Correlation of Complex Random Variables

AI Thread Summary
The discussion addresses the presence of a half factor in the correlation definition of complex random variables, specifically in the formula φ_{zz}(τ) = (1/2)E[z*(t+τ)z(t)]. It is clarified that this factor is not a general rule but serves as a normalization factor, likely due to the independence of the real and imaginary parts of the complex variable z. The general formula for correlation is confirmed to apply to both complex and real-valued cases, where the autocorrelation simplifies to the variance of the variable itself. Participants agree that the half factor is primarily for normalization purposes. The conversation concludes with an acknowledgment of the clarification provided.
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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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