FFT time domain average vs frequency bin average
Let me clarify.


Fourier transform does not represent the histogram of the signal.  Fourier transform is a linear transform that takes signal from time domain (complex function) to frequency domain (another complex function).  It takes a complex function to another complex function.
Fourier transform is linear as the poster above pointed out.
Phase in your samples matters as pointed out above.  If trial-by-trial data varies in phase, then you do not want to average before doing a Fourier transform, but you also do not want to average after Fourier transform.  You want to average after Fourier transform and norm.  I will elaborate below as far as exactly what needs to be done.


The main issue here is that the question is posed wrong.  It is not "should I take the Fourier transform before averaging or after averaging".  Because it makes no difference due to linearity of Fourier transform.  

The correct question to ask is "should I take the amplitude of the Fourier transform before averaging or after averaging".  For this question, the answer is before.

Here are the details. 

Suppose your sampled data is represented by the sequences:

$d_1=d_1[n_1],d_1[n_2],...d_1[n_N]$

$d_2=d_2[n_1],d_2[n_2],...d_2[n_N]$

$d_3=d_3[n_1],d_3[n_2],...d_3[n_N]$

...

$d_M=d_M[n_1],d_M[n_2],...d_M[n_N]$

where $d_1,...d_M$ are data from M trials and $n_1,...n_N$ are sampled time points, then:

$F_1 = \sum_{j=1}^M{|\mathcal{F}\{d_j\}|} \neq  |\mathcal{F}\{\sum_{j=1}^M{d_j}\}| = F_2$

So while the transform $\mathcal{F}$ is linear, $|\mathcal{F}|$ is not.

Furthermore, while $d_j[n_i]$ is real for all $i,j$, $\mathcal{F}\{d_j\}$ is not, but $|\mathcal{F}\{d_j\}|$ is.

As for what you should do, you should take the Fourier transform of individual trials (via FFT), get the amplitude of individual trials, and a average them together.  

Finally, what is $1/f$.  $1/f$ is a short term for the frequency spectrum of "natural" signals (usually people think of images).

When people say there is a large $1/f$ component, it means that the amplitude as a function of frequency looks like $1/f$.  It's totally hand-wavy... probably coming from a biologist :p

The inverse Fourier transform of $1/f$ is some sign function, but that is useless.  It is an imaginary sign function!  Real functions generate symmetric Fourier transform.

In fact saying that the spectrum is $1/f$, tells you something about the signal, but it doesn't let you recover the signal.  All that you know is that $|\mathcal{F}\{x(t)\}| = |1/f|$.  This doesn't let you uniquely determine $x(t)$ because all the phase information is gone, and we know that the structure of a signal relies heavily on its phase.

What does $1/f$ tell you?  Simply that it contains a lot of low frequency and a little high frequency. 

Just as important a question, what does averaging buy you?  and more important is how to interpret the result?  Tune in tomorrow for a more in depth discussion :p