# Physical interpretation of 4th-order correlations

BACKGROUND:

Let's say we have samples of a random process $$X(t)$$ at two different times, $$t_1$$ and $$t_2$$, denoted $$X(t_1), X(t_2)$$. The values of $$X(t)$$ represent some voltage-like quantity (i.e. a voltage, current, or field). Assume $$X(t)$$ is a zero-mean, real-valued wide-sense stationary (WSS) process.

For the second-order statistics, we usually treat the quantity $$\mathbb E[X^2(t)]$$ as the average power.
The expectation $$\mathbb E[X(t_1) X(t_2)]$$ is just the autocorrelation function evaluated at lag $$\tau = t_2-t_1$$.

QUESTION:

Is there some physical interpretation of the 4th-order correlations, i.e. $$\mathbb E[X^3 (t_1)X(t_2)]$$, $$\mathbb E[X^2 (t_1)X^2(t_2)]$$, and $$\mathbb E[X (t_1) X^3(t_2)]$$?

• I'll counter-ask: What's the physical interpretation of the autocorrelation? Nov 19, 2018 at 20:19
• I wonder if this could be considered, conceptually, a duplicate of this question. This was the first thing that came to my mind when trying to infer the physical meaning from the units. In any case, autocorrelation is a product, between two time series of the same unit, so it is that unit squared. So, voltage squared gives you power. But $X^2(t)$ would already be "Power" and the autocorrelation of that would be Power squared...So, not very helpful..What motivated the question?
– A_A
Nov 19, 2018 at 21:41
• @CarlosDanger that's a pretty different question that you're asking!! And the answer is: you can't. I can give you a signal that has pretty much zero autocorrelation, but high $\mathbb E\left[ X^2(t) X^2(t+\tau) \right]$. But then I can give you a signal with the same $\mathbb E\left[ X^2(t) X^2(t+\tau) \right]$, but very different autocorrelation. Nov 19, 2018 at 22:33
• @CarlosDanger PDF $\equiv$ moment generating function ($\equiv$ characteristic function) hence IMHO, in general you cannot compute higher order moments from lower order ones. Nov 19, 2018 at 22:43
• I think you guys are getting too caught up in my example "motivation", so I've deleted it. For now, I'm just looking for the answer to the literal question. Nov 19, 2018 at 22:52