Magnitude of the Fourier Transform of White Noise Say you have two white noise signals with different variation amplitudes A1 and A2 as shown in this beautiful Excel graph:

Ignoring the DC offset as it's been represented here, how do you relate the amplitudes A1 and A2 to the magnitude of the Fourier coefficients after a Fourier transform (as shown in the diagram below)?

In other words, is it possible to relate A1 to Mag1 and A2 to Mag2? Can this even be done analytically or will it require a bit of simulation?
Any help is much appreciated! 
 A: It can be done analytically, but numerical results depend on what conventions you use to define the Fourier transform.
You have $\eta(t)$ a random variate.  I assume that you mean short range correlated
noise, so that $\langle \eta(t) \eta(t') \rangle = \sigma^2 \delta(t-t')$ ($\langle . \rangle$ indicates an ensemble average).
The spectral amplitude is a complex valued random variate
$$z(\omega) = \int e^{i \omega t} \eta(t) dt$$
which has a first moment
$$
\langle z(\omega) \rangle = \int e^{i \omega t} \langle \eta(t) \rangle  dt = \langle \eta \rangle \delta(\omega)
$$
If this is non-zero, you can subtract it off from the noise data; so from here
I'll assume zero mean noise data.
Now, the second moment of this data is given by
$$ 
\langle \lvert z(\omega) \rvert^2 \rangle = \int \int e^{i \omega t } e^{-i \omega t'} \langle \eta(t) \eta(t') \rangle  = \int  \sigma^2 dt
$$
There are a few concepts in this derivation:


*

*Note the interchange in order between doing the integrals, and doing the ensemble averaging; this is almost always mathematically justified.

*The second moment equation comes from explicitly writing out $\lvert z(\omega) \rvert^2=z(\omega) z^*(\omega)$, and then simplifying using the $\delta$ function correlation structure of the noise.

*As a formal, indefinite integral, the final expression is not completely defined; I tend to interpret as the integral from $[-L/2,L/2]$ for some arbitrarily large, but finite $L$. 


The key results are that the expected intensity of the power spectrum:
1. is linearly proportional to the variance of the noise, and
2. grows linearly with the integration time.  
Secondarily, depending on where you put the factor of $2 \pi$ involved in the Fourier transform, you may need to account for it in your noise spectrum.
For discretely sampled data, essentially the same logic applies, but with the integrals replaced by discrete sums.  The key thing in this case is knowing how/where your FFT library applies factors of $1/N$; when doing this kind of thing, I often explicitly check
what is going on by injecting simulated zero-mean, unit-variance Gaussian noise into the spectral processing chain just to verify what comes out.  Once you know the response to unit-variance noise, you scale that result by the observed noise variance.
A: The answer depends whether you have a discrete or continuous signal and wether you have white noise or not:
1) continuous signal (i.e. f(t)) and white noise


*

*No, there is no analytical relationship. This is because the variances/standard deviations A1 and A2 are infinite


2) discrete signal (sampled with sample rate $f_s$) and white noise


*

*The (one sided) noise spectral density is $S_{xx} = 2 A_{1}^2 / f_s$ with $A_1$ the standard deviation of the noise


3) non white noise (with effective bandwidth $B$):


*

*The (one sided) noise spectral density is $S_{xx} = A_{1}^2 / B$ with $A_1$ the standard deviation of the noise


Here the engineering convention is assumed, that the frequency is non-angular and the the noise spectral density is one-sided. This is what spectrum analyzer instruments output. Theorists (also in engineering) usually use the two-sided convention, which just omits the above factor of two.
A: The answer is very simple.  If the amplitude of the noise is multiplied by a factor of D, then the Fourier transform has to be multiplied by the same factor.  (the different conventions make no difference since obviously you are going to use the same conventions for each signal.) 
But I assume you mean power spectrum, not Fourier transform, since that's what spectral analysers measure and it's what your graph looks like.  If the amplitude of the noise is multiplied by a factor of D, its power is multiplied by $D^2$, which gives you the relationship sought.  Note that in this case, the variance of the noise would go up by $D^2$, so, As @Dave says, the power goes up linearly with the variance, like I said.
Worries about infinite variance are irrelevant since white noise is an idealisation, really you probably have some kind of smoothed pink noise, and your graphs from your analyser are clearly finite.  And in mathematical theory, we still distinguish between the Dirac Delta function $\delta$ and its multiples, $D\delta$, anyway, so the analytical relationship holds good for infinity, too.
