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

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.

Answer 1

[michael]

Please have a look on Peter's response here

[Adam Parker]

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.

Answer 2

[seackone]

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..)