Randy Yates writes:
[...]
E[V^2] = E[(X+Y)^2]
= E[X^2 + 2XY + Y^2]
= E[X^2] + 2E[XY] + E[Y^2].

If X and Y are uncorrelated, then E[XY] = 0. Then

E[V^2] = E[X^2] + E[Y^2],

which means that if the two sources are the same power levels, the result
is twice the power, or 3 dB higher.

I should have stated the following:

1. "E[Z]" denotes the expected value of the random variable
underlying the random process Z, where Z is ergodic.

2. We assume X and Y are zero-mean processes.

3. Given assumption 2, E[Z^2] is the variance of Z, which is the
same as the power in the signal.

--RY


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