- #1

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

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

Thanks in advance

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- Thread starter EngWiPy
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- #1

- 1,367

- 61

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

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

Thanks in advance

- #2

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

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

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

Thanks in advance

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

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.

- #3

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

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

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?

- #4

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

The general formula for a correlation is [tex]\frac{Cov(x,y)}{\sqrt{Var(x)Var(y)}}[/tex]. 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].

- #5

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Yes.

The general formula for a correlation is [tex]\frac{Cov(x,y)}{\sqrt{Var(x)Var(y)}}[/tex]. 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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