning is of the phrase: "zero-mean random noise with standard deviation equal to 1"? Also, I want to know why not except zero-mean random noise and standard deviation equal to 1.

Why should I use zero mean and std 1?

An example of the context is from the following paper:

II. OBSERVATION MODEL

We consider an observation model of the form z (x) = y (x) + σ (y (x)) ξ (x), x ∈ X, (1) where X is the set of the sensorís active pixel positions, z is the actual raw-data output, y is the ideal output, ξ is zero-mean random noise with standard deviation equal to 1, and σ is a function y, modulating the standard-deviation of the overall noise component. The function σ (y) is called standard-deviation function or standard-deviation curve. The function σ2 (y) is called variance function. Since E {ξ (x)} = 0 we have E {z (x)} = y (x) and std {z (x)} = σ (E {z (x)}). There are no additional restrictions on the distribution of ξ (x), and different points may have different distributions.

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    $\begingroup$ Hi gmotree: Where does it say that? What's the context? $\endgroup$
    – Time4Tea
    Commented Apr 25, 2015 at 11:34

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Zero mean so that the noise does not present a net disturbance to the system. There's as much positive noise as negative, so they cancel out in the long run. If the mean were not zero, then the noise would appear as an additional dynamic. For example, if the quantity were a force with some random jitter to it, then if the jitter did not have zero mean, the noise would appear as an additional net force on average.

Standard deviation 1 means that the variable has been scaled for convenience. Going back to the force example, one would typically find that the std of the force is some value in Newtons. .01 N, 1 N, 1,000,000 N, whatever, depending on the problem. It is sometimes valuable to rescale the random force by dividing by the standard deviation. This has the effect of making the force dimensionless: Newtons divided by Newtons. It also allows for the calculation to be used for forces at any scale. The solution becomes more easily generalized to other situations. It also creates a notion of "characteristic force". The force is measured with respect to the value of the characteristic force of the system.

There are other ways to define characteristic values, and thus other ways to remove dimensions from the equation. Evidently in the case you cite, the author has decided that the standard deviation of the noise is convenient for his purposes.

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  • $\begingroup$ Thanks Sir, but would you please let me know with more another example? I don't quite a catch about "a net disturbance". is this means the same like some kind of wire noise? Sorry Sir but your answer very professional so please let me know more example. I can't catch all about zero mean. $\endgroup$
    – gmotree
    Commented Apr 25, 2015 at 12:23
  • $\begingroup$ Sir If you can, please let me know more easy example about Zero mean noise. it is not easy to understand. $\endgroup$
    – gmotree
    Commented Apr 26, 2015 at 5:58
  • $\begingroup$ Patience. We're volunteers here. Suppose you want to model a mechanical device. There is some piece that is meant to apply a force to another. You can select a force and the device will provide it. But due to various types of non-ideal behavior (worn bearings, stretching belts, motors, electrical stresses) there is jitter (noise). It's possible that the averge value of the noise is not zero. The noise will add an offset to the desired force. This is both a practical problem and an analytical one. A possible approach in modeling is to subtract the offset so that the noise has zero avg. $\endgroup$
    – garyp
    Commented Apr 26, 2015 at 12:11
  • $\begingroup$ Thanks very much Sir, I get it what are you meaning. But I have some query about z(x) = y(x) + σ(y(x)) ξ(x). As I know, ξ(x) is "zero-mean random noise with standard deviation equal to 1" and The function σ (y) which is called standard-deviation function or standard-deviation curve has multiplied with ξ(x). is this meaning double standard deviation(which mean multiplying each standard deviations)? i'm confusing the meaning and the reason. $\endgroup$
    – gmotree
    Commented Apr 27, 2015 at 10:25

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