# Correlation: A Magnitude or Power quantity?

I am confused about the units we would refer to with a correlation result in that if it would be a magnitude or power quantity (and therefore specifically when using ratios in dB would we use 10Log or 20Log?).

I have since confirmed as detailed in this post that the normalized correlation coefficient $$\rho$$ is a magnitude quantity in that the relationship between SNR and $$\rho$$ repeated here is:

$$\text{SNR} = 10\log_{10}\frac{\rho^2}{1-\rho^2}$$

Where in this case for an SNR relationship to $$\rho$$, the correlation involved is the correlation of $$x(t)$$ being a reference waveform as signal to $$y(t)$$ which is the same signal with added noise. (So that the SNR of $$y(t)$$ is determined in this case).

However I also understand and see directly from the math that the autocorrelation (at $$\tau=0$$ if we are referring to the autocorrelation function) is the variance (scaled by the number of samples), which is a power quantity, and we see the sum of products in the general expression for correlation suggesting a sum of powers. How do we reconcile all this?

Is it that the normalized coefficient through it's normalization process:

$$\rho = \frac {\operatorname*{cov}(x,y)}{\operatorname*{stddev}(x)\operatorname*{stddev}(y)}$$

Has this converted the power quantity of the numerator back to a magnitude quantity, and therefore correlation on its own (the numerator) IS a power quantity, while the correlation coefficient IS a magnitude quantity? If so, I don't yet quite see how. Or is it something else entirely?

• clarification question: in your definition of SNR, what exactly are x & y. Is x the clean signal and y signal+noise ? May 18, 2021 at 13:30
• @Hilmar yes you are correct. May 18, 2021 at 13:48
• Assuming the signal and noise are uncorrelated you and estimate the SNR simply as $P_x/(P_y-P_x)$ . That feels a lot simpler. May 18, 2021 at 21:12

There are different ways to look at this.

In general, it's useful to think of the correlation as a power quantity.

1. If you slog through the units, the correlation will have power-like units: $$V^2$$ for example
2. The Fourier Transform of the autocorrelation is the Power Spectral Density, which is clearly a power-like quantity

So what's happening in your example? Let's assume that we have $$y = x + n$$ where $$x$$ and $$n$$ are mean-free, uncorrelated and have the variances $$P_x = \sigma_x^2 = $$ and $$P_n = \sigma_n^2 = $$. Since they are uncorrelated we have $$P_y = P_x+P_n$$

The SNR would be $$SNR = 10log_{10}\frac{P_x }{P_n}$$

The argument to the log is a ratio of powers.

The covariance is $$cov(x,y) = = P_x$$ which has power-like units. The normalized correlation coefficient comes to be.

$$\rho = \frac{cov(x,y)}{\sqrt{{P_x \cdot P_y }} = \frac{P_x}{\sqrt{P_x \cdot (P_x + P_n) }}$$

$$\rho = \frac{cov(x,y)}{\sqrt{P_x \cdot P_y }} = \frac{P_x}{\sqrt{P_x \cdot (P_x + P_n) }}$$

So that's also a ratio of powers: cross-power to geometric mean of the signal powers. Since it's a ratio it's unitless. Let's square it:

$$\rho^2 =\frac{P_x^2}{P_x \cdot (P_x + P_n) } = \frac{P_x}{P_x + P_n }$$

The most reasonable physical interpretation of this is STILL a ratio powers: signal power to signal power plus noise power. Since the ratio is unitless, the square of ratio is still unitless, so there is no direct conflict here.

