Correlation of Complex Random Variables

EngWiPy
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Hi,

Why there is a half factor in the definition of the correlation of complex random variables, like:

\phi_{zz}(\tau)=\frac{1}{2}\mathbf{E}\left[z^*(t+\tau)z(t)\right]?

Thanks in advance
 
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S_David said:
Hi,

Why there is a half factor in the definition of the correlation of complex random variables, like:

\phi_{zz}(\tau)=\frac{1}{2}\mathbf{E}\left[z^*(t+\tau)z(t)\right]?

Thanks in advance
I don't think that's true as a general rule. For the example you give, an autocorrelation, the general formula would be

\rho_{zz}(\tau)=\frac{\mathbf{E}\left[z^*(t+\tau)z(t)\right]}{\mathbf{E}\left[z^*(t)z(t)\right]}

I'm guessing that in your case, 1/2 is just the normalization factor 1/E[z*z], perhaps because the real and imaginary parts of z are independent with mean square 1.
 
pmsrw3 said:
I don't think that's true as a general rule. For the example you give, an autocorrelation, the general formula would be

\rho_{zz}(\tau)=\frac{\mathbf{E}\left[z^*(t+\tau)z(t)\right]}{\mathbf{E}\left[z^*(t)z(t)\right]}

I'm guessing that in your case, 1/2 is just the normalization factor 1/E[z*z], perhaps because the real and imaginary parts of z are independent with mean square 1.

does this general formula apply to the real-valued case, too?
 
S_David said:
does this general formula apply to the real-valued case, too?
Yes.

The general formula for a correlation is \frac{Cov(x,y)}{\sqrt{Var(x)Var(y)}}. In the case of an autocorrelation, x, and y are the same (except displaced in time, which doesn't affect the variance), so the denominator reduces to Var(x) = E[x^2].
 
pmsrw3 said:
Yes.

The general formula for a correlation is \frac{Cov(x,y)}{\sqrt{Var(x)Var(y)}}. In the case of an autocorrelation, x, and y are the same (except displaced in time, which doesn't affect the variance), so the denominator reduces to Var(x) = E[x^2].

So, 0.5 is just a normalization factor. Ok thanks a lot.

Regards
 
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