When calculating the standard deviation of arrays of complex numbers c, why is std(c) not equal to std(abs(c))?

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Adam Parker
Adam Parker 2016-10-8
回答： seackone 2018-5-1
for example, when both the real and imaginary components are normally distributed:
c = randn(100,1) + i*randn(100,1);
>> std(c)
ans =
1.5367
>> std(abs(c))
ans =
0.7312
Looking at how Matlab calculates these, it appears that the modulus of c is used in both instances. Am I wrong?
Thanks for your help.

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回答（2 个）

michael
michael 2016-10-8
编辑：michael 2016-10-8
Please have a look on Peter's response here
  1 个评论
Adam Parker
Adam Parker 2016-10-8
Michael, I appreciate you taking the time to respond. Thank you.
However, I don't think the Peter's answer gets to the heart of my question - it addresses something slightly different (unless I am missing something?)
My confusion is this: if I look at how the std() function handles a complex number it does the 'squaring' operation by multiplying by the complex conjugate i.e. if c=a+ib then it uses c* x c = a^2+b^2 in its calculation. But this is just the same thing that abs(c) does. So why do you get different answers?
Thanks again.

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seackone
seackone 2018-5-1
Hi, I also want to calculate the standard deviation of a complex number.. At my research, I found your question. But std(c) and std(abs(c)) is not the same!
For example:
A = [1+1i*1 1+1i*2]
The mean of A is:
mean(A) = 1/2 * (1+1i*1 + 1+1i*2) = 1+1i*1.5
Now you can calculate the standard deviation..
But std(abs(A)) is an other equation:
abs(A) = [1.412 2.2361]
and the mean:
mean(abs(A)) = 1.8251
you got a different matrix and a different mean, so the result of the standard deviation isn't the same. If you want, you can calculate the example to the end and the results are:
std(A) = 0.70..
std(abs(A)) = 0.58..
I hope I could help (maybe not you, because the post is more then a year old..)

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