# Covariance and power spectral density of a signal?

I am reading an article on signal processing and I'm not sure of a few things. It defines the covariance function of a signal $y = y(t)$ as

$$\text{cov}(y) = E[y(t)y^\ast(t-k)].$$

And then it says the power spectral density is given by the Fourier transform of this covariance function.

1. Say we have the covariance function $E[XY]$ for two random variables $X$ and $Y$. A positive covariance in this case means that as $X$ increases $Y$ increases. What does it mean in the context of $E[y(t)y^\ast(t-k)]$ as in this case we only have one random variable $y$ which is taken at two different times? What does a positive covariance mean here and what is the significance of variable $k$?
2. If we take the Fourier transform of a signal and look at the resulting plot, a strong peak on this plot at a point $k$ signifies that one of the components of the original signal is a sinusoidal wave corresponding to that $k$. If the power spectral density is the Fourier transform of the covariance what does a strong peak in the power spectral density plot signify?
3. Why do we take the complex conjugate of $y^\ast(t-k)$? Why isn't the covariance $\text{cov}(y) = E[y(t)y(t-k)]$?
• Recall that covariance between $X$ and $Y$ is defined as $\text{Cov}=\mathbb{E}\{(X-\mathbb{E}X)(Y-\mathbb{E}Y)\}$. i.e. you have to subtract the expectations. – Richard Hardy Oct 28 '16 at 17:00

1. In the case in which $y(t)$ is real (to keep it simple) it means that values above the mean of $y(t-k)$ tend to be followed by values above the mean of $Y(t)$; the interpretation is the same as for your $X$ and $Y$, just replace $y(t)$ and $y(t-k)$ in place of $X$ and $Y$.