• in your formula for rho you show the covariance as being equal to the standard deviation of x (and variances throughout should be $\sigma_x^2$ not $\sigma_x$. If you concur can you update as I am trying to follow your otherwise interesting logic May 18, 2021 at 17:19
• Shoot, I mixed up the symbol for variance and standard deviation. Lemme fix it, May 18, 2021 at 17:51
• So rho is a ratio of powers but if we square it it is still a ratio of powers (which is counter-intuitive given a ratio of voltages as expressed in dB using 20Log when squared is a ratio of powers as 10Log. This is still a missing piece for me not yet explained May 19, 2021 at 3:07
• Fair point. Frankly, the definition of $\rho$ feels a little wonky to me. The cross power can be negative (is it still a power then?) and I have no physical interpretation of the "geometric mean" of the powers. There may be a way to interpret this a "scaled magnitude quantity" although it's not obvious to me how to do that. It seems needlessly complicated. Why would you use that instead of just $P_x/(P_y-P_x)$ ? May 19, 2021 at 13:21
• But the rest of your post was great without even mentioning that point and solved a long running mystery for me. In particular, and simply, showing how the SNR was derived. So my real interest is getting to Px/Py as in SNR and using rho to do that. Which you basically showed if you took it one step further for the final rho^2/(1-rho^2). So I appreciate this insight. May 19, 2021 at 13:25

I am adding this which was influenced by Hilmar's answer, and specifically to show that the normalized correlation coefficient, $$\rho$$, when derived from the comparison of a signal to the same signal signal plus noise is indeed a ratio of magnitude quantities and not power quantities. I will derive continuing Hilmar's approach that the result is indeed a ratio of the standard deviations. This is interesting as I also see the logic that correlation itself is a power quantity, especially noting the relationship of power spectral density and the autocorrelation function.

I believe what resolves the confusion is that the Normalized Correlation Coefficient is a magnitude quantity, which is not the same as the Auto-correlation or Cross-correlation functions which I believe are indeed in units of power quantities. The normalization by the standard deviations in the computation for $$\rho$$ which is the only difference must in the process convert the power quantity in the numerator back to a magnitude quantity, as proven by the sequence of equations below (but I don't otherwise grasp that intuitively):

Starting with the relationship between SNR and $$\rho$$ and then confirming this relationship and how it relates to $$\rho$$ itself:

$$SNR = 10\log_{10} \frac{\rho^2}{1-\rho^2} \tag{1} \label{1}$$

$$\rho = \frac{\text{cov}(X,Y)}{\sigma_X \sigma_Y}\tag{2} \label{2}$$

Where

$$X$$: signal, assumed zero-mean

$$Y = X + N$$: signal plus uncorrelated zero-mean noise

$$\text{cov}(X,Y)$$: Covariance of X and Y

$$\sigma_X$$: standard deviation of X

$$\sigma_Y$$: standard deviation of Y

Since the signal and noise are both zero-mean, $$\text{cov}(X,Y) = E(XY)$$:

$$\text{cov}(X,Y) = E(XY)-E(X)E(Y)=E(XY)-0 = E(XY)$$

Since the signal and noise are uncorrelated:

$$E(XY) = E(X(X+N))= E(X^2) + E(X N) = E(X^2) + 0 = E(X^2) = \sigma_X^2 \tag{3} \label{3}$$

Substituting \ref{3} into \ref{2}:

$$\rho = \frac{\sigma_X^2}{\sigma_X \sigma_Y} = \frac{\sigma_X}{\sigma_Y}\tag{4} \label{4}$$

From this we clearly see that the correlation coefficient for the correlation of Signal to Signal+ Noise is simply the ratio of the standard deviation of the two. Standard Deviation is a magnitude quantity, not a power quantity.

We continue to get Hilmar's result above and finally the relationship given in $$\ref{1}$$

$$\rho^2 = \frac{\sigma_X^2}{\sigma_Y^2} = \frac{\sigma_X^2}{\sigma_X^2 + \sigma_N^2}\tag{5} \label{5}$$

and

$$1-\rho^2 = 1-\frac{\sigma_X^2}{\sigma_X^2 + \sigma_N^2} = \frac{\sigma_N^2}{\sigma_X^2 + \sigma_N^2}\tag{6} \label{6}$$

Dividing \ref{5} bu \ref{6} we get the result:

$$\frac{\rho^2 }{1-\rho^2} = \frac{\sigma_X^2}{\sigma_N^2}$$